Recherches sur les fonctions elliptiques.

Researches on the elliptic functions.

Niels Henrik Abel · 1827 · Journal für die reine und angewandte Mathematik (Crelle's Journal), vol. 2, pp. 101–181 · AI draft — Machine output, not yet human-checked. Treat it as a working draft and check anything that matters against the scan. How review levels work →

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Significance

Here Abel turns Legendre's elliptic integral upside down. Legendre had spent decades studying the integral $\displaystyle\int \frac{\partial\theta}{\sqrt{1-c^{2}\sin^{2}\theta}}$ as a function of its upper limit; Abel proposes instead, in the opening pages, to study the inverse function $\Phi\alpha$, defined by $x=\Phi\alpha$ and $\alpha=\displaystyle\int_{0}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}$, together with its companions $f\alpha$ and $F\alpha$.Read directly from the text of this article: the opening pages (pp. 101–103), where Abel reviews Euler, Lagrange and Legendre, takes Legendre's elliptic integrals as his starting point, and introduces "la fonction inverse" $\Phi\alpha$. It is the step that takes the integral $\displaystyle\int \frac{\partial x}{\sqrt{1-x^{2}}}$ to the sine, and it is what Abel builds everything else on. Once the functions are allowed imaginary arguments, as in §. I, they turn out to have two independent periods, $\omega$ and $\varpi$.Read directly from the text of this article: §. I, where the functions are extended to imaginary arguments (equations (7.) and (8.)) and the two periods $\omega$ and $\varpi$ appear. The rest of the article follows the consequences. §. II gives the multiplication formulas, expressing $\Phi(n\alpha)$ rationally in $\Phi\alpha$, $f\alpha$, $F\alpha$. §§. III–V take up the equations that arise when one divides the periods into equal parts, reduce the solution to a single further equation of degree $2n+2$, and solve it in special cases, notably the lemniscatic case $e=c$. For that case Abel states, on p. 146, that the whole circumference of the lemniscate can be divided into $m$ equal parts by ruler and compass for exactly those $m$ for which Gauss had shown it possible for the circle. §§. VI–VII end with other expressions for $\Phi(n\beta)$ and with expansions of the three functions in series and infinite products.Read directly from the text of this article: the section headings of §§. II–VII, the statement on p. 146 of the ruler-and-compass division of the lemniscate (which Abel compares explicitly with Gauss's theorem for the circle), and the series for $\omega/2$ on the final page. The article breaks off with "La continuation dans le cahier suivant": the second part appeared in volume 3 of the same journal (1828) and is not included here.Abel's closing note on p. 181, "La continuation dans le cahier suivant"; the second part is the article of the same title in Crelle's Journal, vol. 3 (1828), pp. 160–190, as recorded in this work's publication entry. Abel died in April 1829, the year in which Jacobi's Fundamenta nova, developed independently, appeared and gave the theory of elliptic functions its other classic foundation.J. J. O'Connor and E. F. Robertson, "Niels Henrik Abel", MacTutor History of Mathematics Archive, University of St Andrews — for Abel's death in April 1829 and his work on elliptic functions in these years. The appearance of Jacobi's Fundamenta nova in 1829 is recorded in this corpus under jacobi-1829-fundamenta-nova.

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Recherches sur les fonctions elliptiques.

(Par M. N. H. Abel.)

Depuis longtems les fonctions logarithmiques, et les fonctions exponentielles et circulaires ont été les seules fonctions transcendantes, qui ont attiré l'attention des géomètres. Ce n'est que dans les derniers tems, qu'on a commencé à en considérer quelques autres. Parmi celles-ci il faut distinguer les fonctions, nommées elliptiques, tant pour leurs belles propriétés analytiques, que pour leur application dans les diverses branches des mathématiques. La première idée de ces fonctions a été donnée par l'immortel Euler, en démontrant, que l'équation séparée \[ \frac{\partial x}{\sqrt{(\alpha+\beta x+\gamma x^{2}+\delta x^{3}+\varepsilon x^{4})}}+\frac{\partial y}{\sqrt{(\alpha+\beta y+\gamma y^{2}+\delta y^{3}+\varepsilon y^{4})}}=0 \tag{1.} \] est intégrable algébriquement. Après Euler, Lagrange y a ajouté quelque chose, en donnant son élégante théorie de la transformation de l'intégrale $\displaystyle\int\frac{R.\partial x}{\sqrt{[(1-p^{2}x^{2})(1-q^{2}x^{2})]}}$, ou $R$ est une fonction rationelle de $x$. Mais le premier et, si je ne me trompe, le seul, qui ait approfondi la nature de ces fonctions, est M. Legendre, qui, d'abord dans un mémoire sur les fonctions elliptiques, et ensuite dans ses excellents exercices de mathématiques, a développé nombre de poropriétésEditorial note: Printed poropriétés instead of propriétés; an apparent misprint. élégantes de ces fonctions, et a montré leur application. Lors de la publication de cet ouvrage, rien n'a été ajouté à la théorie de M. Legendre. Je crois, qu'on ne verra pas ici sans plaisir des recherches ultérieures sur ces fonctions.

En général on comprend sous la dénomination de fonctions elliptiques, toute fonction, comprise dans l'intégrale \[ \int\frac{R\,\partial x}{\sqrt{(\alpha+\beta x+\gamma x^{2}+\delta x^{3}+\varepsilon x^{4})}}, \] où $R$ est une fonction rationelle et $\alpha$, $\beta$, $\gamma$, $\delta$, $\varepsilon$ sont des quantités constantes et réelles. M. Legendre a démontré, que par des substitutions convenables on peut toujours ramener cette intégrale à la forme

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\[ \int\frac{P\,\partial y}{\sqrt{(a+by^{2}+cy^{4})}}, \] où $P$ est une fonction rationelle de $y^{2}$. Par les réductions convenables, cette intégrale peut être ensuite ramenée à la forme \[ \int\frac{A+By^{2}}{C+Dy^{2}}\cdot\frac{\partial y}{\sqrt{(a+by^{2}+cy^{4})}}, \] et celle-ci à: \[ \int\frac{A+B\sin^{2}\theta}{C+D\sin^{2}\theta}\cdot\frac{\partial\theta}{\sqrt{(1-c^{2}\sin^{2}\theta)}}, \] où $c$ est réel, et moindre que l'unité.

De là suit, que toute fonction elliptique peut être réduite à l'une des trois formes: \[ \int\frac{\partial\theta}{\sqrt{(1-c^{2}\sin^{2}\theta)}};\ \int\partial\theta\sqrt{(1-c^{2}\sin^{2}\theta)};\ \int\frac{\partial\theta}{(1+n\sin^{2}\theta)\sqrt{(1-c^{2}\sin^{2}\theta)}}, \] auxquelles M. Legendre donne les noms de fonctions elliptiques de la première, seconde et troisième espèce. Ce sont ces trois fonctions, que M. Legendre a considérées; surtout la première, qui a les propriétés les plus remarquables et les plus simples.

Je me propose, dans ce mémoire, de considérer la fonction inverse, c'est-à-dire la fonction $\Phi\alpha$, déterminée par les équations \[ \begin{gathered} \alpha=\int\frac{\partial\theta}{\sqrt{(1-c^{2}\sin^{2}\theta)}}\ \text{et}\\ \sin\theta=\varphi(\alpha)=x. \end{gathered} \] La dernière équation donne \[ \partial\theta.\sqrt{(1-\sin^{2}\theta)}=\partial.\Phi\alpha=\partial x, \] donc \[ \alpha=\int_{0}\frac{\partial x}{\sqrt{[(1-x^{2})(1-c^{2}x^{2})]}}. \] M. Legendre suppose $c^{2}$ positif, mais j'ai remarqué, que les formules deviennent plus simples, en supposant $c^{2}$ négatif, $=-e^{2}$. De même j'écris pour plus de symmétrie $1-c^{2}x^{2}$ au lieu de $1-x^{2}$. En sorte que la fonction $\Phi\alpha=x$ sera donnée par l'équation \[ \alpha=\int_{0}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}, \] ou bien \[ \Phi'\alpha=\partial\alpha.\sqrt{[(1-c^{2}\Phi^{2}\alpha)(1+e^{2}\Phi^{2}\alpha)]}.\] Editorial note: Printed $\partial\alpha$ before the radical; the factor is spurious, since the derivative is the square root alone. Pour abréger, j'introduis deux autres fonctions de $\alpha$, savoir: \[ f\alpha=\sqrt{(1-c^{2}\Phi^{2}\alpha)};\ F\alpha=\sqrt{(1+e^{2}\Phi^{2}\alpha)}. \]

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Plusieurs propriétés de ces fonctions se déduisent immédiatement des propriétés connues de la fonction elliptique de la première espèce, mais d'autres sont plus cachées. Par ex. on démontre, que les équations $\Phi\alpha=0$, $f\alpha=0$, $F\alpha=0$ ont un nombre infini de racines, qu'on peut trouver toutes. Une des propriétés les plus remarquables est, qu'on peut exprimer rationellement $\Phi(m\alpha)$, $f(m\alpha)$, $F(m\alpha)$ ($m$ étant un nombre entier) en $\Phi\alpha$, $f\alpha$, $F\alpha$. Aussi rien n'est plus facile, que de trouver $\Phi(m\alpha)$, $f(m\alpha)$, $F(m\alpha)$, lorsqu'on connoît $\Phi\alpha$, $f\alpha$, $F\alpha$; mais le problème inverse, savoir de déterminer $\Phi\alpha$, $f\alpha$, $F\alpha$ en $\Phi(m\alpha)$, $f(m\alpha)$, $F(m\alpha)$ est plus difficile, parcequ'il dépend d'une équation d'un degré plus élevé (savoir du degré $m^{2}$).

La solution de cette équation est l'objet principal de ce mémoire. D'abord on fera voir, comment ontEditorial note: Printed ont instead of on; an apparent misprint. peut trouver toutes les racines, au moyen des fonctions $\Phi,f,F$. On traitera ensuite de la solution algébrique de l'equation en question, et on parviendra à ce résultat remarquable, que $\varphi\left(\frac{\alpha}{m}\right)$; $f\left(\frac{\alpha}{m}\right)$; $F\left(\frac{\alpha}{m}\right)$ peuvent être exprimés en $\Phi\alpha$, $f\alpha$, $F\alpha$, au moyen d'une fonction, qui, par rapport à $\alpha$, ne contient d'autres irrationalités que des radicaux. Cela produit une classe très générale d'équations, qui sont resolubles algébriquement. Il est à remarquer, que les expressions des racines contiennent des quantités constantes, qui, en général, ne sont pas exprimables par des quantités algébriques. Ces quantités constantes dépendent d'une équation du degré $\mathrm{m}^{2}-1$. On fera voir comment, au moyen de fonctions algébriques, on peut en ramener la solution à celle d'une équation du degré $\mathrm{m}+1$. On donnera plusieurs expressions des fonctions $\Phi(2n+1)\alpha$; $f(2n+1)\alpha$; $F(2n+1)\alpha$ en fonctions de $\Phi\alpha$, $f\alpha$, $F\alpha$. On en déduira ensuite les valeurs de $\Phi\alpha$, $f\alpha$, $F\alpha$ en fonctions de $\alpha$. On démontrera, que ces fontionsEditorial note: Printed fontions instead of fonctions; an apparent misprint. peuvent être décomposées en un nombre infini de facteurs, et même en une infinité de fractions partielles.

§. I.

Propriétés fondamentales des fonctions $\Phi\alpha$; $f\alpha$; $F\alpha$.

1.

En supposant, que $\Phi\alpha=x$ (1.), on aura en vertu de ce qui précède: \[ \alpha=\int_{0}^{x}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}. \tag{2.} \]

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Par là on voit, que $\alpha$, considéré comme fonction de $x$, est positif depuis $x=0$, jusqu'à $x=\frac{1}{c}$. En faisant donc \[ \frac{\omega}{2}=\int_{0}^{\frac{1}{c}}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}, \tag{3.} \] il est évident, que $\Phi\alpha$ etEditorial note: Printed et instead of est; an apparent misprint. positif et va en augmentant depuis $\alpha=0$ jusqu'à $\alpha=\frac{\omega}{2}$, et qu'on aura \[ \varphi(0)=0,\ \varphi\left(\frac{\omega}{2}\right)=\frac{1}{c}. \tag{4.} \] Parceque $\alpha$ change de signe, lorsqu'on écrit $-x$ à la place de $x$; il en est de même de la fonction $\Phi\alpha$ par rapport à $\alpha$, et par conséquent on aura l'équation \[ \varphi(-\alpha)=-\varphi(\alpha). \tag{5.} \] En mettant dans (1.) $xi$ au lieu de $x$ (ou $i$, pour abréger, représente la quantité imaginaire $\sqrt{-1}$) et désignant la valeur de $\alpha$ par $\beta i$, il viendra \[ xi=\Phi(\beta i)\ \text{et}\ \beta=\int_{0}^{x}\frac{\partial x}{\sqrt{[(1+c^{2}x^{2})(1-e^{2}x^{2})]}}. \tag{6.} \] $\beta$ est réel et positif depuis $x=0$ jusqu'à $x=\frac{1}{e}$, donc en faisant \[ \frac{\varpi}{2}=\int_{0}^{\frac{1}{e}}\frac{\partial x}{\sqrt{[(1-e^{2}x^{2})(1+c^{2}x^{2})]}}, \tag{7.} \] $x$ sera positif, depuis $\beta=$ zéro jusqu'à $\beta=\frac{\varpi}{2}$, c'est-à-dire, la fonction $\frac{1}{i}\Phi(\beta i)$ sera positive entre les mêmes limites. En faisant $\beta=\alpha$ et $y=\frac{\varphi(\alpha i)}{i}$, on a \[ \alpha=\int_{0}^{y}\frac{\partial y}{\sqrt{[(1-e^{2}y^{2})(1+c^{2}y^{2})]}}, \] donc on voit, qu'en supposant $c$ au lieu de $e$ et $e$ au lieu de $\varepsilon$,Editorial note: Printed as a curled e (apparently $\varepsilon$) at the last place; the sense requires $c$ (interchange of $c$ and $e$). \[ \frac{\varphi(\alpha i)}{i}\ \text{se changera en}\ \Phi\alpha. \]

Et parceque \[ \begin{gathered} f\alpha=\sqrt{(1-c^{2}\Phi^{2}\alpha)},\\ F\alpha=\sqrt{(1+e^{2}\Phi^{2}\alpha)}, \end{gathered} \] on voit, que par le changement de $c$ en $e$ et $e$ en $c$, $f(\alpha i)$ et $F(\alpha i)$ rentreront respectivement en $F\alpha$ et $f\alpha$. Enfin les équations (2.) et (7.) font voir, que par la même transformation, $\omega$ et $\varpi$ rentreront respectivement en $\varpi$ et $\omega$.

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Suivant (7.) on aura $x=\frac{1}{e}$ pour $\beta=\frac{\varpi}{2}$, donc en vertu de l'équation $xi=\varphi(\beta i)$, il viendra \[ \varphi\left(\frac{\varpi i}{2}\right)=i.\frac{1}{e}. \tag{8.} \]

2.

En vertu de ce qui précède, on aura les valeurs de $\Phi\alpha$ pour toute valeur réelle de $\alpha$, comprise entre $-\frac{\omega}{2}$ et $+\frac{\omega}{2}$, et pour toute valeur imaginaire de la forme $\beta i$ de cette quantité, si $\beta$ est une quantité continue entre les limites $-\frac{\varpi}{2}$ et $+\frac{\varpi}{2}$. Il s'agit maintenant de trouver la valeur de cette fonction pour une valeur quelconque, réelle ou imaginaire, de la variable. Pour y parvenir, nous allons d'abord établir les propriétés fondamentales des fonctions $\Phi$, $f$ et $F$.

Parcequ'on a \[ \begin{gathered} f^{2}\alpha=1-c^{2}\Phi^{2}\alpha,\\ F^{2}\alpha=1+e^{2}\Phi^{2}\alpha, \end{gathered} \] on aura, en différentiant: \[ \begin{gathered} f\alpha.f'\alpha=-c^{2}\Phi\alpha.\varphi'\alpha,\\ F\alpha.F'\alpha=e^{2}\Phi\alpha\,\varphi'\alpha, \end{gathered} \] Or d'après (2.) on a \[ \varphi'\alpha=\sqrt{[(1-c^{2}\Phi^{2}\alpha)(1+e^{2}\Phi^{2}\alpha)]}=f\alpha.F\alpha, \] donc, en substituant cette valeur de $\Phi'\alpha$ dans les deux équations précédentes, on trouvera, que les fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$ sont liées entre elles par les équations \[ \left\{\begin{aligned} \varphi'\alpha&=f\alpha.F\alpha,\\ f'\alpha&=-c^{2}\Phi\alpha.F\alpha,\\ F'\alpha&=e^{2}\Phi\alpha.f\alpha. \end{aligned}\right. \tag{9.} \] Cela posé, je dis, qu'en désignant par $\alpha$ et $\beta$ deux indéterminées, on aura \[ \left\{\begin{aligned} \Phi(\alpha+\beta)&=\frac{\varphi\alpha.f\beta.F\beta+\varphi\beta.f\alpha.F\alpha}{1+e^{2}c^{2}.\varphi^{2}\alpha.\varphi^{2}\beta},\\ f(\alpha+\beta)&=\frac{f\alpha.f\beta-c^{2}.\varphi\alpha.\varphi\beta.F\alpha.F\beta}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta},\\ F(\alpha+\beta)&=\frac{F\alpha.F\beta+e^{2}.\varphi\alpha.\varphi\beta.f\alpha.f\beta}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta}. \end{aligned}\right. \tag{10.} \]

Ces formules peuvent être déduites sur le champ des propriétés connues des fonctions elliptiques${}^{*)}$; mais on peut aussi les vérifier aisément de la manière suivante.

*) Legendre Exercices de calcul intégral.

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En désignant par $r$ le second membre de la première des équations (10.), on aura, en differentiant par rapport à $\alpha$: \[ \begin{aligned} \left(\frac{\partial r}{\partial\alpha}\right)&=\frac{\{\varphi'\alpha.f\beta.F\beta+\varphi\beta.F\alpha.f'\alpha+\varphi\beta.f\alpha.F'\alpha\}}{1+e^{2}c^{2}.\varphi^{2}\alpha.\varphi^{2}\beta}\\ &\quad-\frac{(\varphi\alpha.f\beta F\beta+\varphi\beta.f\alpha\ F\alpha).2e^{2}c^{2}\varphi\alpha.\varphi^{2}\beta.\varphi'\alpha}{(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)^{2}}. \end{aligned} \] En substituant pour $\Phi'\alpha$, $f'\alpha$, $F'\alpha$ leurs valeurs données par les équations (9.), il viendra \[ \begin{aligned} \left(\frac{\partial r}{\partial\alpha}\right)&=\frac{f\alpha.F\alpha.f\beta.F\beta}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta}-\frac{2e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta.f\alpha.f\beta.F\alpha.F\beta}{(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)^{2}}\\ &\quad+\frac{\varphi\alpha.\varphi\beta.\{1+e^{2}c^{2}\varphi^{2}\alpha\varphi^{2}\beta\}\{-c^{2}F^{2}\alpha+e^{2}f^{2}\alpha\}-2e^{2}c^{2}\varphi\alpha\,\varphi\beta.\varphi^{2}\beta.f^{2}\alpha.F^{2}\alpha}{(1+e^{2}c^{2}\varphi^{2}\alpha\varphi^{2}\beta)^{2}}, \end{aligned} \] d'où, en substituant pour $f^{2}\alpha$ et $F^{2}\alpha$ leurs valeurs: $1-c^{2}\Phi^{2}\alpha$, $1+e^{2}\Phi^{2}\alpha$, et en réduisant, on tire \[ \left(\frac{\partial r}{\partial\alpha}\right)=\frac{(1-e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)\{(e^{2}-c^{2})+\varphi\alpha.\varphi\beta+f\alpha.f\beta.F\alpha.F\beta\}-2e^{2}c^{2}\varphi\alpha\,\varphi\beta(\varphi^{2}\alpha+\varphi^{2}\beta)}{(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)^{2}}. \] Editorial note: Printed with a plus sign (and a stray hairline before it) between $(e^{2}-c^{2})$ and $\varphi\alpha.\varphi\beta$; a product $(e^{2}-c^{2})\varphi\alpha.\varphi\beta$ is expected. Maintenant $\alpha$ et $\beta$ entrent symétriquement dans l'expression de $r$; donc on aura la valeur de $\left(\frac{\partial r}{\partial\beta}\right)$, en permutant $\alpha$ et $\beta$ dans la valeur de $\left(\frac{\partial r}{\partial\alpha}\right)$. Or par cela l'expression de $\left(\frac{\partial r}{\partial\alpha}\right)$ ne change pas de valeur, donc on aura \[ \left(\frac{\partial r}{\partial\alpha}\right)=\left(\frac{\partial r}{\partial\beta}\right). \]

Cette équation aux différentielles partielles fait voir que $r$ est fonction de $\alpha+\beta$; donc on aura \[ r=\psi(\alpha+\beta). \] La forme de la fonction $\psi$ se trouvera, en donnant à $\beta$ une valeur particulière. En supposant par ex. $\beta=0$, et remarquant que $\Phi(0)=0$, $f(0)=1$, $F(0)=1$, les deux valeurs de $r$ deviendront \[ r=\Phi(\alpha)\ \text{et}\ r=\psi(\alpha), \] donc \[ \psi(\alpha)=\Phi(\alpha), \] et de là \[ r=\psi(\alpha+\beta)=\Phi(\alpha+\beta). \] La première des formules (10.) a donc lieu effectivement.

De la même manière on verifira les deux autres formules.

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3.

Des formules (10.) on peut en déduire une foule d'autres. Je vais rapporter quelques-unes des plus remarquables: Pour abréger je fais \[ 1+e^{2}c^{2}\Phi^{2}\alpha.\Phi^{2}\beta=R. \tag{11.} \]

En changeant d'abord le signe de $\beta$, on obtiendra \[ \left\{\begin{aligned} \Phi(\alpha+\beta)+\Phi(\alpha-\beta)&=\frac{2\varphi\alpha.f\beta.F\beta}{R},\\ \Phi(\alpha+\beta)-\Phi(\alpha-\beta)&=\frac{2\varphi\beta.f\alpha.F\alpha}{R},\\ f(\alpha+\beta)+f(\alpha-\beta)&=\frac{2f\alpha.f\beta}{R},\\ f(\alpha+\beta)-f(\alpha-\beta)&=\frac{-2c^{2}.\varphi\alpha.\varphi\beta.F\alpha.F\beta}{R},\\ F(\alpha+\beta)+F(\alpha-\beta)&=\frac{2F\alpha.F\beta}{R},\\ F(\alpha+\beta)-F(\alpha-\beta)&=\frac{2e^{2}.\varphi\alpha.\varphi\beta.f\alpha.f\beta}{R}. \end{aligned}\right. \tag{12.} \]

En formant le produit de $\Phi(\alpha+\beta)$ et $\Phi(\alpha-\beta)$, on trouvera \[ \begin{aligned} \Phi(\alpha+\beta).\Phi(\alpha-\beta)&=\frac{\varphi\alpha.f\beta.F\beta+\varphi\beta.f\alpha.F\alpha}{R}.\frac{\varphi\alpha.f\beta.F\beta-\varphi\beta.f\alpha.F\alpha}{R}\\ &=\frac{\varphi^{2}\alpha.f^{2}\beta.F^{2}\beta-\varphi^{2}\beta.f^{2}\alpha.F^{2}\alpha}{R^{2}}, \end{aligned} \] ou, en substituant les valeurs de $f^{2}\beta$, $F^{2}\beta$, $f^{2}\alpha$, $F^{2}\alpha$ dans $\Phi\beta$ et $\Phi\alpha$: \[ \begin{aligned} \Phi(\alpha+\beta).\Phi(\alpha-\beta)&=\frac{\varphi^{2}\alpha-\varphi^{2}\beta-e^{2}c^{2}\varphi^{2}\alpha.\varphi^{4}\beta+e^{2}c^{2}\varphi^{2}\beta.\varphi^{4}\alpha}{R^{2}}\\ &=\frac{(\varphi^{2}\alpha-\varphi^{2}\beta)(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)}{R^{2}}; \end{aligned} \] or $R=1+e^{2}c^{2}\Phi^{2}\alpha.\Phi^{2}\beta$, donc \[ \Phi(\alpha+\beta).\Phi(\alpha-\beta)=\frac{\varphi^{2}\alpha-\varphi^{2}\beta}{R}. \tag{13.} \] On trouvera de même \[ \left\{\begin{aligned} f(\alpha+\beta).f(\alpha-\beta)&=\frac{f^{2}\alpha-c^{2}\varphi^{2}\beta.F^{2}\alpha}{R}=\frac{f^{2}\beta-c^{2}\varphi^{2}\alpha.F^{2}\beta}{R}\\ &=\frac{1-c^{2}\varphi^{2}\alpha-c^{2}\varphi^{2}\beta-c^{2}e^{2}\varphi^{2}\alpha.\varphi^{2}\beta}{R}=\frac{f^{2}\alpha.f^{2}\beta-c^{2}(c^{2}+e^{2})\varphi^{2}\alpha.\varphi^{2}\beta}{R},\\ F(\alpha+\beta).F(\alpha-\beta)&=\frac{F^{2}\alpha+e^{2}\varphi^{2}\beta.f^{2}\alpha}{R}=\frac{F^{2}\beta+e^{2}\varphi^{2}\alpha.f^{2}\beta}{R}\\ &=\frac{1+e^{2}\varphi^{2}\alpha+e^{2}\varphi^{2}\beta-e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta}{R}=\frac{F^{2}\alpha.F^{2}\beta-e^{2}(c^{2}+e^{2})\varphi^{2}\alpha.\varphi^{2}\beta}{R}. \end{aligned}\right. \tag{14.} \]

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4.

En faisant dans les formules (10.) $\beta=\pm\frac{\omega}{2}$, $\beta=\pm\frac{\varpi}{2}i$, et remarquant, que $f\left(\pm\frac{\omega}{2}\right)=0$, $F\left(\pm\frac{\varpi}{2}i\right)=0$, on aura \[ \left\{\begin{aligned} &\varphi\left(\alpha\pm\frac{\omega}{2}\right)=\pm\varphi\frac{\omega}{2}.\frac{f\alpha}{F\alpha};\ f\left(\alpha\pm\frac{\omega}{2}\right)=\mp\frac{F\frac{\omega}{2}}{\varphi\frac{\omega}{2}}.\frac{\varphi\alpha}{F\alpha};\\ &F\left(\alpha\pm\frac{\omega}{2}\right)=\frac{F\frac{\omega}{2}}{F\alpha};\\ &\varphi\left(\alpha\pm\frac{\varpi}{2}i\right)=\pm\varphi\frac{\varpi}{2}i.\frac{F\alpha}{f\alpha};\ F\left(\alpha\pm\frac{\varpi}{2}i\right)=\mp\frac{f\frac{\varpi}{2}i}{\varphi\frac{\varpi}{2}i}.\frac{\varphi\alpha}{f\alpha};\\ &f\left(\alpha\pm\frac{\varpi}{2}i\right)=\frac{f\frac{\varpi}{2}i}{f\alpha}; \end{aligned}\right. \tag{15.} \] ou bien: \[ \left\{\begin{aligned} &\varphi\left(\alpha\pm\frac{\omega}{2}\right)=\pm\frac{1}{c}.\frac{f\alpha}{F\alpha};\ f\left(\alpha\pm\frac{\omega}{2}\right)=\mp\sqrt{(e^{2}+c^{2})}.\frac{\varphi\alpha}{F\alpha};\\ &F\left(\alpha\pm\frac{\omega}{2}\right)=\frac{\sqrt{(e^{2}+c^{2})}}{c}.\frac{1}{F\alpha};\\ &\varphi\left(\alpha\pm\frac{\varpi}{2}i\right)=\pm\frac{i}{e}.\frac{F\alpha}{f\alpha};\ F\left(\alpha\pm\frac{\varpi}{2}i\right)=\pm i\sqrt{(e^{2}+c^{2})}.\frac{\varphi\alpha}{f\alpha};\\ &f\left(\alpha\pm\frac{\varpi}{2}i\right)=\frac{\sqrt{(e^{2}+c^{2})}}{e}.\frac{1}{f\alpha}. \end{aligned}\right. \tag{16.} \]

De là on tire sur le champ: \[ \left\{\begin{aligned} &\varphi\left(\frac{\omega}{2}+\alpha\right)=\Phi\left(\frac{\omega}{2}-\alpha\right);\ f\left(\frac{\omega}{2}+\alpha\right)=-f\left(\frac{\omega}{2}-\alpha\right);\\ &F\left(\frac{\omega}{2}+\alpha\right)=F\left(\frac{\omega}{2}-\alpha\right);\\ &\varphi\left(\frac{\varpi}{2}i+\alpha\right)=\Phi\left(\frac{\varpi}{2}i-\alpha\right);\ F\left(\frac{\varpi}{2}i+\alpha\right)=-F\left(\frac{\varpi}{2}i-\alpha\right);\\ &f\left(\frac{\varpi}{2}i+\alpha\right)=f\left(\frac{\varpi}{2}i-\alpha\right); \end{aligned}\right. \tag{17.} \] \[ \varphi\left(\alpha\pm\frac{\omega}{2}\right).\varphi\left(\alpha+\frac{\varpi}{2}i\right)=\pm\frac{i}{ce};\ F\left(\alpha\pm\frac{\omega}{2}\right).F\alpha=\frac{b}{c};\ f\left(\alpha\pm\frac{\varpi}{2}i\right).f\alpha=\frac{b}{e}. \tag{18.} \] En faisant $\alpha=\frac{\omega}{2}$ et $\frac{\varpi}{2}i$, on trouve de là \[ \Phi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\ f\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\ F\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}. \]

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Ensuite les équations (17.), en y mettant dans les trois premières $\alpha+\frac{\omega}{2}$ au lieu de $\alpha$, et dans les trois dernières $\alpha+\frac{\varpi}{2}i$ au lieu de $\alpha$, donnent les suivantes \[ \left\{\begin{aligned} &\varphi(\alpha+\omega)=-\Phi\alpha;\ f(\alpha+\omega)=-f\alpha;\ F(\alpha+\omega)=F\alpha;\\ &\varphi(\alpha+\varpi i)=-\Phi\alpha;\ f(\alpha+\varpi i)=f\alpha;\ F(\alpha+\varpi i)=-F\alpha; \end{aligned}\right. \tag{19.} \] et en mettant $\alpha+\omega$ et $\alpha+\varpi i$ au lieu de $\alpha$: \[ \left\{\begin{aligned} &\Phi(2\omega+\alpha)=\Phi\alpha;\ \Phi(2\varpi i+\alpha)=\Phi\alpha;\ \Phi(\omega+\varpi i+\alpha)=\Phi\alpha;\\ &f(2\omega+\alpha)=f\alpha;\ f(\varpi i+\alpha)=f\alpha;\\ &F(\omega+\alpha)=F\alpha;\ F(2\varpi i+\alpha)=F\alpha. \end{aligned}\right. \tag{20.} \]

Ces équations font voir, que les fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$ sont des fonctions périodiques. On en déduira sans peine les suivantes, où $m$ et $n$ sont deux nombres entiers positifs ou négatifs: \[ \left\{\begin{aligned} &\varphi((m+n)\omega+(m-n)\varpi i+\alpha)=\varphi\alpha;\ \varphi((m+n)\omega+(m-n+1)\varpi i+\alpha)=-\varphi\alpha;\\ &f(2m\omega+n\varpi i+\alpha)=f\alpha;\ f((2m+1)\omega+n\varpi i+\alpha)=-f\alpha;\\ &F(m\omega+2n\varpi i+\alpha)=+F\alpha;\ F(m\omega+(2n+1)\varpi i+\alpha)=-F\alpha. \end{aligned}\right. \tag{21.} \] Ces formules peuvent aussi s'écrire comme il suit: \[ \left\{\begin{aligned} &\Phi(m\omega+n\varpi i\pm\alpha)=\pm(-1)^{m+n}.\Phi\alpha,\\ &f(m\omega+n\varpi i\pm\alpha)=(-1)^{m}.f\alpha,\\ &F(m\omega+n\varpi i\pm\alpha)=(-)^{n}.F\alpha. \end{aligned}\right. \tag{22.} \] Editorial note: In the last line of this system the base of the exponent is printed as a bare minus sign, without the digit 1; an apparent misprint for $(-1)^{n}$. On peut remarquer comme cas particuliers: \[ \left\{\begin{aligned} &\Phi(m\omega\pm\alpha)=\pm(-1)^{m}.\Phi\alpha;\ \Phi(n\varpi i\pm\alpha)=\pm(-1)^{n}.\Phi\alpha;\\ &f(m\omega\pm\alpha)=(-1)^{m}.f\alpha;\ f(n\varpi i\pm\alpha)=f\alpha;\\ &F(m\omega\pm\alpha)=F\alpha;\ F(n\varpi i\pm\alpha)=(-1)^{n}.F\alpha. \end{aligned}\right. \tag{22$'$.} \]

5.

Les formules qu'on vient d'établir, font voir qu'on aura les valeurs des fonctions $\Phi\alpha$; $f\alpha$; $F\alpha$ pour toutes les valeurs réelles ou imaginaires de la variable, lorsqu'on les connoît pour les valeurs réelles de cette quantité, comprises entre $\frac{\omega}{2}$ et $-\frac{\omega}{2}$ et pour les valeurs imaginaires de la forme $\beta i$, où $\beta$ est comprise entre $\frac{\varpi}{2}$ et $-\frac{\varpi}{2}$.

En effet, supposons qu'on demande la valeur des fonctions $\Phi(\alpha+\beta i)$, $f(\alpha+\beta i)$, $F(\alpha+\beta i)$, où $\alpha$ et $\beta$ sont des quantités réelles quelconques. En mettant dans les formules (10.) $\beta i$ à la place de $\beta$, il est clair, qu'on aura les trois fonctions dont il s'agit, exprimées par les fonctions: $\Phi\alpha$; $f\alpha$; $F\alpha$; $\Phi(\beta i)$; $f(\beta i)$; $F(\beta i)$. Il ne reste donc, qu'à déterminer ces dernières.

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Or, quelles que soient les valeurs de $\alpha$ et $\beta$, on peut toujours trouver deux nombres entiers $m$ et $n$, tels que $\alpha=m.\omega\pm\alpha'$, $\beta=n\varpi\pm\beta'$, où $\alpha'$ est une quantité comprise entre $0$ et $+\frac{\omega}{2}$, et $\beta'$ entre $0$ et $+\frac{\varpi}{2}$. Donc on aura, en vertu des équations (22.), en substituant les valeurs précédentes de $\alpha$ et $\beta$: \[ \begin{aligned} \Phi(\alpha)&=\varphi(m\omega\pm\alpha')=\pm(-1)^{m}.\Phi\alpha',\\ f(\alpha)&=f(m\omega\pm\alpha')=(-1)^{m}.f\alpha',\\ F(\alpha)&=F(m\omega\pm\alpha')=F\alpha',\\ \Phi(\beta i)&=\varphi(+n\varpi i\pm\beta'i)=\pm(-1)^{n}.\varphi(\beta'i),\\ f(\beta i)&=f(n\varpi i\pm\beta'i)=f(\beta'i),\\ F(\beta i)&=F(n\varpi i\pm\beta'i)=(-1)^{n}.F(\beta'i). \end{aligned} \] Donc les fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$, $\Phi(\beta i)$, $f(\beta i)$, $F(\beta i)$ seront exprimées comme on vient de le dire et par suite aussi les fonctions $\Phi(\alpha+\beta i)$; $f(\alpha+\beta i)$; $F(\alpha+\beta i)$.

Nous avons vu précédemment, que $\Phi\alpha$ est réel depuis $\alpha=-\frac{\omega}{2}$ jusqu'à $\alpha=+\frac{\omega}{2}$, et que $\frac{\varphi(\alpha i)}{i}$ est réel depuis $\alpha=-\frac{\varpi}{2}$ jusqu'à $\alpha=+\frac{\varpi}{2}$.

Donc en vertu des équations (22.) il est clair:

1) que $\Phi(\alpha)$ et $\frac{\varphi(\alpha i)}{i}$ sont réels pour toute valeur réelle de $\alpha$; $\Phi\alpha$ est compris entre $-\frac{1}{c}$ et $+\frac{1}{c}$, et $\frac{\varphi(\alpha i)}{i}$ entre $-\frac{1}{e}$ et $+\frac{1}{e}$;

2) que $\Phi(\alpha)$ s'évanouit pour $\alpha=m\omega$, et $\frac{\varphi(\alpha i)}{i}$ pour $\alpha=m\varpi$, $m$ étant un nombre entier positif ou négatif; mais $\Phi(\alpha)$ n'est pas nul pour aucune autre valeur réelle de $\alpha$.

En remarquant, que $f\alpha=\sqrt{(1-c^{2}\Phi^{2}\alpha)}$, $F\alpha=\sqrt{(1+e^{2}\Phi^{2}\alpha)}$, il suit de ce que nous venons de dire:

1) que les fonctions $f(\alpha)$; $F(\alpha)$; $f(\alpha i)$; $F(\alpha i)$ sont réelles pour toute valeur de $\alpha$;

2) que $f(\alpha)$ est compris entre les limites $-1$ et $+1$ et $F(\alpha)$ entre les limites $+1$ et $+\sqrt{\left(1+\frac{e^{2}}{c^{2}}\right)}$, ensorte que $F\alpha$ est positif pour toute valeur réelle de $\alpha$;

3) que $f(\alpha i)$ est positif et compris entre les limites $+1$ et $\sqrt{\left(1+\frac{c^{2}}{e^{2}}\right)}$ et $F(\alpha i)$ entre les limites $-1$ et $+1$ pour toute valeur réelle de $\alpha$;

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4) que $f(\alpha)$ s'évanouit pour $\alpha=(m+\frac{1}{2})\omega$ et $F(i\alpha)$ pour $\alpha=(m+\frac{1}{2})\varpi$; mais pour nulle autre valeur de $\alpha$.

On remarquera ce qui suit comme corollaires qu'on tire des formules (22.):

1) Soit $\alpha=0$. Dans ce cas, en remarquant, que $\Phi(0)=0$, $f(0)=1$, $F(0)=1$, on aura \[ \left\{\begin{aligned} \varphi(m\omega+n\varpi i)&=0,\\ f(m\omega+n\varpi i)&=(-1)^{m},\\ F(m\omega+n\varpi i)&=-1. \end{aligned}\right. \tag{23.} \] Editorial note: The last line is printed as $-1$ over a blotted impression; by formula (22.) one would expect $(-1)^{n}$.

2) Soit $\alpha=\frac{\omega}{2}$. En vertu des équations: \[ \Phi\left(\frac{\omega}{2}\right)=\frac{1}{c};\ f\left(\frac{\omega}{2}\right)=0;\ F\left(\frac{\omega}{2}\right)=\frac{b}{c}; \] on aura \[ \left\{\begin{aligned} \Phi((m+\tfrac{1}{2})\omega+n\varpi i)&=(-1)^{m+n}.\frac{1}{c},\\ f((m+\tfrac{1}{2})\omega+n\varpi i)&=0,\\ F((m+\tfrac{1}{2})\omega+n\varpi i)&=(-1)^{n}.\frac{b}{c}. \end{aligned}\right. \tag{24.} \]

3) Soit $\alpha=\frac{\varpi}{2}i$. En vertu des équations \[ \Phi\left(\frac{\varpi}{2}i\right)=\frac{i}{e};\ f\left(\frac{\varpi}{2}i\right)=\frac{b}{e};\ F\left(\frac{\varpi}{2}i\right)=0; \] on aura: \[ \left\{\begin{aligned} \Phi(m\omega+(n+\tfrac{1}{2})\varpi i)&=(-1)^{m+n}\frac{i}{e},\\ f(m\omega+(n+\tfrac{1}{2})\varpi i)&=(-1)^{m}.\frac{b}{e},\\ F(m\omega+(n+\tfrac{1}{2})\varpi i)&=0. \end{aligned}\right. \tag{25.} \]

4) Soit $\alpha=\frac{\omega}{2}+\frac{\varpi}{2}i$. En vertu des équations ci dessus on aura \[ \left\{\begin{aligned} \Phi((m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i)&=\frac{1}{0},\\ f((m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i)&=\frac{1}{0},\\ F((m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i)&=\frac{1}{0}. \end{aligned}\right. \tag{26.} \]

6.

Les équations (23.), (24.), (25.) font voir, que la fonction $\Phi(\alpha)$ s'évanouit toutes les fois que $\alpha$ est de la forme $\alpha=m\omega+n\varpi i$; que $f\alpha$ s'évanouit toutes les fois que $\alpha$ est de la forme $\alpha=(m+\frac{1}{2})\omega+n\varpi i$, et que $F\alpha$ s'évanouit toutes les fois que $\alpha$ est de la forme $\alpha=m\omega+(n+\frac{1}{2})\varpi i$.

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Or je dis que pour toute autre valeur de $\alpha$, les fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$ auront nécessairement une valeur différente de zéro.

Supposons en effet, qu'on ait \[ \Phi(\alpha+\beta i)=0, \] $\alpha$ et $\beta$ étant des quantités réelles. En vertu de la première des formules (10.), cette équation peut s'écrire comme il suit: \[ \frac{\varphi\alpha.f(\beta i).F(\beta i)+\varphi(\beta i).f\alpha.F\alpha}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}(\beta i)}=0. \]

Maintenant les quantités $\Phi\alpha$, $f(\beta i)$, $F(\beta i)$ sont réelles et $\Phi(\beta i)$ est de la forme $i.A$, où $A$ est réel; donc cette équation ne peut pas subsister à moins qu'on n'ait séparément: \[ \Phi(\alpha).f(\beta i).F(\beta i)=0;\ \Phi(\beta i).f\alpha.F\alpha=0. \] Ces équations ne peuvent être satisfaites, que de deux manières, savoir en faisant \[ \Phi(\alpha)=0,\ \Phi(\beta i)=0, \] ou \[ f(\beta i).F(\beta i)=0,\ f\alpha.F\alpha=0. \]

Les deux premières équations donnent $\alpha=m\omega$; $\beta=n\varpi$.

Les deux dernières, en remarquant que $F\alpha$ et $f(\beta i)$ ne peuvent jamais s'évanouir, donnent \[ f\alpha=0,\ F(\beta i)=0, \] d'où \[ \alpha=(m+\tfrac{1}{2})\omega,\ \beta=(n+\tfrac{1}{2})\varpi. \] Mais pour ces valeurs de $\alpha$ et $\beta$ la valeur de $\Phi(\alpha+\beta i)$ deviendra infinie; donc les seules valeurs de $\alpha$ et $\beta$ sont $\alpha=m\omega$ et $\beta=n\varpi$, et par conséquent toutes les racines de l'équation \[ \Phi(x)=0, \] peuvent être représentées par \[ x=m\omega+n\varpi i. \tag{27.} \] De la même manière on trouvera, que toutes les racines de l'équation \[ f(x)=0, \] peuvent être représentées par \[ x=(m+\tfrac{1}{2})\omega+n\varpi i, \tag{28.} \] et celles de l'équation \[ F(x)=0, \] par \[ x=m\omega+(n+\tfrac{1}{2})\varpi i. \tag{29.} \]

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7.

Les formules (26.) font voir, qu'on satisfait aux trois équations \[ \Phi(x)=\frac{1}{0},\ f(x)=\frac{1}{0},\ F(x)=\frac{1}{0}, \] en donnant à $x$ une des valeurs de la forme \[ x=(m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i. \tag{30.} \]

Or on peut démontrer que les équations en question n'ont pas d'autres racines.

En effet, ayant \[ \begin{gathered} F(x)=\frac{b}{c}.\frac{1}{F\left(x-\frac{\omega}{2}\right)};\ \Phi(x)=\frac{-i}{ec}.\frac{1}{\varphi\left(x-\frac{\omega}{2}-\frac{\varpi}{2}i\right)};\\ f(x)=\frac{b}{e}.\frac{1}{f\left(x-\frac{\varpi}{2}i\right)}; \end{gathered} \] les équations en question entraîneront celles-ci: \[ \varphi\left(x-\frac{\omega}{2}-\frac{\varpi}{2}i\right)=0;\ f\left(x-\frac{\varpi}{2}i\right)=0;\ F\left(x-\frac{\omega}{2}\right)=0, \] mais en vertu de ce qu'on vient de voir dans le numéro précédent, les équations donnent respectivement: \[ \begin{gathered} x-\frac{\omega}{2}-\frac{\varpi}{2}i=m\omega+n\varpi i;\ x-\frac{\varpi}{2}i=(m+\tfrac{1}{2})\omega+n\varpi i;\\ x-\frac{\omega}{2}=m\omega+(n+\tfrac{1}{2})\varpi i; \end{gathered} \] c'est-à-dire: on aura pour les trois équations: \[ x=(m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i, \] c. q. f. d.

8.

Ayant trouvé comme ci dessus toutes les racines des équations \[ \begin{gathered} \Phi(x)=0;\ f(x)=0;\ F(x)=0;\\ \Phi(x)=\frac{1}{0};\ f(x)=\frac{1}{0};\ F(x)=\frac{1}{0}; \end{gathered} \] je vais maintenant chercher les racines des équations plus générales. \[ \Phi(x)=\Phi a,\ f(x)=fa,\ F(x)=Fa, \] où $a$ est une quantité quelconque réelle ou imaginaire.

Considérons d'abord l'équation \[ \Phi(x)-\Phi a=0. \] En faisant dans la seconde des formules (12.) \[ \alpha=\frac{x+a}{2},\ \beta=\frac{x-a}{2}, \] on trouvera \[ \Phi x-\Phi a=\frac{2\varphi\left(\frac{x-a}{2}\right).f\left(\frac{x+a}{2}\right).F\left(\frac{x+a}{2}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{x+a}{2}\right).\varphi^{2}\left(\frac{x-a}{2}\right)}=0. \]

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Cette équation ne peut subsister que dans l'un des cinq cas suivans:

1. si $\Phi\left(\frac{x-a}{2}\right)=0$, d'où $x=a+2m\omega+2n\varpi i$,

2. si $f\left(\frac{x+a}{2}\right)=0$, d'où $x=-a+(2m+1)\omega+2n\varpi i$,

3. si $F\left(\frac{x+a}{2}\right)=0$, d'où $x=-a+2m\omega+(2n+1)\varpi i$,

4. si $\Phi\left(\frac{x-a}{2}\right)=\frac{1}{0}$, d'où $x=a+(2m+1)\omega+(2n+1)\varpi i$,

5. si $\Phi\left(\frac{x+a}{2}\right)=\frac{1}{0}$, d'où $x=-a+(2m+1)\omega+(2n+1)\varpi i$.

La résolution de ces cinq équations est contenue dans les formules (27.), (28.), (29.).

Des valeurs trouvées de $x$ il faut rejeter celles, que donne la formule \[ x=-a+(2m+1)\omega+(2n+1)\varpi i, \] car une telle valeur de $x$ donne en vertu de (22.): \[ \Phi x=-\Phi a, \] tandis qu'on doit avoir $\Phi x=\Phi a$; mais les autres valeurs de $x$, exprimées par les quatre premières formules, peuvent être admises. Elles sont, comme on voit, contenues dans la seule formule: \[ x=(-1)^{m+n}.a+m\omega+n\varpi i. \tag{31.} \] Telle est donc l'expression générale de toutes les racines de l'équation \[ \Phi x=\Phi a. \] De la même manière on trouvera, que toutes les racines de l'équation \[ fx=fa \] sont représentées par la formule \[ x=\pm a+2m\omega+n\varpi i, \tag{32.} \] et toutes celles de l'équation \[ Fx=Fa, \] par la formule \[ x=\pm a+m\omega+2n\varpi i. \tag{33.} \]

§. II.

Formules, qui donnent les valeurs de $\Phi(n\alpha)$, $f(n\alpha)$, $F(n\alpha)$ exprimées en fonctions rationelles de $\Phi\alpha$, $f\alpha$, $F\alpha$.

9.

Reprenons les formules (12.). En faisant dans la $1{}^{\mathrm{e}}$, $3{}^{\mathrm{e}}$ et $5{}^{\mathrm{e}}$ $\alpha=n\beta$, il viendra:

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\[ \left\{\begin{aligned} \Phi(n+1)\beta&=-\Phi(n-1)\beta+\frac{2\varphi(n\beta).f\beta.F\beta}{R},\\ f(n+1)\beta&=-f(n-1)\beta+\frac{2f(n\beta).f\beta}{R},\\ F(n+1)\beta&=-F(n-1)\beta+\frac{2F(n\beta).F\beta}{R}, \end{aligned}\right. \tag{34.} \] où $R=1+c^{2}e^{2}.\Phi^{2}(n\beta).\Phi^{2}\beta$.

Ces formules donnent la valeur de $\Phi(n+1)\beta$ en $\Phi(n-1)\beta$ et $\Phi(n\beta)$; celle de $f(n+1)\beta$ en $f(n-1)\beta$ et $f(n\beta)$, et celle de $F(n+1)\beta$ en $F(n-1)\beta$ et $F(n\beta)$. Donc en faisant successivement $n=1$, 2, 3…, on trouvera successivement les valeurs des fonctions: \[ \begin{gathered} \Phi(2\beta);\ \Phi(3\beta);\ \Phi(4\beta)\ .....\ \Phi(n\beta),\\ f(2\beta);\ f(3\beta);\ f(4\beta)\ .....\ f(n\beta),\\ F(2\beta);\ F(3\beta);\ F(4\beta)\ .....\ F(n\beta) \end{gathered} \] exprimées en fonctions rationelles des trois quantités \[ \Phi\beta;\ f\beta;\ F\beta. \] En faisant p. ex. $n=1$, on aura: \[ \left\{\begin{aligned} \Phi(2\beta)&=\frac{2.\varphi\beta.f\beta.F\beta}{1+e^{2}c^{2}\varphi^{4}\beta},\\ f(2\beta)&=-1+\frac{2f^{2}\beta}{1+e^{2}c^{2}\varphi^{4}\beta},\\ F(2\beta)&=-1+\frac{2F^{2}\beta}{1+e^{2}c^{2}\varphi^{4}\beta}. \end{aligned}\right. \tag{35.} \]

Les fonctions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ étant des fonctions rationelles de $\Phi\beta$, $f\beta$, $F\beta$, on peut toujours les réduire à la forme $\frac{P}{Q}$, où $P$ et $Q$ sont des fonctions entières de $\Phi\beta$, $f\beta$, $F\beta$. De même il est clair, que le dénominateur $Q$ aura la même valeur pour les trois fonctions, que l'on considère. Soit donc \[ \Phi(n\beta)=\frac{P_n}{Q_n},\ f(n\beta)=\frac{P'_n}{Q_n},\ F(n\beta)=\frac{P''_n}{Q_n}, \tag{35$'$.} \] on aura également \[ \begin{gathered} \Phi(n+1)\beta=\frac{P_{n+1}}{Q_{n+1}},\ f(n+1)\beta=\frac{P'_{n+1}}{Q_{n+1}},\ F(n+1)\beta=\frac{P''_{n+1}}{Q_{n+1}},\\ \Phi(n-1)\beta=\frac{P_{n-1}}{Q_{n-1}},\ f(n-1)\beta=\frac{P'_{n-1}}{Q_{n-1}},\ F(n-1)\beta=\frac{P''_{n-1}}{Q_{n-1}}. \end{gathered} \] En substituant ces valeurs, la première des formules (34.) deviendra: \[ \frac{P_{n+1}}{Q_{n+1}}=-\frac{P_{n-1}}{Q_{n-1}}+\frac{2f\beta.F\beta.\dfrac{P_n}{Q_n}}{1+c^{2}e^{2}\varphi^{2}\beta.\dfrac{P_n^{2}}{Q_n^{2}}}, \]

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ou bien \[ \frac{P_{n+1}}{Q_{n+1}}=\frac{-P_{n-1}.(Q_n^{2}+c^{2}e^{2}\varphi^{2}\beta.P_n^{2})+2P_n.Q_n.Q_{n-1}.f\beta.F\beta}{Q_{n-1}.(Q_n^{2}+e^{2}c^{2}\varphi^{2}\beta.P_n^{2})}. \]

En égalant les numérateurs et les dénominateurs de ces deux fractions, on aura \[ P_{n+1}=-P_{n-1}(Q_n^{2}+c^{2}e^{2}\Phi^{2}\beta.P_n^{2})+2f\beta.F\beta.P_n.Q_n.Q_{n-1}, \tag{36.} \] \[ Q_{n+1}=Q_{n-1}(Q_n^{2}+e^{2}c^{2}\Phi^{2}\beta.P_n^{2}). \tag{37.} \] La seconde et la troisième des équations (34.) donneront de la même manière: \[ P'_{n+1}=-P'_{n-1}.(Q_n^{2}+c^{2}e^{2}\Phi^{2}\beta.P_n^{2})+2f\beta\,P'_n.Q_n.Q_{n-1}, \tag{38.} \] \[ P''_{n+1}=-P''_{n-1}.(Q_n^{2}+c^{2}e^{2}\Phi^{2}\beta.P_n^{2})+2F\beta.P''_n.Q_n.Q_{n-1}. \tag{39.} \] En faisant dans ces quatre formules $n=1$, 2, 3 ....., et remarquant, qu'on aura: \[ \begin{gathered} Q_0=1;\ Q_1=1;\ P_0=0;\ P_1=\Phi\beta;\\ P'_0=1;\ P'_1=f\beta;\ P''_0=1;\ P''_1=F\beta; \end{gathered} \] on trouvera successivement les fonctions entières $Q_n$, $P_n$, $P'_n$, $P''_n$, pour toutes les valeurs de $n$.

Soient pour abréger: \[ \Phi\beta=x,\ f\beta=y,\ F\beta=z\ \text{et} \tag{40.} \] \[ R_n=Q_n^{2}+e^{2}c^{2}x^{2}P_n^{2}, \tag{41.} \] les formules précédentes donneront: \[ \left\{\begin{aligned} Q_{n+1}&=Q_{n-1}.R_n,\\ P_{n+1}&=-P_{n-1}.R_n+2yz.P_n.Q_n.Q_{n-1},\\ P'_{n+1}&=-P'_{n-1}.R_n+2y.P'_n.Q_n.Q_{n-1},\\ P''_{n+1}&=-P''_{n-1}.R_n+2z.P''_n.Q_n.Q_{n-1}. \end{aligned}\right. \tag{42.} \] En posant $n=1$, 2, on aura: \[ \left\{\begin{aligned} R_1&=Q_1^{2}+e^{2}c^{2}x^{2}P_1^{2}=1+e^{2}c^{2}x^{4},\\ Q_2&=Q_0R_1=1+e^{2}c^{2}x^{4},\\ P_2&=-P_0R_1+2yzP_1.Q_1.Q_0=2xyz,\\ P'_2&=-P'_0R_1+2yP'_1.Q_1.Q_0=-1-e^{2}c^{2}x^{4}+2y^{2},\\ P''_2&=-P''_0R_1+2zP''_1.Q_1.Q_0=-1-e^{2}c^{2}x^{4}+2z^{2}. \end{aligned}\right. \tag{43.} \] \[ \left\{\begin{aligned} R_2&=Q_2^{2}+e^{2}c^{2}x^{2}.P_2^{2}=(1+e^{2}c^{2}x^{4})^{2}+e^{2}c^{2}x^{2}.4x^{2}y^{2}z^{2},\\ Q_3&=Q_1R_2=R_2,\\ P_3&=-P_1R_2+2yzP_2Q_2Q_1=-xR_2+2y^{2}z^{2}x.Q_2\\ &=x\{2y^{2}z^{2}.Q_2-R_2\},\\ P'_3&=-P'_1.R_2+2y.P'_2.Q_2Q_1=-y.R_2+2yP'_2Q_2\\ &=y\{2Q_2P'_2-R_2\},\\ P''_3&=z\{2Q_2P''_2-R_2\}. \end{aligned}\right. \tag{44.} \]

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En continuant de cette sorte, et en remarquant que $y^{2}=1-c^{2}x^{2}$, $z^{2}=1+e^{2}x^{2}$, on verra aisément, que les quantités: \[ Q_n,\ \frac{P_{2n}}{xyz},\ \frac{P_{2n+1}}{x},\ P'_{2n},\ \frac{P'_{2n+1}}{y},\ P''_{2n},\ \frac{P''_{2n+1}}{z} \] sont des fonctions entières des trois quantités $x^{2}$, $y^{2}$, $z^{2}$ et par conséquent aussi de l'une de ces quantités quelconque pour une valeur entière quelconque de $n$.

Cela fait voir que les expressions de $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ seront de la forme suivante: \[ \left\{\begin{aligned} \Phi(2n\beta)&=\Phi\beta.f\beta.F\beta.T, & \Phi(2n+1)\beta&=\Phi\beta.T',\\ f(2n\beta)&=T_1, & f(2n+1)\beta&=f\beta.T'',\\ F(2n\beta)&=T_2, & F(2n+1)\beta&=F\beta.T''', \end{aligned}\right. \tag{45.} \] où $T$ etc. représentent des fonctions rationelles des quantités $(\Phi\beta)^{2}$, $(f\beta)^{2}$, $(F\beta)^{2}$.

§. III.

Résolution des équations \[ \Phi(n\beta)=\frac{P_n}{Q_n},\ f(n\beta)=\frac{P'_n}{Q_n},\ F(n\beta)=\frac{P''_n}{Q_n}. \]

10.

Suivant ce qu'on a vu, les fonctions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ s'expriment rationellement en $x$, $y$, $z$. Le cas contraire n'a pas lieu, car les équations (35.) sont en général d'un degré très élevé. Elles ont par cette raison un certain nombre de racines. Nous allons voir, comment on peut aisément exprimer toutes ces racines au moyen des fonctions $\Phi$, $f$, $F$.

A. Considérons d'abord l'équation $\Phi(n\beta)=\frac{P_n}{Q_n}$, ou $Q_n.\Phi(n\beta)=P_n$, et cherchons toutes les valeurs de $x$.

Il faut distinguer deux cas, selon que $n$ est pair ou impair:

1) Si $n$ est un nombre pair.

D'après ce qu'on a vu dans le paragraphe précédent (45.), on aura dans ce cas \[ \Phi(2n\beta)=\psi(x^{2}).x.y.z, \] c'est-à-dire, en vertu des formules \[ y=\sqrt{(1-c^{2}x^{2})},\ z=\sqrt{(1+e^{2}x^{2})}, \] \[ \Phi(2n\beta)=x\psi(x^{2}).\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}. \] Donc l'équation en $x$ deviendra, \[ \Phi^{2}(2n\beta)=x^{2}(\psi(x^{2}))^{2}.(1-c^{2}x^{2})(1+e^{2}x^{2}). \]

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En désignant le second nombre par $\theta(x^{2})$, on aura \[ \Phi^{2}(2n\beta)=\theta(x^{2}). \] $\Phi\beta$ étant une des valeurs de $x$, on aura \[ \Phi^{2}(2n\beta)=\theta(\Phi^{2}\beta), \tag{46.} \] équation, qui a lieu, quelle que soit la valeur de $\beta$. Pour trouver les autres valeurs de $x$, soit $x=\Phi\alpha$ une racine quelconque, on doit avoir \[ \Phi^{2}(2n\beta)=\theta(\Phi^{2}\alpha). \] Or, en mettant dans (46.) $\alpha$ au lieu de $\beta$, il viendra \[ \Phi^{2}(2n\alpha)=\theta(\Phi^{2}\alpha),\ \text{donc:} \] \[ \Phi^{2}(2n\beta)=\Phi^{2}(2n\alpha); \tag{47.} \] équation, qui revient à ces deux-ci: \[ \Phi(2n\alpha)=\Phi(2n\beta)\ \text{et}\ \Phi(2n\alpha)=-\Phi(2n\beta). \] La première donne en vertu de (31.) \[ 2n\alpha=2n\beta.(-1)^{m+\mu}+m\omega+\mu\varpi i, \] où $m$ et $\mu$ sont deux nombres entiers quelconques, positifs ou négatifs, zéro y compris.

La seconde donne les mêmes valeurs de $2m\alpha$Editorial note: Printed $2m\alpha$ instead of $2n\alpha$; an apparent misprint., mais de signe contraire, comme il est aisé de voir, en l'écrivant comme suit: \[ \Phi(-2n\alpha)=\Phi(2n\beta). \] Toute valeur de $2n\alpha$, qui satisfait à l'équation (47.), peut donc être représentée par \[ 2n\alpha=\pm(2n\beta.(-1)^{m+\mu}+m\omega+\mu\varpi i). \] De là on tire la valeur de $\alpha$, en divisant par $2n$, savoir: \[ \alpha=\pm\left((-1)^{m+\mu}.\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right). \] Ayant la valeur de $\alpha$, on aura \[ \Phi\alpha=\pm\Phi\left((-1)^{m+\mu}.\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right)=x. \tag{48.} \] Donc toutes les valeurs de $x$ sont contenues dans cette expression, et on les trouvera, en donnant aux nombres $m$ et $\mu$ toutes les valeurs entières depuis $-\infty$ jusqu'à $+\infty$. Or pour avoir toutes celles qui sont différentes entre elles, il suffit de donner à $m$ et $\mu$ des valeurs entières moindres que $2n$. En effet, quels que soient ces nombres, on peut toujours les supposer réduits a la forme: \[ m=2n.k+m',\ \mu=2n.k'+\mu', \] où $k$, $k'$ sont des nombres entiers, et $m'$, $\mu'$, des nombres entiers moindres que $2n$. En substituant ces valeurs dans l'expression de $x$, elle deviendra:

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\[ x=\pm\Phi\left\{(-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i+k\omega+k'\varpi i\right\}, \] or en vertu de (22.) cette expression se réduit à \[ x=\pm\Phi\left\{(-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i\right\}. \tag{49.} \] Cette valeur de $x$ est de la même forme que la précédente (48.), $m$ et $\mu$ seulement sont remplacés par $m'$ et $\mu'$, qui, tous les deux, sont positifs et moindres que $2n$; donc on obtiendra toutes les valeurs différentes de $x$, en donnant seulement à $m$ et $\mu$ toutes les valeurs entières depuis zéro jusqu'à $2n$ excl. Toutes ces valeurs sont nécessairement différentes entre elles. En effet, supposons par ex. qu'on ait \[ \begin{gathered} \pm\Phi\left\{(-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i\right\}\\ =\pm\Phi\left\{(-1)^{m+\mu}\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right\}, \end{gathered} \] il s'en suivrait, d'après (31.): \[ (-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i=\pm\left\{(-1)^{m+\mu}.\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right\}+k\omega+k'\varpi i, \] $k$ et $k'$ étant des entiers.

Cette équation donne: \[ \mu'=k'.2n\pm\mu,\ m'=k.2n\pm m,\ (-1)^{m'+\mu'}=\pm(-1)^{m+\mu}. \] Les deux premières équations ne peuvent pas subsister à moins que $k'=1$, $k=1$, $\mu'=2n-\mu$, $m'=2n-m$, et alors la dernière deviendra: \[ (-1)^{4n-m-\mu}=-(-1)^{m+\mu}, \] d'où l'on tire: \[ (-1)^{2m+2\mu}=-1, \] résultat absurde.

Donc toutes les valeurs de $x$, contenues dans la formule (48.), sont différentes entre elles, si $m$ et $\mu$ sont positifs et moindres que $m$Editorial note: Printed $m$ instead of $2n$; an apparent misprint..

Le nombre total des valeurs de $x$ est, comme il est aisé de voir, égal à $2(2n)^{2}=8n^{2}$, or l'équation $\Phi^{2}(2n\beta)=\theta(x^{2})$ ne peut pas avoir des racines égales, car dans ce cas on auroit $\frac{\partial.\theta(x^{2})}{\partial x}=0$, ce qui donneroit pour $x$ une valeur indépendante de $\beta$. Donc le degré de l'équation $\Phi^{2}(2n\beta)=\theta(x^{2})$ est égal au nombre des racines, c'est-à-dire à $8n^{2}$. Si par ex. $n=1$, on aura l'équation \[ \Phi^{2}(2\beta)=\theta(x^{2})=\frac{4x^{2}(1-c^{2}x^{2})(1+e^{2}x^{2})}{(1+e^{2}c^{2}x^{4})^{2}}, \] ou bien \[ (1+e^{2}c^{2}x^{4})^{2}.\Phi^{2}(2\beta)=4x^{2}(1-c^{2}x^{2})(1+e^{2}x^{2}), \]

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et d'après la formule (48.) les racines de cette équation, au nombre de huit, seront: \[ \begin{gathered} x=\pm\Phi\beta,\ x=\pm\Phi\left(-\beta+\frac{\omega}{2}\right),\\ x=\pm\Phi\left(-\beta+\frac{\varpi}{2}i\right),\ x=\pm\Phi\left(\beta+\frac{\omega}{2}+\frac{\varpi}{2}i\right). \end{gathered} \]

2) Si $n$ est un nombre impair $=2n+1$.

Dans ce cas $\frac{P_{2n+1}}{Q_{2n+1}}$ est, comme nous l'avons vu, une fonction rationelle de $x$, et par conséquent l'équation en $x$ sera: \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \tag{50.} \] Précisément comme dans le cas précédent, on trouvera, que toutes les racines de cette équation peuvent être représentées par \[ x=\Phi\left((-1)^{m+\mu}.\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right), \tag{51.} \] où il faut donner à $m$ et $\mu$ toutes les valeurs entières depuis $-n$ jusqu'à $+n$ incl. Donc le nombre des racines différentes est $(2n+1)^{2}$. C'est aussi le degré de l'équation en question. On peut aussi exprimer les racines par \[ x=(-1)^{m+\mu}.\Phi\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right). \] Si par ex. $n=1$, on aura une équation du degré $3^{2}=9$.

La formule (51.) donne pour $x$ les 9 valeurs suivantes: \[ \begin{gathered} \Phi(\beta);\\ \Phi\left(-\beta-\frac{\omega}{3}\right);\\ \Phi\left(-\beta+\frac{\omega}{3}\right);\\ \Phi\left(-\beta-\frac{\varpi}{3}i\right);\\ \Phi\left(-\beta+\frac{\varpi}{3}i\right);\\ \Phi\left(\beta-\frac{\omega}{3}-\frac{\varpi}{3}i\right);\\ \Phi\left(\beta-\frac{\omega}{3}+\frac{\varpi}{3}i\right);\\ \Phi\left(\beta+\frac{\omega}{3}-\frac{\varpi}{3}i\right);\\ \Phi\left(\beta+\frac{\omega}{3}+\frac{\varpi}{3}i\right); \end{gathered} \]

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B. Considérons maintenant l'équation \[ f(n\beta)=\frac{P'_n}{Q_n} \tag{52.} \] et cherchons les valeurs de $y$, qui satisfont à cette équation. La fonction $\frac{P'_n}{Q_n}$ étant, comme on a vu plus haut, rationelle en $y$: l'équation en $y$, en faisant $\frac{P'_n}{Q_n}=\psi(y)$, sera \[ f(n\beta)=\psi(y). \] Une des racines de cette équation est $y=f\beta$, donc, quelle que soit la leurEditorial note: Printed “leur” instead of “valeur”; the first syllable of the word is missing at the line break. de $\beta$: \[ f(n\beta)=\psi(f\beta). \tag{53.} \] Pour trouver les autres valeurs de $y$, soit $\alpha$ une nouvelle inconnue, telle que $y=f\alpha$, on aura \[ f(n\beta)=\psi(f\alpha); \] or, en vertu de (53.), le second membre est égal à $f(n\alpha)$, donc pour déterminer $\alpha$, on aura l'équation: \[ f(n\alpha)=f(n\beta). \] En vertu de (32.) cette équation donne pour expression générale de $n\alpha$: \[ n\alpha=\pm n\beta+2m\omega+\mu\varpi i, \] $m$ et $\mu$ étant deux nombres entiers positifs ou négatifs, zéro y compris.

De là on tire \[ \alpha=\pm\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i, \] et par conséquent: \[ f\alpha=f\left(\pm\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right)=y. \] C'est la valeur générale de $y$. Maintenant pour avoir les valeurs différentes de $y$, je dis, qu'il suffit de prendre $\beta$ avec le signe $+$ et de donner à $m$ et $\mu$ toutes les valeurs entières, moindres que $n$. En effet, comme on a $f(+\alpha)=f(-\alpha)$, on aura d'abord: \[ f\left(-\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right)=f\left(\beta-\frac{2m}{n}\omega-\frac{\mu}{n}\varpi i\right). \] Donc on peut toujours dans l'expression de $y$ prendre $\beta$ avec le signe $+$. Ainsi toutes les valeurs de $y$ sont contenues dans l'expression \[ y=f\left(\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right). \tag{54.} \]

Maintenant quels que soient les nombres $m$ et $\mu$, on peut toujours supposer, \[ m=k.n+m',\ \mu'=k'.n+\mu', \] Editorial note: Printed $\mu'$ on the left of the second equation instead of $\mu$; an apparent misprint.

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où $k$, $k'$, $m'$, $\mu'$ sont des nombres entiers, les deux derniers étant en même temps positifs et moindres que $n$.

En substituant, il viendra \[ y=f\left(\beta+\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i+2k\omega+k'\varpi i\right). \] Or, en vertu de (22.) le second membre de cette équation est égal à \[ f\left(\beta+\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i\right)=y, \tag{55.} \] quantité de la même forme, que le second membre de (54.); seulement $m'$ et $\mu'$ sont positifs et moindres que $n$. Donc etc.

En donnant à $m$ et $\mu$ toutes les valeurs possibles, moindres que $n$, on trouvera un nombre $n^{2}$ de valeurs de $y$. Or, en général toutes ces quantités sont différentes entre elles. En effet, supposons par ex. \[ f\left(\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right)=f\left(\beta+\frac{2\mu'}{n}+\frac{\mu'}{n}\varpi i\right), \] Editorial note: In the right member the print has $2\mu'/n$ where $2m'\omega/n$ is expected; the $\omega$ is absent. An apparent misprint. on aura en vertu de (32.), en désignant par $k$, $k'$ deux nombres entiers: \[ \beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i=\pm\left(\beta+\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i\right)+2k\omega+k'\varpi i. \] Puisque $\beta$ peut avoir une valeur irrationelle quelconque, il est clair que cette équation ne peut pas subsister à moins qu'on ne préfère dans le second membre le signe supérieur. Alors il viendra \[ \frac{2m}{n}\omega+\frac{\mu}{n}\varpi i=\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i+2k\omega+k'\varpi i, \] d'où l'on tire, en égalant les parties réelles et les parties imaginaires: \[ m=m'+kn;\ \mu=\mu'+k'n, \] équations absurdes, en remarquant que les nombres $m$, $m'$, $\mu$ et $\mu'$ sont tous positifs et inférieurs à $n$. Donc en général l'équation \[ f(n\beta)=\psi(y) \] a un nombre $n^{2}$ de racines différentes entre elles et non un plus grand nombre. Or généralement toutes les racines de cette équation sont différentes entres elles. En effet, si deux d'entre elles étoient égales, on auroit à la fois: \[ f(n\beta)=\psi(y)\ \text{et}\ 0=\psi'(y), \] et cela est impossible, si l'on remarque que les coéfficiens de $y$ dans $\psi(y)$ ne contiennent pas $\beta$. Donc généralement l'équation (52.) est nécessairement du degré $n^{2}$.

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C. L'équation \[ F(n\beta)=\frac{P''_n}{Q_n}, \tag{56.} \] étant traitée absolument de la même manière par rapport à $z$, que l'équation $f(n\beta)=\frac{P'_n}{Q_n}$ l'a été par rapport à $y$, donne pour expression générale des valeurs de $z$: \[ z=F\left(\beta+\frac{m}{n}\omega+\frac{2\mu}{n}\varpi i\right), \tag{57.} \] où $m$ et $\mu$ sont entiers, positifs et moindres que $n$. Le nombre des valeurs de $z$ est $n^{2}$, et elles sont en général toutes différentes entre elles.

Donc généralement l'équation (56.) est du degré $n^{2}$.

11.

Nous avons trouvé ci-dessus toutes les racines des équations \[ \varphi(n\beta)=\frac{P_n}{Q_n},\ f(n\beta)=\frac{P'_n}{Q_n},\ F(n\beta)=\frac{P''_n}{Q_n}, \] racines, qui sont exprimées par les formules (48.), (51.), (54.), (57.). Toutes ces racines sont différentes entre elles, excepté les cas de valeurs particulières de $\beta$; mais pour ces valeurs, les racines différentes sont contenues dans les mêmes formules. — Dans ce dernier cas un certain nombre des valeurs des quantités $x$, $y$, $z$ seront égales; mais il est clair, que toutes les valeurs égales ou inégales seront néanmoins les racines des équations, dont il s'agit. Cela se fait voir en faisant converger $\beta$ vers une valeur particulière, qui donne pour $x$, ou $y$, ou $z$ des valeurs égales.

En faisant dans la formule (48.) $\beta=\frac{\alpha}{2n}$, on aura l'équation \[ \varphi^{2}\alpha=\frac{P_{2n}^{2}}{Q_{2n}^{2}},\ \text{dont les racines sont}\ x=\pm\varphi\left((-1)^{m+\mu}\frac{\alpha}{2n}+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right); \tag{58.} \] et où $m$ et $\mu$ ont toutes les valeurs entières et positives moindres que $2n$.

En faisant de même dans la formule (50.) $\beta=\frac{\alpha}{2n+1}$, on aura \[ \Phi\alpha=\frac{P_{2n+1}}{Q_{2n+1}},\ \text{dont les racines sont} \] \[ x=(-1)^{m+\mu}.\Phi\left(\frac{\alpha}{2n+1}+\frac{m\omega+\mu\varpi i}{2n+1}\right), \tag{59.} \] $m$ et $\mu$ ayant pour valeurs tous les nombres entiers depuis $-n$ jusqu'à $+n$.

Enfin en faisant dans (52.), (56.) $\beta=\frac{\alpha}{n}$, on aura l'équation \[ f\alpha=\frac{P'_n}{Q_n},\ \text{dont les racines sont} \] \[ y=f\left(\frac{\alpha}{n}+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right), \tag{60.} \]

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et l'équation \[ F\alpha=\frac{P''_n}{Q_n},\ \text{dont les racines sont} \] \[ z=F\left(\frac{\alpha}{n}+\frac{m}{n}\omega+\frac{2\mu}{n}\varpi i\right), \tag{61.} \] où $m$ et $\mu$ sont renfermés entre les limites 0 et $n-1$ incl. Si $n$ est impair $=2n+1$, on peut aussi supposer \[ \begin{gathered} y=(-1)^{m}.f\left(\frac{\alpha}{2n+1}+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right),\\ z=(-1)^{\mu}.F\left(\frac{\alpha}{2n+1}+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right), \end{gathered} \] $m$ et $\mu$ ayant toutes les valeurs entières de $-n$ à $+n$.

Dans toutes ces équations la quantité $\alpha$ peut avoir une valeur quelconque.

Comme cas particuliers on doit remarquer les suivans:

1) En faisant dans (58.) et (59.) $\alpha=0$, on aura les équations \[ \left\{\begin{aligned} &P_{2n}^{2}=0,\ \text{dont les racines sont}\ x=\pm\Phi\left(\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right)\\ &\text{(les limites de $m$ et $\mu$ étant 0 et $2n-1$),}\\ &P_{2n+1}=0,\ \text{dont les racines sont}\ x=\Phi\left(\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right)\\ &\text{(les limites de $m$ et $\mu$ étant $-n$ et $+n$).} \end{aligned}\right. \tag{62.} \]

2) En faisant dans (60.) $\alpha=\frac{\omega}{2}$ et dans (61.) $\alpha=\frac{\varpi}{2}i$, et remarquant, que $f\left(\frac{\omega}{2}\right)=0$, $F\left(\frac{\varpi}{2}i\right)=0$, on obtiendra les deux équations: \[ P'_n=0,\ \text{dont les racines sont}\ y=f\left((2m+\tfrac{1}{2})\frac{\omega}{n}+\frac{\mu}{n}\varpi i\right) \tag{63.} \] \[ P''_n=0,\ \text{dont les racines sont}\ z=F\left(\frac{m}{n}\omega+(2\mu+\tfrac{1}{2})\frac{\varpi i}{n}\right) \tag{64.} \] (les limites de $m$ et $\mu$ étant 0 et $n-1$).

3) En faisant dans (58.) $\alpha=\frac{\omega}{2}+\frac{\varpi}{2}i$, et en remarquant, que $\varphi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}$, on aura l'équation \[ Q_{2n}^{2}=0, \] dont les racines seront: \[ x=\pm\Phi\left((m+\tfrac{1}{2}(-1)^{m+\mu})\frac{\omega}{2n}+(\mu+\tfrac{1}{2}(-1)^{m+\mu}).\frac{\varpi i}{2n}\right). \]

Les valeurs de $x$, doivent être égales par paires, et l'on verra aisement, que les valeurs inégales peuvent être représentées par \[ x=\Phi\left((m+\tfrac{1}{2})\frac{\omega}{2n}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n}\right), \tag{65.} \]

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en donnant à $m$ et $\mu$ toutes les valeurs entières depuis 0 à $2n-1$. Donc ce sont les racines de l'équation \[ Q_{2n}=0,\ \text{par rapport à } x. \]

En faisant de même dans (59.) $\alpha=\frac{\omega}{2}+\frac{\varpi}{2}i$, on aura l'équation \[ Q_{2n+1}=0, \] dont les racines seront: \[ \left\{\begin{aligned} x&=(-1)^{m+\mu}.\varphi\left((m+\tfrac{1}{2})\frac{\omega}{2n+1}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n+1}\right),\\ y&=(-1)^{m}.f\left((m+\tfrac{1}{2})\frac{\omega}{2n+1}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n+1}\right),\\ z&=(-1)^{\mu}.F\left((m+\tfrac{1}{2})\frac{\omega}{2n+1}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n+1}\right), \end{aligned}\right. \tag{66.} \] $m$ et $\mu$ ayant pour valeurs tous les nombres entiers de $-n$ à $+n$.

Parmi les valeurs de $x$, $y$, $z$, il faut remarquer celle, qui répond à $m=n$, $\mu=n$. Alors on a \[ \begin{aligned} x&=(-1)^{n}.\varphi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\\ y&=(-1)^{n}.f\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\\ z&=(-1)^{n}.F\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}. \end{aligned} \]

Ces valeurs infinies font voir que l'équation $Q_{2n+1}=0$ est d'un degré moindre d'une unité que celui des équations, dont elle sort. En écartant ces valeurs, les restantes, en nombre de $(2n+1)^{2}-1$, seront les racines de l'équation $Q_{2n+1}=0$.

§. IV.

Résolution algébrique des équations \[ \Phi\alpha=\frac{P_{2n+1}}{Q_{2n+1}},\quad f\alpha=\frac{P'_{2n+1}}{Q_{2n+1}},\quad F\alpha=\frac{P''_{2n+1}}{Q_{2n+1}}. \]

12.

Nous avons vu dans le §. précédent, comment on peut exprimer aisément les racines des équations en question au moyen des fonctions $\Phi$, $f$, $F$. Nous allons maintenant en déduire la résolution de ces mêmes équations, ou la determination des fonctions $\varphi\left(\frac{\alpha}{n}\right)$, $f\left(\frac{\alpha}{n}\right)$, $F\left(\frac{\alpha}{n}\right)$, en fonctions de $\varphi\alpha$, $f\alpha$, $F\alpha$.

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Comme on a \[ \varphi\left(\frac{\alpha}{m\mu}\right)=\varphi\left(\frac{1}{m}\left(\frac{\alpha}{\mu}\right)\right), \] on peut supposer que $n$ est un nombre premier. D'abord nous considérerons le cas, où $n=2$, et ensuite celui, où $n$ est un nombre impair.

A. Expressions des fonctions $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$.

13.

Les valeurs de $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$ peuvent être trouvées très facilement de la manière suivante. En supposant dans les formules (35.) $\beta=\frac{\alpha}{2}$, et faisant \[ x=\varphi\left(\frac{\alpha}{2}\right),\ y=f\left(\frac{\alpha}{2}\right),\ z=F\left(\frac{\alpha}{2}\right), \] il viendra: \[ f(\alpha)=\frac{y^{2}-c^{2}x^{2}z^{2}}{1+e^{2}c^{2}x^{4}},\ F(\alpha)=\frac{z^{2}+e^{2}y^{2}x^{2}}{1+e^{2}c^{2}x^{4}}, \] ou bien, en substituant les valeurs de $y^{2}$ et $z^{2}$ en $x^{2}$: \[ f(\alpha)=\frac{1-2c^{2}x^{2}-c^{2}e^{2}x^{4}}{1+e^{2}c^{2}x^{4}},\ F(\alpha)=\frac{1+2e^{2}x^{2}-e^{2}c^{2}x^{4}}{1+e^{2}c^{2}x^{4}}. \] Ces équations donnent \[ \begin{aligned} 1+f\alpha&=\frac{2(1-c^{2}x^{2})}{1+e^{2}c^{2}x^{4}},\ 1-f\alpha=\frac{2c^{2}x^{2}(1+e^{2}x^{2})}{1+e^{2}c^{2}x^{4}},\\ F\alpha-1&=\frac{2e^{2}x^{2}(1-c^{2}x^{2})}{1+e^{2}c^{2}x^{4}},\ F\alpha+1=\frac{2(1+e^{2}x^{2})}{1+e^{2}c^{2}x^{4}}, \end{aligned} \] Editorial note: A blot resembling a 2 is printed after the minus sign in the numerator of the first fraction; the sense requires $2(1-c^{2}x^{2})$, as in the later fractions. d'où \[ \frac{F\alpha-1}{1+f\alpha}=e^{2}.x^{2},\ \frac{1-f\alpha}{F\alpha+1}=c^{2}x^{2}, \] et de là, en remarquant, que $y^{2}=1-c^{2}x^{2}$, $z^{2}=1+e^{2}x^{2}$: \[ z^{2}=\frac{F\alpha+f\alpha}{1+f\alpha},\ y^{2}=\frac{F\alpha+f\alpha}{1+F\alpha}. \]

De ces équations on tire, en extrayant la racine carrée, et remplaçant $x$, $y$, $z$ par leurs valeurs $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$: \[ \left\{\begin{aligned} \varphi\left(\frac{\alpha}{2}\right)&=\frac{1}{c}\sqrt{\left(\frac{1-f\alpha}{1+F\alpha}\right)}=\frac{1}{e}\sqrt{\left(\frac{F\alpha-1}{f\alpha+1}\right)},\\ f\left(\frac{\alpha}{2}\right)&=\sqrt{\left(\frac{F\alpha+f\alpha}{1+F\alpha}\right)},\ F\left(\frac{\alpha}{2}\right)=\sqrt{\left(\frac{F\alpha+f\alpha}{1+f\alpha}\right)}. \end{aligned}\right. \tag{67.} \] Telles sont les formes les plus simples qu'on peut donner aux valeurs des fonctions $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$. De cette manière on peut exprimer

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algébriquement $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$ en $f\alpha$, $F\alpha$. De la même manière $\varphi\left(\frac{\alpha}{4}\right)$, $f\left(\frac{\alpha}{4}\right)$, $F\left(\frac{\alpha}{4}\right)$ s'exprimeront en $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$, et ainsi de suite. Donc en général les fonctions $\varphi\left(\frac{\alpha}{2^{n}}\right)$, $f\left(\frac{\alpha}{2^{n}}\right)$, $F\left(\frac{\alpha}{2^{n}}\right)$ peuvent être exprimées au moyen d'extractions de racines carrées en fonctions des trois quantités $\varphi\alpha$, $f\alpha$, $F\alpha$.

Pour appliquer les formules trouvées ci-dessus pour la bissection à un exemple, supposons $\alpha=\frac{\omega}{2}$.

Alors on aura $f\left(\frac{\omega}{2}\right)=0$, $F\left(\frac{\omega}{2}\right)=\frac{\sqrt{(e^{2}+c^{2})}}{c}$, donc en substituant: \[ \varphi\left(\frac{\omega}{4}\right)=\frac{1}{c}\sqrt{\left(\frac{1}{1+\frac{1}{c}\sqrt{(e^{2}+c^{2})}}\right)}=\frac{1}{e}.\sqrt{\left(\frac{1}{c}\sqrt{(e^{2}+c^{2})}-1\right)}, \] \[ f\left(\frac{\omega}{4}\right)=\sqrt{\left(\frac{\frac{1}{c}\sqrt{(e^{2}+c^{2})}}{1+\frac{1}{c}\sqrt{(e^{2}+c^{2})}}\right)}, \] \[ F\left(\frac{\omega}{4}\right)=\sqrt{\left(\frac{1}{c}\sqrt{(e^{2}+c^{2})}\right)}, \] ou bien \[ \begin{aligned} \varphi\left(\frac{\omega}{4}\right)&=\frac{1}{\sqrt{(c^{2}+c\sqrt{(e^{2}+c^{2})})}}=\frac{\sqrt{(\sqrt{(e^{2}+c^{2})}-c)}}{e\sqrt{c}},\\ f\left(\frac{\omega}{4}\right)&=\frac{\sqrt[4]{(e^{2}+c^{2})}}{\sqrt{(c+\sqrt{(e^{2}+c^{2})})}}=\frac{1}{e}.\sqrt{(e^{2}+c^{2}-c\sqrt{(e^{2}+c^{2})})}\\ F\left(\frac{\omega}{4}\right)&=\sqrt[4]{\left(1+\frac{e^{2}}{c^{2}}\right)}=\sqrt{\left[F\left(\frac{\omega}{2}\right)\right]}. \end{aligned} \]

B. Expressions des fonctions $\varphi\left(\frac{\alpha}{2n+1}\right)$, $f\left(\frac{\alpha}{2n+1}\right)$, $F\left(\frac{\alpha}{2n+1}\right)$, en fonctions algébriques des quantités $\varphi\alpha$, $f\alpha$, $F\alpha$.

14.

Pour trouver les valeurs de $\varphi\left(\frac{\alpha}{2n+1}\right)$, $f\left(\frac{\alpha}{2n+1}\right)$, $F\left(\frac{\alpha}{2n+1}\right)$ en $\Phi\alpha$, $f\alpha$, $F\alpha$, il faut résoudre les équations \[ \Phi\alpha=\frac{P_{2n+1}}{Q_{2n+1}},\ f\alpha=\frac{P'_{2n+1}}{Q_{2n+1}},\ F\alpha=\frac{P''_{2n+1}}{Q_{2n+1}}, \] qui toutes sont du degré $(2n+1)^{2}$. Nous allons voir, qu'il est toujours possible, d'effectuer algébriquement cette résolution.

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Soient \[ \varphi_{1}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\varphi\left(\beta+\frac{2m\omega}{2n+1}\right) \tag{68.} \] et \[ \psi\beta=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\ \psi_{1}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}\varphi_{1}\left(\beta-\frac{2\mu\varpi i}{2n+1}\right), \tag{69.} \] où $\theta$ est une racine imaginaire quelconque de l'équation $\theta^{2n+1}-1=0$. Cela posé, je dis que les deux quantités \[ \psi\beta.\psi_{1}\beta\ \text{et}\ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}, \] pourront être exprimées rationellement en $\varphi(2n+1)\beta$.

D'abord en écrivant $\varphi_{1}\beta$ comme suit: \[ \begin{aligned} \varphi_{1}\beta&=\varphi\beta.\overset{n}{\underset{1}{\Sigma}}.\left\{\varphi\left(\beta+\frac{2m\omega}{2n+1}\right)+\varphi\left(\beta-\frac{2m\omega}{2n+1}\right)\right\}\\ &=\varphi\beta.\overset{n}{\underset{1}{\Sigma}}(-1)^{m}.\frac{2\varphi\beta.f\left(\frac{2m\omega}{2n+1}\right).F\left(\frac{2m\omega}{2n+1}\right)}{1+e^{2}c^{2}\varphi^{2}\left(\frac{2m\omega}{2n+1}\right).\varphi^{2}\beta}, \end{aligned} \] Editorial note: Printed with a point after $\varphi\beta$ before the sum in both lines, where the sense requires a plus sign; the first line shows a blot over the point and a further point after the sum sign. on voit, que $\varphi_{1}\beta$ peut s'exprimer rationellement en $\varphi\beta$. Soit donc $\varphi_{1}\beta=\chi(\varphi\beta)$, on a de même: \[ \begin{aligned} \varphi_{1}\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)&=\chi\left(\varphi\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)\right)\\ &=\chi\left(\frac{\varphi\beta.f\left(\frac{2\mu\varpi i}{2n+1}\right).F\left(\frac{2\mu\varpi i}{2n+1}\right)\pm\varphi\left(\frac{2\mu\varpi i}{2n+1}\right).f\beta.F\beta}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{2\mu\varpi i}{2n+1}\right).\varphi^{2}\beta}\right), \end{aligned} \] ou bien, en faisant $\Phi\beta=x$: \[ f\left(\frac{2\mu\varpi i}{2n+1}\right).F\left(\frac{2\mu\varpi i}{2n+1}\right)=a,\ \varphi\left(\frac{2\mu\varpi i}{2n+1}\right)=b, \] et en substituant pour $f\beta$ et $F\beta$ leurs valeurs $\sqrt{(1-c^{2}x^{2})}$ et $\sqrt{(1+e^{2}x^{2})}$: \[ \varphi_{1}\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)=\chi\left(\frac{ax\pm b.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}{1+e^{2}c^{2}b^{2}.x^{2}}\right); \] or, $\chi$ désignant une fonction rationelle, le second membre de cette équation peut se mettre sous la forme \[ R_{\mu}\pm R'_{\mu}.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}, \] où $R_{\mu}$ et $R'_{\mu}$ sont des fonctions rationelles de $x$.

Donc on a \[ \varphi_{1}\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)=R_{\mu}\pm R'_{\mu}.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}. \] En substituant dans les expressions de $\psi\beta$ et $\psi_{1}\beta$, il viendra:

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\[ \left\{\begin{aligned} \psi\beta&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R_{\mu}+\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}.\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R'_{\mu},\\ \psi_{1}\beta&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R_{\mu}-\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}.\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R'_{\mu}. \end{aligned}\right. \tag{70.} \] Editorial note: The plus sign inside the parentheses before $\theta$ is printed overstruck with a short horizontal bar throughout (70.); kept here as a plain plus.

Maintenant $R_{\mu}$ et $R'_{\mu}$ étant des fonctions rationelles de $x$, les quantités $\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R_{\mu}$ et $\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R'_{\mu}$ le sont également. En élevant donc $\psi\beta$ et $\psi_{1}\beta$ à la $(2n+1)^{\text{ième}}$ puissance: les deux quantités $(\psi\beta)^{2n+1}$ et $(\psi_{1}\beta)^{2n+1}$ pourront se mettre sous la forme: \[ \begin{aligned} (\psi\beta)^{2n+1}&=t+t'.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]},\\ (\psi_{1}\beta)^{2n+1}&=t-t'.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}, \end{aligned} \] $t$ et $t'$ étant des fonctions rationelles de $x$. En prenant la somme des valeurs de $(\psi\beta)^{2n+1}$ et $(\psi_{1}\beta)^{2n+1}$, on aura \[ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}=2t. \]

Donc la quantité $(\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}$ peut être exprimée rationellement en $x$. Il en est de même du produit $\psi\beta.\psi_{1}\beta$, comme on voit par les équations (70.).

Donc on peut faire \[ \left\{\begin{aligned} \psi\beta.\psi_{1}\beta&=\lambda(x),\\ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}&=\lambda_{1}(x), \end{aligned}\right. \tag{71.} \] $\lambda(x)$ et $\lambda_{1}(x)$ désignant des fonctions rationelles de $x$. Maintenant ces fonctions ont la propriété, de ne pas changer de valeur, lorsqu'on met à la place de $x$ une autre racine quelconque de l'équation \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \]

Considérons d'abord la fonction $\lambda(x)$. En remettant la valeur de $x=\Phi\beta$, on aura \[ \psi\beta.\psi_{1}\beta=\lambda(\Phi\beta), \] d'où l'on tire, en mettant $\beta+\frac{2k\omega}{2n+1}+\frac{2k'\varpi i}{2n+1}$ au lieu de $\beta$: \[ \lambda\left(\varphi\left(\beta+\frac{2k\omega}{2n+1}+\frac{2k'\varpi i}{2n+1}\right)\right)=\psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right).\psi_{1}\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right). \] Cela posé, en remarquant, que: \[ \overset{+n}{\underset{-n}{\Sigma}}\psi(m+k)=\overset{+n}{\underset{-n}{\Sigma}}_{m}\psi(m)+\overset{k}{\underset{1}{\Sigma}}_{m}.(\psi(m+n)-\psi(m-n-1)), \tag{72.} \] on aura, en faisant dans l'expression de $\psi\beta$, $\beta=\beta+\frac{2k\omega}{2n+1}$:

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\[ \begin{aligned} \varphi_{1}\left(\beta+\frac{2k\omega}{2n+1}\right)&=\overset{+n}{\underset{-n}{\Sigma}}_{m}\varphi\left(\beta+\frac{2(k+m)}{2n+1}\omega\right)\\ &=\varphi_{1}\beta+\overset{k}{\underset{1}{\Sigma}}_{m}\left\{\varphi\left(\beta+\frac{2(m+n)\omega}{2n+1}\right)-\varphi\left(\beta+\frac{2(m-n-1)\omega}{2n+1}\right)\right\}, \end{aligned} \] or: $\varphi\left(\beta+\frac{2(m-n-1)}{2n+1}\omega\right)=\varphi\left(\beta+\frac{2(m+n)}{2n+1}\omega-2\omega\right)=\varphi\left(\beta+\frac{2(m+n)}{2n+1}\omega\right)$, donc: \[ \varphi_{1}\left(\beta+\frac{2k\omega}{2n+1}\right)=\varphi_{1}\beta. \tag{73.} \]

En mettant dans l'expression de $\psi\beta$, $\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}$ au lieu de $\beta$, on trouvera \[ \psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right), \] or en vertu de (73.) on a: \[ \varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}\right), \] donc \[ \psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}\right). \] En vertu de (72.) on a \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}\right)\\ &=\theta^{-k'}.\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right)+\overset{k'}{\underset{1}{\Sigma}}_{\mu}.\theta^{n+\mu-k'}.\varphi_{1}\left(\beta+\frac{2(\mu+n)\varpi i}{2n+1}\right)\\ &\quad-\overset{k'}{\underset{1}{\Sigma}}_{\mu}\theta^{\mu-n-1-k'}.\varphi_{1}\left(\beta+\frac{2(\mu-n-1)\varpi i}{2n+1}\right), \end{aligned} \] donc, en remarquant que $\theta^{n+\mu-k'}=\theta^{\mu-n-1-k'}$ et \[ \varphi_{1}\left(\beta+\frac{2(\mu-n-1)\varpi i}{2n+1}\right)=\varphi_{1}\left(\beta+\frac{2(\mu+n)\varpi i}{2n+1}-2\varpi i\right)=\varphi_{1}\left(\beta+\frac{2(\mu+n)\varpi i}{2n+1}\right), \] il viendra \[ \psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\theta^{-k'}.\psi\beta. \tag{74.} \] De la même manière on trouvera aussi \[ \psi_{1}\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\theta^{k'}.\psi_{1}\beta. \] Ces deux équations donneront \[ \begin{aligned} &\psi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right).\psi_{1}\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)=\psi\beta.\psi_{1}\beta,\\ &\left\{\psi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right\}^{2n+1}+\left\{\psi_{1}\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right\}^{2n+1}=(\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}. \end{aligned} \]

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En vertu de ces équations on obtiendra, en mettant dans les valeurs de $\lambda(\Phi\beta)$ et $\lambda_{1}(\Phi\beta)$, $\beta+\frac{2k\omega+2k'\varpi i}{2n+1}$ au lieu de $\beta$: \[ \begin{aligned} \lambda(\Phi\beta)&=\lambda\left[\Phi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right],\\ \lambda_{1}(\Phi\beta)&=\lambda_{1}\left[\Phi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right]. \end{aligned} \] Or, $\Phi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)$ exprime une racine quelconque de l'équation \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \]

Donc comme nous avons dit, les fonctions $\lambda(x)$ et $\lambda_{1}(x)$ auront les mêmes valeurs, quelle que soit la racine, qu'on met à la place de $x$.

Soient donc $x_{0}$, $x_{1}$, $x_{2}....x_{2n+1}$ ces racines, on aura \[ \begin{aligned} \lambda(x)&=\frac{1}{2n+1}\{\lambda(x_{0})+\lambda(x_{1})+....+\lambda(x_{2n})\},\\ \lambda_{1}(x)&=\frac{1}{2n+1}\{\lambda_{1}(x_{0})+\lambda_{1}(x_{1})+....+\lambda_{1}(x_{2n})\}. \end{aligned} \] Editorial note: The list of roots is printed as ending with $x_{2n+1}$, while the sums run only to $x_{2n}$. Or le second membre de ces équations est une fonction rationelle et symétrique des racines de l'équation $\Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}$, donc $\lambda(x)$ et $\lambda_{1}(x)$ pourront s'exprimer rationellement en $\Phi(2n+1)\beta$. En faisant \[ \lambda(x)=B,\ \lambda_{1}(x)=2A, \] les équations (71.) donneront \[ (\psi\beta)^{2n+1}.(\psi_{1}\beta)^{2n+1}=B^{2n+1};\ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}=2A, \] d'où l'on tire \[ \psi\beta=\sqrt[2n+1]{[A+\sqrt{(A^{2}-B^{2n+1})}]}=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right). \tag{75.} \]

15.

Ayant trouvé la valeur de $\psi\beta$, on en déduira facilement celles de $\Phi_{1}\beta$.

En effet, en prenant pour $\theta$ successivement toutes les racines imaginaires de l'équation $\theta^{2n+1}-1=0$, et désignant les valeurs correspondantes de $A$ et $B$ par $A_{1}$, $B_{1}$, $A_{2}$, $B_{2}$ etc., on obtiendra: \[ \begin{aligned} \sqrt[2n+1]{[A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}]}&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta_{1}^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\\ \sqrt[2n+1]{[A_{2}+\sqrt{(A_{2}^{2}-B_{2}^{2n+1})}]}&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta_{2}^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\\ \cdots\cdots\cdots&\cdots\cdots\cdots\\ \sqrt[2n+1]{[A_{2n}+\sqrt{(A_{2n}^{2}-B_{2n}^{2n+1})}]}&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta_{2n}^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right). \end{aligned} \]

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De même on connoît la somme des racines: \[ \overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\varphi\left(\beta+\frac{2m\omega}{2n+1}+\frac{2\mu\varpi i}{2n+1}\right)=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right), \] qui est égale à $(2n+1)\varphi(2n+1)\beta$, comme nous le verrons dans la suite. En ajoutant ces équations membre à membre, après avoir multiplié la première par $\theta_{0}^{-k}$, la seconde par $\theta_{1}^{-k}$, la troisième par $\theta_{2}^{-k}....$ et la $(2n)^{\text{ième}}$ par $\theta_{2n}^{-k}$, il viendra: \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Sigma}}_{\mu}.(\theta_{0}^{\mu-k}+\theta_{1}^{\mu-k}+\theta_{2}^{\mu-k}+....+\theta_{2n}^{\mu-k}+1).\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right)\\ &=(2n+1).\varphi(2n+1)\beta+\overset{2n}{\underset{1}{\Sigma}}_{\mu}\theta_{\mu}^{-k}.\sqrt[2n+1]{[A_{\mu}+\sqrt{(A_{\mu}^{2}-B_{\mu}^{2n+1})}]}; \end{aligned} \] Editorial note: In the first sum the term $\theta_{0}^{\mu-k}$ and the final $1$ are both printed, so 1 is counted twice; the next sum is printed without $\theta_{0}^{\mu-k}$. or la somme \[ 1+\theta_{1}^{\mu-k}+\theta_{2}^{\mu-k}+....+\theta_{2n}^{\mu-k} \] se réduit à zéro pour toutes les valeurs de $k$, excepté pour $k=\mu$. Dans ce cas elle devient égale à $2n+1$. Donc le premier membre de l'équation précédente devient \[ (2n+1).\varphi_{1}\left(\beta+\frac{2k\varpi i}{2n+1}\right), \] donc, en substituant et divisant par $(2n+1)$, on a: \[ \begin{aligned} \varphi_{1}\left(\beta+\frac{2k\varpi i}{2n+1}\right)=&\\ \varphi(2n+1)\beta+\frac{1}{2n+1}.&\left\{\theta_{1}^{-k}.\sqrt[2n+1]{[A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}]}+\theta_{2}^{-k}\sqrt[2n+1]{[A_{2}+\sqrt{(A_{2}^{2}-B_{2}^{2n+1})}]}+....\right.\\ &\left.....+\theta_{2n}^{-k}\sqrt[2n+1]{[A_{2n}+\sqrt{(A_{2n}^{2}-B_{2n}^{2n+1})}]}\right\}. \end{aligned} \tag{76.} \] Pour $k=0$, on a: \[ \begin{aligned} \varphi_{1}\beta=\varphi(2n+1)\beta+\frac{1}{2n+1}&\left\{\sqrt[2n+1]{[A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}]}+\sqrt[2n+1]{[A_{2}+\sqrt{(A_{2}^{2}-B_{2}^{2n+1})}]}+\ldots\right.\\ &\left.....+\sqrt[2n+1]{[A_{2n}+\sqrt{(A_{2n}^{2}-B_{2n}^{2n+1})}]}\right\}. \end{aligned} \tag{77.} \]

16.

Ayant ainsi trouvé la valeur de $\varphi_{1}\beta$, il s'agit, d'en tirer celle de $\varphi\beta$. Or cela peut se faire aisément comme suit:

Soit \[ \psi_{2}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}.\varphi\left(\beta+\frac{2m\omega}{2n+1}\right);\ \psi_{3}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}.\varphi\left(\beta-\frac{2m\omega}{2n+1}\right), \tag{78.} \] on a \[ \varphi\left(\beta\pm\frac{2m\omega}{2n+1}\right)=\frac{\varphi\beta.f\left(\frac{2m\omega}{2n+1}\right).F\left(\frac{2m\omega}{2n+1}\right)\pm f\beta.F\beta.\varphi\left(\frac{2m\omega}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{2m\omega}{2n+1}\right).\varphi^{2}\beta}. \]

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De là suit, qu'on peut faire \[ \psi_{2}\beta=r+f\beta.F\beta.s;\ \psi_{3}\beta=r-f\beta.F\beta.s, \] où $r$ et $s$ sont des fonctions rationelles de $\Phi\beta$.

De là on tire \[ \left\{\begin{aligned} \psi_{2}\beta.\psi_{3}\beta&=\chi(\Phi\beta),\\ (\psi_{2}\beta)^{2n+1}+(\psi_{3}\beta)^{2n+1}&=\chi_{1}(\Phi\beta), \end{aligned}\right. \tag{79.} \] $\chi(\Phi\beta)$ et $\chi_{1}(\Phi\beta)$ étant deux fonctions rationelles de $\Phi\beta$.

Cela posé je dis, que $\chi(\Phi\beta)$ et $\chi_{1}(\Phi\beta)$ pourront s'exprimer rationellement en $\Phi_{1}\beta$.

On a vu, que \[ \varphi_{1}\beta=\varphi\beta+\overset{n}{\underset{1}{\Sigma}}_{m}.\left\{\frac{2\varphi\beta.f\left(\frac{2m\omega}{2n+1}\right).F\left(\frac{2m\omega}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{2m\omega}{2n+1}\right)\varphi^{2}\beta}\right\}. \tag{80.} \] En faisant $\Phi\beta=x$, on aura une équation en $x$ du degré $(2n+1)$. Une racine de cette équation est $x=\Phi\beta$; or, en mettant $\beta+\frac{2k\omega}{2n+1}$ au lieu de $\beta$, $\Phi_{1}\beta$ ne change pas de valeur, donc $x=\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)$ sera une racine, quel que soit le nombre entier $k$. Or, en donnant à $k$ toutes les valeurs entières de $-n$ jusqu'á $+n$, $\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)$ prendra $2n+1$ valeurs différentes, donc ces $2n+1$ quantités seront précisément les $2n+1$ racines de l'équation en $x$.

Cela posé, en mettant $\beta+\frac{2k\omega}{2n+1}$ au lieu de $\beta$ dans l'expression de $\psi_{2}\beta$, il viendra en vertu de (72.): \[ \begin{aligned} \psi_{2}\left(\beta+\frac{2k\omega}{2n+1}\right)&=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}\Phi\left(\beta+\frac{2(k+m)\omega}{2n+1}\right)\\ &=\theta^{-k}.\psi_{2}3+\overset{k}{\underset{1}{\Sigma}}_{m}\theta^{m+n-k}.\Phi\left(\beta+\frac{2(m+n)\omega}{2n+1}\right)\\ &\quad-\overset{k}{\underset{1}{\Sigma}}_{m}\theta^{m-n-1-k}.\Phi\left(\beta+\frac{2(m-n-1)\omega}{2n+1}\right), \end{aligned} \] Editorial note: Printed $\psi_{2}3$ with a digit three instead of $\psi_{2}\beta$; an apparent misprint. donc en ayant attention, que $\theta^{m+n-k}=\theta^{m-n-1-k}$ et $\Phi\left(\beta+\frac{2(m-n-1)}{2n+1}\omega\right)=\Phi\left(\beta+\frac{2(m+n)\omega}{2n+1}\right)$, il en résultera \[ \psi_{2}\left(\beta+\frac{2k\omega}{2n+1}\right)=\theta^{-k}.\psi_{2}\beta. \tag{81.} \]

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De même on aura \[ \psi_{3}\left(\beta+\frac{2k\omega}{2n+1}\right)=\theta^{+k}.\psi_{3}\beta. \] On voit en vertu de ces relations, que les équations, qui donnent les valeurs des fonctions $\chi(\Phi\beta)$ et $\chi_{1}(\Phi\beta)$Editorial note: The second printed argument looks like a damaged or misprinted $\beta$., conduisent à ces deux égalités: \[ \begin{aligned} \chi_{1}\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right]&=\chi_{1}(\Phi\beta),\\ \chi\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right]&=\chi(\Phi\beta). \end{aligned} \] De là on tire \[ \begin{aligned} \chi(\Phi\beta)&=\frac{1}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{k}.\chi\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right],\\ \chi_{1}(\Phi\beta)&=\frac{1}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{k}.\chi_{1}\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right]. \end{aligned} \]

Or, ces valeurs de $\chi(\Phi\beta)$ et $\chi_{1}(\Phi\beta)$ sont des fonctions rationelles et symétriques de toutes les racines de l'équation (80.). Donc elles peuvent être exprimées rationellement par les coefficiens de la même équation, c'est-à-dire rationellement en $\Phi_{1}\beta$.

Soit \[ \chi_{1}(\Phi\beta)=2C,\ \chi(\Phi\beta)=D, \] Editorial note: The subscripts of both printed $\chi$ are damaged and hard to read; they are read here as $\chi_1$ for $2C$ and $\chi$ for $D$, as (79.) requires. les équations (79.) donneront \[ \psi_{2}\beta=\sqrt[2n+1]{[C+\sqrt{(C^{2}-D^{2n+1})}]}, \] d'où, en remettant la valeur de $\psi_{2}\beta$: \[ \sqrt[2n+1]{[C+\sqrt{(C^{2}-D^{2n+1})}]}=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}.\Phi\left(\beta+\frac{2m\omega}{2n+1}\right). \tag{82.} \] De là on tire, en mettant $\theta_{\mu}$ au lieu de $\theta$, et désignant les valeurs correspondantes de $C$ et $D$ par $C_{\mu}$ et $D_{\mu}$: \[ \theta_{\mu}^{-k}\sqrt[2n+1]{[C_{\mu}+\sqrt{(C_{\mu}^{2}-D_{\mu}^{2n+1})}]}=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta_{\mu}^{m-k}.\Phi\left(\beta+\frac{2m\omega}{2n+1}\right). \] En y joignant l'équation \[ \Phi_{1}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\Phi\left(\beta+\frac{2m\omega}{2n+1}\right), \] on en tirera facilement: \[ (2n+1).\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)=\Phi_{1}\beta+\overset{2n}{\underset{1}{\Sigma}}_{\mu}\theta_{\mu}^{-k}.\sqrt[2n+1]{[C_{\mu}+\sqrt{(C_{\mu}^{2}-D_{\mu}^{2n+1})}]}. \tag{83.} \] En supposant $k=0$, il viendra: \[ \Phi\beta=\frac{1}{2n+1}\left\{\Phi_{1}\beta+\sqrt[2n+1]{[C_{1}+\sqrt{(C_{1}^{2}-D_{1}^{2n+1})}]}+\ldots+\sqrt[2n+1]{[C_{2n}+\sqrt{(C_{2n}^{2}-D_{2n}^{2n+1})}]}\right\}. \tag{84.} \]

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Cette équation donne $\Phi\beta$ en fonction algébrique de $\Phi_{1}\beta$, or précédemment nous avons trouvé $\Phi_{1}\beta$ en fonction algébrique de $\Phi(2n+1)\beta$. Donc en mettant $\frac{\alpha}{2n+1}$ au lieu de $\beta$, on aura $\Phi\left(\frac{\alpha}{2n+1}\right)$ en fonction algébrique de $\Phi\alpha$.

Par une analyse toute semblable on trouvera $f\left(\frac{\alpha}{2n+1}\right)$ en $f\alpha$ et $F\left(\frac{\alpha}{2n+1}\right)$ en $F\alpha$.

17.

Les valeurs, que nous venons de trouver des quantités $\Phi_{1}\beta$ et $\Phi\beta$, la première en $\Phi(2n+1)\beta$ et la seconde en $\Phi_{1}\beta$, contiennent chacune la somme de $2n$ radicaux différentes du $(2n+1)$ degré. Il en résultera pour $\Phi\beta$, $\Phi_{1}\beta....$ un nombre $(2n+1)^{2n}$ de valeurs, tandisque chacune de ces quantités est la racine d'une équation du $(2n+1)^{\text{ème}}$ degré. Or on peut donner aux expressions de $\Phi\beta$ et $\Phi_{1}\beta$ une telle forme, que le nombre des valeurs de ces quantités soit précisément égal à $2n+1$.

Pour cela soit \[ \theta=\cos\frac{2\pi}{2n+1}+i\sin\frac{2\pi}{2n+1}, \] on peut faire \[ \theta_{1}=\theta,\ \theta_{2}=\theta^{2},\ \theta_{3}=\theta^{3}....\theta_{2n}=\theta^{2n}. \]

Soient de même \[ \left\{\begin{aligned} \psi^{k}(\beta)&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{k\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\\ \psi_{1}^{k}(\beta)&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{k\mu}.\Phi_{1}\left(\beta-\frac{2\mu\varpi i}{2n+1}\right), \end{aligned}\right. \tag{85.} \] on aura en vertu de (74.) \[ \begin{aligned} \psi^{k}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{-k\nu}.\psi^{k}(\beta),\\ \psi_{1}^{k}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{+k\nu}.\psi_{1}^{k}(\beta),\\ \psi^{1}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{-\nu}.\psi^{1}(\beta),\\ \psi_{1}^{1}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{+\nu}.\psi_{1}^{1}(\beta). \end{aligned} \]

Soit maintenant \[ \left\{\begin{aligned} \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k}}+\frac{\psi_{1}^{k}(\beta)}{(\psi_{1}^{1}\beta)^{k}}&=P(\Phi\beta),\\ \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k-2n-1}}+\frac{\psi_{1}^{k}(\beta)}{\psi_{1}^{1}(\beta)^{k-2n-1}}&=Q(\Phi\beta), \end{aligned}\right. \tag{86.} \]

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$P(\Phi\beta)$ et $Q(\Phi\beta)$ seront des fonctions rationelles de $\Phi\beta$; or, en mettant $\beta+\frac{2m\omega+2\mu\varpi i}{2n+1}$ au lieu de $\beta$, il est clair en vertu des formules, que $P$ et $Q$ ne changent pas de valeurs; donc on aura \[ P(\Phi\beta)=\frac{1}{(2n+1)^{2}}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}.P\left[\Phi\left(\beta+\frac{2m\omega+2\mu\varpi i}{2n+1}\right)\right]; \] or, le second membre étant une fonction symétrique et rationelle des racines de l'équation $\Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}$, $P(\Phi\beta)$ pourra s'exprimer rationellement en $\Phi(2n+1)\beta$. Il en est de même de $Q(\Phi\beta)$. Connoissant ces deux quantités, les équations (86.) donneront \[ \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k}}\left\{1-\left(\frac{\psi_{1}^{1}\beta}{\psi_{1}^{1}\beta}\right)^{2n+1}\right\}=P(\Phi\beta)-\frac{Q(\varphi\beta)}{(\psi_{1}^{1}\beta)^{2n+1}}, \] or \[ \begin{aligned} \psi_{1}^{1}\beta&=A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})},\\ \psi_{1}^{1}\beta&=A_{1}-\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}, \end{aligned} \] Editorial note: In the first of these two equations the subscript of the printed $\psi$ is damaged; it should read $\psi^{1}\beta$ without subscript, and the quotient in the preceding display should likewise have $\psi^{1}\beta$ in the numerator. donc \[ \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k}}.2\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}=Q(\Phi\beta)-[A_{1}-\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}].P(\Phi\beta). \] Donc on aura \[ \psi^{k}(\beta)=(\psi^{1}\beta)^{k}.\{F_{k}+H_{k}.\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}\}, \] où $F_{k}$ et $H_{k}$ sont des fonctions rationelles de $\Phi(2n+1)\beta$. En remplaçant $A_{1}$ et $B_{1}$ et substituant les valeurs de $\psi^{k}(\beta)$ et $(\psi^{1}\beta)^{k}$, il viendra: \[ \sqrt[2n+1]{[A_{k}+\sqrt{(A_{k}^{2}-B_{k}^{2n+1})}]}=[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{k}{2n+1}}[F_{k}+H_{k}.\sqrt{(A^{2}-B^{2n+1})}], \] donc la valeur de $\Phi_{1}\beta$ deviendra: \[ \begin{aligned} \Phi_{1}\beta=\Phi(2n+1)\beta+\frac{1}{2n+1}.&\left\{[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{1}{2n+1}}\right.\\ &+[F_{2}+H_{2}\sqrt{(A^{2}-B_{2n}^{2n+1})}]\,[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{2}{2n+1}}\\ &\left.+....[F_{2n}+H_{2n}\sqrt{(A^{2}-B^{2n+1})}]\,[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{2n}{2n+1}}\right\}. \end{aligned} \tag{87.} \] Editorial note: In the second line the printed $B$ under the radical carries the subscript $2n$, which the first and third lines lack; an apparent misprint. Par un procédé tout semblable on trouvera \[ \begin{aligned} \Phi\beta=\frac{1}{2n+1}&\left\{\Phi_{1}\beta+[C+\sqrt{(C^{2}-D^{2n+1})}]^{\frac{1}{2n+1}}\right.\\ &+[K_{2}+L_{2}\sqrt{(C^{2}-D^{2n+1})}]\,[C+\sqrt{(C^{2}-D^{2n+1})}]^{\frac{2}{2n+1}}\\ &\left.+....[K_{2n}+L_{2n}\sqrt{(C^{2}-D^{2n+1})}]\,[C+\sqrt{(C^{2}-D^{2n+1})}]^{\frac{2n}{2n+1}}\right\}, \end{aligned} \tag{88.} \] où $K_{2}$, $L_{2}$, $K_{3}$, $L_{3}....K_{2n}$, $L_{2n}$ sont des fonctions rationelles de $\Phi_{1}\beta$.

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Ces expressions de $\Phi_{1}\beta$ et $\Phi\beta$ n'ont que $2n+1$ valeurs différentes, qu'on obtiendra en attribuant aux radicaux leurs $2n+1$ valeurs. — Il suit de notre analyse, qu'on peut prendre $\sqrt{(A^{2}-B^{2n+1})}$ et $\sqrt{(C^{2}-D^{2n+1})}$ avec tel signe qu'on voudra.

18.

La valeur, que nous avons trouvée pour $\Phi(\beta)$ ou $\Phi\left(\frac{\alpha}{2n+1}\right)$ contient, encore outre la fonction $\Phi\alpha$, les suivantes: \[ \begin{gathered} e,\ c,\ \theta,\\ \Phi\left(\frac{m\omega}{2n+1}\right),\ \Phi\left(\frac{m\varpi i}{2n+1}\right),\ f\left(\frac{m\omega}{2n+1}\right),\\ f\left(\frac{m\varpi i}{2n+1}\right),\ F\left(\frac{m\omega}{2n+1}\right),\ F\left(\frac{m\varpi i}{2n+1}\right), \end{gathered} \] pour des valeurs quelconques de $m$ depuis $1$ jusqu'à $2n$. Maintenant quelle que soit la valeur de $m$, on peut toujours exprimer algébriquement $\Phi\left(\frac{m\omega}{2n+1}\right)$, $f\left(\frac{m\omega}{2n+1}\right)$, $F\left(\frac{m\omega}{2n+1}\right)$ en $\Phi\left(\frac{\omega}{2n+1}\right)$, et $\Phi\left(\frac{\mu\varpi i}{2n+1}\right)$, $f\left(\frac{\mu\varpi i}{2n+1}\right)$, $F\left(\frac{\mu\varpi i}{2n+1}\right)$ en $\Phi\left(\frac{\varpi i}{2n+1}\right)$. Tout est donc connu dans l'expression de $\Phi\left(\frac{\alpha}{2n+1}\right)$, excepté les deux quantités indépendantes de $\alpha$; $\Phi\left(\frac{\omega}{2n+1}\right)$, $\Phi\left(\frac{\varpi i}{2n+1}\right)$. — Ces quantités dépendent seulement de $c$ et $e$, et elles peuvent être trouvées par la résolution d'une équation du degré $(2n+1)^{2}-1$, savoir de l'équation $\frac{P_{2n+1}}{x}=0$. — Nous allons voir dans le paragraphe suivant, comment on peut en ramener la résolution à celle d'équations moins élevées.

§. V.

C. Sur l'équation $P_{2n+1}=0$.

19.

L'expression, que nous venons de trouver pour $\Phi\left(\frac{\alpha}{2n+1}\right)$ contiendra, comme nous avons vu, les deux quantités constantes $\Phi\left(\frac{\omega}{2n+1}\right)$ et $\Phi\left(\frac{\varpi i}{2n+1}\right)$. On trouvera ces quantités en résolvant l'équation \[ P_{2n+1}=0, \] dont les racines seront représentées par \[ x=\Phi\left(\frac{m\omega+\mu\varpi i}{2n+1}\right), \]

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où $m$ et $\mu$ pourront être tous les nombres entiers depuis $-n$ jusqu'à $+n$. Une de ces racines, qui répond à $m=0$, $\mu=0$, est égale à zéro. Donc $P_{2n+1}$ est divisible par $x$. En écartant ce facteur, on aura une équation \[ R=0,\ \text{du degré }(2n+1)^{2}-1. \] En faisant $x^{2}=r$, l'équation (90.) $R=0$, en $r$, sera du degré $\frac{(2n+1)^{2}-1}{2}=2.n(n+1)$, et les racines de cette équation seront \[ r=\Phi^{2}\left(\frac{m\omega\pm\mu\varpi i}{2n+1}\right), \tag{91.} \] $\mu$ et $m$ ayant toutes les valeurs positives, au dessous de $n$, en faisant abstraction de la racine zéro.

Nous allons voir maintenant, comment on peut ramener la résolution de l'équation $R=0$ à celle de deux équations, l'une du degré $n$ et l'autre du degré $2n+2$.

D'abord, je dis, qu'on peut représenter toutes les valeurs de $r$ par \[ \Phi^{2}\left(\frac{m\omega}{2n+1}\right)\ \text{et}\ \Phi^{2}\left(m\frac{\mu\omega+\varpi i}{2n+1}\right)\ldots\ldots, \tag{92.} \] en donnant à $\mu$ toutes les valeurs entières depuis zéro jusqu'à $2n$, et à $m$ toutes celles depuis $1$ jusqu'à $n$.

En effet $\Phi^{2}\left(\frac{m\omega}{2n+1}\right)$ représente d'abord un nombre $n$ de valeurs de $r$; or les autres peuvent être représentées par $\Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)$. Soit, pour le démontrer, $m\mu=(2n+1)k+m'$, où $m'$ est un nombre entier compris entre les limites $-n$ et $+n$. En substituant, on aura \[ \begin{aligned} \Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)&=\Phi^{2}\left(k.\omega+\frac{m'\omega+\mu\varpi i}{2n+1}\right)\\ &=\Phi^{2}\left(\frac{m'\omega+\mu\varpi i}{2n+1}\right)=\Phi^{2}\left(\frac{-m'\omega-\mu\varpi i}{2n+1}\right). \end{aligned} \] $\Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)$ est donc une valeur de $r$; maintenant à chaque valeur de $m$, répond une valeur différente de $m'$. Car si l'on avoit \[ m_{1}\mu=(2n+1)k_{1}+m', \] il s'en suivrait \[ (m_{1}-m)\mu=(2n+1).(k-k_{1}), \] ce qui est impossible, en remarquant que $2n+1$ est un nombre premier. Donc $\Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)$ combiné avec $\Phi^{2}\left(\frac{m\omega}{2n+1}\right)$ représente toutes les valeurs de $r$.

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Cela posé, soit \[ \begin{aligned} &\left[r-\Phi^{2}\left(\frac{\omega}{2n+1}\right)\right]\left[r-\Phi^{2}\left(\frac{2\omega}{2n+1}\right)\right]\ldots\left[r-\Phi^{2}\left(\frac{n\omega}{2n+1}\right)\right]\\ &=r^{n}+p_{n-1}.r^{n-1}+p_{n-2}.r^{n-2}+\ldots\ldots+p_{1}.r+p_{0}. \end{aligned} \tag{93.} \] Les quantités $p_{0}$, $p_{1}$, $\ldots p_{n-1}$, seront des fonctions rationelles et symétriques de $\Phi^{2}\left(\frac{\omega}{2n+1}\right)$, $\Phi^{2}\left(\frac{2\omega}{2n+1}\right)\ldots\Phi^{2}\left(\frac{n\omega}{2n+1}\right)$; or ces fonctions peuvent être trouvées au moyen d'une équation de degré $2n+2$.

Soit $p$ une fonction rationelle et symétrique quelconque de $\Phi^{2}\left(\frac{\omega'}{2n+1}\right)$, $\Phi^{2}\left(\frac{2\omega'}{2n+1}\right)\ldots\Phi^{2}\left(\frac{n\omega'}{2n+1}\right)$, où $\omega'$ désigne la quantité $m\omega+\mu\varpi i$.

Par les formules que nous avons données plus haut pour exprimer $\Phi(n\beta)$ en $\Phi\beta$, il est clair, qu'on peut exprimer $\Phi^{2}\left(m'.\frac{\omega'}{2n+1}\right)$ en fonction rationelle de $\Phi^{2}\left(\frac{\omega'}{2n+1}\right)$. Donc on peut faire \[ p=\psi.\left[\Phi^{2}\left(\frac{\omega'}{2n+1}\right)\right]=\theta\left[\Phi^{2}\left(\frac{\omega'}{2n+1}\right),\ \Phi^{2}\left(\frac{2\omega'}{2n+1}\right),\ \ldots\ \Phi^{2}\left(\frac{n\omega'}{2n+1}\right)\right], \tag{94.} \] $\theta$ désignant une fonction symétrique et rationelle. En mettant $\nu\omega'$ au lieu de $\omega'$, il viendra \[ \psi\left[\Phi^{2}\left(\frac{\nu\omega'}{2n+1}\right)\right]=\theta\left[\Phi^{2}\left(\frac{\nu\omega'}{2n+1}\right),\ \Phi^{2}\left(\frac{2\nu\omega'}{2n+1}\right),\ \ldots\ \Phi^{2}\left(\frac{n\nu\omega'}{2n+1}\right)\right], \tag{95.} \] or, en faisant: \[ a.\nu=(2n+1).k_{a}'+k_{a}, \] où $k_{a}$ est entier et compris entre $-n$ et $+n$, la serie \[ k_{1},\ k_{2}\ \ldots\ldots\ k_{n} \] aura au signe près les mêmes termes que celle-ci: \[ 1,\ 2,\ 3\ \ldots\ldots\ n; \] donc il est clair, que le second membre de l'équation (95.) aura la même valeur que $p$. Donc: \[ \psi\left[\Phi^{2}\left(\frac{\nu\omega'}{2n+1}\right)\right]=\psi\left[\Phi^{2}\left(\frac{\omega'}{2n+1}\right)\right], \tag{96.} \] Editorial note: The exponent of the second Phi in (96.) is damaged in the print and appears as a raised dot; read as the exponent 2 of the first member. équation, qui en faisant $\omega'=\omega$ et $\omega'=m\omega+\varpi i$, donnera ces deux-ci: \[ \left\{\begin{aligned} &\psi\left[\Phi^{2}\left(\frac{\nu\omega}{2n+1}\right)\right]=\psi.\left[\Phi^{2}\left(\frac{\omega}{2n+1}\right)\right],\\ &\psi\left[\Phi^{2}\left(\nu.\frac{m\omega+\varpi i}{2n+1}\right)\right]=\psi\left[\Phi^{2}\left(\frac{m\omega+\varpi i}{2n+1}\right)\right], \end{aligned}\right. \tag{97.} \] ou bien, en faisant, pour abréger, \[ \Phi^{2}\left(\frac{\nu\omega}{2n+1}\right)=r_{\nu},\ \Phi^{2}\left(\nu.\frac{m\omega+\varpi i}{2n+1}\right)=r_{\nu,m}, \tag{98.} \]

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il viendra: \[ \psi.r_{\nu}=\psi r_{1};\ \psi r_{\nu,m}=\psi r_{1,m}. \tag{99.} \]

Cela posé, soit \[ \left\{\begin{aligned} &(p-\psi r_{1})(p-\psi r_{1,0})(p-\psi r_{1,1})(p-\psi r_{1,2})\ldots\ (p-\psi r_{1,2n})\\ &=q_{0}+q_{1}.p+q_{2}.p^{2}+\ldots q_{2n+1}.p^{2n+1}+p^{2n}. \end{aligned}\right. \tag{100.} \] Editorial note: The exponent of the last term is partly obscured by a smudge and reads $2n$; the degree $2n+2$ is expected from (104.).

Je dis, qu'on peut exprimer les coefficiens $q_{0}$, $q_{1}$ etc. rationellement en $e$ et $c$.

D'abord en vertu des formules connues on peut exprimer rationellement ces coefficiens en $t_{1}$, $t_{2}\ldots t_{2n}$, si l'on fait, pour abréger, \[ t_{k}=(\psi r_{1})^{k}+(\psi r_{1,0})^{k}+(\psi r_{1,1})^{k}+\ldots+(\psi r_{1,2n})^{k}. \tag{101.} \]

Il s'agit donc de trouver les quantités $t_{1}$, $t_{2}\ldots$, or cela se pourra aisément au moyen des relations (99.). En effet, en y faisant successivement $\nu=1$, $2\ldots n$, après avoir élevé les deux membres à la $k^{\text{ème}}$ puissance, on en tirera sur le champ: \[ \left\{\begin{aligned} (\psi r_{1})^{k}&=\frac{1}{n}\left[(\psi r_{1})^{k}+(\psi r_{2})^{k}+\ldots+(\psi r_{n})^{k}\right],\\ (\psi r_{1,m})^{k}&=\frac{1}{n}\left[(\psi r_{1,m})^{k}+(\psi r_{2,m})^{k}+\ldots+(\psi r_{n,m})^{k}\right]. \end{aligned}\right. \tag{102.} \] Donc en mettant pour $m$ tous les nombres entiers $0$, $1$, $\ldots$ $2n$, et ensuite substituant dans l'expression de $t_{k}$, il viendra: \[ \left\{\begin{aligned} n.t_{k}&=(\psi r_{1})^{k}+(\psi r_{2})^{k}+\ldots\ldots+(\psi r_{n})^{k}\\ &+(\psi r_{1,0})^{k}+(\psi r_{2,0})^{k}+\ldots\ldots+(\psi r_{n,0})^{k}\\ &+(\psi r_{1,1})^{k}+(\psi r_{2,1})^{k}+\ldots\ldots+(\psi r_{n,1})^{k}\\ &+\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &+(\psi r_{1,2n})^{k}+(\psi r_{2,2n})^{k}+\ldots\ldots+(\psi r_{n,2n})^{k}. \end{aligned}\right. \tag{103.} \]

Cette valeur de $t_{k}$ est, comme on voit, une fonction rationelle et symétrique des $2n+2$ quantités $r_{1}$, $r_{2}$, $\ldots r_{n}$; $r_{1,0}$, $r_{2,0}$, $\ldots r_{n,0}\ldots$ $r_{1,2n}$, $r_{2,2n}\ldots r_{n,2n}$, qui sont les $2n+2$ racines de l'équation $R=0$. Donc, comme on sait, $t_{k}$ pourra s'exprimer rationellement par les coefficiens de cette équation, et par suite en fonction rationelle de $e$ et $c$. Ayant ainsi trouvé les quantités $t_{k}$, on en tire les valeurs de $q_{0}$, $q_{1}$, $\ldots$ $\ldots q_{2n+1}$; qui seront également des fonctions rationelles de $e$ et $c$.

20.

Cela posé, en supposant \[ 0=q_{0}+q_{1}.p+q_{2}.p^{2}+\ldots\ldots+q_{2n+1}.p^{2n+1}+p^{2n+2}, \tag{104.} \] on aura une équation du $(2n+2)^{\text{ème}}$ degré, dont les racines seront \[ \psi r_{1};\ \psi r_{1,0};\ \psi r_{1,1};\ \psi r_{1,2}\ \ldots\ldots\ \psi r_{1,2n}. \]

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La fonction $\psi r_{1}$, c'est-à-dire, une fonction quelconque rationelle et symétrique des racines $r_{1}$, $r_{2}$, $r_{3}$, $\ldots\ldots r_{n}$ pourra donc être trouvée au moyen d'une équation du degré $2n+2$.

Donc on aura de cette manière les coefficiens $p_{0}$, $p_{1}$, $\ldots\ldots p_{n-1}$, en résolvant un nombre $n$ d'équations, chacune du $(2n+2)^{\text{ème}}$ degré.

Ayant déterminé $p_{0}$, $p_{1}$, $\ldots\ldots$, on aura, en résolvant l'équation \[ 0=p_{0}+p_{1}r+\ldots\ldots+p_{n-1}.r^{n-1}+r^{n}, \tag{105.} \] la valeur des quantités \[ r_{1},\ r_{2},\ \ldots\ldots r_{n};\ r_{1,0},\ r_{2,0},\ \ldots\ldots r_{n,0};\ r_{1,1},\ r_{2,1},\ \ldots\ldots r_{n,1}\ \text{etc. etc.,} \] dont la première est égale à $\Phi^{2}\left(\frac{\omega}{2n+1}\right)$. Donc la détermination de cette quantité, ou bien la résolution de l'équation $R=0$, qui est du degré $(2n+2).n$, est réduite à celle d'équations du degré $(2n+2)$ et $n$.

Mais on peut encore simplifier le procédé précédent. En effet, comme nous le verrons, pour avoir les quantités $p_{0}$, $p_{1}$, $\ldots\ldots$, il suffit de connoître l'une quelconque d'entre elles, et alors on peut exprimer les autres rationellement par celle-là.

Soient généralement $p$, $q$ deux fonctions rationelles et symétriques des quantités $r_{1}$, $r_{2}$, $\ldots\ldots r_{n}$, on peut faire, comme nous l'avons vu: \[ p=\psi r_{1};\ q=\theta r_{1}, \] $\psi r_{1}$ et $\theta r_{1}$ désignent deux fonctions rationelles de $r_{1}$, qui ont cette propriété: de rester les mêmes, si l'on change $r_{1}$ en une autre quelconque des quantités $r_{1}$, $r_{2}$, $\ldots\ldots r_{n}$.

Supposons maintenant: \[ s_{k}=(\psi r_{1})^{k}.\theta r_{1}+(\psi r_{1,0})^{k}.\theta r_{1,0}+(\psi r_{1,1})^{k}.\theta r_{1,1}+\ldots\ldots+(\psi r_{1,2n})^{k}.\theta r_{1,2n}, \] je dis que $s_{k}$ pourra être exprimé rationellement en $e$ et $c$.

En effet, on a \[ (\psi r_{1})^{k}.\theta r_{1}=(\psi r_{\nu})^{k}.\theta r_{\nu}=\frac{1}{n}\left\{(\psi r_{1})^{k}.\theta r_{1}+(\psi r_{2})^{k}.\theta r_{2}+\ldots\ldots+(\psi r_{n})^{k}.\theta r_{n}\right\}, \] \[ (\psi r_{1,m})^{k}.\theta r_{1,m}=(\psi r_{\nu,m})^{k}.\theta r_{\nu,m}=\frac{1}{n}\left\{(\psi r_{1,m})^{k}.\theta r_{1,m}+(\psi r_{2,m})^{k}.\theta r_{2,m}+\ldots+(\psi r_{n,m})^{k}.\theta r_{n,m}\right\}. \] En faisant $m=0$, $1$, $2$, $\ldots\ldots 2n$, et substituant dans l'expression de $s_{k}$, on verra, que $s_{k}$ sera une fonction rationelle et symétrique des racines $r_{0}$, $r_{1}$, $\ldots\ldots r_{1,0}$ etc., etc. de l'équation $R=0$; donc $s_{k}$ pourra s'exprimer rationellement en $e$ et $c$.

Connoissant $s_{k}$, on obtiendra, en faisant $k=0$, $1$, $2$, $\ldots\ldots 2n$, $2n+1$ équations, desquelles on tirera aisément la valeur de $\theta r_{1}$ en fonction

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rationelle de $\psi_{1}r$.Editorial note: Printed $\psi_{1}r$ with the subscript on the psi; $\psi r_{1}$ is expected, as on the previous page. Donc, une fonction de la forme $p$ étant donnée, on peut exprimer une autre fonction quelconque de la même forme en fonction rationelle de $p$. Donc, comme nous l'avons dit, on peut exprimer les coefficiens $p_{0}$, $p_{1}$, $\ldots\ldots p_{n-1}$ rationellement par l'un quelconque d'entre eux. Donc enfin, pour en avoir les valeurs, il suffit de résoudre une seule équation du degré $2n+2$, et par conséquent, pour avoir les racines de l'équation $R=0$, il suffit de résoudre une équation du degré $2n+2$, et $2n+2$ équations du degré $n$.

21.

Maintenant, parmi les équations, dont dépend la détermination des quantités $\varphi\left(\frac{\omega}{2m+1}\right)$; $\varphi\left(\frac{\varpi i}{2n+1}\right)$,Editorial note: Printed $2m+1$ in the first denominator; $2n+1$ is apparently expected. celles du degré $n$ peuvent être résolues algébriquement. Le procédé, par lequel nous allons effectuer cette résolution est entièrement semblable à celui, qui est dû à M. Gauss pour la résolution de l'équation \[ \theta^{2n+1}-1=0. \] Soit proposée l'équation \[ 0=p_{1}+p_{1}.r+p_{2}.r^{2}+\ldots\ldots+p_{n-1}.r^{n-1}+r^{n}, \tag{106.} \] Editorial note: The first coefficient is printed $p_{1}$; $p_{0}$ is expected. dont les racines sont: \[ \Phi^{2}\left(\frac{\omega'}{2n+1}\right);\ \Phi^{2}\left(\frac{2\omega'}{2n+1}\right);\ \ldots\ldots\ \Phi^{2}\left(\frac{n\omega'}{2n+1}\right), \] où $\omega'$ a une des valeurs de $\omega$, $m\omega+\varpi i$. Désignons par $\alpha$ une des racines primitives du nombre $2n+1$, c'est-à-dire, un nombre entier tel, que $\mu=2n$$\,+1$Editorial note: The characters after $2n$ are overstruck by a heavy bar and hard to read; $\mu=2n$ is expected. est le nombre le plus petit, qui rende $\alpha^{\mu}-1$ divisible par $2n+1$, je dis, que les racines de l'équation (106.) peuvent aussi être représentées par \[ \Phi^{2}(\varepsilon),\ \Phi^{2}(\alpha\varepsilon),\ \Phi^{2}(\alpha^{2}\varepsilon),\ \Phi^{2}(\alpha^{3}\varepsilon)\ldots\ \Phi^{2}(\alpha^{n-1}.\varepsilon), \tag{107.} \] où $\varepsilon=\frac{\omega'}{2n+1}$.

Soit \[ \alpha^{m}=(2n+1)k_{m}\pm a_{m}, \] où $k$ est entier et $a_{m}$ entier, positif et moindre que $n+1$, je dis, que les termes de la série \[ 1,\ a_{1},\ a_{2},\ \ldots\ldots\ a_{n-1} \] seront tous différens entre eux.

En effet, si l'on a \[ a_{m}=a_{\mu}, \] il en résulte,

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\[ \begin{aligned} \text{ou }\ \alpha^{m}-\alpha^{\mu}&=(2n+1)(k_{m}-k_{\mu}),\\ \text{ou }\ \alpha^{m}+\alpha^{\mu}&=(2n+1)(k_{m}+k_{\mu}). \end{aligned} \] Il faut donc que l'une des quantités $\alpha^{m}-\alpha^{\mu}$, $\alpha^{m}+\alpha^{\mu}$ soit divisible par $2n+1$; or soit posé $m>\mu$, ce qui est permis, il faut que $\alpha^{m-\mu}-1$ ou $\alpha^{m-\mu}+1$ soit divisible par $n$,Editorial note: Printed $n$ here; $2n+1$ is apparently expected. or cela est impossible, car $m-\mu$ est moindre que $n$.

Donc les quantités $1$, $a_{1}$, $a_{2}\ldots a_{n-1}$ sont différentes entre elles et par conséquent elles coincident, mais dans un ordre différent, avec les nombres $1$, $2$, $3$, $4\ldots n$.

Donc, en remarquant que \[ \varphi^{2}\{[(2n+1)k_{m}\pm a_{m}]\varepsilon\}=\varphi^{2}(a_{m}\varepsilon), \] on voit que les quantités (107.) sont les mêmes que celles-ci: \[ \Phi^{2}(\varepsilon),\ \Phi^{2}(2\varepsilon)\ \ldots\ \Phi^{2}(n\varepsilon), \] c'est-à-dire les racines de l'équation (106.) c. q. f. d.

Il y a encore à remarquer, qu'ayant \[ \alpha^{n}=(2n+1)k_{n}-1, \] on aura \[ \alpha^{n+m}=(2n+1)k_{n}\alpha^{m}-\alpha^{m}, \] donc \[ a_{n+m}=\pm a_{m} \] et \[ \Phi^{2}(\alpha^{n+m}\varepsilon)=\Phi^{2}(\alpha^{m}\varepsilon). \] Cela posé, soit $\theta$ une racine imaginaire quelconque de l'équation \[ \theta^{n}-1=0 \] et \[ \psi(\varepsilon)=\Phi^{2}(\varepsilon)+\Phi^{2}(\alpha\varepsilon)\theta+\Phi^{2}(\alpha^{2}\varepsilon)\theta^{2}+\ldots+\Phi^{2}(\alpha^{n-1}\varepsilon)\theta^{n-1}. \tag{108.} \] En vertu de ce que nous avons vu précédemment, le second membre de cette équation peut être transformé en une fonction rationelle de $\Phi^{2}(\varepsilon)$. Faisons \[ \psi(\varepsilon)=\chi[\Phi^{2}(\varepsilon)]. \tag{109.} \] En mettant dans la première expression de $\psi(\varepsilon)$, $\alpha^{m}\varepsilon$ au lieu de $\varepsilon$, il viendra: \[ \begin{aligned} \psi(\alpha^{m}\varepsilon)&=\varphi^{2}(\alpha^{m}\varepsilon)+\varphi^{2}(\alpha^{m+1}\varepsilon).\theta+\varphi^{2}(\alpha^{m+2}\varepsilon).\theta^{2}+\ldots\\ &\ldots+\Phi^{2}(\alpha^{n-1}\varepsilon).\theta^{n-m-1}+\Phi^{2}(\alpha^{n}\varepsilon).\theta^{n-m}+\ldots+\Phi^{2}(\alpha^{n+m-1}\varepsilon)\theta^{n-1}, \end{aligned} \] mais nous avons vu, que $\Phi^{2}(\alpha^{n+m}\varepsilon)=\Phi^{2}(\alpha^{m}\varepsilon)$; donc: \[ \begin{aligned} \psi(\alpha^{m}\varepsilon)&=\theta^{n-m}.\Phi^{2}(\varepsilon)+\theta^{n-m+1}.\Phi^{2}(\alpha\varepsilon)+\theta^{n-m+2}.\Phi^{2}(\alpha^{2}\varepsilon)+\ldots\\ &\ldots+\theta^{n-1}.\Phi^{2}(\alpha^{m-1}\varepsilon)+\Phi^{2}(\alpha^{m}\varepsilon)+\theta.\Phi^{2}(\alpha^{m+1}\varepsilon)+\ldots+\theta^{n-m-1}.\Phi^{2}(\alpha^{n-1}\varepsilon). \end{aligned} \]

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En multipliant par $\theta^{m}$, le second membre deviendra égal a $\psi(\varepsilon)$, donc: \[ \psi(\alpha^{m}\varepsilon)=\theta^{-m}.\psi(\varepsilon), \tag{110.} \] ou bien: \[ \psi(\varepsilon)=\theta^{m}.\chi[\Phi^{2}(\alpha^{m}\varepsilon)], \] d'où, en élevant les deux à la $n^{\text{ème}}$ puissance, on tire en vertu de la relation $\theta^{+mn}=1$: \[ [\psi(\varepsilon)]^{n}=\{\chi[\Phi^{2}(\alpha^{m}\varepsilon)]\}^{n}. \tag{111.} \]

Cette formule donne, en faisant successivement $m=0$, $1$, $2$, $3\ldots\ldots n-1$, $n$ équations, qui, ajoutées membres à membres donneront la suivante: \[ n[\psi(\varepsilon)]^{n}=\{\chi[\Phi^{2}(\varepsilon)]\}^{n}+\{\chi[\Phi^{2}(\alpha\varepsilon)]\}^{n}+\{\chi[\Phi^{2}(\alpha^{2}\varepsilon)]\}^{n}+\ldots+\{\chi[\Phi^{2}(\alpha^{n-1}\varepsilon)]\}^{n}; \tag{112.} \] or, le second membre de cette équation est une fonction rationelle et symétrique des quantités $\Phi^{2}(\varepsilon)$, $\Phi^{2}(\alpha\varepsilon)\ldots\Phi^{2}(\alpha^{n-1}\varepsilon)$, c'est-à-dire des racines de l'équation (106.); donc $(\psi\varepsilon)^{n}$ peut être exprimé en fonction rationelle de $p_{0}$, $p_{1}\ldots p_{n-1}$, par conséquent en fonction rationelle de l'une quelconque de ces quantités. Soit $v$ la valeur de $\psi(\varepsilon)$, on aura \[ \sqrt[n]{v}=\Phi^{2}(\varepsilon)+\theta.\Phi^{2}(\alpha\varepsilon)+\theta^{2}.\Phi^{2}(\alpha^{2}\varepsilon)+\ldots+\theta^{n-1}.\Phi^{2}(\alpha^{n-1}\varepsilon). \tag{113.} \] Cela posé, soit $\theta=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. Les racines imaginaires de l'équation $\theta^{n}-1$ peuvent être représentées par \[ \theta^{1},\ \theta^{2}\ \ldots\ldots\ \theta^{n-1}. \] Donc en faisant successivement $\theta$ égal à chacune de ces racines et désignant les valeurs correspondantes de $v$ par $v_{1}$, $v_{2}\ldots v_{n-1}$, il viendra \[ \begin{aligned} \sqrt[n]{v_{1}}&=\Phi^{2}(\varepsilon)+\theta.\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\theta^{n-1}.\Phi^{2}(\alpha^{n-1}\varepsilon),\\ \sqrt[n]{v_{2}}&=\Phi^{2}(\varepsilon)+\theta^{2}.\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\theta^{2n-2}.\Phi^{2}(\alpha^{n-1}\varepsilon)\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ \sqrt[n]{v_{n-1}}&=\Phi^{2}(\varepsilon)+\theta^{n-1}.\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\theta^{(n-1)^{2}}.\Phi^{2}(\alpha^{n-1}\varepsilon), \end{aligned} \] En combinant ces équations avec la suivante: \[ -p_{n-1}=\Phi^{2}(\varepsilon)+\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\Phi^{2}(\alpha^{n-1}\varepsilon), \] on en tire aisément: \[ \begin{aligned} \Phi^{2}(\alpha^{m}\varepsilon)=+\frac{1}{n}&\left\{-p_{n-1}+\theta^{-m}.\sqrt[n]{v_{1}}+\theta^{-2m}.\sqrt[n]{v_{2}}+\theta^{-3m}.\sqrt[n]{v_{3}}+\ldots\right.\\ &\left.\ldots+\theta^{-(n-1)m}\sqrt[n]{v_{n-1}}\right\}, \end{aligned} \tag{114.} \] et pour $m=0$: \[ \Phi^{2}(\varepsilon)=\frac{1}{n}\left\{-p_{n-1}+\sqrt[n]{v_{1}}+\sqrt[n]{v_{2}}+\ldots+\sqrt[n]{v_{n-1}}\right\}. \tag{115.} \]

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22.

Toutes les racines de l'équation (106.) sont contenues dans la formule (115.), mais puisque leur nombre n'est que $n$, il reste encore à donner à $\Phi^{2}(\varepsilon)$ une forme, qui ne contienne pas des racines étrangères à la question. Or cela se fait aisément, comme suit:

Soit \[ s_{k}=\frac{\sqrt[n]{v_{k}}}{(\sqrt[n]{v_{1}})^{k}}. \] En posant ici $\varepsilon$ au lieu de $\alpha^{m}\varepsilon$, $\sqrt[n]{v_{k}}$ se changera en $\theta^{-km}.\sqrt[n]{v_{k}}$, et $v_{1}$ en $\theta^{-m}.v_{1}$, donc $s_{k}$ se changera en: \[ \frac{\theta^{-km}.\sqrt[n]{v_{k}}}{(\theta^{-m}\sqrt[n]{v_{1}})^{k}}=\frac{\sqrt[n]{v_{k}}}{(\sqrt[n]{v_{1}})^{k}}. \]

La fonction $s_{k}$, comme on voit, ne change pas de valeur, en mettant $\alpha^{m}\varepsilon$ au lieu de $\varepsilon$. Or $s_{k}$ est une fonction rationelle de $\Phi^{2}(\varepsilon)$. Donc, en désignant $s_{k}$ par $\lambda[\Phi^{2}(\varepsilon)]$, on aura \[ s_{k}=\lambda[\Phi^{2}(\alpha^{m}\varepsilon)], \] quel que soit le nombre entier $m$. De là on tirera de la même manière, et comme nous avons trouvé $(\psi\varepsilon)^{n}$, la valeur de $s_{k}$, en fonction rationelle de l'une des quantités $p_{0}$, $p_{1}$, $....p_{n-1}$. Connoissant $s_{k}$, on a: \[ \sqrt[n]{v_{k}}=s_{k}.(\sqrt[n]{v_{1}})^{k}. \] Donc, en mettant $v$ au lieu de $v_{1}$, l'expression de $\Phi^{2}(\alpha^{m}\varepsilon)$ deviendra: \[ \Phi^{2}(\alpha^{m}\varepsilon)=\frac{1}{n}\left\{-p_{n-1}+\theta^{-m}.v^{\frac{1}{n}}+s_{2}\theta^{-2m}.v^{\frac{2}{n}}+....+s_{n-1}\theta^{-(n-1)m}.v^{\frac{n-1}{n}}\right\}, \tag{116.} \] pour $m=0$: \[ \Phi^{2}(\varepsilon)=\frac{1}{n}\left\{-p_{n-1}+v^{\frac{1}{n}}+s_{2}.v^{\frac{2}{n}}+s_{3}.v^{\frac{3}{n}}+....+s_{n-1}.v^{\frac{n-1}{n}}\right\}. \tag{117.} \] Cette valeur n'a que $n$ valeurs différentes, qui répondent aux $n$ valeurs de $v^{\frac{1}{n}}$. Donc en dernier lieu la résolution de l'équation $P_{2n+1}=0$ est réduite à celle d'une seule équation du degré $2n+2$; mais cette équation ne paraît pas en général être résoluble algébriquement. Néanmoins on peut la résoudre complètement dans plusieurs cas particuliers, p. ex., lorsque $e=c$, $e=c\sqrt{3}$, $e=c(2\pm\sqrt{3})$ etc. Dans le cours de ce mémoire je m'occuperai de ces cas, dont le premier surtout est remarquable, tant par la simplicité de la solution, que par sa belle application dans la géométrie.

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En effet entre autres je suis parvenu à ce théorème:

„On peut diviser la circonférence entière de la lemniscate par la règle et le compas seuls, en $m$ parties égales, si $m$ est de la forme $2^{n}$ ou $2^{n}+1$, le dernier nombre étant en même temps premier; ou bien si $m$ est un produit de plusieurs nombres de ces deux formes.”

Ce théorème est, comme on voit, précisément le même que celui de M. Gauss, relativement au cercle.

§. VI.

Expressions diverses des fonctions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$.

23.

En faisant usage des formules connues, qui donnent les valeurs des coefficiens d'une équation algébrique en fonction des racines, on peut tirer plusieurs expressions des fonctions $\varphi(n\beta)$, $f(n\beta)$, $F(n\beta)$ des formules du paragraphe précédent.

Je vais considérer les plus remarquables.

Pour abréger les formules, je me servirai des notations suivantes: Je désignerai:

1) Par $\overset{k'}{\underset{k}{\Sigma}}_{m}\psi(m)$ la somme, et par $\overset{k'}{\underset{k}{\Pi}}_{m}\psi(m)$ le produit de toutes les quantités de la forme $\psi(m)$, qu'on obtiendra, en donnant à $m$ toutes les valeurs entières, depuis $k$ jusqu'à $k'$, les limites $k$ et $k'$ y compris.

2) Par $\overset{k'}{\underset{k}{\Sigma}}_{m}\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(m,\mu)$ la somme, et par $\overset{k'}{\underset{k}{\Pi}}_{m}\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(m,\mu)$ le produit de toutes les quantités de la forme $\psi(m,\mu)$, qu'on obtiendra, en donnant à $m$ toutes les valeurs entières de $k$ à $k'$, et à $\mu$ les valeurs entières de $\nu$ à $\nu'$, en y comprenant toujours les limites.

D'après cela il est clair, qu'on aura: \[ \overset{k'}{\underset{k}{\Sigma}}_{m}\psi(m)=\psi(k)+\psi(k+1)+.....\psi(k'), \tag{119.} \] Editorial note: No equation number 118. is printed; the numbering passes from 117. to 119. \[ \overset{k'}{\underset{k}{\Pi}}_{m}\psi(m)=\psi(k).\psi(k+1)+.....\psi(k'), \tag{120.} \] Editorial note: A plus sign is printed before the row of dots in the product; a multiplication dot is expected. \[ \overset{k'}{\underset{k}{\Sigma}}_{m}\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(k,\mu)+\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(k+1,\mu)+.....+\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(k',\mu), \tag{121.} \] \[ \overset{k'}{\underset{k}{\Pi}}_{m}\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(m,\mu)=\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(k,\mu).\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(k+1,\mu).........\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(k',\mu). \tag{122.} \]

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Cela posé, considérons les équations \[ \left\{\begin{aligned} \Phi(2n+1)\beta&=\frac{P_{2n+1}}{Q_{2n+1}},\\ f(2n+1)\beta&=\frac{P'_{2n+1}}{Q_{2n+1}},\\ F(2n+1)\beta&=\frac{P''_{2n+1}}{Q_{2n+1}}. \end{aligned}\right. \tag{123.} \]

Nous avons vu, que $P_{2n+1}$ est une fonction rationelle de $x$ du degré $(2n+1)^{2}$ et de la forme $x.\psi(x^{2})$. De même $P'_{2n+1}$ et $F''_{2n+1}$ sont des fonctions de cette même forme, la première de $y$ et la seconde de $z$.Editorial note: Printed $F''_{2n+1}$ where $P''_{2n+1}$ is expected. Enfin $Q_{2n+1}$ est une fonction qui, exprimée indistinctement en $x$, $y$ ou $z$, sera du degré $(2n+1)^{2}-1$, et contiendra seulement des puissances paires. Donc on aura \[ \begin{aligned} P_{2n+1}&=A.x^{(2n+1)^{2}}+......+B.x;&\quad P'_{2n+1}&=A'.y^{(2n+1)^{2}}+......+B'.y;\\ P''_{2n+1}&=A''.z^{(2n+1)^{2}}+......+B''.z;&\quad Q_{2n+1}&=C.x^{(2n+1)^{2}-1}+....+D;\\ Q_{2n+1}&=C'.y^{(2n+1)^{2}-1}+....+D';&\quad Q_{2n+1}&=C''.z^{(2n+1)^{2}-1}+....+D''. \end{aligned} \]

En substituant ces valeurs dans (123.), il viendra \[ \begin{aligned} &\{A.x^{(2n+1)^{2}}+.....+B.x\}-\varphi(2n+1)\beta.\{C.x^{(2n+1)^{2}-1}+....+D\},\\ &\{A'.y^{(2n+1)^{2}}+.....+B'.y\}-f(2n+1)\beta.\{C'.y^{(2n+1)^{2}-1}+....+D'\},\\ &\{A''.z^{(2n+1)^{2}}+.....+B''.z\}-F(2n+1)\beta.\{C''.z^{(2n+1)^{2}-1}+....+D''\}. \end{aligned} \]

Dans la première de ces équations $A$ est le coefficient du premier terme, $-\varphi(2n+1)\beta.C$ celui du second et $-\varphi(2n+1)\beta.D$ le dernier terme. Donc $\frac{C}{A}\varphi(2n+1)\beta$ est égal à la somme et $\pm\frac{D}{A}\varphi(2n+1)\beta$ égal au produit des racines de l'équation dont il s'agit, équation, qui est la même, que celle-ci: \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \tag{124.} \]

Donc en remarquant que $A$, $C$ et $D$ (et en général tous les coefficiens) sont indépendans de $\beta$, on voit que $\Phi(2n+1)\beta$ est (d'un coefficient constant près) égal à la somme et au produit de toutes les racines de l'équation (124.).

De la même manière on voit que $f(2n+1)\beta$ et $F(2n+1)\beta$ sont respectivement égaux au produit ou à la somme des racines des équations \[ f(2n+1)\beta=\frac{P'_{2n+1}}{Q_{2n+1}},\ F(2n+1)\beta=\frac{P''_{2n+1}}{Q_{2n+1}}, \]

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en ayant attention de multiplier le résultat par un coefficient constant, choisi convenablement.

Maintenant les racines des équations (123.) d'après le No. 11. sont respectivement: \[ \begin{aligned} x&=(-1)^{m+\mu}.\Phi\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right),\\ y&=(-1)^{m}.f\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right),\\ z&=(-1)^{\mu}.F\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right), \end{aligned} \] où les limites de $m$ et $\mu$ sont $-n$ et $+n$.

Donc en vertu de ce qu'on vient de voir, et en faisant usage des notations adoptées, on aura les formules suivantes: \[ \left\{\begin{aligned} \Phi(2n+1)\beta&=A.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m+\mu}.\Phi\left(\beta+\frac{m.\omega+\mu\varpi i}{2n+1}\right),\\ f(2n+1)\beta&=A'.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m}.f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ F(2n+1)\beta&=A''.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{\mu}.F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right);\\ \Phi(2n+1)\beta&=B.\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}.\Phi\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ f(2n+1)\beta&=B'.\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}.f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ F(2n+1)\beta&=B''.\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}.F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right). \end{aligned}\right. \tag{125.} \]

Pour déterminer les quantités constantes $A$, $A'$, $A''$, $B$, $B'$, $B''$, il faudra donner à $\beta$ une valeur particulière. Ainsi en faisant dans les trois premières formules $\beta=\frac{\omega}{2}+\frac{\varpi}{2}i$, après avoir divisé les deux membres par $\Phi\beta$, il viendra, en remarquant que $\Phi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}$: \[ \left.\begin{aligned} A&=\frac{\varphi(2n+1)\beta}{\varphi\beta}\\ A'&=\frac{f(2n+1)\beta}{f\beta}\\ A''&=\frac{F(2n+1)\beta}{f\beta} \end{aligned}\right\}\ \text{pour }\beta=\frac{\omega}{2}+\frac{\varpi}{2}i. \] Editorial note: The denominator of the third fraction is printed $f\beta$; $F\beta$ is apparently expected.

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Soit $\beta=\frac{\omega}{2}+\frac{\varpi}{2}i+\alpha$, on a: \[ \left.\begin{aligned} A&=\frac{\varphi\left((2n+1)\alpha+n\omega+n\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{\varphi\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}\\ &=\frac{\varphi\left((2n+1)\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{\varphi\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}=\frac{\varphi\alpha}{\varphi(2n+1)\alpha},\\ A'&=\frac{f\left((2n+1)\alpha+n\omega+n\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{f\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}\\ &=(-1)^{n}.\frac{f\left((2n+1)\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{f\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}=(-1)^{n}.\frac{f\left(\alpha+\frac{\omega}{2}\right)}{f\left((2n+1)\alpha+\frac{\omega}{2}\right)},\\ A''&=\frac{F\left((2n+1)\alpha+n\omega+n\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{F\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}\\ &=(-1)^{n}.\frac{F\left((2n+1)\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{F\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}=(-1)^{n}.\frac{F\left(\alpha+\frac{\varpi}{2}i\right)}{F\left((2n+1)\alpha+\frac{\varpi}{2}i\right)}. \end{aligned}\right\}\ \text{pour }\alpha=0. \]

Ces expressions de $A$, $A'$, $A''$ deviendront de la forme $\frac{0}{0}$, en faisant $\alpha=0$, donc on trouvera d'après les règles connues: \[ A=\frac{1}{2n+1},\quad A'=A''=\frac{(-1)^{n}}{2n+1}. \] D'après cela les trois premières formules deviendront: \[ \left\{\begin{aligned} \Phi(2n+1)\beta&=\frac{1}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m+\mu}.\varphi\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ f(2n+1)\beta&=\frac{(-1)^{n}}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m}.f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ F(2n+1)\beta&=\frac{(-1)^{n}}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{\mu}.F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right). \end{aligned}\right. \tag{126.} \]

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Pour avoir la valeur des constantes $B$, $B'$, $B''$, je remarque, qu'on aura: \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}\psi(m,\mu)=\psi(0,0).\overset{n}{\underset{1}{\Pi}}_{m}\psi(m,0).\overset{n}{\underset{1}{\Pi}}_{\mu}\psi(0,\mu)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\psi(m,\mu).\psi(-m,-\mu).\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\psi(m,-\mu).\psi(-m,\mu). \end{aligned} \tag{127.} \]

En appliquant cette transformation aux formules (126.), divisant la première par $\Phi\beta$, la seconde par $f\beta$ et la troisième par $F\beta$, faisant ensuite dans la première $\beta=0$, dans la seconde $\beta=\frac{\omega}{2}$ et dans la troisième $\beta=\frac{\varpi}{2}i$, et remarquant que $\frac{\varphi(2n+1)\beta}{\varphi\beta}=2n+1$, pour $\beta=0$, que $\frac{f(2n+1)\beta}{f\beta}=2n+1$, pour $\beta=\frac{\omega}{2}$, et que $\frac{F(2n+1)\beta}{F\beta}=2n+1$, pour $\beta=\frac{\varpi}{2}i$: on trouvera: \[ \left\{\begin{aligned} (2n+1)&=B.\overset{n}{\underset{1}{\Pi}}_{m}\varphi^{2}\left(\frac{m\omega}{2n+1}\right).\overset{n}{\underset{1}{\Pi}}_{\mu}\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right).\varphi^{2}\left(\frac{m\omega-\mu\varpi i}{2n+1}\right),\\ (2n+1)&=B'.\overset{n}{\underset{1}{\Pi}}_{m}f^{2}\left(\frac{\omega}{2}+\frac{m\omega}{2n+1}\right).\overset{n}{\underset{1}{\Pi}}_{\mu}f^{2}\left(\frac{\omega}{2}+\frac{\mu\varpi i}{2n+1}\right)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}f^{2}\left(\frac{\omega}{2}+\frac{m\omega+\mu\varpi i}{2n+1}\right).f^{2}\left(\frac{\omega}{2}+\frac{m\omega-\mu\varpi i}{2n+1}\right),\\ (2n+1)&=B''.\overset{n}{\underset{1}{\Pi}}_{m}F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right).\overset{n}{\underset{1}{\Pi}}_{\mu}F^{2}\left(\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right).F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right). \end{aligned}\right. \tag{128.} \]

En tirant de ces équations les valeurs de $B$, $B'$, $B''$, et les substituant ensuite dans les formules transformées, il viendra:

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\[ \left\{\begin{aligned} &\Phi(2n+1)\beta=\\ &(2n+1)\Phi\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{\varphi\left(\beta+\frac{m\omega}{2n+1}\right).\varphi\left(\beta-\frac{m\omega}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega}{2n+1}\right)}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{\varphi\left(\beta+\frac{\mu\varpi i}{2n+1}\right).\varphi\left(\beta-\frac{\mu\varpi i}{2n+1}\right)}{\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right)}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{\varphi\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right).\varphi\left(\beta-\frac{m\omega+\mu\varpi i}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)}.\frac{\varphi\left(\beta+\frac{m\omega-\mu\varpi i}{2n+1}\right).\varphi\left(\beta-\frac{m\omega-\mu\varpi i}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega-\mu\varpi i}{2n+1}\right)},\\ &f(2n+1)\beta=\\ &(2n+1)f\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{f\left(\beta+\frac{m\omega}{2n+1}\right).f\left(\beta-\frac{m\omega}{2n+1}\right)}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega}{2n+1}\right)}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{f\left(\beta+\frac{\mu\varpi i}{2n+1}\right).f\left(\beta-\frac{\mu\varpi i}{2n+1}\right)}{f^{2}\left(\frac{\omega}{2}+\frac{\mu\varpi i}{2n+1}\right)}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right).f\left(\beta-\frac{m\omega+\mu\varpi i}{2n+1}\right)}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega+\mu\varpi i}{2n+1}\right)}.\frac{f\left(\beta+\frac{m\omega-\mu\varpi i}{2n+1}\right).f\left(\beta-\frac{m\omega-\mu\varpi i}{2n+1}\right)}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega-\mu\varpi i}{2n+1}\right)},\\ &F(2n+1)\beta=\\ &(2n+1)F\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{F\left(\beta+\frac{m\omega}{2n+1}\right).F\left(\beta-\frac{m\omega}{2n+1}\right)}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{F\left(\beta+\frac{\mu\varpi i}{2n+1}\right).F\left(\beta-\frac{\mu\varpi i}{2n+1}\right)}{F^{2}\left(\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right).F\left(\beta-\frac{m\omega+\mu\varpi i}{2n+1}\right)}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}.\frac{F\left(\beta+\frac{m\omega-\mu\varpi i}{2n+1}\right).F\left(\beta-\frac{m\omega-\mu\varpi i}{2n+1}\right)}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}. \end{aligned}\right. \tag{129.} \]

On peut donner à ces formules des formes plus simples, en faisant usage des formules suivantes: \[ \frac{\varphi(\beta+\alpha).\varphi(\beta-\alpha)}{\varphi^{2}\alpha}=-\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\alpha}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}}, \] \[ \frac{f(\beta+\alpha).f(\beta-\alpha)}{f^{2}\left(\frac{\omega}{2}+\alpha\right)}=-\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\alpha\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}}, \] \[ \frac{F^{2}(\beta+\alpha).F(\beta-\alpha)}{F^{2}\left(\frac{\varpi}{2}i+\alpha\right)}=-\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\alpha\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\alpha\right)}}, \] Editorial note: The numerator of the third left-hand fraction is printed $F^{2}(\beta+\alpha).F(\beta-\alpha)$; $F(\beta+\alpha).F(\beta-\alpha)$ is apparently expected.

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qu'on verifiera aisément au moyen des formules (13.), (16.), (18.).

En vertu de ces formules il est clair, qu'on peut mettre les équations (129.) sous la forme: \[ \left\{\begin{aligned} &\Phi(2n+1)\beta=\\ &(2n+1)\Phi\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}},\\ &f(2n+1)\beta=\\ &(2n+1)f\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{m\omega}{2n+1}+\frac{\omega}{2}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}},\\ &F(2n+1)\beta=\\ &(2n+1)F\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}. \end{aligned}\right. \tag{130.} \]

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Ces formules donnent, comme on voit, les valeurs de $\Phi(2n+1)\beta$, $f(2n+1)\beta$ et $F(2n+1)\beta$, exprimées respectivement en fonctions rationelles de $\Phi\beta$, $f\beta$ et $F\beta$, sous la forme de produits.

Nous donnerons encore les valeurs de $f(2n+1)\beta$, $F(2n+1)\beta$ sous une autre forme, qui sera utile dans la suite.

On a $f^{2}\beta=1-c^{2}\varphi^{2}\beta$, donc \[ 1-\frac{f^{2}\beta}{f^{2}\alpha}=\frac{c^{2}(\varphi^{2}\beta-\varphi^{2}\alpha)}{f^{2}\alpha}=\frac{c^{2}}{f^{2}\alpha}\Phi^{2}\beta-\frac{c^{2}\varphi^{2}\alpha}{f^{2}\alpha} \] et \[ 1-\frac{f^{2}\beta}{f^{2}\left(\frac{\varpi}{2}i+\alpha\right)}=\frac{c^{2}\left[\varphi^{2}\beta-\varphi^{2}\left(\frac{\varpi}{2}i+\alpha\right)\right]}{f^{2}\left(\frac{\varpi}{2}i+\alpha\right)}, \] or en vertu de (18.) on a: \[ f^{2}\left(\frac{\varpi}{2}i+\alpha\right)=\frac{e^{2}+c^{2}}{e^{2}}.\frac{1}{f^{2}\alpha}; \] donc: \[ \frac{1-\frac{f^{2}\beta}{f^{2}\alpha}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\varpi}{2}i+\alpha\right)}}=\frac{1}{f^{4}\alpha}.\frac{e^{2}+c^{2}}{e^{2}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\alpha}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\varpi}{2}i+\alpha\right)}}. \] On trouvera de même: \[ \frac{1-\frac{F^{2}\beta}{F^{2}\alpha}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\alpha\right)}}=\frac{1}{F^{4}\alpha}.\frac{e^{2}+c^{2}}{e^{2}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\alpha}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\alpha\right)}}. \]

En vertu de ces formules, et en faisant $\beta=0$, pour déterminer le facteur constant, il est clair, qu'on peut écrire les expressions de $f(2n+1)\beta$, $F(2n+1)\beta$, comme suit:

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\[ \left\{\begin{aligned} &f(2n+1)\beta=f\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{m\omega}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}},\\ &F(2n+1)\beta=F\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}. \end{aligned}\right. \tag{130$'$.} \] Editorial note: In the numerator of the second product of the first line, the print sets an $i$ after $\omega/2$; there is apparently no $i$ there, by analogy with the first product. Reproduced as printed.

Dans ce paragraphe nous n'avons considéré les fonctions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$, que dans le cas des valeurs impaires de $n$. On pourroit trouver des expressions analogues de ces fonctions pour des valeurs paires de $n$; mais comme il n'y a dans cela aucune difficulté et que d'ailleurs les formules, aux quelles nous sommes parvenus, sont celles qui nous seront le plus utiles dans la suite, je ne m'en occuperai pas.

§. VII.

Développement des fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$ en séries et en produits infinis.

24.

En faisant dans les formules du paragraphe précédent $\beta=\frac{\alpha}{2n+1}$, on obtiendra des expressions des fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$, qui, à cause du nombre indéterminé $n$, peuvent être variées d'une infinité de manières.

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Parmi toutes les formules, qu'on obtiendra ainsi, celles qui résultent de la supposition de $n$ infini, sont les plus remarquables. Alors les fonctions $\Phi$, $f$, $F$ disparaîtront des valeurs de $\Phi\alpha$, $f\alpha$, $F\alpha$, et on obtiendra pour ces fonctions des expressions algébriques, mais composées d'une infinité de termes. Pour avoir ces expressions, il faut faire dans les formules (126.), (130.) $\beta=\frac{\alpha}{2n+1}$, et ensuite chercher la limite du second membre de ces équations pour des valeurs toujours croissantes de $n$. Pour abréger, soit $v$ une quantité, dont la limite est zéro pour des valeurs toujours croissantes de $n$. Cela posé considérons successivement les trois formules (126.).

En faisant dans la première des formules (126.) $\beta=\frac{\alpha}{2n+1}$, et remarquant que \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta(m,\mu)=\theta(0,0)+\overset{n}{\underset{1}{\Sigma}}_{m}\theta(m,0)+\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(0,\mu)\\ &\quad+\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}\{\theta(m,\mu)+\theta(-m,-\mu)+\theta(m,-\mu)+\theta(-m,\mu)\}, \end{aligned} \tag{131.} \] il est clair, qu'on peut mettre la formule, dont il s'agit, sous la forme: \[ \begin{aligned} \Phi\alpha={}&\frac{1}{2n+1}.\varphi\left(\frac{\alpha}{2n+1}\right)+\frac{1}{2n+1}\overset{n}{\underset{1}{\Sigma}}_{m}(-1)^{m}\left\{\Phi\left(\frac{\alpha+m\omega}{2n+1}\right)+\Phi\left(\frac{\alpha-m\omega}{2n+1}\right)\right\}\\ &+\frac{1}{2n+1}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{\mu}\left\{\Phi\left(\frac{\alpha+\mu\varpi i}{2n+1}\right)+\Phi\left(\frac{\alpha-\mu\varpi i}{2n+1}\right)\right\}-\frac{i}{ec}.\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(n-m,n-\mu)\\ &+\frac{i}{ec}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(n-m,n-\mu), \end{aligned} \tag{132.} \] où l'on a fait, pour abréger, \[ \left\{\begin{aligned} \psi(m,\mu)&=\frac{1}{2n+1}.\left\{\frac{1}{\varphi\left(\frac{\alpha+(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}+\frac{1}{\varphi\left(\frac{\alpha-(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}\right\},\\ \psi_{1}(m,\mu)&=\frac{1}{2n+1}.\left\{\frac{1}{\varphi\left(\frac{\alpha+(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}+\frac{1}{\varphi\left(\frac{\alpha-(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}\right\}. \end{aligned}\right. \tag{133.} \]

Maintenant, en remarquant, que \[ \Phi\left(\frac{\alpha+m\omega}{2n+1}\right)+\Phi\left(\frac{\alpha-m\omega}{2n+1}\right)=\frac{2\varphi\left(\frac{\alpha}{2n+1}\right).f\left(\frac{m\omega}{2n+1}\right).F\left(\frac{m\omega}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{m\omega}{2n+1}\right).\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}=\frac{A_{m}}{2n+1}, \] \[ \Phi\left(\frac{\alpha+\mu\varpi i}{2n+1}\right)+\Phi\left(\frac{\alpha-\mu\varpi i}{2n+1}\right)=\frac{2\varphi\left(\frac{\alpha}{2n+1}\right).f\left(\frac{\mu\varpi i}{2n+1}\right).F\left(\frac{\mu\varpi i}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right).\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}=\frac{B_{\mu}}{2n+1}, \]

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où $A_{m}$ et $B_{\mu}$ sont des quantités finies, la partie de l'équation (132.) jusqu'au membre, qui a le signe $-$, prendra la forme: \[ \frac{1}{2n+1}.\varphi\left(\frac{\alpha}{2n+1}\right)+\frac{1}{(2n+1)^{2}}.\overset{n}{\underset{1}{\Sigma}}_{m}(-1)^{m}(A_{m}+B_{\mu}); \] or la limite de cette quantité est évidemment zéro; donc, en prenant la limite de la formule (132.), on aura: \[ \begin{aligned} \Phi\alpha&=-\frac{i}{ec}\ \text{lim.}\ \overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(n-m,n-\mu)\\ &\quad+\frac{i}{ec}\ \text{lim.}\ \overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(n-m,n-\mu), \end{aligned} \] ou bien: \[ \begin{aligned} \Phi\alpha&=-\frac{i}{ec}\ \text{lim.}\ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu)\\ &\quad+\frac{i}{ec}\ \text{lim.}\ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(m,\mu). \end{aligned} \tag{134.} \]

Il suffit de connoître l'une de ces limites, car on aura l'autre, en changeant seulement le signe de $i$.

Cherchons la limite de \[ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu). \] Pour cela, il faut essayer de mettre la quantité précédente sous la forme \[ P+v, \] où $P$ est indépendant de $n$, et $v$ une quantité, qui a zéro pour limite; car alors la quantité $P$ sera précisement la limite, dont il s'agit.

25.

Considérons d'abord l'expression: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\psi(m,\mu). \]

Soit \[ \theta(m,\mu)=\frac{2\alpha}{\alpha^{2}-(m\omega+\mu\varpi i)^{2}}, \tag{135.} \] et faisons \[ \psi(m,\mu)-\theta(m,\mu)=\frac{2\alpha}{(2n+1)^{2}}.R_{\mu}, \tag{136.} \] on aura: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\psi(m,\mu)-\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)=2\alpha.\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{n}.\frac{R_{\mu}}{(2n+1)^{2}}. \tag{137.} \] Editorial note: The exponent of the factor $-1$ in the last sum is apparently printed $n$; $\mu$ is expected, as in the two sums on the left.

Cela posé, je dis que le second membre de cette équation est une quantité de la forme $\frac{v}{2n+1}$.

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D'après (12.), (13.) on aura \[ \frac{1}{\varphi(\beta+\varepsilon)}+\frac{1}{\varphi(\beta-\varepsilon)}=\frac{\varphi(\beta+\varepsilon)+\varphi(\beta-\varepsilon)}{\varphi(\beta+\varepsilon).\varphi(\beta-\varepsilon)}=\frac{2\varphi\beta.f\varepsilon.F\varepsilon}{\varphi^{2}\beta-\varphi^{2}\varepsilon}, \] donc en faisant $\beta=\frac{\alpha}{2n+1}$ et $\varepsilon=\frac{m\omega+\mu\varpi i}{2n+1}=\frac{\varepsilon_{\mu}}{2n+1}$ et $f\varepsilon.F\varepsilon=\theta\varepsilon$, on a: \[ \psi(m,\mu)=\frac{1}{2n+1}.\frac{2\varphi\left(\frac{\alpha}{2n+1}\right).\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}. \] Maintenant on a: \[ \varphi\left(\frac{\alpha}{2n+1}\right)=\frac{\alpha}{2n+1}+\frac{A\alpha^{3}}{(2n+1)^{3}}, \] donc: \[ \psi(m,\mu)=\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}.\left\{\frac{2A\alpha^{3}}{(2n+1)^{4}}+\frac{2\alpha}{(2n+1)^{2}}\right\}, \] et par conséquent: \[ \begin{aligned} \psi(m,\mu)-\theta(m,\mu)&=\frac{2\alpha}{(2n+1)^{2}}\left\{\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}-\frac{1}{\left(\frac{\alpha}{2n+1}\right)^{2}-\left(\frac{\varepsilon_{\mu}}{2n+1}\right)^{2}}\right\}\\ &\quad+\frac{2A\alpha^{3}}{(2n+1)^{4}}.\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}. \end{aligned} \] Donc la valeur de $R_{\mu}$ deviendra \[ R_{\mu}=\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}.\left\{1+\frac{A\alpha^{2}}{(2n+1)^{2}}\right\}-\frac{1}{\left(\frac{\alpha}{2n+1}\right)^{2}-\left(\frac{\varepsilon_{\mu}}{2n+1}\right)^{2}}. \tag{138.} \]

Cela posé, il y a deux cas à considérer, savoir si $\frac{\varepsilon_{\mu}}{2n+1}$ a zéro pour limite ou non.

$a)$ Si $\frac{\varepsilon_{\mu}}{2n+1}$ a zéro pour limite, on aura: \[ \begin{aligned} \varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)&=\frac{\varepsilon_{\mu}^{2}}{(2n+1)^{2}}+\frac{B_{\mu}.\varepsilon_{\mu}^{4}}{(2n+1)^{4}},\\ \theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)&=\sqrt{\left[1-c^{2}\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)\right]}.\sqrt{\left[1+e^{2}\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)\right]}=1+\frac{C_{\mu}.\varepsilon_{\mu}^{2}}{(2n+1)^{2}},\\ \varphi^{2}\left(\frac{\alpha}{2n+1}\right)&=\frac{\alpha^{2}}{(2n+1)^{2}}+\frac{D.\alpha^{4}}{(2n+1)^{4}}, \end{aligned} \] ou $B_{\mu}$, $C_{\mu}$, $D$ ont des limites finies; donc en substituant:

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\[ \begin{aligned} R_{\mu}&=A\alpha^{2}.\frac{\frac{1}{\varepsilon_{\mu}^{2}}+\frac{C_{\mu}}{(2n+1)^{2}}}{\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}-1+\frac{D\alpha^{4}}{(2n+1)^{2}.\varepsilon_{\mu}^{2}}-B_{\mu}\frac{\varepsilon_{\mu}^{2}}{(2n+1)^{2}}}\\ &\quad+\frac{C_{\mu}.\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}-\frac{D\alpha^{4}}{\varepsilon_{\mu}^{4}}-C_{\mu}+B_{\mu}}{\left(1-\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}\right)^{2}-\left(1-\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}\right)\left\{\frac{D\alpha^{4}}{(2n+1)^{2}.\varepsilon_{\mu}^{2}}-B_{\mu}.\frac{\varepsilon_{\mu}^{2}}{(2n+1)^{2}}\right\}}, \end{aligned} \tag{139.} \] or soit que $\varepsilon_{\mu}$ soit fini ou infini, il est clair, que cette quantité convergera toujours vers une quantité finie pour des valeurs toujours croissantes de $n$. Donc on aura \[ R_{\mu}=r_{\mu}+v_{\mu}, \tag{140.} \] où $r_{\mu}$ est une quantité finie indépendante de $n$.

$b)$ Si $\frac{\varepsilon_{\mu}}{2n+1}$ a pour limite une quantité finie, il est clair, qu'en nommant cette limite $\delta_{\mu}$, on aura: \[ R_{\mu}=-\frac{\theta(\delta_{\mu})}{\varphi^{2}(\delta_{\mu})}+\frac{1}{\delta_{\mu}^{2}}+v_{\mu}^{1}. \tag{141.} \] Cela posé, considérons l'expression $\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}$. On a \[ \begin{aligned} \overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}&=\frac{1}{(2n+1)^{2}}.\{R_{0}-R_{1}+R_{2}-R_{3}+....\\ &\quad....+(-1)^{\nu-1}.R_{\nu-1}+(-1)^{\nu+1}[R_{\nu+1}-R_{\nu+1}+R_{\nu+2}-R_{\nu+3}...+(-1)^{n-\nu-1}.R_{n-1}]\}. \end{aligned} \tag{142.} \] Editorial note: The first term inside the square bracket is printed $R_{\nu+1}$; $R_{\nu}$ is apparently expected. Supposons d'abord, que $\frac{\varepsilon_{\mu}}{2n+1}$ ait pour limite une quantité finie, quelle que soit la valeur de $\mu$. Alors en remarquant, que \[ \delta_{\mu+1}=\delta_{\mu}, \] on aura \[ R_{\mu}-R_{\mu+1}=v_{\mu}^{1}-v_{\mu+1}^{1}, \] donc: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{1}{(2n+1)^{2}}(v_{0}^{1}-v_{1}^{1}+v_{2}^{1}-v_{3}^{1}+....+v_{k-2}^{1}-v_{k-1}^{1})+\frac{B}{(2n+1)^{2}}, \] où $k=n$ ou $n-1$, selon que $n$ est pair ou impair. La quantité $B$ a toujours pour limite une quantité finie, savoir $B=0$, si $n$ est pair, et $B=R_{n-1}$, si $n$ est impair.

Maintenant on sait, qu'une somme telle que \[ v_{0}^{1}-v_{1}^{1}+v_{2}^{1}-....+v_{k-2}^{1}-v_{k-1}^{1}, \] peut être mise sous la forme $k.v$, $v$ ayant zéro pour limite. Donc en substituant:

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\[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{k.v+B}{(2n+1)^{2}}; \] or, $k$ étant égal à $n$ ou à $n-1$, et $B$ fini, la limite de $\frac{k.v+B}{2n+1}$ sera zéro, donc: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{v}{2n+1}. \tag{143.} \]

Supposons maintenant, que $\frac{m}{2n+1}$ a zéro pour limite. Alors $\frac{\varepsilon_{\mu}}{2n+1}$ a également zéro pour limite, à moins qu'en même tems $\frac{\mu}{2n+1}$ n'ait pour limite une quantité finie. Soit dans ce cas $\nu$ le nombre entier immédiatement inferieur à $\sqrt{n}$, et considérons la somme \[ R_{0}-R_{1}+R_{2}-R_{3}+....+(-1)^{\nu-1}R_{\nu-1}. \] En supposant, que $\mu$ est un des nombres $0$, $1$, $....\nu$, il est clair, que $\frac{\varepsilon_{\mu}}{(2n+1)}=\frac{m\omega+\mu\varpi i}{2n+1}$ a zéro pour limite; donc, selon ce qu'on a vu, $R_{\mu}$ sera une quantité finie, et par conséquent \[ R_{0}-R_{1}+R_{2}-....+(-1)^{\nu-1}R_{\nu-1}=\nu.R, \] où $R$ est également une quantité finie.

Considérons maintenant la somme \[ (-1)^{\nu}\{R_{\nu}-R_{\nu+1}+R_{\nu+2}-....+(-1)^{n-\nu-1}.R_{n-1}\}. \]

Si $\frac{\varepsilon_{\mu}}{2n+1}$ a pour limite une quantité différente de zéro, on a, comme on verra: \[ R_{\mu}-R_{\mu+1}=v_{\mu}^{1}-v_{\mu+1}^{1}; \] si au contraire $\frac{\varepsilon_{\mu}}{2n+1}$ a pour limite zéro, on a: \[ R_{\mu}=v_{\mu}+v_{\mu}^{1}; \] or, si en même tems $\mu>\sqrt{n}$, il est clair, qu'en vertu de la valeur de $R_{\mu}$, \[ v_{\mu}=B_{\mu}-C_{\mu}; \] or il est clair, que $B_{\mu}$ et $C_{\mu}$, tous deux, ont pour limites des quantités indépendantes de $\mu$, donc en nommant ces limites $B$ et $C$, on aura: \[ R_{\mu}=B-C+v_{\mu}, \] et par suite, aussi dans ce cas, \[ R_{\mu}-R_{\mu+1}=v_{\mu}-v_{\mu+1}. \] Editorial note: The last subscript is printed indistinctly, apparently $v_{\mu+1}$ with the $\mu$ damaged in print.

Donc comme dans le cas où $\frac{\varepsilon_{\mu}}{2n+1}$ auroit une limite différente de zéro pour toutes les valeurs de $\mu$, on démontrera que \[ \frac{(-1)^{\nu}}{(2n+1)^{2}}\{R_{\nu}-R_{\nu+1}+....+(-1)^{n-\nu-1}R_{n-1}\}=\frac{v}{(2n+1)}. \]

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Maintenant en combinant les équations ci-dessus, on en tirera \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{1}{(2n+1)^{2}}.(\nu R)+\frac{v}{2n+1}; \] or $\frac{v}{2n+1}$ a zéro pour limite, donc \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{v}{2n+1}. \] Donc cette formule a toujours lieu, et par conséquent la formule (137.) deviendra: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\psi(m,\mu)-\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)=\frac{v}{2n+1}. \tag{144.} \] Cela posé, il s'agit de mettre $\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)$ sous la forme $P+\frac{v}{2n+1}$. Or c'est ce qu'on peut faire comme il suit. On a: \[ \left\{\begin{aligned} &\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)=\overset{\frac{1}{0}}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)-\overset{\frac{1}{0}}{\underset{n}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)\ \text{et}\\ &\overset{\frac{1}{0}}{\underset{n}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)=(-1)^{n}\{\theta(m,n)-\theta(m,n+1)+\theta(m,n+2)-....\text{etc.}\}. \end{aligned}\right. \tag{145.} \] Or d'après une formule connue on a: \[ \begin{aligned} &\theta(m,n)-\theta(m,n+1)+\theta(m,n+2)-......\\ &=\tfrac{1}{2}\theta(m,n)+A\frac{\partial\theta(m,n)}{\partial n}+B.\frac{\partial^{3}\theta(m,n)}{\partial^{3}n}+......, \end{aligned} \] où $A$, $B$$....$ sont des nombres; or \[ \theta(m,n)=\frac{2\alpha}{\alpha^{2}-(m\omega+n\varpi i)^{2}}, \] donc en substituant: \[ \theta(m,n)-\theta(m,n+1)+....=\frac{\alpha}{\alpha^{2}-(m\omega+n\varpi i)^{2}}+\frac{4A\alpha\varpi i.n}{\{\alpha^{2}-(m\omega+n\varpi i)^{2}\}^{2}}+.... \] Editorial note: The exponent outside the braces in the denominator of the second term is printed indistinctly; $2$ is read here. De là il suit, que \[ \theta(m,n)-\theta(m,n+1)+....=\frac{\alpha}{\varpi^{2}.n^{2}}+\frac{v}{n^{2}}=\frac{v}{2n+1}. \] Donc en vertu des équations (145.) \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)=\overset{\frac{1}{0}}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)+\frac{v}{2n+1}, \] et par conséquent \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\psi(m,\mu)=\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)+\frac{v}{2n+1}. \tag{146.} \]

26.

Ayant transformé de cette sorte la quantité $\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\psi(m,\mu)$, on tire de l'équation (146.):

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\[ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}\psi(m,\mu)=\overset{n-1}{\underset{0}{\Sigma}}(-1)^{m}.\varrho_{m}+\overset{n-1}{\underset{0}{\Sigma}}.\frac{v_{m}}{2n+1}, \tag{147.} \] en faisant \[ \varrho_{m}=\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu); \tag{148.} \] or \[ \overset{n-1}{\underset{0}{\Sigma}}.\frac{v_{m}}{2n+1}=\frac{v_{0}+v_{1}+v_{2}\ldots v_{n-1}}{2n+1}=\frac{nv}{2n+1}=\frac{v}{2}, \] $v$ ayant zéro pour limite. Donc l'équation (147.) donnera, en faisant $n$ infini: \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu)=\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho_{m}. \tag{149.} \] De même, si l'on fait, pour abréger: \[ \left\{\begin{aligned}\theta_{1}(m,\mu)&=\frac{2\alpha}{\alpha^{2}-(m\omega-\mu\varpi i)^{2}},\\ \varrho'_{m}&=\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta_{1}(m,\mu),\end{aligned}\right. \tag{150.} \] on aura: \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(m,\mu)=\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho'_{m}. \tag{151.} \] Ayant trouvé ces deux quantités, dont l'expression de $\Phi\alpha$ est composée, on aura, en substituant: \[ \Phi\alpha=-\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho_{m}+\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho'_{m}=\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}\{\varrho'_{m}-\varrho_{m}\}, \] ou bien, en remettant les valeurs de $\varrho'_{m}$ et $\varrho_{m}$, \[ \Phi\alpha=\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}-\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}}\right\}\right\}. \tag{152.} \] Maintenant \[ \frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega\pm(\mu+\tfrac{1}{2})\varpi i]^{2}}=\frac{1}{\alpha-(m+\tfrac{1}{2})\omega\mp(\mu+\tfrac{1}{2})\varpi i}+\frac{1}{\alpha+(m+\tfrac{1}{2})\omega\pm(\mu+\tfrac{1}{2})\varpi i}, \] donc: \[ \begin{aligned}&\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}-\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}}\\ &=+\frac{(2\mu+1)\varpi i}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{(2\mu+1)\varpi i}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}},\end{aligned} \] donc l'expression de $\Phi\alpha$ deviendra sous une forme réelle: \[ \Phi\alpha=\frac{1}{ec}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\left\{\frac{(2\mu+1)\varpi}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{(2\mu+1)\varpi}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right\}, \tag{153.} \] c'est-à-dire, on aura:

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\[ \begin{aligned}\Phi\alpha&=\frac{\varpi}{ec}(\delta_{0}-\delta_{1}+\delta_{2}-\delta_{3}+....+(-1)^{m}\delta_{m}-....)\\ &-\frac{\varpi}{ec}(\delta'_{0}-\delta'_{1}+\delta'_{2}-\delta'_{3}+....+(-1)^{m}\delta'_{m}-....),\end{aligned} \tag{154.} \] ou bien: \[ \left\{\begin{aligned}\delta_{m}&=\frac{1}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+\frac{\varpi^{2}}{4}}-\frac{3}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+\frac{9\varpi^{2}}{4}}+\frac{5}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+\frac{25\varpi^{2}}{4}}-\ldots\\ \delta'_{\mu}&=\frac{1}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+\frac{\varpi^{2}}{4}}-\frac{3}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+\frac{9\varpi^{2}}{4}}+\frac{5}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+\frac{25\varpi^{2}}{4}}-\ldots\end{aligned}\right. \tag{155.} \]

Si l'on commence la recherche de la limite de la fonction $\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu)$ par celle de $\overset{n-1}{\underset{0}{\Sigma}}(-1)^{m}\psi(m,\mu)$ au lieu de celle de $\overset{n-1}{\underset{0}{\Sigma}}(-1)^{\mu}\psi(m,\mu)$, comme nous l'avons fait, on trouvera au lieu de la formule (53.)Editorial note: Printed (53.); the formula meant is presumably (153.), an apparent misprint. la suivante: \[ \Phi\alpha=\frac{1}{ec}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{(2\mu+1)\varpi}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{(2\mu+1)\varpi}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right\}, \tag{156.} \] c'est-à-dire: \[ \begin{aligned}\Phi\alpha&=\frac{\varpi}{ec}(\varepsilon_{0}-3\varepsilon_{1}+5\varepsilon_{2}-7\varepsilon_{3}+....+(-1)^{\mu}.(2\mu+1)\varepsilon_{\mu}-....)\\ &-\frac{\varpi}{ec}(\varepsilon'_{0}-3\varepsilon'_{1}+5\varepsilon'_{2}-7\varepsilon'_{3}+....+(-1)^{\mu}.(2\mu+1)\varepsilon_{\mu}-....),\end{aligned} \tag{157.} \] Editorial note: In the second row the last term is printed with $\varepsilon_{\mu}$ without the prime; $\varepsilon'_{\mu}$ is expected, an apparent misprint. où \[ \left\{\begin{aligned}\varepsilon_{\mu}&=\frac{1}{\left(\alpha-\frac{\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{1}{\left(\alpha-\frac{3\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}+\frac{1}{\left(\alpha-\frac{5\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-....\\ \varepsilon'_{\mu}&=\frac{1}{\left(\alpha+\frac{\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{1}{\left(\alpha+\frac{3\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}+\frac{1}{\left(\alpha+\frac{5\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-....\end{aligned}\right. \tag{158.} \]

27.

Cherchons maintenant l'expression de $f\alpha$, au moyen de la deuxième des formules (126.). En y faisant $\beta=\frac{\alpha}{2n+1}$, et remarquant, qu'alors la limite des quantités contenues dans les deux premières lignes, devient égale à zéro, on aura: \[ \begin{aligned}f\alpha&=\text{lim.}(-1)^{n}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m}.\psi(n-m,n-\mu)\\ &+\text{lim.}(-1)^{n}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m}.\psi_{1}(n-m,n-\mu),\end{aligned} \tag{159.} \]

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ou l'on a fait pour abréger: \[ \begin{aligned}\psi(n-m,n-\mu)&=\frac{1}{2n+1}\left\{f\left(\frac{\alpha+m\omega+\mu\varpi i}{2n+1}\right)+f\left(\frac{\alpha-m\omega-\mu\varpi i}{2n+1}\right)\right\},\\ \psi_{1}(n-m,n-\mu)&=\frac{1}{2n+1}\left\{f\left(\frac{\alpha+m\omega-\mu\varpi i}{2n+1}\right)+f\left(\frac{\alpha-m\omega+\mu\varpi i}{2n+1}\right)\right\}.\end{aligned} \] Maintenant on a: \[ f(\beta-\varepsilon)+f(\beta-\varepsilon)=\frac{2f\beta.f\varepsilon}{1+e^{2}c^{2}\varphi^{2}\varepsilon.\varphi^{2}\beta}=\frac{f\varepsilon}{e^{2}c^{2}\varphi^{2}\varepsilon}.\frac{f\beta}{\varphi^{2}\beta+\frac{1}{e^{2}c^{2}\varphi^{2}\varepsilon}}. \] Editorial note: Printed $f(\beta-\varepsilon)$ in both terms; the first term is presumably $f(\beta+\varepsilon)$, an apparent misprint.

Soit \[ \varepsilon=\frac{m\omega+\mu\varpi i}{2n+1}, \] on aura: \[ \begin{aligned}\frac{1}{\varphi\varepsilon}&=-iec.\Phi\left(\frac{\omega}{2}+\frac{\varpi}{2}i-\varepsilon\right)=-iec.\Phi\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right),\\ \frac{f\varepsilon}{\varphi\varepsilon}&=-\frac{i\sqrt{(e^{2}+c^{2})}}{F\left(\varepsilon-\frac{\varpi}{2}i\right)}=-i\sqrt{(e^{2}+c^{2})}.\frac{c}{\sqrt{(e^{2}+c^{2})}}.F\left(\varepsilon-\frac{\omega}{2}-\frac{\varpi}{2}i\right)\\ &=-ci.F.\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right).\end{aligned} \] Donc on aura, en substituant et mettant $m$ et $\mu$ respectivement au lieu de $n-m$ et $n-\mu$: \[ \psi(m,\mu)=\frac{1}{e}.\frac{2\varphi\left(\frac{(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right).F\left(\frac{(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}{(2n+1)\left[\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)\right]}. \] On aura la valeur de $\psi_{1}(m,\mu)$, en changeant seulement le signe de $i$. En faisant maintenant \[ \theta(m,\mu)=\frac{(2m+1)\omega+(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}} \] et \[ \theta_{1}(m,\mu)=\frac{(2m+1)\omega-(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}, \] et cherchant ensuite la limite de la fonction \[ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m}.\psi(m,\mu), \] de la même manière que précédemment, on trouvera: \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m}.\psi(m,\mu)=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\theta(m,\mu)\right\} \] et \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m}.\psi_{1}(m,\mu)=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\theta_{1}(m,\mu)\right\}; \]

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donc en substituant dans (159.), et remettant les valeurs de $\theta(m,\mu)$ et $\theta_{1}(m,\mu)$, on a: \[ f\alpha=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\left\{\frac{(2m+1)\omega+(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}}+\frac{(2m+1)\omega-(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}\right\}. \tag{160.} \]

La quantité renfermée entre les crochets peut aussi se mettre sous la forme: \[ \frac{2[\alpha-(m+\tfrac{1}{2})\omega]}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{2[\alpha+(m+\tfrac{1}{2})\omega]}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}, \] donc aussi: \[ f\alpha=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\frac{2[\alpha-(m+\tfrac{1}{2})\omega]}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\frac{2[\alpha+(m+\tfrac{1}{2})\omega]}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right\}. \tag{161.} \]

On aura de la même manière: \[ F(\alpha)=\frac{1}{c}.\overset{\infty}{\underset{0}{\Sigma}}_{m}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{(2\mu+1)\varpi}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})\varpi^{2}}+\overset{\infty}{\underset{0}{\Sigma}}(-1)^{\mu}.\frac{(2\mu+1)\varpi}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})\varpi^{2}}\right\}. \tag{162.} \] Editorial note: Both denominators are printed with $(\mu+\tfrac{1}{2})\varpi^{2}$; the square on the bracket, $(\mu+\tfrac{1}{2})^{2}\varpi^{2}$, is expected as in (161.), an apparent misprint.

28.

Venons maintenant aux formules (130.). Pour trouver la valeur du second membre, après avoir fait $\beta=\frac{\alpha}{2n+1}$, et supposé $n$ infini, nous allons d'abord chercher la limite de l'expression suivante: \[ t=\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}, \tag{163.} \] où $k$ et $l$ sont deux quantités indépendantes de $n$, $m$, $\mu$.

En prenant le logarithme, et faisant, pour abréger: \[ \psi(m,\mu)=\log\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}, \tag{164.} \]

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on aura: \[ \log t=\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}.\psi(m,\mu). \tag{165.} \]

Considérons d'abord l'expression: $\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)$. Soit \[ \theta(m,\mu)=\log\left\{\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}}\right\}, \tag{166.} \] on aura: \[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}.\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}. \] Cela posé, je dis que le second membre de cette équation est pour toute valeur de $m$ et $\mu$ de la forme: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{v}{(2n+1)^{2}}. \]

Pour démontrer cela, il faut distinguer deux cas, si la limite de $\frac{m\omega+\mu\varpi i}{2n+1}$ est une quantité différente de zéro, et si elle est égale à zéro.

$a)$ Dans le premier cas on aura, en nommant $a$ la limite, dont il s'agit: \[ \begin{aligned}\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)&=\Phi^{2}a+v,\\ \varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)&=\Phi^{2}a+v',\\ \varphi^{2}\left(\frac{\alpha}{2n+1}\right)&=\frac{\alpha^{2}}{(2n+1)^{2}}+\frac{v'}{(2n+1)^{2}},\end{aligned} \] donc: \[ \begin{aligned}1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}&=1-\frac{\alpha^{2}}{(2n+1)^{2}\varphi^{2}a}+\frac{v}{(2n+1)^{2}},\\ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}&=1-\frac{\alpha^{2}}{(2n+1)^{2}.\varphi^{2}a}+\frac{v'}{(2n+1)^{2}}.\end{aligned} \]

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On a de même \[ \begin{aligned}1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}&=1-\frac{\alpha^{2}}{(2n+1)^{2}\left\{\frac{m\omega+\mu\varpi i+k}{(2n+1)}\right\}}=1-\frac{\alpha^{2}}{(2n+1)^{2}a^{2}}+\frac{v}{(2n+1)^{2}},\\ 1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}&=1-\frac{\alpha^{2}}{(2n+1)^{2}a^{2}}+\frac{v'}{(2n+1)^{2}}.\end{aligned} \] Editorial note: In the first line the braced quotient in the middle denominator is printed without a square; a square is expected, as in the neighbouring terms.

En substituant ces valeurs, l'expression de $\psi(m,\mu)-\theta(m,\mu)$ prendra la forme: \[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1-\frac{v}{(2n+1)^{2}}}{1-\frac{v'}{(2n+1)^{2}}}.\frac{1-\frac{v_{1}}{(2n+1)^{2}}}{1-\frac{v'_{1}}{(2n+1)^{2}}}\right\}, \] les quantités $v$, $v'$, $v_{1}$, $v'_{1}$ ayant toutes zéro pour limite.

On a donc: \[ \log\left(1-\frac{v}{(2n+1)^{2}}\right)=\frac{v}{(2n+1)^{2}}\ \text{etc.}, \] et par conséquent: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{v}{(2n+1)^{2}}. \]

$b)$ Si la limite de la quantité $\frac{m\omega+\mu\varpi i}{2n+1}$ est égale à zéro, on aura: \[ \begin{aligned}\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)&=\frac{(m\omega+\mu\varpi i+k)^{2}}{(2n+1)^{2}}+A.\frac{(m\omega+\mu\varpi i+k)^{4}}{(2n+1)^{4}}+....,\\ \varphi^{2}\left(\frac{\alpha}{2n+1}\right)&=\frac{\alpha^{2}}{(2n+1)^{2}}+A'.\frac{\alpha^{4}}{(2n+1)^{4}}+....,\end{aligned} \] donc: \[ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}=1-\frac{\alpha^{2}+A'.\frac{\alpha^{4}}{(2n+1)^{2}}+.....}{(m\omega+\mu\varpi i+k)^{2}+A.\frac{(m\omega+\mu\varpi i+k)^{4}}{(2n+1)^{2}}}. \]

Si maintenant $m\omega+\mu\varpi i$ ne va pas indéfiniment en augmentant avec $n$, on aura: \[ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+m\varpi i+k}{2n+1}\right)}=1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}+\frac{B}{(2n+1)^{2}}, \] Editorial note: Printed $m\varpi i$ in the argument of the denominator function instead of $\mu\varpi i$; an apparent misprint. de même: \[ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}=1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}+\frac{C}{(2n+1)^{2}}, \] Editorial note: The argument of the denominator function is printed with $+k$ (the glyph is blotted) where $+l$ is expected, as in the right-hand side. donc dans ce cas:

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\[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1-\frac{B'}{(2n+1)^{2}}}{1-\frac{C'}{(2n+1)^{2}}}\right\}, \] $B'$ et $C'$ ayant des limites finies, ou bien: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{D}{(2n+1)^{2}}, \] la limite de $D$ étant également une quantité finie.

Si au contraire la quantité $m\omega+\mu\varpi i$ augmente indéfiniment avec $n$, on a: \[ \begin{aligned}\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}&=1+\frac{\alpha^{2}}{(2n+1)^{2}}.\left\{\frac{A-A'\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}{1+A\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)^{2}}\right\}\\ &\times\frac{1}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}};\end{aligned} \] or les quantités $\frac{1}{m\omega+\mu\varpi i+k}$, $\frac{m\omega+\mu\varpi i+k}{2n+1}$ ont zéro pour limite; donc la quantité précédente sera de la forme: \[ 1+\frac{\alpha^{2}}{(2n+1)^{2}}.A'', \] $A''$ ayant une quantité finie pour limite. En changeant $k$ en $l$, et désignant la valeur correspondante de $A''$ par $A''_{1}$, la valeur de $\psi(m,\mu)-\theta(m,\mu)$ deviendra: \[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1+\frac{\alpha^{2}}{(2n+1)^{2}}A''}{1+\frac{\alpha^{2}}{(2n+1)^{2}}A''_{1}}\right\}=\frac{\alpha^{2}(A''-A''_{1})}{(2n+1)^{2}}+\frac{v}{(2n+1)^{2}}. \] Maintenant la limite de $A''$ est la même que celle de $A$; or il est clair, que cette dernière limite est indépendante de $k$, $m$, $\mu$ (elle est en effet égale au coefficient de $\alpha^{4}$ dans le développement de $\Phi^{2}\alpha$). Donc on aura: \[ A''=M+v, \] et en changeant $k$ en $l$: \[ A''_{1}=M+v', \] d'où $A''-A''_{1}=v-v'=v$. Donc $A''-A''_{1}$ a zéro pour limite, et par conséquent on a: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{v}{(2n+1)^{2}}. \]

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Donc nous avons démontré, qu'en faisant \[ \psi(m,\mu)-\theta(m,\mu)=\frac{A_{m,\mu}}{(2n+1)^{2}}, \tag{167.} \] la limite de $A_{m,\mu}$ sera égale à zéro toutes les fois que $m\omega+\mu\varpi i$ augmente indéfiniment avec $n$, et qu'elle sera égale à une quantité finie dans le cas contraire.

29.

Cela posé, considérons la quantité \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu). \] En substituant la valeur de $\psi(m,\mu)$, il viendra: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)+\frac{1}{(2n+1)^{2}}.\overset{n}{\underset{1}{\Sigma}}_{\mu}(A_{m,\mu}). \tag{168.} \] Soit $\nu$ le plus grand nombre entier, contenu dans $\sqrt{n}$, on peut faire: \[ \begin{aligned}\overset{n}{\underset{1}{\Sigma}}_{\mu}A_{m,\mu}&=A_{m,1}+A_{m,2}+.....+A_{m,\nu}\\ &+A_{m,\nu+1}+A_{m,\nu+2}+.....+A_{m,n}.\end{aligned} \]

Or, d'après la nature des quantités $A_{m,\mu}$, la somme contenue dans la première ligne sera égale à $\nu.A_{m}$, et la seconde égale à $A'_{m}(n-\nu-1)$, où $A_{m}$ est une quantité finie et $A'_{m}$ une quantité, qui a zéro pour limite, donc: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}A_{m,\mu}=\nu.A_{m}+(n-\nu-1)A'_{m}=(2n+1).B_{m}, \] où \[ B_{m}=\frac{\nu}{2n+1}A_{m}+\frac{n-\nu-1}{2n+1}A'_{m}. \] Donc la quantité $B_{m}$ a zéro pour limite, en remarquant, que $\nu$ ne surpasse pas $\sqrt{n}$.

Par là l'expression de $\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)$ se change en: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)+\frac{B_{m}}{2n+1}. \tag{169.} \]

Pour avoir la limite de $\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)$, j'écris \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)-\overset{\infty}{\underset{n+1}{\Sigma}}_{\mu}\theta(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)-\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n). \] Or on peut trouver la valeur de $\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)$ comme suit.

page 169

On a: \[ \begin{aligned}\theta(m.\,\mu+n)&=\log\left\{\frac{1-\frac{\alpha^{2}}{[m\omega+(\mu+n)\varpi i+k]^{2}}}{1-\frac{\alpha^{2}}{[m\omega+(\mu+n)\varpi i+l]^{2}}}\right\}\\ &=\alpha^{2}\left\{\frac{1}{[m\omega+(\mu+n)\varpi i+l]^{2}}-\frac{1}{[m\omega+(\mu+n)\varpi i+k]^{2}}\right\}\\ &-\tfrac{1}{2}\alpha^{4}\left\{\frac{1}{[m\omega+(\mu+n)\varpi i+l]^{2}}-\frac{1}{[m\omega+(\mu+n)\varpi i+k]^{2}}\right\}\\ &+\ \text{etc.}\end{aligned} \] Editorial note: Printed $\theta(m.\,\mu+n)$ with a full stop instead of a comma between the arguments; an apparent misprint. De là on tire: \[ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)=\frac{\alpha^{2}}{n}.\Sigma\frac{1}{n}.\theta\left(\frac{\mu}{n}\right)-\frac{\alpha^{4}}{2n^{3}}.\Sigma\frac{1}{n}.\theta_{1}\left(\frac{\mu}{n}\right)+...., \] où \[ \begin{gathered}\theta\left(\frac{\mu}{n}\right)=\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{2}}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{2}},\\ \theta_{1}\left(\frac{\mu}{n}\right)=\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{4}}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{4}},\\ \text{etc.}\end{gathered} \] Or on sait que la limite de \[ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\frac{1}{n}\theta\left(\frac{\mu}{n}\right)\ \text{est égale à}\ \int_{0}^{x}\theta(x)\,\partial x, \] donc: \[ \begin{gathered}\overset{\mu}{\underset{1}{\Sigma}}_{\mu}\frac{1}{n}\theta\left(\frac{\mu}{n}\right)=\int_{0}^{x}\theta(x)\,\partial x+v,\\ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\frac{1}{n}\theta_{1}\left(\frac{\mu}{n}\right)=\int_{0}^{x}\theta_{1}(x)\,\partial x+v_{1},\\ \text{etc},\end{gathered} \] et par conséquent, en substituant: \[ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)=\frac{\alpha^{2}}{n}.\int_{0}^{x}\theta(x)-\frac{\alpha^{4}}{2n^{3}}.\int_{0}^{x}\theta_{1}(x)+....+\frac{v\alpha^{2}}{n}-\frac{v_{1}\alpha^{3}}{2n^{3}}+.... \] Editorial note: The integrals are printed without the differential $\partial x$, and the last fraction has $\alpha^{3}$ where $\alpha^{4}$ is expected; both apparent misprints, reproduced as printed. or \[ \begin{gathered}\theta(x)=\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+x\varpi i\right)^{2}}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+x\varpi i\right)^{2}}\\ \text{etc.}\end{gathered} \] donc on aura:

page 170

\[ \begin{aligned}\int_{0}^{x}\theta(x)\,\partial x&=\frac{1}{\varpi i}.\left\{\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+x\varpi i\right)}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+x\varpi i\right)}\right\}\\ &-\frac{1}{\varpi i}.\left\{\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i\right)}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i\right)}\right\}\\ &=\frac{1}{\varpi i}.\frac{k-l}{n}.\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+x\varpi i\right)\left(\frac{m\omega+k}{n}+\varpi i+x\varpi i\right)}\\ &-\frac{1}{\varpi i}.\frac{k-l}{n}.\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i\right)\left(\frac{m\omega+k}{n}+\varpi i\right)}.\end{aligned} \]

La limite de cette expression de $\displaystyle\int_{0}^{x}\theta(x)\,\partial x$ est zéro pour une valeur quelconque de $x$. De même on trouvera, que la limite de $\displaystyle\int_{0}^{x}\theta_{1}(x)\,\partial x$ est zéro, donc: \[ \begin{aligned}\overset{\mu}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)&=\frac{\alpha^{2}}{n}.v-\frac{\alpha^{4}}{2n^{3}}.v'+\frac{\alpha^{6}}{3n^{5}}.v''-....\\ &=\frac{\alpha^{2}}{2n+1}.\left\{v-\frac{\alpha^{2}}{2n^{2}}v'+\frac{\alpha^{4}}{3n^{4}}v''-....\right\}\frac{2n+1}{n}=\frac{v}{2n+1},\end{aligned} \] donc aussi, en faisant $\mu=\infty$: \[ \overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)=\frac{v}{2n+1}, \] d'où: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu+n)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)-\frac{v}{2n+1},\ \text{et} \] \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)+\frac{v_{m}}{2n+1}, \tag{170.} \] $v_{m}$ ayant zéro pour limite. De là on tire \[ \overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{n}{\underset{1}{\Sigma}}_{m}\left\{\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)\right\}+\overset{n}{\underset{1}{\Sigma}}\frac{v_{m}}{2n+1}. \] En prenant la limite des deux membres et remarquant, que \[ \overset{n}{\underset{1}{\Sigma}}\frac{v_{m}}{2n+1}=\frac{v_{1}+v_{2}+.....v_{m}}{2n+1}=v, \] on aura: \[ \text{lim.}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{m}\left\{\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)\right\}. \tag{171.} \] En remettant les valeurs de $\psi(m,\mu)$ et $\theta(m,\mu)$, et passant des logarithmes aux nombres, on en tire:

page 171

\[ \text{lim.}\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}=\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}}\right)\right\}. \tag{172.} \]

Par une analyse toute semblable à la précédente, mais plus simple, on trouvera de même: \[ \text{lim.}\overset{n}{\underset{1}{\Pi}}_{m}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+l}{2n+1}\right)}}\right\}=\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\frac{1-\frac{\alpha^{2}}{(m\omega+k)^{2}}}{1-\frac{\alpha^{2}}{(m\omega+l)^{2}}}\right\}, \tag{173.} \] \[ \text{lim.}\overset{n}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{\mu\varpi i+k}{2n+1}\right)}}\right\}=\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\alpha^{2}}{(\mu\varpi i+k)^{2}}}{1-\frac{\alpha^{2}}{(\mu\varpi i+l)^{2}}}\right\}. \tag{174.} \] Editorial note: In the left side of the last equation the lower fraction is printed with $k$ in the argument of the denominator function, as the upper one; presumably $l$ was meant, as in the right side. An apparent misprint.

30.

Maintenant rien n'est plus facile, que de trouver les valeurs de $\Phi\alpha$, $f\alpha$, $F\alpha$.

Considérons d'abord la première formule (130.). On a: \[ \begin{aligned}e^{2}c^{2}\Phi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)&=-\frac{1}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}-\frac{\omega}{2}-\frac{\varpi}{2}i\right)}\\ &=-\frac{1}{\varphi^{2}\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)},\end{aligned} \] donc:

page 172

\[ \begin{aligned}\overset{n}{\underset{1}{\Pi}}_{m}\overset{\mu}{\underset{1}{\Pi}}\left\{\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)\varphi^{2}\beta}\right\}&=\frac{\overset{n}{\underset{1}{\Pi}}_{m}\overset{\mu}{\underset{1}{\Pi}}.\left\{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)}\right\}}{\overset{n}{\underset{1}{\Pi}}_{m}\overset{\mu}{\underset{1}{\Pi}}.\left\{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}\right\}}\\ &=\overset{n}{\underset{1}{\Pi}}_{m}\overset{\mu}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i}{2n+1}\right)}}\right\}.\end{aligned} \]

Cela posé, si l'on fait $\beta=\frac{\alpha}{2n+1}$ et qu'on suppose $n$ infini, il viendra, en faisant usage des formules (172.), (173.), (174.), et remarquant que $\alpha$ est la limite de $(2n+1)\varphi\left(\frac{\alpha}{2n+1}\right)$ et égale à $\alpha$: \[ \begin{aligned}\Phi\alpha&=\alpha\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{1-\frac{\alpha^{2}}{(m\omega)^{2}}\right\}.\overset{\infty}{\underset{1}{\Pi}}\left\{1+\frac{\alpha^{2}}{(\mu\varpi)^{2}}\right\}\\ &\times\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i)^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right).\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{(m\omega-\mu\varpi i)^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega-(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\right\}.\end{aligned} \tag{175.} \]

Les deux formules (130.) donneront de la même manière, en faisant $\beta=\frac{\alpha}{2n+1}$, et remarquant, que $f(0)=1$, $F(0)=1$: \[ \begin{aligned}f\alpha&=\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+\mu\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\left(\frac{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega-\mu\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega-(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\right\}\\ &\times\overset{\infty}{\underset{0}{\Pi}}_{\mu}\left\{1-\frac{\alpha^{2}}{(m+\tfrac{1}{2})^{2}\omega^{2}}\right\},\end{aligned} \tag{176.} \] \[ \begin{aligned}F\alpha&=\overset{\infty}{\underset{0}{\Pi}}_{\mu}\left\{1+\frac{\alpha^{2}}{(\mu+\tfrac{1}{2})\varpi^{2}}\right\}.\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{[m\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\left(\frac{1-\frac{\alpha^{2}}{[m\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\right\}.\end{aligned} \tag{177.} \] Editorial note: The two factors of the last product are printed identically, where the earlier formulas have opposite signs; an apparent misprint, reproduced as printed. In the preceding formulas the factor $(m+\tfrac{1}{2})^{2}\omega^{2}$ of (176.) carries its square, and the product index there is printed $\mu$ although $m$ occurs; the factor $(\mu+\tfrac{1}{2})\varpi^{2}$ of (177.) is printed without a square on the bracket. All reproduced as printed.

On peut aussi donner une forme réelle aux expressions précédentes comme suit:

page 173

\[ \begin{aligned} \Phi\alpha &= \alpha.\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(1+\frac{\alpha^{2}}{\mu^{2}\varpi^{2}}\right).\overset{\infty}{\underset{1}{\Pi}}_{m}\left(1-\frac{\alpha^{2}}{m^{2}\omega^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\overset{\infty}{\underset{1}{\Pi}}_{\mu}\frac{1+\frac{(\alpha+m\omega)^{2}}{\mu^{2}\varpi^{2}}}{1+\frac{[\alpha+(m-\tfrac{1}{2})\omega]^{2}}{\mu^{2}\varpi^{2}}}.\frac{1+\frac{(\alpha-m\omega)^{2}}{\mu^{2}\varpi^{2}}}{1+\frac{[\alpha-(m-\tfrac{1}{2})\omega]^{2}}{\mu^{2}\varpi^{2}}}.\left(\frac{1+\frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{\mu^{2}\varpi^{2}}}{1+\frac{m^{2}\omega^{2}}{\mu^{2}\varpi^{2}}}\right)^{2}, \end{aligned} \tag{178.} \] \[ \begin{aligned} f\alpha &= \overset{\infty}{\underset{1}{\Pi}}_{m}\left(1-\frac{\alpha^{2}}{(m-\tfrac{1}{2})^{2}\omega^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\overset{\infty}{\underset{1}{\Pi}}_{\mu}\frac{1+\frac{[\alpha+(m-\tfrac{1}{2})\omega]^{2}}{\mu^{2}\varpi^{2}}}{1+\frac{[\alpha+(m-\tfrac{1}{2})\omega]^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}.\frac{1+\frac{[\alpha-(m-\tfrac{1}{2})\omega]^{2}}{\mu^{2}\varpi^{2}}}{1+\frac{[\alpha-(m-\tfrac{1}{2})\omega]^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}.\left(\frac{1+\frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}{1+\frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{\mu^{2}\varpi^{2}}}\right)^{2}, \end{aligned} \tag{179.} \] \[ \begin{aligned} F\alpha &= \overset{\infty}{\underset{0}{\Pi}}_{\mu}\left(1+\frac{\alpha^{2}}{(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\overset{\infty}{\underset{1}{\Pi}}_{\mu}\frac{1+\frac{(\alpha+m\omega)^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}{1+\frac{[\alpha+(m-\tfrac{1}{2})\omega]^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}.\frac{1+\frac{(\alpha-m\omega)^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}{1+\frac{[\alpha-(m-\tfrac{1}{2})\omega]^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}.\left(\frac{1+\frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}{1+\frac{m^{2}\omega^{2}}{(\mu-\tfrac{1}{2})^{2}\varpi^{2}}}\right)^{2}. \end{aligned} \tag{180.} \]

Ces transformations s'opèrent aisément au moyen de la formule: \[ \begin{aligned} \left(1-\frac{\alpha^{2}}{(a+bi)^{2}}\right)\left(1-\frac{\alpha^{2}}{(a-bi)^{2}}\right) &= \left(1+\frac{\alpha}{a+bi}\right)\left(1+\frac{\alpha}{a-bi}\right)\left(1-\frac{\alpha}{a+bi}\right)\left(1-\frac{\alpha}{a-bi}\right)\\ &= \frac{(\alpha+a)^{2}+b^{2}}{a^{2}+b^{2}}.\frac{(\alpha-a)^{2}+b^{2}}{a^{2}+b^{2}} = \left(1+\frac{(\alpha+a)^{2}}{b^{2}}\right)\left(1+\frac{(\alpha-a)^{2}}{b^{2}}\right).\frac{1}{\left(1+\frac{a^{2}}{b^{2}}\right)^{2}}. \end{aligned} \]

31.

Dans ce qui précède nous sommes parvenus à deux espèces d'expressions des fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$: les unes donnent ces fonctions décomposées en fractions partielles, dont la totalité forme des séries infinies doubles, les autres donnent ces mêmes fonctions décomposées en un nombre infini de facteurs, dont chacun est à son tour composé d'une infinité de facteurs.

Or on peut beaucoup simplifier les formules précédentes au moyen des fonctions exponentielles et circulaires. C'est ce que nous allons voir par ce qui suit:

Considérons d'abord les équations (178.), (179.), (180.). En vertu des formules connues, on a:

page 174

\[ \frac{\sin y}{y} = \overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(1-\frac{y^{2}}{\mu^{2}\pi^{2}}\right);\ \cos y = \overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(1-\frac{y^{2}}{(\mu-\tfrac{1}{2})^{2}\pi^{2}}\right); \] donc: \[ \overset{\infty}{\underset{1}{\Pi}}_{\mu}\frac{\left(1-\frac{z^{2}}{\mu^{2}\pi^{2}}\right)}{\left(1-\frac{y^{2}}{(\mu-\tfrac{1}{2})\pi^{2}}\right)} = \frac{\sin z}{z\cos y};\ \overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{z^{2}}{(\mu-\tfrac{1}{2})\pi^{2}}}{1-\frac{y^{2}}{(\mu-\tfrac{1}{2})\pi^{2}}}\right) = \frac{\cos z}{\cos y}. \] Editorial note: In both products the bracket $(\mu-\tfrac{1}{2})$ under $\pi^{2}$ is printed without the square (the denominator of the first product, and the numerator and denominator of the second); an apparent misprint for $(\mu-\tfrac{1}{2})^{2}\pi^{2}$, reproduced as printed.

En vertu de cette formule il est clair, que les expressions de $\Phi\alpha$, $f\alpha$, $F\alpha$ peuvent être mises sous la forme: \[ \begin{aligned} \Phi\alpha &= \frac{\varpi}{\pi}.\frac{\sin\left(\alpha\frac{\pi}{\varpi}i\right)}{i}.\overset{\infty}{\underset{1}{\Pi}}_{m}\left(1-\frac{\alpha^{2}}{m^{2}\omega^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\left(\frac{\sin(\alpha+m\omega)\frac{\pi}{\varpi}i.\sin(\alpha-m\omega)\frac{\pi}{\varpi}i.\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}{\cos[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\sin^{2}m\omega\frac{\pi}{\varpi}i}.\frac{\left(m\omega\frac{\pi}{\varpi}i\right)^{2}}{(\alpha+m\omega)(\alpha-m\omega).\frac{\pi^{2}}{\omega^{2}}i^{2}}\right), \end{aligned} \] Editorial note: In the denominator of the second fraction the factor $\frac{\pi^{2}}{\omega^{2}}$ is printed with $\omega$, where $\varpi$ is expected (compare the factor $\frac{\pi}{\varpi}$ elsewhere in the line); an apparent misprint, reproduced as printed. \[ \begin{aligned} f\alpha &= \overset{\alpha}{\underset{1}{\Pi}}_{m}\left(1-\frac{\alpha^{2}}{(m-\tfrac{1}{2})^{2}\omega^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\left(\text{tang}[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\text{tang}[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cot^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i.\frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{\alpha^{2}-(m-\tfrac{1}{2})^{2}\omega^{2}}\right), \end{aligned} \] Editorial note: In the first line of the expression for $f\alpha$ the upper limit of the product is printed as $\alpha$, where $\infty$ is expected; an apparent misprint, reproduced as printed. \[ F\alpha = \cos\left(\alpha\frac{\pi}{\varpi}i\right).\overset{\infty}{\underset{1}{\Pi}}_{m}\left(\frac{\cos(\alpha+m\omega)\frac{\pi}{\varpi}i.\cos(\alpha-m\omega)\frac{\pi}{\varpi}i.\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}{\cos[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos^{2}m\omega.\frac{\pi}{\varpi}i}\right). \]

On trouvera des expressions réelles, en substituant au lieu des fonctions circulaires leurs expressions en fonctions exponentielles.

On a: \[ \begin{gathered} \sin(a-b).\sin(a+b) = \sin^{2}a-\sin^{2}b,\\ \cos(a+b).\cos(a-b) = \cos^{2}a-\sin^{2}b, \end{gathered} \] donc: \[ \frac{\sin(\alpha+m\omega)\frac{\pi}{\varpi}i.\sin(\alpha-m\omega)\frac{\pi}{\varpi}i}{\sin^{2}m\omega\frac{\pi}{\varpi}i} = -\left\{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}m\omega\frac{\pi}{\varpi}i}\right\}, \] \[ \frac{\cos[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i} = 1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}, \]

page 175

\[ \text{tang}[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\text{tang}[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cot^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i = -\frac{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}. \]

D'après cela et en remarquant, que \[ \frac{m^{2}\omega^{2}}{\alpha^{2}-m^{2}\omega^{2}} = -\frac{1}{1-\frac{\alpha^{2}}{m^{2}\omega^{2}}}\ \text{et} \] \[ \frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{\alpha^{2}-(m-\tfrac{1}{2})^{2}\omega^{2}} = -\frac{1}{1-\frac{\alpha^{2}}{(m-\tfrac{1}{2})^{2}\omega^{2}}}, \] il est clair, qu'on aura: \[ \Phi\alpha = \frac{\pi}{\varpi}.\frac{\sin\left(\frac{\alpha}{\varpi}\pi i\right)}{i}.\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}m\omega\frac{\pi}{\varpi}i}}{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}, \tag{181.} \] Editorial note: The factor before the sine is printed as the fraction with $\pi$ over $\varpi$, whereas the preceding expression has $\varpi$ over $\pi$; an apparent misprint, reproduced as printed. \[ f\alpha = \overset{\infty}{\underset{0}{\Pi}}_{m}.\frac{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}(m+\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m+\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}, \tag{182.} \] \[ F\alpha = \cos\left(\frac{\alpha}{\varpi}\pi i\right).\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\frac{\sin^{2}\frac{\alpha}{\varpi}\pi i}{\cos^{2}m\frac{\omega}{\varpi}\pi i}}{1-\frac{\sin^{2}\frac{\alpha}{\varpi}\pi i}{\cos^{2}(m-\tfrac{1}{2})\frac{\omega}{\varpi}\pi i}}. \tag{183.} \]

page 176

En substituant au lieu des cosinus et sinus d'arcs imaginaires leurs valeurs en quantités exponentielles, ces formules deviendront: \[ \Phi\alpha = \tfrac{1}{2}.\frac{\varpi}{\pi}\left(h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}\right).\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{m\frac{\omega}{\varpi}\pi}-h^{-m\frac{\omega}{\varpi}\pi}}\right)^{2}}{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}+h^{-(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}, \tag{184.} \] \[ f\alpha = \overset{\infty}{\underset{0}{\Pi}}_{m}.\frac{1-\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}-h^{-(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}+h^{-(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}, \tag{185.} \] \[ F\alpha = \tfrac{1}{2}\left(e^{\frac{\alpha}{\varpi}\pi}+e^{-\frac{\alpha}{\varpi}\pi}\right).\overset{\infty}{\underset{1}{\Pi}}_{m}\frac{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{m\frac{\omega}{\varpi}\pi}+h^{-m\frac{\omega}{\varpi}\pi}}\right)^{2}}{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}+h^{-(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}, \tag{186.} \] où $h$ est le nombre 2,1718218..... Editorial note: The value of $h$ is printed as 2,1718218; the base of the natural logarithms is 2,7182818; an apparent misprint, reproduced as printed. Editorial note: In the first factor of the last formula the base is printed as $e$ in the sum, where the other exponentials of the page use $h$; an apparent misprint, reproduced as printed.

On peut encore transformer cette formule de la manière suivante:

Si l'on remplace $\alpha$ par $\alpha i$, on aura les valeurs de $\Phi(\alpha i)$, $f(\alpha i)$, $F(\alpha i)$. En changeant maintenant $c$ en $e$ et $e$ en $c$, les quantités: \[ \omega;\ \varpi;\ \Phi(\alpha i);\ f(\alpha i);\ F(\alpha i) \] se changeront respectivement en: \[ \varpi;\ \omega;\ i\Phi\alpha;\ F\alpha;\ f\alpha, \] donc les formules précédentes donneront: \[ \Phi\alpha = \frac{\omega}{\pi}.\sin\frac{\alpha\pi}{\omega}.\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1+\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{m\varpi\pi}{\omega}}-h^{-\frac{m\varpi\pi}{\omega}}\right)^{2}}}{1-\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{(2m-1)\varpi\pi}{2\omega}}+h^{-\frac{(2m-1)\varpi\pi}{2\omega}}\right)^{2}}}, \tag{187.} \]

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\[ F\alpha = \overset{\infty}{\underset{0}{\Pi}}_{m}\frac{1+\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{(2m+1)\varpi\pi}{2\omega}}-h^{-\frac{(2m+1)\varpi\pi}{2\omega}}\right)^{2}}}{1-\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{(2m+1)\varpi\pi}{2\omega}}+h^{-\frac{(2m+1)\varpi\pi}{2\omega}}\right)^{2}}}, \tag{188.} \] \[ f\alpha = \cos\left(\frac{\alpha\pi}{\omega}\right).\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{m\varpi\pi}{\omega}}+h^{-\frac{m\varpi\pi}{\omega}}\right)^{2}}}{1-\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{(2m+1)\varpi\pi}{2\omega}}+h^{-\frac{(2m+1)\varpi\pi}{2\omega}}\right)^{2}}}, \tag{189.} \]

32.

Considérons maintenant les formules (160.), (161.), (162.).

On a: \[ \overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{(2\mu+1)\pi}{y^{2}+(\mu+\tfrac{1}{2})^{2}\pi^{2}} = \frac{2}{h^{y}+h^{-y}}, \] donc, en faisant \[ y = [\alpha\pm(m+\tfrac{1}{2})\omega]\frac{\pi}{\varpi}: \] \[ \overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{(2\mu+1)\varpi}{[\alpha\pm(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}} = \frac{2\pi}{\varpi}.\frac{1}{h^{[\alpha\pm(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha\pm(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}. \] En vertu de cette formule il est aisé de voir, que les expressions (153.), (162.) de $\Phi\alpha$ et $F\alpha$ deviendront: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{1}{h^{[\alpha-(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha-(m-\frac{1}{2})\omega]\frac{\pi}{\varpi}}}-\frac{1}{h^{[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}\right\}, \tag{190.} \] Editorial note: In the first denominator the second exponent is printed with $m-\tfrac{1}{2}$ where the first has $m+\tfrac{1}{2}$; an apparent misprint, reproduced as printed. \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}\left\{\frac{1}{h^{[\alpha-(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha-(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}+\frac{1}{h^{[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}\right\}. \tag{191.} \]

Les expressions précédentes de $\Phi\alpha$, $F\alpha$, peuvent être mises encore sous beaucoup d'autres formes: je vais rappeler les plus remarquables. D'abord en réunissant les termes du second membre, on trouvera:

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\[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{\left(h^{\frac{\alpha\pi}{\varpi}}-h^{-\frac{\alpha\pi}{\varpi}}\right)\left(h^{(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}-h^{-(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}\right)}{h^{\frac{2\alpha\pi}{\varpi}}+h^{-\frac{2\alpha\pi}{\varpi}}+h^{(2m+1)\frac{\omega\pi}{\varpi}}+h^{-(2m+1)\frac{\omega\pi}{\varpi}}}\right\}, \tag{192.} \] \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}.\left\{\frac{\left(h^{\frac{\alpha\pi}{\varpi}}+h^{-\frac{\alpha\pi}{\varpi}}\right)\left(h^{(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}+h^{-(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}\right)}{h^{\frac{2\alpha\pi}{\varpi}}+h^{-\frac{2\alpha\pi}{\varpi}}+h^{(2m+1)\frac{\omega\pi}{\varpi}}+h^{-(2m+1)\frac{\omega\pi}{\varpi}}}\right\}. \tag{193.} \] Si, pour abréger, on suppose: \[ h^{\frac{\alpha\pi}{\varpi}} = \varepsilon\ \text{et}\ h^{\frac{\omega\pi}{\varpi}} = r^{2}, \tag{194.} \] ces formules, en développant le second membre, deviendront: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}\left(\varepsilon-\frac{1}{\varepsilon}\right).\left\{\frac{r-\frac{1}{r}}{r^{2}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{2}}}-\frac{r^{3}-\frac{1}{r^{3}}}{r^{6}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{6}}}+\frac{r^{5}-\frac{1}{r^{5}}}{r^{10}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{10}}}-\ldots\right\}, \tag{195.} \] \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\left(\varepsilon+\frac{1}{\varepsilon}\right).\left\{\frac{r+\frac{1}{r}}{r^{2}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{2}}}+\frac{r^{3}+\frac{1}{r^{3}}}{r^{6}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{6}}}+\frac{r^{5}+\frac{1}{r^{5}}}{r^{10}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{10}}}-\ldots\right\}. \tag{196.} \]

En mettant $\alpha i$ au lieu de $\alpha$ dans les formules (192.), (193.), changeant ensuite $c$ en $e$ et $e$ en $c$, et remarquant, que les quantités \[ \omega,\ \varpi,\ \Phi(\alpha i),\ F(\alpha i),\ h^{\alpha i\frac{\pi}{\varpi}}-h^{-\alpha i\frac{\pi}{\varpi}},\ h^{\alpha i\frac{\pi}{\varpi}}+h^{\alpha i\frac{\pi}{\varpi}}, \] Editorial note: In the last quantity the sign of the second exponent is printed without a minus sign; an apparent misprint for $h^{-\alpha i\frac{\pi}{\varpi}}$, reproduced as printed. se changeront respectivement en: \[ \varpi,\ \omega,\ i\Phi(\alpha),\ f\alpha,\ 2i.\sin\alpha\frac{\pi}{\omega},\ 2\cos\alpha\frac{\pi}{\omega}, \] il viendra: \[ \Phi\alpha = \frac{4}{ec}.\frac{\pi}{\omega}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{\sin\frac{\alpha\pi}{\omega}.\left(h^{(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}-h^{-(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}\right)}{h^{(2m+1)\frac{\varpi\pi}{\omega}}+2\cos2\alpha\frac{\pi}{\omega}+h^{-(2m+1)\frac{\varpi\pi}{\omega}}}\right\}, \tag{197.} \] \[ f\alpha = \frac{4}{e}.\frac{\pi}{\omega}.\overset{\infty}{\underset{0}{\Sigma}}_{m}.\left\{\frac{\cos\frac{\alpha\pi}{\omega}.\left(h^{(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}+h^{-(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}\right)}{h^{(2m+1)\frac{\varpi\pi}{\omega}}+2\cos2\alpha\frac{\pi}{\omega}+h^{-(2m+1)\frac{\varpi\pi}{\omega}}}\right\}. \tag{198.} \] En faisant pour abréger \[ h^{\frac{\varpi\pi}{2\omega}} = \varrho, \tag{199.} \]

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et développant, on obtiendra: \[ \varphi\left(\alpha\frac{\omega}{2}\right) = \frac{4}{ec}.\frac{\pi}{\omega}.\sin\left(\alpha\frac{\pi}{2}\right).\left\{\frac{\varrho-\frac{1}{\varrho}}{\varrho^{2}+2\cos(\alpha\pi)+\frac{1}{\varrho^{2}}}-\frac{\varrho^{3}-\frac{1}{\varrho^{3}}}{\varrho^{6}+2\cos(\alpha\pi)+\frac{1}{\varrho^{6}}}+\frac{\varrho^{5}-\frac{1}{\varrho^{5}}}{\varrho^{10}+2\cos(\alpha\pi)+\frac{1}{\varrho^{10}}}-....\right\}, \tag{200.} \] \[ f\left(\alpha\frac{\omega}{2}\right) = \frac{4}{e}.\frac{\pi}{\omega}.\cos\left(\alpha\frac{\pi}{2}\right).\left\{\frac{\varrho+\frac{1}{\varrho}}{\varrho^{2}+2\cos(\alpha\pi)+\frac{1}{\varrho^{2}}}+\frac{\varrho^{3}+\frac{1}{\varrho^{3}}}{\varrho^{6}+2\cos(\alpha\pi)+\frac{1}{\varrho^{6}}}+\frac{\varrho^{5}+\frac{1}{\varrho^{5}}}{\varrho^{10}+2\cos(\alpha\pi)+\frac{1}{\varrho^{10}}}+....\right\}. \tag{201.} \]

En substituant dans les formules (190.), (191.) au lieu de $h^{\alpha\frac{\pi}{\varpi}}$ et $h^{\frac{\omega\pi}{2\varpi}}$ leurs valeurs $\varepsilon$ et $r$, il viendra: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-m).\left\{\frac{1}{\varepsilon.r^{-(2m+1)}+\varepsilon^{-1}.r^{2m+1}}-\frac{1}{\varepsilon.r^{2m+1}+\varepsilon^{-1}.r^{-(2m+1)}}\right\}, \tag{202.} \] Editorial note: The factor is printed as $(-m)$ where $(-1)^{m}$ is expected; an apparent misprint, reproduced as printed. \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}\left\{\frac{1}{\varepsilon.r^{-2m-1}+\varepsilon^{-1}.r^{2m+1}}+\frac{1}{\varepsilon.r^{2m+1}+\varepsilon^{-1}.r^{-2m-1}}\right\}. \tag{203.} \]

En supposant maintenant $\alpha<\frac{\omega}{2}$, on aura: \[ \frac{1}{\varepsilon.r^{-2m-1}+\varepsilon^{-1}.r^{2m+1}} = \frac{\varepsilon^{-1}.r^{-2m-1}}{1+\varepsilon^{2}.r^{-4m-2}} = \varepsilon.r^{-2m-1}-\varepsilon^{3}.r^{-6m-3}+\varepsilon^{5}.r^{-10m-5}-...., \] Editorial note: In the first fraction the numerator is printed $\varepsilon^{-1}.r^{-2m-1}$ where $\varepsilon.r^{-2m-1}$ is expected (the expansion beside it begins with $\varepsilon.r^{-2m-1}$); an apparent misprint, reproduced as printed. \[ \frac{1}{\varepsilon.r^{2m+1}+\varepsilon^{-1}.r^{-2m-1}} = \frac{\varepsilon^{-1}.r^{-2m-1}}{1+\varepsilon^{-2}.r^{-4m-2}} = \varepsilon^{-1}.r^{-2m-1}-\varepsilon^{-3}.r^{-6m-3}+\varepsilon^{-5}.r^{-10m-5}-...., \] donc: \[ \begin{aligned} \Phi\alpha &= \frac{2}{ec}.\frac{\pi}{\varpi}.\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\left\{(\varepsilon-\varepsilon^{-1}).r^{-2m-1}-(\varepsilon^{3}-\varepsilon^{-3}).r^{-6m-3}+(\varepsilon^{5}-\varepsilon^{-5})\,r^{-10m-5}-....\right\}\right\}\\ &= \frac{2}{ec}.\frac{\pi}{\varpi}.\left\{(\varepsilon-\varepsilon^{-1}).\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{-2m-1}-(\varepsilon^{3}-\varepsilon^{-3}).\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{-6m-3}+....\right\}. \end{aligned} \] Or: \[ \overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{-2m-1} = r^{-1}-r^{-3}+r^{-5}-.... = \frac{r^{-1}}{1+r^{-2}} = \frac{r}{r^{2}+1}, \] \[ \overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{3m-3} = r^{-3}-r^{-6}+r^{-9}-.... = \frac{r^{-3}}{1+r^{-6}} = \frac{r^{3}}{r^{6}+1} \] Editorial note: The exponent in the summand is printed $3m-3$ and the series exponents as $-3$, $-6$, $-9$; the sum is evidently meant to be that of $(-1)^{m}.r^{-6m-3}$, with series $r^{-3}-r^{-9}+r^{-15}-\ldots$; apparent misprints, reproduced as printed. etc., donc: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\left\{\frac{\varepsilon-\varepsilon^{-1}}{r+r^{-1}}-\frac{\varepsilon^{3}-\varepsilon^{-3}}{r^{3}+r^{-3}}+\frac{\varepsilon^{5}-\varepsilon^{-5}}{r^{5}+r^{-5}}-.....\right\}. \tag{204.} \] De la même manière on trouvera: \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\left\{\frac{\varepsilon+\varepsilon^{-1}}{r-r^{-1}}-\frac{\varepsilon^{3}+\varepsilon^{-3}}{r^{3}-r^{-3}}+\frac{\varepsilon^{5}+\varepsilon^{-5}}{r^{5}-r^{-5}}-.....\right\}. \tag{205.} \]

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En mettant $\alpha\frac{\varpi}{2}i$ au lieu de $\alpha$, et changeant ensuite $e$ en $c$ et $c$ en $e$, \[ \varepsilon;\ \varpi;\ \Phi\left(\alpha\frac{\varpi}{2}i\right);\ F\left(\alpha\frac{\varpi}{2}i\right);\ r;\ \varepsilon^{m}+\varepsilon^{-m};\ \varepsilon^{m}-\varepsilon^{-m} \] se changent en: \[ \varpi;\ \omega;\ i\varphi\left(\alpha\frac{\omega}{2}\right);\ f\left(\alpha\frac{\omega}{2}\right);\ \varrho;\ 2\cos\left(m\alpha\frac{\pi}{2}\right);\ 2i\sin\left(m\alpha\frac{\pi}{2}\right); \] donc: \[ \varphi\left(\alpha\frac{\omega}{2}\right) = \frac{4}{ec}.\frac{\pi}{\omega}.\left\{\frac{\sin\left(\alpha\frac{\pi}{2}\right)}{\varrho+\frac{1}{\varrho}}-\frac{\sin\left(3\alpha\frac{\pi}{2}\right)}{\varrho^{3}+\frac{1}{\varrho^{3}}}+\frac{\sin\left(5\alpha\frac{\pi}{2}\right)}{\varrho^{5}+\frac{1}{\varrho^{5}}}-....\right\}, \tag{206.} \] \[ f\left(\alpha\frac{\omega}{2}\right) = \frac{4}{e}.\frac{\pi}{\omega}.\left\{\frac{\cos\left(\alpha\frac{\pi}{2}\right)}{\varrho-\frac{1}{\varrho}}-\frac{\cos\left(3\alpha\frac{\pi}{2}\right)}{\varrho^{3}-\frac{1}{\varrho^{3}}}+\frac{\cos\left(5\alpha\frac{\pi}{2}\right)}{\varrho^{5}-\frac{1}{\varrho^{5}}}-....\right\}. \tag{207.} \]

Ces quatre dernières formules offrent d'expressions très simples des fonctions $\Phi\alpha$, $f\alpha$, $F\alpha$. Par les différentiations et intégrations on peut en déduire une foule d'autres plus ou moins remarquables.

33.

Dans le cas, $e=c$, les formules précédentes prennent une forme plus simple, à cause de la relation $\omega=\varpi$, qui a lieu dans ce cas. Soit pour plus de simplicité $e=c=1$. On a: \[ r = h^{\frac{\omega\pi}{2\varpi}} = h^{\frac{\pi}{2}},\ \varrho = h^{\frac{\varpi\pi}{2\omega}} = h^{\frac{\pi}{2}}, \] donc en substituant, et faisant dans (204.), (205.) $\alpha=\alpha\frac{\varpi}{2}$, il vient: \[ \varphi\left(\alpha\frac{\omega}{2}\right) = 2\frac{\pi}{\omega}\left\{\frac{h^{\frac{\alpha\pi}{2}}-h^{-\frac{\alpha\pi}{2}}}{h^{\frac{\pi}{2}}+h^{-\frac{\pi}{2}}}-\frac{h^{\frac{3\alpha\pi}{2}}-h^{-\frac{3\alpha\pi}{2}}}{h^{\frac{3\pi}{2}}+h^{-\frac{3\pi}{2}}}+\frac{h^{\frac{5\alpha\pi}{2}}-h^{-\frac{5\alpha\pi}{2}}}{h^{\frac{5\pi}{2}}+h^{-\frac{5\pi}{2}}}-.....\right\}, \] \[ F\left(\alpha\frac{\omega}{2}\right) = 2\frac{\pi}{\omega}\left\{\frac{h^{\frac{\alpha\pi}{2}}+h^{-\frac{\alpha\pi}{2}}}{h^{\frac{\pi}{2}}-h^{-\frac{\pi}{2}}}-\frac{h^{\frac{3\alpha\pi}{2}}+h^{-\frac{3\alpha\pi}{2}}}{h^{\frac{3\pi}{2}}-h^{-\frac{3\pi}{2}}}+\frac{h^{\frac{5\alpha\pi}{2}}+h^{-\frac{5\alpha\pi}{2}}}{h^{\frac{5\pi}{2}}-h^{-\frac{5\pi}{2}}}-.....\right\}, \] \[ \varphi\left(\alpha\frac{\omega}{2}\right) = \frac{4\pi}{\omega}\left\{\sin\left(\alpha\frac{\pi}{2}\right).\frac{h^{\frac{\pi}{2}}}{1+h^{\pi}}-\sin\left(3\alpha\frac{\pi}{2}\right).\frac{h^{\frac{3\pi}{2}}}{1+h^{3\pi}}+\sin\left(5\alpha\frac{\pi}{2}\right).\frac{h^{\frac{5\pi}{2}}}{1+h^{5\pi}}-....\right\}, \] \[ f\left(\alpha\frac{\omega}{2}\right) = \frac{4\pi}{\omega}\left\{\cos\left(\alpha\frac{\pi}{2}\right).\frac{h^{\frac{\pi}{2}}}{h^{\pi}-1}-\cos\left(3\alpha\frac{\pi}{2}\right).\frac{h^{\frac{3\pi}{2}}}{h^{3\pi}-1}+\sin\left(5\alpha\frac{\pi}{2}\right).\frac{h^{\frac{5\pi}{2}}}{h^{5\pi}-1}-....\right\}. \] Editorial note: In the last display the third term has a sine, where the pattern of the first two terms requires a cosine; an apparent misprint, reproduced as printed.

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Les fonctions $\varphi$, $f$, $F$ sont déterminées par les équations \[ \begin{gathered} \alpha.\frac{\omega}{2} = \int_{0}^{x}\frac{\partial x}{\sqrt{(1-x^{4})}};\ \frac{\omega}{2} = \int_{0}^{1}\frac{\partial x}{\sqrt{(1-x^{4})}};\\ x = \varphi\left(\alpha\frac{\omega}{2}\right) = \sqrt{(1-x^{2})} = f\left(\alpha\frac{\omega}{2}\right) = \sqrt{(1+x^{2})} = F\left(\alpha\frac{\omega}{2}\right). \end{gathered} \]

Si dans les deux dernières formules on fait $\alpha=0$, et qu'on remarque, qu'alors la valeur de $\dfrac{\varphi\left(\alpha\frac{\omega}{2}\right)}{\sin\left(\alpha\frac{\pi}{2}\right)}$ est $=\frac{\omega}{\pi}$, et celle de $\dfrac{\sin\left(m\alpha\frac{\pi}{2}\right)}{\sin\left(\alpha\frac{\pi}{2}\right)}=m$, on trouvera: \[ \frac{\omega}{2} = \frac{\pi}{2}.\left\{\frac{h^{\frac{\pi}{2}}}{h^{\pi}-1}-\frac{h^{\frac{3\pi}{2}}}{h^{3\pi}-1}+\frac{h^{\frac{5\pi}{2}}}{h^{5\pi}-1}-\ldots\right\} = \int_{0}^{1}\frac{\partial x}{\sqrt{(1-x^{4})}}, \] \[ \frac{\omega^{2}}{4} = \pi^{2}.\left\{\frac{h^{\frac{\pi}{2}}}{h^{\pi}+1}-3.\frac{h^{\frac{3\pi}{2}}}{h^{3\pi}+1}+5.\frac{h^{\frac{5\pi}{2}}}{h^{5\pi}+1}-\ldots\right\} = \left(\int_{0}^{1}\frac{\partial x}{\sqrt{(1-x^{4})}}\right)^{2}. \]

(La continuation dans le cahier suivant.)

↓ .tex ↓ PDF Produced by claude-sonnet-5-5, effort: medium (2026-10-07)

page 101

Researches on elliptic functions.

(By Mr. N. H. Abel.)

For a long time the logarithmic functions, and the exponential and circular functions, were the only transcendental functions that attracted the attention of the geometers. It is only in recent times that one has begun to consider some others. Among these one must distinguish the functions called elliptic, as much for their beautiful analytical properties as for their application in the various branches of mathematics. The first idea of these functions was given by the immortal Euler, in demonstrating that the separated equation \[ \frac{\partial x}{\sqrt{(\alpha+\beta x+\gamma x^{2}+\delta x^{3}+\varepsilon x^{4})}}+\frac{\partial y}{\sqrt{(\alpha+\beta y+\gamma y^{2}+\delta y^{3}+\varepsilon y^{4})}}=0 \tag{1.} \] is algebraically integrable. After Euler, Lagrange added something to it, in giving his elegant theory of the transformation of the integral $\displaystyle\int\frac{R.\partial x}{\sqrt{[(1-p^{2}x^{2})(1-q^{2}x^{2})]}}$, where $R$ is a rational function of $x$. But the first and, if I am not mistaken, the only one who has gone deeply into the nature of these functions is Mr. Legendre, who, first in a memoir on elliptic functions, and then in his excellent exercises of mathematics, has developed a number of elegant propertiesEditorial note: Printed poropriétés instead of propriétés; an apparent misprint. of these functions, and has shown their application. At the time of the publication of that work, nothing had been added to the theory of Mr. Legendre. I believe that further researches on these functions will not be seen here without pleasure.

In general one understands under the denomination of elliptic functions every function contained in the integral \[ \int\frac{R\,\partial x}{\sqrt{(\alpha+\beta x+\gamma x^{2}+\delta x^{3}+\varepsilon x^{4})}}, \] where $R$ is a rational function and $\alpha$, $\beta$, $\gamma$, $\delta$, $\varepsilon$ are constant and real quantities. Mr. Legendre has demonstrated that by suitable substitutions one can always reduce this integral to the form

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\[ \int\frac{P\,\partial y}{\sqrt{(a+by^{2}+cy^{4})}}, \] where $P$ is a rational function of $y^{2}$. By suitable reductions, this integral can then be brought to the form \[ \int\frac{A+By^{2}}{C+Dy^{2}}\cdot\frac{\partial y}{\sqrt{(a+by^{2}+cy^{4})}}, \] and this to: \[ \int\frac{A+B\sin^{2}\theta}{C+D\sin^{2}\theta}\cdot\frac{\partial\theta}{\sqrt{(1-c^{2}\sin^{2}\theta)}}, \] where $c$ is real, and less than unity.

From this it follows that every elliptic function can be reduced to one of the three forms: \[ \int\frac{\partial\theta}{\sqrt{(1-c^{2}\sin^{2}\theta)}};\ \int\partial\theta\sqrt{(1-c^{2}\sin^{2}\theta)};\ \int\frac{\partial\theta}{(1+n\sin^{2}\theta)\sqrt{(1-c^{2}\sin^{2}\theta)}}, \] to which Mr. Legendre gives the names of elliptic functions of the first, second and third species. It is these three functions that Mr. Legendre has considered; above all the first, which has the most remarkable and the simplest properties.

I propose, in this memoir, to consider the inverse function, that is, the function $\Phi\alpha$, determined by the equations \[ \begin{gathered} \alpha=\int\frac{\partial\theta}{\sqrt{(1-c^{2}\sin^{2}\theta)}}\ \text{and}\\ \sin\theta=\varphi(\alpha)=x. \end{gathered} \] The last equation gives \[ \partial\theta.\sqrt{(1-\sin^{2}\theta)}=\partial.\Phi\alpha=\partial x, \] hence \[ \alpha=\int_{0}\frac{\partial x}{\sqrt{[(1-x^{2})(1-c^{2}x^{2})]}}. \] Mr. Legendre supposes $c^{2}$ positive, but I have remarked that the formulas become simpler on supposing $c^{2}$ negative, $=-e^{2}$. Likewise, for the sake of greater symmetry, I write $1-c^{2}x^{2}$ in place of $1-x^{2}$. So that the function $\Phi\alpha=x$ will be given by the equation \[ \alpha=\int_{0}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}, \] or else \[ \Phi'\alpha=\partial\alpha.\sqrt{[(1-c^{2}\Phi^{2}\alpha)(1+e^{2}\Phi^{2}\alpha)]}.\] Editorial note: Printed $\partial\alpha$ before the radical; the factor is spurious, since the derivative is the square root alone. For brevity, I introduce two other functions of $\alpha$, namely: \[ f\alpha=\sqrt{(1-c^{2}\Phi^{2}\alpha)};\ F\alpha=\sqrt{(1+e^{2}\Phi^{2}\alpha)}. \]

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Several properties of these functions are deduced immediately from the known properties of the elliptic function of the first species, but others are more hidden. For example, one demonstrates that the equations $\Phi\alpha=0$, $f\alpha=0$, $F\alpha=0$ have an infinite number of roots, which one can find all. One of the most remarkable properties is that one can express $\Phi(m\alpha)$, $f(m\alpha)$, $F(m\alpha)$ ($m$ being an integer) rationally in terms of $\Phi\alpha$, $f\alpha$, $F\alpha$. Also nothing is easier than to find $\Phi(m\alpha)$, $f(m\alpha)$, $F(m\alpha)$, when one knows $\Phi\alpha$, $f\alpha$, $F\alpha$; but the inverse problem, namely to determine $\Phi\alpha$, $f\alpha$, $F\alpha$ in terms of $\Phi(m\alpha)$, $f(m\alpha)$, $F(m\alpha)$, is more difficult, because it depends on an equation of a higher degree (namely of the degree $m^{2}$).

The solution of this equation is the principal object of this memoir. First it will be shown how oneEditorial note: Printed ont instead of on; an apparent misprint. can find all the roots, by means of the functions $\Phi,f,F$. Next the algebraic solution of the equation in question will be treated, and one will arrive at this remarkable result, that $\varphi\left(\frac{\alpha}{m}\right)$; $f\left(\frac{\alpha}{m}\right)$; $F\left(\frac{\alpha}{m}\right)$ can be expressed in terms of $\Phi\alpha$, $f\alpha$, $F\alpha$, by means of a function which, with respect to $\alpha$, contains no other irrationalities than radicals. This produces a very general class of equations which are algebraically soluble. It is to be remarked that the expressions of the roots contain constant quantities which, in general, are not expressible by algebraic quantities. These constant quantities depend on an equation of the degree $\mathrm{m}^{2}-1$. It will be shown how, by means of algebraic functions, the solution of it can be reduced to that of an equation of the degree $\mathrm{m}+1$. Several expressions will be given of the functions $\Phi(2n+1)\alpha$; $f(2n+1)\alpha$; $F(2n+1)\alpha$ in terms of $\Phi\alpha$, $f\alpha$, $F\alpha$. From these will then be deduced the values of $\Phi\alpha$, $f\alpha$, $F\alpha$ in terms of $\alpha$. It will be demonstrated that these functionsEditorial note: Printed fontions instead of fonctions; an apparent misprint. can be decomposed into an infinite number of factors, and even into an infinity of partial fractions.

§. I.

Fundamental properties of the functions $\Phi\alpha$; $f\alpha$; $F\alpha$.

1.

Supposing that $\Phi\alpha=x$ (1.), we shall have by virtue of what precedes: \[ \alpha=\int_{0}^{x}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}. \tag{2.} \]

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Hence one sees that $\alpha$, considered as a function of $x$, is positive from $x=0$ to $x=\frac{1}{c}$. Making therefore \[ \frac{\omega}{2}=\int_{0}^{\frac{1}{c}}\frac{\partial x}{\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}, \tag{3.} \] it is evident that $\Phi\alpha$ andEditorial note: Printed et instead of est; an apparent misprint. positive and goes on increasing from $\alpha=0$ to $\alpha=\frac{\omega}{2}$, and that we shall have \[ \varphi(0)=0,\ \varphi\left(\frac{\omega}{2}\right)=\frac{1}{c}. \tag{4.} \] Because $\alpha$ changes sign when one writes $-x$ in place of $x$, the same is true of the function $\Phi\alpha$ with respect to $\alpha$, and consequently we shall have the equation \[ \varphi(-\alpha)=-\varphi(\alpha). \tag{5.} \] Putting in (1.) $xi$ in place of $x$ (where $i$, for brevity, represents the imaginary quantity $\sqrt{-1}$) and denoting the value of $\alpha$ by $\beta i$, there will result \[ xi=\Phi(\beta i)\ \text{and}\ \beta=\int_{0}^{x}\frac{\partial x}{\sqrt{[(1+c^{2}x^{2})(1-e^{2}x^{2})]}}. \tag{6.} \] $\beta$ is real and positive from $x=0$ to $x=\frac{1}{e}$, hence making \[ \frac{\varpi}{2}=\int_{0}^{\frac{1}{e}}\frac{\partial x}{\sqrt{[(1-e^{2}x^{2})(1+c^{2}x^{2})]}}, \tag{7.} \] $x$ will be positive from $\beta=$ zero to $\beta=\frac{\varpi}{2}$, that is, the function $\frac{1}{i}\Phi(\beta i)$ will be positive between the same limits. Making $\beta=\alpha$ and $y=\frac{\varphi(\alpha i)}{i}$, we have \[ \alpha=\int_{0}^{y}\frac{\partial y}{\sqrt{[(1-e^{2}y^{2})(1+c^{2}y^{2})]}}, \] hence one sees that, on supposing $c$ in place of $e$ and $e$ in place of $\varepsilon$,Editorial note: Printed as a curled e (apparently $\varepsilon$) at the last place; the sense requires $c$ (interchange of $c$ and $e$). \[ \frac{\varphi(\alpha i)}{i}\ \text{will be changed into}\ \Phi\alpha. \]

And because \[ \begin{gathered} f\alpha=\sqrt{(1-c^{2}\Phi^{2}\alpha)},\\ F\alpha=\sqrt{(1+e^{2}\Phi^{2}\alpha)}, \end{gathered} \] one sees that by the change of $c$ into $e$ and $e$ into $c$, $f(\alpha i)$ and $F(\alpha i)$ will return respectively into $F\alpha$ and $f\alpha$. Finally the equations (2.) and (7.) show that by the same transformation, $\omega$ and $\varpi$ will return respectively into $\varpi$ and $\omega$.

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According to (7.) we shall have $x=\frac{1}{e}$ for $\beta=\frac{\varpi}{2}$, hence by virtue of the equation $xi=\varphi(\beta i)$, there will result \[ \varphi\left(\frac{\varpi i}{2}\right)=i.\frac{1}{e}. \tag{8.} \]

2.

By virtue of what precedes, we shall have the values of $\Phi\alpha$ for every real value of $\alpha$ lying between $-\frac{\omega}{2}$ and $+\frac{\omega}{2}$, and for every imaginary value of this quantity of the form $\beta i$, if $\beta$ is a quantity continuous between the limits $-\frac{\varpi}{2}$ and $+\frac{\varpi}{2}$. It is now a question of finding the value of this function for any value whatever, real or imaginary, of the variable. To attain this, we shall first establish the fundamental properties of the functions $\Phi$, $f$ and $F$.

Because we have \[ \begin{gathered} f^{2}\alpha=1-c^{2}\Phi^{2}\alpha,\\ F^{2}\alpha=1+e^{2}\Phi^{2}\alpha, \end{gathered} \] we shall have, on differentiating: \[ \begin{gathered} f\alpha.f'\alpha=-c^{2}\Phi\alpha.\varphi'\alpha,\\ F\alpha.F'\alpha=e^{2}\Phi\alpha\,\varphi'\alpha, \end{gathered} \] Now by (2.) we have \[ \varphi'\alpha=\sqrt{[(1-c^{2}\Phi^{2}\alpha)(1+e^{2}\Phi^{2}\alpha)]}=f\alpha.F\alpha, \] hence, on substituting this value of $\Phi'\alpha$ in the two preceding equations, we shall find that the functions $\Phi\alpha$, $f\alpha$, $F\alpha$ are connected with one another by the equations \[ \left\{\begin{aligned} \varphi'\alpha&=f\alpha.F\alpha,\\ f'\alpha&=-c^{2}\Phi\alpha.F\alpha,\\ F'\alpha&=e^{2}\Phi\alpha.f\alpha. \end{aligned}\right. \tag{9.} \] This being premised, I say that, denoting by $\alpha$ and $\beta$ two indeterminates, we shall have \[ \left\{\begin{aligned} \Phi(\alpha+\beta)&=\frac{\varphi\alpha.f\beta.F\beta+\varphi\beta.f\alpha.F\alpha}{1+e^{2}c^{2}.\varphi^{2}\alpha.\varphi^{2}\beta},\\ f(\alpha+\beta)&=\frac{f\alpha.f\beta-c^{2}.\varphi\alpha.\varphi\beta.F\alpha.F\beta}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta},\\ F(\alpha+\beta)&=\frac{F\alpha.F\beta+e^{2}.\varphi\alpha.\varphi\beta.f\alpha.f\beta}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta}. \end{aligned}\right. \tag{10.} \]

These formulas can be deduced at once from the known properties of the elliptic functions${}^{*)}$; but one can also verify them easily in the following manner.

*) Legendre Exercices de calcul intégral.

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Denoting by $r$ the second member of the first of the equations (10.), we shall have, on differentiating with respect to $\alpha$: \[ \begin{aligned} \left(\frac{\partial r}{\partial\alpha}\right)&=\frac{\{\varphi'\alpha.f\beta.F\beta+\varphi\beta.F\alpha.f'\alpha+\varphi\beta.f\alpha.F'\alpha\}}{1+e^{2}c^{2}.\varphi^{2}\alpha.\varphi^{2}\beta}\\ &\quad-\frac{(\varphi\alpha.f\beta F\beta+\varphi\beta.f\alpha\ F\alpha).2e^{2}c^{2}\varphi\alpha.\varphi^{2}\beta.\varphi'\alpha}{(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)^{2}}. \end{aligned} \] On substituting for $\Phi'\alpha$, $f'\alpha$, $F'\alpha$ their values given by the equations (9.), there will result \[ \begin{aligned} \left(\frac{\partial r}{\partial\alpha}\right)&=\frac{f\alpha.F\alpha.f\beta.F\beta}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta}-\frac{2e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta.f\alpha.f\beta.F\alpha.F\beta}{(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)^{2}}\\ &\quad+\frac{\varphi\alpha.\varphi\beta.\{1+e^{2}c^{2}\varphi^{2}\alpha\varphi^{2}\beta\}\{-c^{2}F^{2}\alpha+e^{2}f^{2}\alpha\}-2e^{2}c^{2}\varphi\alpha\,\varphi\beta.\varphi^{2}\beta.f^{2}\alpha.F^{2}\alpha}{(1+e^{2}c^{2}\varphi^{2}\alpha\varphi^{2}\beta)^{2}}, \end{aligned} \] whence, on substituting for $f^{2}\alpha$ and $F^{2}\alpha$ their values: $1-c^{2}\Phi^{2}\alpha$, $1+e^{2}\Phi^{2}\alpha$, and on reducing, we derive \[ \left(\frac{\partial r}{\partial\alpha}\right)=\frac{(1-e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)\{(e^{2}-c^{2})+\varphi\alpha.\varphi\beta+f\alpha.f\beta.F\alpha.F\beta\}-2e^{2}c^{2}\varphi\alpha\,\varphi\beta(\varphi^{2}\alpha+\varphi^{2}\beta)}{(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)^{2}}. \] Editorial note: Printed with a plus sign (and a stray hairline before it) between $(e^{2}-c^{2})$ and $\varphi\alpha.\varphi\beta$; a product $(e^{2}-c^{2})\varphi\alpha.\varphi\beta$ is expected. Now $\alpha$ and $\beta$ enter symmetrically into the expression of $r$; hence we shall have the value of $\left(\frac{\partial r}{\partial\beta}\right)$ by permuting $\alpha$ and $\beta$ in the value of $\left(\frac{\partial r}{\partial\alpha}\right)$. Now by this the expression of $\left(\frac{\partial r}{\partial\alpha}\right)$ does not change its value, hence we shall have \[ \left(\frac{\partial r}{\partial\alpha}\right)=\left(\frac{\partial r}{\partial\beta}\right). \]

This equation in partial differentials shows that $r$ is a function of $\alpha+\beta$; hence we shall have \[ r=\psi(\alpha+\beta). \] The form of the function $\psi$ will be found by giving to $\beta$ a particular value. Supposing for example $\beta=0$, and remarking that $\Phi(0)=0$, $f(0)=1$, $F(0)=1$, the two values of $r$ will become \[ r=\Phi(\alpha)\ \text{and}\ r=\psi(\alpha), \] hence \[ \psi(\alpha)=\Phi(\alpha), \] and from this \[ r=\psi(\alpha+\beta)=\Phi(\alpha+\beta). \] The first of the formulas (10.) therefore holds good indeed.

In the same manner one will verify the two other formulas.

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3.

From the formulas (10.) one can deduce a multitude of others. I shall set down some of the most remarkable: For brevity I make \[ 1+e^{2}c^{2}\Phi^{2}\alpha.\Phi^{2}\beta=R. \tag{11.} \]

On changing first the sign of $\beta$, we shall obtain \[ \left\{\begin{aligned} \Phi(\alpha+\beta)+\Phi(\alpha-\beta)&=\frac{2\varphi\alpha.f\beta.F\beta}{R},\\ \Phi(\alpha+\beta)-\Phi(\alpha-\beta)&=\frac{2\varphi\beta.f\alpha.F\alpha}{R},\\ f(\alpha+\beta)+f(\alpha-\beta)&=\frac{2f\alpha.f\beta}{R},\\ f(\alpha+\beta)-f(\alpha-\beta)&=\frac{-2c^{2}.\varphi\alpha.\varphi\beta.F\alpha.F\beta}{R},\\ F(\alpha+\beta)+F(\alpha-\beta)&=\frac{2F\alpha.F\beta}{R},\\ F(\alpha+\beta)-F(\alpha-\beta)&=\frac{2e^{2}.\varphi\alpha.\varphi\beta.f\alpha.f\beta}{R}. \end{aligned}\right. \tag{12.} \]

On forming the product of $\Phi(\alpha+\beta)$ and $\Phi(\alpha-\beta)$, we shall find \[ \begin{aligned} \Phi(\alpha+\beta).\Phi(\alpha-\beta)&=\frac{\varphi\alpha.f\beta.F\beta+\varphi\beta.f\alpha.F\alpha}{R}.\frac{\varphi\alpha.f\beta.F\beta-\varphi\beta.f\alpha.F\alpha}{R}\\ &=\frac{\varphi^{2}\alpha.f^{2}\beta.F^{2}\beta-\varphi^{2}\beta.f^{2}\alpha.F^{2}\alpha}{R^{2}}, \end{aligned} \] or, on substituting the values of $f^{2}\beta$, $F^{2}\beta$, $f^{2}\alpha$, $F^{2}\alpha$ in $\Phi\beta$ and $\Phi\alpha$: \[ \begin{aligned} \Phi(\alpha+\beta).\Phi(\alpha-\beta)&=\frac{\varphi^{2}\alpha-\varphi^{2}\beta-e^{2}c^{2}\varphi^{2}\alpha.\varphi^{4}\beta+e^{2}c^{2}\varphi^{2}\beta.\varphi^{4}\alpha}{R^{2}}\\ &=\frac{(\varphi^{2}\alpha-\varphi^{2}\beta)(1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta)}{R^{2}}; \end{aligned} \] now $R=1+e^{2}c^{2}\Phi^{2}\alpha.\Phi^{2}\beta$, hence \[ \Phi(\alpha+\beta).\Phi(\alpha-\beta)=\frac{\varphi^{2}\alpha-\varphi^{2}\beta}{R}. \tag{13.} \] We shall find in the same way \[ \left\{\begin{aligned} f(\alpha+\beta).f(\alpha-\beta)&=\frac{f^{2}\alpha-c^{2}\varphi^{2}\beta.F^{2}\alpha}{R}=\frac{f^{2}\beta-c^{2}\varphi^{2}\alpha.F^{2}\beta}{R}\\ &=\frac{1-c^{2}\varphi^{2}\alpha-c^{2}\varphi^{2}\beta-c^{2}e^{2}\varphi^{2}\alpha.\varphi^{2}\beta}{R}=\frac{f^{2}\alpha.f^{2}\beta-c^{2}(c^{2}+e^{2})\varphi^{2}\alpha.\varphi^{2}\beta}{R},\\ F(\alpha+\beta).F(\alpha-\beta)&=\frac{F^{2}\alpha+e^{2}\varphi^{2}\beta.f^{2}\alpha}{R}=\frac{F^{2}\beta+e^{2}\varphi^{2}\alpha.f^{2}\beta}{R}\\ &=\frac{1+e^{2}\varphi^{2}\alpha+e^{2}\varphi^{2}\beta-e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}\beta}{R}=\frac{F^{2}\alpha.F^{2}\beta-e^{2}(c^{2}+e^{2})\varphi^{2}\alpha.\varphi^{2}\beta}{R}. \end{aligned}\right. \tag{14.} \]

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4.

Making in the formulas (10.) $\beta=\pm\frac{\omega}{2}$, $\beta=\pm\frac{\varpi}{2}i$, and remarking that $f\left(\pm\frac{\omega}{2}\right)=0$, $F\left(\pm\frac{\varpi}{2}i\right)=0$, we shall have \[ \left\{\begin{aligned} &\varphi\left(\alpha\pm\frac{\omega}{2}\right)=\pm\varphi\frac{\omega}{2}.\frac{f\alpha}{F\alpha};\ f\left(\alpha\pm\frac{\omega}{2}\right)=\mp\frac{F\frac{\omega}{2}}{\varphi\frac{\omega}{2}}.\frac{\varphi\alpha}{F\alpha};\\ &F\left(\alpha\pm\frac{\omega}{2}\right)=\frac{F\frac{\omega}{2}}{F\alpha};\\ &\varphi\left(\alpha\pm\frac{\varpi}{2}i\right)=\pm\varphi\frac{\varpi}{2}i.\frac{F\alpha}{f\alpha};\ F\left(\alpha\pm\frac{\varpi}{2}i\right)=\mp\frac{f\frac{\varpi}{2}i}{\varphi\frac{\varpi}{2}i}.\frac{\varphi\alpha}{f\alpha};\\ &f\left(\alpha\pm\frac{\varpi}{2}i\right)=\frac{f\frac{\varpi}{2}i}{f\alpha}; \end{aligned}\right. \tag{15.} \] or else: \[ \left\{\begin{aligned} &\varphi\left(\alpha\pm\frac{\omega}{2}\right)=\pm\frac{1}{c}.\frac{f\alpha}{F\alpha};\ f\left(\alpha\pm\frac{\omega}{2}\right)=\mp\sqrt{(e^{2}+c^{2})}.\frac{\varphi\alpha}{F\alpha};\\ &F\left(\alpha\pm\frac{\omega}{2}\right)=\frac{\sqrt{(e^{2}+c^{2})}}{c}.\frac{1}{F\alpha};\\ &\varphi\left(\alpha\pm\frac{\varpi}{2}i\right)=\pm\frac{i}{e}.\frac{F\alpha}{f\alpha};\ F\left(\alpha\pm\frac{\varpi}{2}i\right)=\pm i\sqrt{(e^{2}+c^{2})}.\frac{\varphi\alpha}{f\alpha};\\ &f\left(\alpha\pm\frac{\varpi}{2}i\right)=\frac{\sqrt{(e^{2}+c^{2})}}{e}.\frac{1}{f\alpha}. \end{aligned}\right. \tag{16.} \]

From this we derive at once: \[ \left\{\begin{aligned} &\varphi\left(\frac{\omega}{2}+\alpha\right)=\Phi\left(\frac{\omega}{2}-\alpha\right);\ f\left(\frac{\omega}{2}+\alpha\right)=-f\left(\frac{\omega}{2}-\alpha\right);\\ &F\left(\frac{\omega}{2}+\alpha\right)=F\left(\frac{\omega}{2}-\alpha\right);\\ &\varphi\left(\frac{\varpi}{2}i+\alpha\right)=\Phi\left(\frac{\varpi}{2}i-\alpha\right);\ F\left(\frac{\varpi}{2}i+\alpha\right)=-F\left(\frac{\varpi}{2}i-\alpha\right);\\ &f\left(\frac{\varpi}{2}i+\alpha\right)=f\left(\frac{\varpi}{2}i-\alpha\right); \end{aligned}\right. \tag{17.} \] \[ \varphi\left(\alpha\pm\frac{\omega}{2}\right).\varphi\left(\alpha+\frac{\varpi}{2}i\right)=\pm\frac{i}{ce};\ F\left(\alpha\pm\frac{\omega}{2}\right).F\alpha=\frac{b}{c};\ f\left(\alpha\pm\frac{\varpi}{2}i\right).f\alpha=\frac{b}{e}. \tag{18.} \] Making $\alpha=\frac{\omega}{2}$ and $\frac{\varpi}{2}i$, we find from this \[ \Phi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\ f\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\ F\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}. \]

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Next the equations (17.), on putting in the first three $\alpha+\frac{\omega}{2}$ in place of $\alpha$, and in the last three $\alpha+\frac{\varpi}{2}i$ in place of $\alpha$, give the following \[ \left\{\begin{aligned} &\varphi(\alpha+\omega)=-\Phi\alpha;\ f(\alpha+\omega)=-f\alpha;\ F(\alpha+\omega)=F\alpha;\\ &\varphi(\alpha+\varpi i)=-\Phi\alpha;\ f(\alpha+\varpi i)=f\alpha;\ F(\alpha+\varpi i)=-F\alpha; \end{aligned}\right. \tag{19.} \] and on putting $\alpha+\omega$ and $\alpha+\varpi i$ in place of $\alpha$: \[ \left\{\begin{aligned} &\Phi(2\omega+\alpha)=\Phi\alpha;\ \Phi(2\varpi i+\alpha)=\Phi\alpha;\ \Phi(\omega+\varpi i+\alpha)=\Phi\alpha;\\ &f(2\omega+\alpha)=f\alpha;\ f(\varpi i+\alpha)=f\alpha;\\ &F(\omega+\alpha)=F\alpha;\ F(2\varpi i+\alpha)=F\alpha. \end{aligned}\right. \tag{20.} \]

These equations show that the functions $\Phi\alpha$, $f\alpha$, $F\alpha$ are periodic functions. One will deduce from them without difficulty the following, where $m$ and $n$ are two integers positive or negative: \[ \left\{\begin{aligned} &\varphi((m+n)\omega+(m-n)\varpi i+\alpha)=\varphi\alpha;\ \varphi((m+n)\omega+(m-n+1)\varpi i+\alpha)=-\varphi\alpha;\\ &f(2m\omega+n\varpi i+\alpha)=f\alpha;\ f((2m+1)\omega+n\varpi i+\alpha)=-f\alpha;\\ &F(m\omega+2n\varpi i+\alpha)=+F\alpha;\ F(m\omega+(2n+1)\varpi i+\alpha)=-F\alpha. \end{aligned}\right. \tag{21.} \] These formulas can also be written as follows: \[ \left\{\begin{aligned} &\Phi(m\omega+n\varpi i\pm\alpha)=\pm(-1)^{m+n}.\Phi\alpha,\\ &f(m\omega+n\varpi i\pm\alpha)=(-1)^{m}.f\alpha,\\ &F(m\omega+n\varpi i\pm\alpha)=(-)^{n}.F\alpha. \end{aligned}\right. \tag{22.} \] Editorial note: In the last line of this system the base of the exponent is printed as a bare minus sign, without the digit 1; an apparent misprint for $(-1)^{n}$. One can remark as particular cases: \[ \left\{\begin{aligned} &\Phi(m\omega\pm\alpha)=\pm(-1)^{m}.\Phi\alpha;\ \Phi(n\varpi i\pm\alpha)=\pm(-1)^{n}.\Phi\alpha;\\ &f(m\omega\pm\alpha)=(-1)^{m}.f\alpha;\ f(n\varpi i\pm\alpha)=f\alpha;\\ &F(m\omega\pm\alpha)=F\alpha;\ F(n\varpi i\pm\alpha)=(-1)^{n}.F\alpha. \end{aligned}\right. \tag{22$'$.} \]

5.

The formulas which we have just established show that we shall have the values of the functions $\Phi\alpha$; $f\alpha$; $F\alpha$ for all the real or imaginary values of the variable, when we know them for the real values of this quantity lying between $\frac{\omega}{2}$ and $-\frac{\omega}{2}$ and for the imaginary values of the form $\beta i$, where $\beta$ lies between $\frac{\varpi}{2}$ and $-\frac{\varpi}{2}$.

In effect, suppose that the value of the functions $\Phi(\alpha+\beta i)$, $f(\alpha+\beta i)$, $F(\alpha+\beta i)$ is asked, where $\alpha$ and $\beta$ are any real quantities whatever. On putting in the formulas (10.) $\beta i$ in place of $\beta$, it is clear that we shall have the three functions in question expressed by the functions: $\Phi\alpha$; $f\alpha$; $F\alpha$; $\Phi(\beta i)$; $f(\beta i)$; $F(\beta i)$. It remains therefore only to determine these latter.

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Now, whatever the values of $\alpha$ and $\beta$ may be, one can always find two integral numbers $m$ and $n$, such that $\alpha=m.\omega\pm\alpha'$, $\beta=n\varpi\pm\beta'$, where $\alpha'$ is a quantity comprised between $0$ and $+\frac{\omega}{2}$, and $\beta'$ between $0$ and $+\frac{\varpi}{2}$. Therefore, by virtue of equations (22.), on substituting the preceding values of $\alpha$ and $\beta$, there will result: \[ \begin{aligned} \Phi(\alpha)&=\varphi(m\omega\pm\alpha')=\pm(-1)^{m}.\Phi\alpha',\\ f(\alpha)&=f(m\omega\pm\alpha')=(-1)^{m}.f\alpha',\\ F(\alpha)&=F(m\omega\pm\alpha')=F\alpha',\\ \Phi(\beta i)&=\varphi(+n\varpi i\pm\beta'i)=\pm(-1)^{n}.\varphi(\beta'i),\\ f(\beta i)&=f(n\varpi i\pm\beta'i)=f(\beta'i),\\ F(\beta i)&=F(n\varpi i\pm\beta'i)=(-1)^{n}.F(\beta'i). \end{aligned} \] Therefore the functions $\Phi\alpha$, $f\alpha$, $F\alpha$, $\Phi(\beta i)$, $f(\beta i)$, $F(\beta i)$ will be expressed as has just been said, and consequently also the functions $\Phi(\alpha+\beta i)$; $f(\alpha+\beta i)$; $F(\alpha+\beta i)$.

We have seen previously that $\Phi\alpha$ is real from $\alpha=-\frac{\omega}{2}$ to $\alpha=+\frac{\omega}{2}$, and that $\frac{\varphi(\alpha i)}{i}$ is real from $\alpha=-\frac{\varpi}{2}$ to $\alpha=+\frac{\varpi}{2}$.

Therefore, by virtue of equations (22.), it is clear:

1) that $\Phi(\alpha)$ and $\frac{\varphi(\alpha i)}{i}$ are real for every real value of $\alpha$; $\Phi\alpha$ is comprised between $-\frac{1}{c}$ and $+\frac{1}{c}$, and $\frac{\varphi(\alpha i)}{i}$ between $-\frac{1}{e}$ and $+\frac{1}{e}$;

2) that $\Phi(\alpha)$ vanishes for $\alpha=m\omega$, and $\frac{\varphi(\alpha i)}{i}$ for $\alpha=m\varpi$, $m$ being a positive or negative integral number; but $\Phi(\alpha)$ is not zero for any other real value of $\alpha$.

On remarking that $f\alpha=\sqrt{(1-c^{2}\Phi^{2}\alpha)}$, $F\alpha=\sqrt{(1+e^{2}\Phi^{2}\alpha)}$, it follows from what we have just said:

1) that the functions $f(\alpha)$; $F(\alpha)$; $f(\alpha i)$; $F(\alpha i)$ are real for every value of $\alpha$;

2) that $f(\alpha)$ is comprised between the limits $-1$ and $+1$ and $F(\alpha)$ between the limits $+1$ and $+\sqrt{\left(1+\frac{e^{2}}{c^{2}}\right)}$, so that $F\alpha$ is positive for every real value of $\alpha$;

3) that $f(\alpha i)$ is positive and comprised between the limits $+1$ and $\sqrt{\left(1+\frac{c^{2}}{e^{2}}\right)}$ and $F(\alpha i)$ between the limits $-1$ and $+1$ for every real value of $\alpha$;

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4) that $f(\alpha)$ vanishes for $\alpha=(m+\frac{1}{2})\omega$ and $F(i\alpha)$ for $\alpha=(m+\frac{1}{2})\varpi$; but for no other value of $\alpha$.

The following will be remarked as corollaries drawn from the formulas (22.):

1) Let $\alpha=0$. In this case, on remarking that $\Phi(0)=0$, $f(0)=1$, $F(0)=1$, we shall have \[ \left\{\begin{aligned} \varphi(m\omega+n\varpi i)&=0,\\ f(m\omega+n\varpi i)&=(-1)^{m},\\ F(m\omega+n\varpi i)&=-1. \end{aligned}\right. \tag{23.} \] Editorial note: The last line is printed as $-1$ over a blotted impression; by formula (22.) one would expect $(-1)^{n}$.

2) Let $\alpha=\frac{\omega}{2}$. By virtue of the equations: \[ \Phi\left(\frac{\omega}{2}\right)=\frac{1}{c};\ f\left(\frac{\omega}{2}\right)=0;\ F\left(\frac{\omega}{2}\right)=\frac{b}{c}; \] we shall have \[ \left\{\begin{aligned} \Phi((m+\tfrac{1}{2})\omega+n\varpi i)&=(-1)^{m+n}.\frac{1}{c},\\ f((m+\tfrac{1}{2})\omega+n\varpi i)&=0,\\ F((m+\tfrac{1}{2})\omega+n\varpi i)&=(-1)^{n}.\frac{b}{c}. \end{aligned}\right. \tag{24.} \]

3) Let $\alpha=\frac{\varpi}{2}i$. By virtue of the equations \[ \Phi\left(\frac{\varpi}{2}i\right)=\frac{i}{e};\ f\left(\frac{\varpi}{2}i\right)=\frac{b}{e};\ F\left(\frac{\varpi}{2}i\right)=0; \] we shall have: \[ \left\{\begin{aligned} \Phi(m\omega+(n+\tfrac{1}{2})\varpi i)&=(-1)^{m+n}\frac{i}{e},\\ f(m\omega+(n+\tfrac{1}{2})\varpi i)&=(-1)^{m}.\frac{b}{e},\\ F(m\omega+(n+\tfrac{1}{2})\varpi i)&=0. \end{aligned}\right. \tag{25.} \]

4) Let $\alpha=\frac{\omega}{2}+\frac{\varpi}{2}i$. By virtue of the equations above we shall have \[ \left\{\begin{aligned} \Phi((m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i)&=\frac{1}{0},\\ f((m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i)&=\frac{1}{0},\\ F((m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i)&=\frac{1}{0}. \end{aligned}\right. \tag{26.} \]

6.

The equations (23.), (24.), (25.) show that the function $\Phi(\alpha)$ vanishes whenever $\alpha$ is of the form $\alpha=m\omega+n\varpi i$; that $f\alpha$ vanishes whenever $\alpha$ is of the form $\alpha=(m+\frac{1}{2})\omega+n\varpi i$, and that $F\alpha$ vanishes whenever $\alpha$ is of the form $\alpha=m\omega+(n+\frac{1}{2})\varpi i$.

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Now I say that for every other value of $\alpha$, the functions $\Phi\alpha$, $f\alpha$, $F\alpha$ will necessarily have a value different from zero.

Suppose in fact that we have \[ \Phi(\alpha+\beta i)=0, \] $\alpha$ and $\beta$ being real quantities. By virtue of the first of the formulas (10.), this equation may be written as follows: \[ \frac{\varphi\alpha.f(\beta i).F(\beta i)+\varphi(\beta i).f\alpha.F\alpha}{1+e^{2}c^{2}\varphi^{2}\alpha.\varphi^{2}(\beta i)}=0. \]

Now the quantities $\Phi\alpha$, $f(\beta i)$, $F(\beta i)$ are real and $\Phi(\beta i)$ is of the form $i.A$, where $A$ is real; therefore this equation cannot subsist unless we have separately: \[ \Phi(\alpha).f(\beta i).F(\beta i)=0;\ \Phi(\beta i).f\alpha.F\alpha=0. \] These equations can be satisfied only in two ways, namely by making \[ \Phi(\alpha)=0,\ \Phi(\beta i)=0, \] or \[ f(\beta i).F(\beta i)=0,\ f\alpha.F\alpha=0. \]

The first two equations give $\alpha=m\omega$; $\beta=n\varpi$.

The last two, on remarking that $F\alpha$ and $f(\beta i)$ can never vanish, give \[ f\alpha=0,\ F(\beta i)=0, \] whence \[ \alpha=(m+\tfrac{1}{2})\omega,\ \beta=(n+\tfrac{1}{2})\varpi. \] But for these values of $\alpha$ and $\beta$ the value of $\Phi(\alpha+\beta i)$ will become infinite; therefore the only values of $\alpha$ and $\beta$ are $\alpha=m\omega$ and $\beta=n\varpi$, and consequently all the roots of the equation \[ \Phi(x)=0, \] may be represented by \[ x=m\omega+n\varpi i. \tag{27.} \] In the same manner we shall find that all the roots of the equation \[ f(x)=0, \] may be represented by \[ x=(m+\tfrac{1}{2})\omega+n\varpi i, \tag{28.} \] and those of the equation \[ F(x)=0, \] by \[ x=m\omega+(n+\tfrac{1}{2})\varpi i. \tag{29.} \]

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7.

The formulas (26.) show that the three equations \[ \Phi(x)=\frac{1}{0},\ f(x)=\frac{1}{0},\ F(x)=\frac{1}{0}, \] are satisfied by giving to $x$ one of the values of the form \[ x=(m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i. \tag{30.} \]

Now it can be demonstrated that the equations in question have no other roots.

In fact, having \[ \begin{gathered} F(x)=\frac{b}{c}.\frac{1}{F\left(x-\frac{\omega}{2}\right)};\ \Phi(x)=\frac{-i}{ec}.\frac{1}{\varphi\left(x-\frac{\omega}{2}-\frac{\varpi}{2}i\right)};\\ f(x)=\frac{b}{e}.\frac{1}{f\left(x-\frac{\varpi}{2}i\right)}; \end{gathered} \] the equations in question will entail these: \[ \varphi\left(x-\frac{\omega}{2}-\frac{\varpi}{2}i\right)=0;\ f\left(x-\frac{\varpi}{2}i\right)=0;\ F\left(x-\frac{\omega}{2}\right)=0, \] but by virtue of what we have just seen in the preceding number, these equations give respectively: \[ \begin{gathered} x-\frac{\omega}{2}-\frac{\varpi}{2}i=m\omega+n\varpi i;\ x-\frac{\varpi}{2}i=(m+\tfrac{1}{2})\omega+n\varpi i;\\ x-\frac{\omega}{2}=m\omega+(n+\tfrac{1}{2})\varpi i; \end{gathered} \] that is: we shall have for the three equations: \[ x=(m+\tfrac{1}{2})\omega+(n+\tfrac{1}{2})\varpi i, \] Q. E. D.

8.

Having found as above all the roots of the equations \[ \begin{gathered} \Phi(x)=0;\ f(x)=0;\ F(x)=0;\\ \Phi(x)=\frac{1}{0};\ f(x)=\frac{1}{0};\ F(x)=\frac{1}{0}; \end{gathered} \] I shall now seek the roots of the more general equations. \[ \Phi(x)=\Phi a,\ f(x)=fa,\ F(x)=Fa, \] where $a$ is any quantity whatever, real or imaginary.

Let us consider first the equation \[ \Phi(x)-\Phi a=0. \] On making in the second of the formulas (12.) \[ \alpha=\frac{x+a}{2},\ \beta=\frac{x-a}{2}, \] we shall find \[ \Phi x-\Phi a=\frac{2\varphi\left(\frac{x-a}{2}\right).f\left(\frac{x+a}{2}\right).F\left(\frac{x+a}{2}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{x+a}{2}\right).\varphi^{2}\left(\frac{x-a}{2}\right)}=0. \]

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This equation can subsist only in one of the five following cases:

1. if $\Phi\left(\frac{x-a}{2}\right)=0$, whence $x=a+2m\omega+2n\varpi i$,

2. if $f\left(\frac{x+a}{2}\right)=0$, whence $x=-a+(2m+1)\omega+2n\varpi i$,

3. if $F\left(\frac{x+a}{2}\right)=0$, whence $x=-a+2m\omega+(2n+1)\varpi i$,

4. if $\Phi\left(\frac{x-a}{2}\right)=\frac{1}{0}$, whence $x=a+(2m+1)\omega+(2n+1)\varpi i$,

5. if $\Phi\left(\frac{x+a}{2}\right)=\frac{1}{0}$, whence $x=-a+(2m+1)\omega+(2n+1)\varpi i$.

The solution of these five equations is contained in the formulas (27.), (28.), (29.).

Of the values found for $x$, those must be rejected which the formula \[ x=-a+(2m+1)\omega+(2n+1)\varpi i, \] gives, for such a value of $x$ gives by virtue of (22.): \[ \Phi x=-\Phi a, \] whereas we must have $\Phi x=\Phi a$; but the other values of $x$, expressed by the first four formulas, may be admitted. They are, as one sees, contained in the single formula: \[ x=(-1)^{m+n}.a+m\omega+n\varpi i. \tag{31.} \] Such then is the general expression of all the roots of the equation \[ \Phi x=\Phi a. \] In the same manner we shall find that all the roots of the equation \[ fx=fa \] are represented by the formula \[ x=\pm a+2m\omega+n\varpi i, \tag{32.} \] and all those of the equation \[ Fx=Fa, \] by the formula \[ x=\pm a+m\omega+2n\varpi i. \tag{33.} \]

§. II.

Formulas which give the values of $\Phi(n\alpha)$, $f(n\alpha)$, $F(n\alpha)$ expressed in rational functions of $\Phi\alpha$, $f\alpha$, $F\alpha$.

9.

Let us resume the formulas (12.). On making in the $1{}^{\mathrm{e}}$, $3{}^{\mathrm{e}}$ and $5{}^{\mathrm{e}}$ $\alpha=n\beta$, there will result:

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\[ \left\{\begin{aligned} \Phi(n+1)\beta&=-\Phi(n-1)\beta+\frac{2\varphi(n\beta).f\beta.F\beta}{R},\\ f(n+1)\beta&=-f(n-1)\beta+\frac{2f(n\beta).f\beta}{R},\\ F(n+1)\beta&=-F(n-1)\beta+\frac{2F(n\beta).F\beta}{R}, \end{aligned}\right. \tag{34.} \] where $R=1+c^{2}e^{2}.\Phi^{2}(n\beta).\Phi^{2}\beta$.

These formulas give the value of $\Phi(n+1)\beta$ in terms of $\Phi(n-1)\beta$ and $\Phi(n\beta)$; that of $f(n+1)\beta$ in terms of $f(n-1)\beta$ and $f(n\beta)$, and that of $F(n+1)\beta$ in terms of $F(n-1)\beta$ and $F(n\beta)$. Therefore, on making successively $n=1$, 2, 3…, we shall find successively the values of the functions: \[ \begin{gathered} \Phi(2\beta);\ \Phi(3\beta);\ \Phi(4\beta)\ .....\ \Phi(n\beta),\\ f(2\beta);\ f(3\beta);\ f(4\beta)\ .....\ f(n\beta),\\ F(2\beta);\ F(3\beta);\ F(4\beta)\ .....\ F(n\beta) \end{gathered} \] expressed in rational functions of the three quantities \[ \Phi\beta;\ f\beta;\ F\beta. \] On making e.g. $n=1$, we shall have: \[ \left\{\begin{aligned} \Phi(2\beta)&=\frac{2.\varphi\beta.f\beta.F\beta}{1+e^{2}c^{2}\varphi^{4}\beta},\\ f(2\beta)&=-1+\frac{2f^{2}\beta}{1+e^{2}c^{2}\varphi^{4}\beta},\\ F(2\beta)&=-1+\frac{2F^{2}\beta}{1+e^{2}c^{2}\varphi^{4}\beta}. \end{aligned}\right. \tag{35.} \]

The functions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ being rational functions of $\Phi\beta$, $f\beta$, $F\beta$, they can always be reduced to the form $\frac{P}{Q}$, where $P$ and $Q$ are integral functions of $\Phi\beta$, $f\beta$, $F\beta$. Likewise it is clear that the denominator $Q$ will have the same value for the three functions under consideration. Let then \[ \Phi(n\beta)=\frac{P_n}{Q_n},\ f(n\beta)=\frac{P'_n}{Q_n},\ F(n\beta)=\frac{P''_n}{Q_n}, \tag{35$'$.} \] we shall likewise have \[ \begin{gathered} \Phi(n+1)\beta=\frac{P_{n+1}}{Q_{n+1}},\ f(n+1)\beta=\frac{P'_{n+1}}{Q_{n+1}},\ F(n+1)\beta=\frac{P''_{n+1}}{Q_{n+1}},\\ \Phi(n-1)\beta=\frac{P_{n-1}}{Q_{n-1}},\ f(n-1)\beta=\frac{P'_{n-1}}{Q_{n-1}},\ F(n-1)\beta=\frac{P''_{n-1}}{Q_{n-1}}. \end{gathered} \] On substituting these values, the first of the formulas (34.) will become: \[ \frac{P_{n+1}}{Q_{n+1}}=-\frac{P_{n-1}}{Q_{n-1}}+\frac{2f\beta.F\beta.\dfrac{P_n}{Q_n}}{1+c^{2}e^{2}\varphi^{2}\beta.\dfrac{P_n^{2}}{Q_n^{2}}}, \]

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or else \[ \frac{P_{n+1}}{Q_{n+1}}=\frac{-P_{n-1}.(Q_n^{2}+c^{2}e^{2}\varphi^{2}\beta.P_n^{2})+2P_n.Q_n.Q_{n-1}.f\beta.F\beta}{Q_{n-1}.(Q_n^{2}+e^{2}c^{2}\varphi^{2}\beta.P_n^{2})}. \]

On equating the numerators and the denominators of these two fractions, we shall have \[ P_{n+1}=-P_{n-1}(Q_n^{2}+c^{2}e^{2}\Phi^{2}\beta.P_n^{2})+2f\beta.F\beta.P_n.Q_n.Q_{n-1}, \tag{36.} \] \[ Q_{n+1}=Q_{n-1}(Q_n^{2}+e^{2}c^{2}\Phi^{2}\beta.P_n^{2}). \tag{37.} \] The second and the third of the equations (34.) will give in the same manner: \[ P'_{n+1}=-P'_{n-1}.(Q_n^{2}+c^{2}e^{2}\Phi^{2}\beta.P_n^{2})+2f\beta\,P'_n.Q_n.Q_{n-1}, \tag{38.} \] \[ P''_{n+1}=-P''_{n-1}.(Q_n^{2}+c^{2}e^{2}\Phi^{2}\beta.P_n^{2})+2F\beta.P''_n.Q_n.Q_{n-1}. \tag{39.} \] On making in these four formulas $n=1$, 2, 3 ....., and remarking that we shall have: \[ \begin{gathered} Q_0=1;\ Q_1=1;\ P_0=0;\ P_1=\Phi\beta;\\ P'_0=1;\ P'_1=f\beta;\ P''_0=1;\ P''_1=F\beta; \end{gathered} \] we shall find successively the integral functions $Q_n$, $P_n$, $P'_n$, $P''_n$, for all values of $n$.

Let, for brevity: \[ \Phi\beta=x,\ f\beta=y,\ F\beta=z\ \text{and} \tag{40.} \] \[ R_n=Q_n^{2}+e^{2}c^{2}x^{2}P_n^{2}, \tag{41.} \] the preceding formulas will give: \[ \left\{\begin{aligned} Q_{n+1}&=Q_{n-1}.R_n,\\ P_{n+1}&=-P_{n-1}.R_n+2yz.P_n.Q_n.Q_{n-1},\\ P'_{n+1}&=-P'_{n-1}.R_n+2y.P'_n.Q_n.Q_{n-1},\\ P''_{n+1}&=-P''_{n-1}.R_n+2z.P''_n.Q_n.Q_{n-1}. \end{aligned}\right. \tag{42.} \] On putting $n=1$, 2, we shall have: \[ \left\{\begin{aligned} R_1&=Q_1^{2}+e^{2}c^{2}x^{2}P_1^{2}=1+e^{2}c^{2}x^{4},\\ Q_2&=Q_0R_1=1+e^{2}c^{2}x^{4},\\ P_2&=-P_0R_1+2yzP_1.Q_1.Q_0=2xyz,\\ P'_2&=-P'_0R_1+2yP'_1.Q_1.Q_0=-1-e^{2}c^{2}x^{4}+2y^{2},\\ P''_2&=-P''_0R_1+2zP''_1.Q_1.Q_0=-1-e^{2}c^{2}x^{4}+2z^{2}. \end{aligned}\right. \tag{43.} \] \[ \left\{\begin{aligned} R_2&=Q_2^{2}+e^{2}c^{2}x^{2}.P_2^{2}=(1+e^{2}c^{2}x^{4})^{2}+e^{2}c^{2}x^{2}.4x^{2}y^{2}z^{2},\\ Q_3&=Q_1R_2=R_2,\\ P_3&=-P_1R_2+2yzP_2Q_2Q_1=-xR_2+2y^{2}z^{2}x.Q_2\\ &=x\{2y^{2}z^{2}.Q_2-R_2\},\\ P'_3&=-P'_1.R_2+2y.P'_2.Q_2Q_1=-y.R_2+2yP'_2Q_2\\ &=y\{2Q_2P'_2-R_2\},\\ P''_3&=z\{2Q_2P''_2-R_2\}. \end{aligned}\right. \tag{44.} \]

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On continuing in this manner, and remarking that $y^{2}=1-c^{2}x^{2}$, $z^{2}=1+e^{2}x^{2}$, one will readily see that the quantities: \[ Q_n,\ \frac{P_{2n}}{xyz},\ \frac{P_{2n+1}}{x},\ P'_{2n},\ \frac{P'_{2n+1}}{y},\ P''_{2n},\ \frac{P''_{2n+1}}{z} \] are integral functions of the three quantities $x^{2}$, $y^{2}$, $z^{2}$ and consequently also of any one whatever of these quantities, for any integral value whatever of $n$.

This shows that the expressions of $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ will be of the following form: \[ \left\{\begin{aligned} \Phi(2n\beta)&=\Phi\beta.f\beta.F\beta.T, & \Phi(2n+1)\beta&=\Phi\beta.T',\\ f(2n\beta)&=T_1, & f(2n+1)\beta&=f\beta.T'',\\ F(2n\beta)&=T_2, & F(2n+1)\beta&=F\beta.T''', \end{aligned}\right. \tag{45.} \] where $T$ etc. represent rational functions of the quantities $(\Phi\beta)^{2}$, $(f\beta)^{2}$, $(F\beta)^{2}$.

§. III.

Solution of the equations \[ \Phi(n\beta)=\frac{P_n}{Q_n},\ f(n\beta)=\frac{P'_n}{Q_n},\ F(n\beta)=\frac{P''_n}{Q_n}. \]

10.

According to what has been seen, the functions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ are expressed rationally in $x$, $y$, $z$. The converse does not hold, for the equations (35.) are in general of a very high degree. They have for this reason a certain number of roots. We shall see how one can readily express all these roots by means of the functions $\Phi$, $f$, $F$.

A. Let us consider first the equation $\Phi(n\beta)=\frac{P_n}{Q_n}$, or $Q_n.\Phi(n\beta)=P_n$, and let us seek all the values of $x$.

Two cases must be distinguished, according as $n$ is even or odd:

1) If $n$ is an even number.

From what has been seen in the preceding paragraph (45.), we shall have in this case \[ \Phi(2n\beta)=\psi(x^{2}).x.y.z, \] that is, by virtue of the formulas \[ y=\sqrt{(1-c^{2}x^{2})},\ z=\sqrt{(1+e^{2}x^{2})}, \] \[ \Phi(2n\beta)=x\psi(x^{2}).\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}. \] Therefore the equation in $x$ will become, \[ \Phi^{2}(2n\beta)=x^{2}(\psi(x^{2}))^{2}.(1-c^{2}x^{2})(1+e^{2}x^{2}). \]

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On denoting the second member by $\theta(x^{2})$, we shall have \[ \Phi^{2}(2n\beta)=\theta(x^{2}). \] $\Phi\beta$ being one of the values of $x$, we shall have \[ \Phi^{2}(2n\beta)=\theta(\Phi^{2}\beta), \tag{46.} \] an equation which holds, whatever the value of $\beta$ may be. To find the other values of $x$, let $x=\Phi\alpha$ be any root whatever; we must have \[ \Phi^{2}(2n\beta)=\theta(\Phi^{2}\alpha). \] Now, on putting in (46.) $\alpha$ in place of $\beta$, there will result \[ \Phi^{2}(2n\alpha)=\theta(\Phi^{2}\alpha),\ \text{therefore:} \] \[ \Phi^{2}(2n\beta)=\Phi^{2}(2n\alpha); \tag{47.} \] an equation which comes to these two: \[ \Phi(2n\alpha)=\Phi(2n\beta)\ \text{and}\ \Phi(2n\alpha)=-\Phi(2n\beta). \] The first gives by virtue of (31.) \[ 2n\alpha=2n\beta.(-1)^{m+\mu}+m\omega+\mu\varpi i, \] where $m$ and $\mu$ are any two integral numbers, positive or negative, zero included.

The second gives the same values of $2m\alpha$Editorial note: Printed $2m\alpha$ instead of $2n\alpha$; an apparent misprint., but of contrary sign, as it is easy to see, on writing it as follows: \[ \Phi(-2n\alpha)=\Phi(2n\beta). \] Every value of $2n\alpha$ which satisfies the equation (47.) can therefore be represented by \[ 2n\alpha=\pm(2n\beta.(-1)^{m+\mu}+m\omega+\mu\varpi i). \] From this we draw the value of $\alpha$, on dividing by $2n$, namely: \[ \alpha=\pm\left((-1)^{m+\mu}.\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right). \] Having the value of $\alpha$, we shall have \[ \Phi\alpha=\pm\Phi\left((-1)^{m+\mu}.\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right)=x. \tag{48.} \] Therefore all the values of $x$ are contained in this expression, and they will be found on giving to the numbers $m$ and $\mu$ all the integral values from $-\infty$ to $+\infty$. Now to have all those which are different from one another, it suffices to give to $m$ and $\mu$ integral values less than $2n$. In fact, whatever these numbers may be, one can always suppose them reduced to the form: \[ m=2n.k+m',\ \mu=2n.k'+\mu', \] where $k$, $k'$ are integral numbers, and $m'$, $\mu'$, integral numbers less than $2n$. On substituting these values in the expression of $x$, it will become:

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\[ x=\pm\Phi\left\{(-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i+k\omega+k'\varpi i\right\}, \] now by virtue of (22.) this expression reduces to \[ x=\pm\Phi\left\{(-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i\right\}. \tag{49.} \] This value of $x$ is of the same form as the preceding (48.), $m$ and $\mu$ only being replaced by $m'$ and $\mu'$, which, both of them, are positive and less than $2n$; therefore we shall obtain all the different values of $x$, on giving only to $m$ and $\mu$ all the integral values from zero up to $2n$ excl. All these values are necessarily different from one another. In fact, suppose for example that we have \[ \begin{gathered} \pm\Phi\left\{(-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i\right\}\\ =\pm\Phi\left\{(-1)^{m+\mu}\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right\}, \end{gathered} \] it would follow, according to (31.): \[ (-1)^{m'+\mu'}\beta+\frac{m'}{2n}\omega+\frac{\mu'}{2n}\varpi i=\pm\left\{(-1)^{m+\mu}.\beta+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right\}+k\omega+k'\varpi i, \] $k$ and $k'$ being integers.

This equation gives: \[ \mu'=k'.2n\pm\mu,\ m'=k.2n\pm m,\ (-1)^{m'+\mu'}=\pm(-1)^{m+\mu}. \] The first two equations cannot subsist unless $k'=1$, $k=1$, $\mu'=2n-\mu$, $m'=2n-m$, and then the last will become: \[ (-1)^{4n-m-\mu}=-(-1)^{m+\mu}, \] whence we draw: \[ (-1)^{2m+2\mu}=-1, \] an absurd result.

Therefore all the values of $x$, contained in the formula (48.), are different from one another, if $m$ and $\mu$ are positive and less than $m$Editorial note: Printed $m$ instead of $2n$; an apparent misprint..

The total number of the values of $x$ is, as it is easy to see, equal to $2(2n)^{2}=8n^{2}$, now the equation $\Phi^{2}(2n\beta)=\theta(x^{2})$ cannot have equal roots, for in that case we should have $\frac{\partial.\theta(x^{2})}{\partial x}=0$, which would give for $x$ a value independent of $\beta$. Therefore the degree of the equation $\Phi^{2}(2n\beta)=\theta(x^{2})$ is equal to the number of the roots, that is to $8n^{2}$. If for example $n=1$, we shall have the equation \[ \Phi^{2}(2\beta)=\theta(x^{2})=\frac{4x^{2}(1-c^{2}x^{2})(1+e^{2}x^{2})}{(1+e^{2}c^{2}x^{4})^{2}}, \] or else \[ (1+e^{2}c^{2}x^{4})^{2}.\Phi^{2}(2\beta)=4x^{2}(1-c^{2}x^{2})(1+e^{2}x^{2}), \]

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and, according to formula (48.), the roots of this equation, eight in number, will be: \[ \begin{gathered} x=\pm\Phi\beta,\ x=\pm\Phi\left(-\beta+\frac{\omega}{2}\right),\\ x=\pm\Phi\left(-\beta+\frac{\varpi}{2}i\right),\ x=\pm\Phi\left(\beta+\frac{\omega}{2}+\frac{\varpi}{2}i\right). \end{gathered} \]

2) If $n$ is an odd number $=2n+1$.

In this case $\frac{P_{2n+1}}{Q_{2n+1}}$ is, as we have seen, a rational function of $x$, and consequently the equation in $x$ will be: \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \tag{50.} \] Precisely as in the preceding case, one will find that all the roots of this equation may be represented by \[ x=\Phi\left((-1)^{m+\mu}.\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right), \tag{51.} \] where one must give to $m$ and $\mu$ all integral values from $-n$ to $+n$ inclusive. Hence the number of different roots is $(2n+1)^{2}$. This is also the degree of the equation in question. One may also express the roots by \[ x=(-1)^{m+\mu}.\Phi\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right). \] If for example $n=1$, one will have an equation of the degree $3^{2}=9$.

The formula (51.) gives for $x$ the following 9 values: \[ \begin{gathered} \Phi(\beta);\\ \Phi\left(-\beta-\frac{\omega}{3}\right);\\ \Phi\left(-\beta+\frac{\omega}{3}\right);\\ \Phi\left(-\beta-\frac{\varpi}{3}i\right);\\ \Phi\left(-\beta+\frac{\varpi}{3}i\right);\\ \Phi\left(\beta-\frac{\omega}{3}-\frac{\varpi}{3}i\right);\\ \Phi\left(\beta-\frac{\omega}{3}+\frac{\varpi}{3}i\right);\\ \Phi\left(\beta+\frac{\omega}{3}-\frac{\varpi}{3}i\right);\\ \Phi\left(\beta+\frac{\omega}{3}+\frac{\varpi}{3}i\right); \end{gathered} \]

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B. Let us now consider the equation \[ f(n\beta)=\frac{P'_n}{Q_n} \tag{52.} \] and seek the values of $y$ which satisfy this equation. The function $\frac{P'_n}{Q_n}$ being, as has been seen above, rational in $y$: the equation in $y$, on putting $\frac{P'_n}{Q_n}=\psi(y)$, will be \[ f(n\beta)=\psi(y). \] One of the roots of this equation is $y=f\beta$, therefore, whatever be the valueEditorial note: Printed “leur” instead of “valeur”; the first syllable of the word is missing at the line break. of $\beta$: \[ f(n\beta)=\psi(f\beta). \tag{53.} \] To find the other values of $y$, let $\alpha$ be a new unknown, such that $y=f\alpha$; there will be \[ f(n\beta)=\psi(f\alpha); \] now, by virtue of (53.), the second member is equal to $f(n\alpha)$, therefore to determine $\alpha$ one will have the equation: \[ f(n\alpha)=f(n\beta). \] By virtue of (32.) this equation gives for the general expression of $n\alpha$: \[ n\alpha=\pm n\beta+2m\omega+\mu\varpi i, \] $m$ and $\mu$ being two integral numbers, positive or negative, zero included.

From this we derive \[ \alpha=\pm\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i, \] and consequently: \[ f\alpha=f\left(\pm\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right)=y. \] This is the general value of $y$. Now, to have the different values of $y$, I say that it suffices to take $\beta$ with the sign $+$ and to give to $m$ and $\mu$ all integral values less than $n$. Indeed, since one has $f(+\alpha)=f(-\alpha)$, one will have first: \[ f\left(-\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right)=f\left(\beta-\frac{2m}{n}\omega-\frac{\mu}{n}\varpi i\right). \] Therefore one may always take $\beta$ with the sign $+$ in the expression of $y$. Thus all the values of $y$ are contained in the expression \[ y=f\left(\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right). \tag{54.} \]

Now, whatever the numbers $m$ and $\mu$ may be, one may always suppose, \[ m=k.n+m',\ \mu'=k'.n+\mu', \] Editorial note: Printed $\mu'$ on the left of the second equation instead of $\mu$; an apparent misprint.

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where $k$, $k'$, $m'$, $\mu'$ are integral numbers, the last two being at the same time positive and less than $n$.

On substituting, there will result \[ y=f\left(\beta+\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i+2k\omega+k'\varpi i\right). \] Now, by virtue of (22.), the second member of this equation is equal to \[ f\left(\beta+\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i\right)=y, \tag{55.} \] a quantity of the same form as the second member of (54.); only $m'$ and $\mu'$ are positive and less than $n$. Therefore etc.

On giving to $m$ and $\mu$ all possible values less than $n$, one will find a number $n^{2}$ of values of $y$. Now, in general all these quantities are different from one another. Indeed, suppose for example \[ f\left(\beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right)=f\left(\beta+\frac{2\mu'}{n}+\frac{\mu'}{n}\varpi i\right), \] Editorial note: In the right member the print has $2\mu'/n$ where $2m'\omega/n$ is expected; the $\omega$ is absent. An apparent misprint. one will have by virtue of (32.), denoting by $k$, $k'$ two integral numbers: \[ \beta+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i=\pm\left(\beta+\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i\right)+2k\omega+k'\varpi i. \] Since $\beta$ may have any irrational value whatever, it is clear that this equation cannot subsist unless one takes in the second member the upper sign. Then there will result \[ \frac{2m}{n}\omega+\frac{\mu}{n}\varpi i=\frac{2m'}{n}\omega+\frac{\mu'}{n}\varpi i+2k\omega+k'\varpi i, \] whence one derives, on equating the real parts and the imaginary parts: \[ m=m'+kn;\ \mu=\mu'+k'n, \] absurd equations, on observing that the numbers $m$, $m'$, $\mu$ and $\mu'$ are all positive and less than $n$. Therefore in general the equation \[ f(n\beta)=\psi(y) \] has a number $n^{2}$ of roots different from one another, and not a greater number. Now in general all the roots of this equation are different from one another. Indeed, if two of them were equal, one would have at once: \[ f(n\beta)=\psi(y)\ \text{and}\ 0=\psi'(y), \] and this is impossible, if one observes that the coefficients of $y$ in $\psi(y)$ do not contain $\beta$. Therefore in general the equation (52.) is necessarily of the degree $n^{2}$.

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C. The equation \[ F(n\beta)=\frac{P''_n}{Q_n}, \tag{56.} \] being treated in absolutely the same manner with respect to $z$ as the equation $f(n\beta)=\frac{P'_n}{Q_n}$ has been with respect to $y$, gives for the general expression of the values of $z$: \[ z=F\left(\beta+\frac{m}{n}\omega+\frac{2\mu}{n}\varpi i\right), \tag{57.} \] where $m$ and $\mu$ are integral, positive and less than $n$. The number of values of $z$ is $n^{2}$, and they are in general all different from one another.

Therefore in general the equation (56.) is of the degree $n^{2}$.

11.

We have found above all the roots of the equations \[ \varphi(n\beta)=\frac{P_n}{Q_n},\ f(n\beta)=\frac{P'_n}{Q_n},\ F(n\beta)=\frac{P''_n}{Q_n}, \] roots which are expressed by the formulas (48.), (51.), (54.), (57.). All these roots are different from one another, except in the cases of particular values of $\beta$; but for these values the different roots are contained in the same formulas. — In this latter case a certain number of the values of the quantities $x$, $y$, $z$ will be equal; but it is clear that all the values, equal or unequal, will nevertheless be the roots of the equations in question. This is made evident by making $\beta$ converge towards a particular value which gives for $x$, or $y$, or $z$ equal values.

On putting in the formula (48.) $\beta=\frac{\alpha}{2n}$, one will have the equation \[ \varphi^{2}\alpha=\frac{P_{2n}^{2}}{Q_{2n}^{2}},\ \text{whose roots are}\ x=\pm\varphi\left((-1)^{m+\mu}\frac{\alpha}{2n}+\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right); \tag{58.} \] and where $m$ and $\mu$ have all integral and positive values less than $2n$.

On putting likewise in the formula (50.) $\beta=\frac{\alpha}{2n+1}$, one will have \[ \Phi\alpha=\frac{P_{2n+1}}{Q_{2n+1}},\ \text{whose roots are} \] \[ x=(-1)^{m+\mu}.\Phi\left(\frac{\alpha}{2n+1}+\frac{m\omega+\mu\varpi i}{2n+1}\right), \tag{59.} \] $m$ and $\mu$ having for values all integral numbers from $-n$ to $+n$.

Finally, on putting in (52.), (56.) $\beta=\frac{\alpha}{n}$, one will have the equation \[ f\alpha=\frac{P'_n}{Q_n},\ \text{whose roots are} \] \[ y=f\left(\frac{\alpha}{n}+\frac{2m}{n}\omega+\frac{\mu}{n}\varpi i\right), \tag{60.} \]

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and the equation \[ F\alpha=\frac{P''_n}{Q_n},\ \text{whose roots are} \] \[ z=F\left(\frac{\alpha}{n}+\frac{m}{n}\omega+\frac{2\mu}{n}\varpi i\right), \tag{61.} \] where $m$ and $\mu$ are enclosed between the limits 0 and $n-1$ inclusive. If $n$ is odd $=2n+1$, one may also suppose \[ \begin{gathered} y=(-1)^{m}.f\left(\frac{\alpha}{2n+1}+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right),\\ z=(-1)^{\mu}.F\left(\frac{\alpha}{2n+1}+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right), \end{gathered} \] $m$ and $\mu$ having all integral values from $-n$ to $+n$.

In all these equations the quantity $\alpha$ may have any value whatever.

As particular cases one must notice the following:

1) On putting in (58.) and (59.) $\alpha=0$, one will have the equations \[ \left\{\begin{aligned} &P_{2n}^{2}=0,\ \text{whose roots are}\ x=\pm\Phi\left(\frac{m}{2n}\omega+\frac{\mu}{2n}\varpi i\right)\\ &\text{(the limits of $m$ and $\mu$ being 0 and $2n-1$),}\\ &P_{2n+1}=0,\ \text{whose roots are}\ x=\Phi\left(\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right)\\ &\text{(the limits of $m$ and $\mu$ being $-n$ and $+n$).} \end{aligned}\right. \tag{62.} \]

2) On putting in (60.) $\alpha=\frac{\omega}{2}$ and in (61.) $\alpha=\frac{\varpi}{2}i$, and observing that $f\left(\frac{\omega}{2}\right)=0$, $F\left(\frac{\varpi}{2}i\right)=0$, one will obtain the two equations: \[ P'_n=0,\ \text{whose roots are}\ y=f\left((2m+\tfrac{1}{2})\frac{\omega}{n}+\frac{\mu}{n}\varpi i\right) \tag{63.} \] \[ P''_n=0,\ \text{whose roots are}\ z=F\left(\frac{m}{n}\omega+(2\mu+\tfrac{1}{2})\frac{\varpi i}{n}\right) \tag{64.} \] (the limits of $m$ and $\mu$ being 0 and $n-1$).

3) On putting in (58.) $\alpha=\frac{\omega}{2}+\frac{\varpi}{2}i$, and observing that $\varphi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}$, one will have the equation \[ Q_{2n}^{2}=0, \] whose roots will be: \[ x=\pm\Phi\left((m+\tfrac{1}{2}(-1)^{m+\mu})\frac{\omega}{2n}+(\mu+\tfrac{1}{2}(-1)^{m+\mu}).\frac{\varpi i}{2n}\right). \]

The values of $x$ must be equal in pairs, and one will readily see that the unequal values may be represented by \[ x=\Phi\left((m+\tfrac{1}{2})\frac{\omega}{2n}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n}\right), \tag{65.} \]

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on giving to $m$ and $\mu$ all integral values from 0 to $2n-1$. Therefore these are the roots of the equation \[ Q_{2n}=0,\ \text{with respect to } x. \]

On putting likewise in (59.) $\alpha=\frac{\omega}{2}+\frac{\varpi}{2}i$, one will have the equation \[ Q_{2n+1}=0, \] whose roots will be: \[ \left\{\begin{aligned} x&=(-1)^{m+\mu}.\varphi\left((m+\tfrac{1}{2})\frac{\omega}{2n+1}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n+1}\right),\\ y&=(-1)^{m}.f\left((m+\tfrac{1}{2})\frac{\omega}{2n+1}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n+1}\right),\\ z&=(-1)^{\mu}.F\left((m+\tfrac{1}{2})\frac{\omega}{2n+1}+(\mu+\tfrac{1}{2})\frac{\varpi i}{2n+1}\right), \end{aligned}\right. \tag{66.} \] $m$ and $\mu$ having for values all integral numbers from $-n$ to $+n$.

Among the values of $x$, $y$, $z$, one must notice that which answers to $m=n$, $\mu=n$. Then one has \[ \begin{aligned} x&=(-1)^{n}.\varphi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\\ y&=(-1)^{n}.f\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0},\\ z&=(-1)^{n}.F\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}. \end{aligned} \]

These infinite values show that the equation $Q_{2n+1}=0$ is of a degree lower by one unit than that of the equations from which it comes. Setting aside these values, the remaining ones, $(2n+1)^{2}-1$ in number, will be the roots of the equation $Q_{2n+1}=0$.

§. IV.

Algebraic resolution of the equations \[ \Phi\alpha=\frac{P_{2n+1}}{Q_{2n+1}},\quad f\alpha=\frac{P'_{2n+1}}{Q_{2n+1}},\quad F\alpha=\frac{P''_{2n+1}}{Q_{2n+1}}. \]

12.

We have seen in the preceding §. how one may readily express the roots of the equations in question by means of the functions $\Phi$, $f$, $F$. We are now going to deduce from this the resolution of these same equations, or the determination of the functions $\varphi\left(\frac{\alpha}{n}\right)$, $f\left(\frac{\alpha}{n}\right)$, $F\left(\frac{\alpha}{n}\right)$, in terms of the functions $\varphi\alpha$, $f\alpha$, $F\alpha$.

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As one has \[ \varphi\left(\frac{\alpha}{m\mu}\right)=\varphi\left(\frac{1}{m}\left(\frac{\alpha}{\mu}\right)\right), \] one may suppose that $n$ is a prime number. First we shall consider the case where $n=2$, and afterwards that where $n$ is an odd number.

A. Expressions of the functions $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$.

13.

The values of $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$ may be found very easily in the following manner. Supposing in the formulas (35.) $\beta=\frac{\alpha}{2}$, and putting \[ x=\varphi\left(\frac{\alpha}{2}\right),\ y=f\left(\frac{\alpha}{2}\right),\ z=F\left(\frac{\alpha}{2}\right), \] there will result: \[ f(\alpha)=\frac{y^{2}-c^{2}x^{2}z^{2}}{1+e^{2}c^{2}x^{4}},\ F(\alpha)=\frac{z^{2}+e^{2}y^{2}x^{2}}{1+e^{2}c^{2}x^{4}}, \] or else, on substituting the values of $y^{2}$ and $z^{2}$ in $x^{2}$: \[ f(\alpha)=\frac{1-2c^{2}x^{2}-c^{2}e^{2}x^{4}}{1+e^{2}c^{2}x^{4}},\ F(\alpha)=\frac{1+2e^{2}x^{2}-e^{2}c^{2}x^{4}}{1+e^{2}c^{2}x^{4}}. \] These equations give \[ \begin{aligned} 1+f\alpha&=\frac{2(1-c^{2}x^{2})}{1+e^{2}c^{2}x^{4}},\ 1-f\alpha=\frac{2c^{2}x^{2}(1+e^{2}x^{2})}{1+e^{2}c^{2}x^{4}},\\ F\alpha-1&=\frac{2e^{2}x^{2}(1-c^{2}x^{2})}{1+e^{2}c^{2}x^{4}},\ F\alpha+1=\frac{2(1+e^{2}x^{2})}{1+e^{2}c^{2}x^{4}}, \end{aligned} \] Editorial note: A blot resembling a 2 is printed after the minus sign in the numerator of the first fraction; the sense requires $2(1-c^{2}x^{2})$, as in the later fractions. whence \[ \frac{F\alpha-1}{1+f\alpha}=e^{2}.x^{2},\ \frac{1-f\alpha}{F\alpha+1}=c^{2}x^{2}, \] and from this, observing that $y^{2}=1-c^{2}x^{2}$, $z^{2}=1+e^{2}x^{2}$: \[ z^{2}=\frac{F\alpha+f\alpha}{1+f\alpha},\ y^{2}=\frac{F\alpha+f\alpha}{1+F\alpha}. \]

From these equations one derives, on extracting the square root, and replacing $x$, $y$, $z$ by their values $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$: \[ \left\{\begin{aligned} \varphi\left(\frac{\alpha}{2}\right)&=\frac{1}{c}\sqrt{\left(\frac{1-f\alpha}{1+F\alpha}\right)}=\frac{1}{e}\sqrt{\left(\frac{F\alpha-1}{f\alpha+1}\right)},\\ f\left(\frac{\alpha}{2}\right)&=\sqrt{\left(\frac{F\alpha+f\alpha}{1+F\alpha}\right)},\ F\left(\frac{\alpha}{2}\right)=\sqrt{\left(\frac{F\alpha+f\alpha}{1+f\alpha}\right)}. \end{aligned}\right. \tag{67.} \] Such are the simplest forms that can be given to the values of the functions $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$. In this manner one can express

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algebraically $\varphi\left(\frac{\alpha}{2}\right)$, $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$ in $f\alpha$, $F\alpha$. In the same manner $\varphi\left(\frac{\alpha}{4}\right)$, $f\left(\frac{\alpha}{4}\right)$, $F\left(\frac{\alpha}{4}\right)$ will be expressed in $f\left(\frac{\alpha}{2}\right)$, $F\left(\frac{\alpha}{2}\right)$, and so on. Therefore in general the functions $\varphi\left(\frac{\alpha}{2^{n}}\right)$, $f\left(\frac{\alpha}{2^{n}}\right)$, $F\left(\frac{\alpha}{2^{n}}\right)$ may be expressed by means of extractions of square roots in terms of the three quantities $\varphi\alpha$, $f\alpha$, $F\alpha$.

To apply the formulas found above for the bisection to an example, let us suppose $\alpha=\frac{\omega}{2}$.

Then one will have $f\left(\frac{\omega}{2}\right)=0$, $F\left(\frac{\omega}{2}\right)=\frac{\sqrt{(e^{2}+c^{2})}}{c}$, therefore on substituting: \[ \varphi\left(\frac{\omega}{4}\right)=\frac{1}{c}\sqrt{\left(\frac{1}{1+\frac{1}{c}\sqrt{(e^{2}+c^{2})}}\right)}=\frac{1}{e}.\sqrt{\left(\frac{1}{c}\sqrt{(e^{2}+c^{2})}-1\right)}, \] \[ f\left(\frac{\omega}{4}\right)=\sqrt{\left(\frac{\frac{1}{c}\sqrt{(e^{2}+c^{2})}}{1+\frac{1}{c}\sqrt{(e^{2}+c^{2})}}\right)}, \] \[ F\left(\frac{\omega}{4}\right)=\sqrt{\left(\frac{1}{c}\sqrt{(e^{2}+c^{2})}\right)}, \] or else \[ \begin{aligned} \varphi\left(\frac{\omega}{4}\right)&=\frac{1}{\sqrt{(c^{2}+c\sqrt{(e^{2}+c^{2})})}}=\frac{\sqrt{(\sqrt{(e^{2}+c^{2})}-c)}}{e\sqrt{c}},\\ f\left(\frac{\omega}{4}\right)&=\frac{\sqrt[4]{(e^{2}+c^{2})}}{\sqrt{(c+\sqrt{(e^{2}+c^{2})})}}=\frac{1}{e}.\sqrt{(e^{2}+c^{2}-c\sqrt{(e^{2}+c^{2})})}\\ F\left(\frac{\omega}{4}\right)&=\sqrt[4]{\left(1+\frac{e^{2}}{c^{2}}\right)}=\sqrt{\left[F\left(\frac{\omega}{2}\right)\right]}. \end{aligned} \]

B. Expressions of the functions $\varphi\left(\frac{\alpha}{2n+1}\right)$, $f\left(\frac{\alpha}{2n+1}\right)$, $F\left(\frac{\alpha}{2n+1}\right)$, in algebraic functions of the quantities $\varphi\alpha$, $f\alpha$, $F\alpha$.

14.

To find the values of $\varphi\left(\frac{\alpha}{2n+1}\right)$, $f\left(\frac{\alpha}{2n+1}\right)$, $F\left(\frac{\alpha}{2n+1}\right)$ in $\Phi\alpha$, $f\alpha$, $F\alpha$, one must resolve the equations \[ \Phi\alpha=\frac{P_{2n+1}}{Q_{2n+1}},\ f\alpha=\frac{P'_{2n+1}}{Q_{2n+1}},\ F\alpha=\frac{P''_{2n+1}}{Q_{2n+1}}, \] which are all of the degree $(2n+1)^{2}$. We shall see that it is always possible to effect this resolution algebraically.

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Let \[ \varphi_{1}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\varphi\left(\beta+\frac{2m\omega}{2n+1}\right) \tag{68.} \] and \[ \psi\beta=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\ \psi_{1}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}\varphi_{1}\left(\beta-\frac{2\mu\varpi i}{2n+1}\right), \tag{69.} \] where $\theta$ is any imaginary root whatever of the equation $\theta^{2n+1}-1=0$. This being premised, I say that the two quantities \[ \psi\beta.\psi_{1}\beta\ \text{and}\ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}, \] may be expressed rationally in $\varphi(2n+1)\beta$.

First, on writing $\varphi_{1}\beta$ as follows: \[ \begin{aligned} \varphi_{1}\beta&=\varphi\beta.\overset{n}{\underset{1}{\Sigma}}.\left\{\varphi\left(\beta+\frac{2m\omega}{2n+1}\right)+\varphi\left(\beta-\frac{2m\omega}{2n+1}\right)\right\}\\ &=\varphi\beta.\overset{n}{\underset{1}{\Sigma}}(-1)^{m}.\frac{2\varphi\beta.f\left(\frac{2m\omega}{2n+1}\right).F\left(\frac{2m\omega}{2n+1}\right)}{1+e^{2}c^{2}\varphi^{2}\left(\frac{2m\omega}{2n+1}\right).\varphi^{2}\beta}, \end{aligned} \] Editorial note: Printed with a point after $\varphi\beta$ before the sum in both lines, where the sense requires a plus sign; the first line shows a blot over the point and a further point after the sum sign. one sees that $\varphi_{1}\beta$ can be expressed rationally in $\varphi\beta$. Let then $\varphi_{1}\beta=\chi(\varphi\beta)$; one has likewise: \[ \begin{aligned} \varphi_{1}\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)&=\chi\left(\varphi\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)\right)\\ &=\chi\left(\frac{\varphi\beta.f\left(\frac{2\mu\varpi i}{2n+1}\right).F\left(\frac{2\mu\varpi i}{2n+1}\right)\pm\varphi\left(\frac{2\mu\varpi i}{2n+1}\right).f\beta.F\beta}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{2\mu\varpi i}{2n+1}\right).\varphi^{2}\beta}\right), \end{aligned} \] or else, on putting $\Phi\beta=x$: \[ f\left(\frac{2\mu\varpi i}{2n+1}\right).F\left(\frac{2\mu\varpi i}{2n+1}\right)=a,\ \varphi\left(\frac{2\mu\varpi i}{2n+1}\right)=b, \] and on substituting for $f\beta$ and $F\beta$ their values $\sqrt{(1-c^{2}x^{2})}$ and $\sqrt{(1+e^{2}x^{2})}$: \[ \varphi_{1}\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)=\chi\left(\frac{ax\pm b.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}}{1+e^{2}c^{2}b^{2}.x^{2}}\right); \] now, $\chi$ denoting a rational function, the second member of this equation may be put under the form \[ R_{\mu}\pm R'_{\mu}.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}, \] where $R_{\mu}$ and $R'_{\mu}$ are rational functions of $x$.

Therefore one has \[ \varphi_{1}\left(\beta\pm\frac{2\mu\varpi i}{2n+1}\right)=R_{\mu}\pm R'_{\mu}.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}. \] On substituting in the expressions of $\psi\beta$ and $\psi_{1}\beta$, there will result:

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Now $R_{\mu}$ and $R'_{\mu}$ being rational functions of $x$, the quantities $\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R_{\mu}$ and $\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(+\theta)^{\mu}.R'_{\mu}$ are so likewise. On raising then $\psi\beta$ and $\psi_{1}\beta$ to the $(2n+1)^{\text{th}}$ power: the two quantities $(\psi\beta)^{2n+1}$ and $(\psi_{1}\beta)^{2n+1}$ may be put under the form: \[ \begin{aligned} (\psi\beta)^{2n+1}&=t+t'.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]},\\ (\psi_{1}\beta)^{2n+1}&=t-t'.\sqrt{[(1-c^{2}x^{2})(1+e^{2}x^{2})]}, \end{aligned} \] $t$ and $t'$ being rational functions of $x$. On taking the sum of the values of $(\psi\beta)^{2n+1}$ and $(\psi_{1}\beta)^{2n+1}$, one will have \[ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}=2t. \]

Therefore the quantity $(\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}$ can be expressed rationally in $x$. The same holds of the product $\psi\beta.\psi_{1}\beta$, as one sees by the equations (70.).

Therefore one may put \[ \left\{\begin{aligned} \psi\beta.\psi_{1}\beta&=\lambda(x),\\ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}&=\lambda_{1}(x), \end{aligned}\right. \tag{71.} \] $\lambda(x)$ and $\lambda_{1}(x)$ denoting rational functions of $x$. Now these functions have the property of not changing in value when one puts in place of $x$ any other root whatever of the equation \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \]

Let us consider first the function $\lambda(x)$. On restoring the value of $x=\Phi\beta$, one will have \[ \psi\beta.\psi_{1}\beta=\lambda(\Phi\beta), \] whence one derives, on putting $\beta+\frac{2k\omega}{2n+1}+\frac{2k'\varpi i}{2n+1}$ in place of $\beta$: \[ \lambda\left(\varphi\left(\beta+\frac{2k\omega}{2n+1}+\frac{2k'\varpi i}{2n+1}\right)\right)=\psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right).\psi_{1}\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right). \] This being premised, observing that: \[ \overset{+n}{\underset{-n}{\Sigma}}\psi(m+k)=\overset{+n}{\underset{-n}{\Sigma}}_{m}\psi(m)+\overset{k}{\underset{1}{\Sigma}}_{m}.(\psi(m+n)-\psi(m-n-1)), \tag{72.} \] one will have, on putting in the expression of $\psi\beta$, $\beta=\beta+\frac{2k\omega}{2n+1}$:

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\[ \begin{aligned} \varphi_{1}\left(\beta+\frac{2k\omega}{2n+1}\right)&=\overset{+n}{\underset{-n}{\Sigma}}_{m}\varphi\left(\beta+\frac{2(k+m)}{2n+1}\omega\right)\\ &=\varphi_{1}\beta+\overset{k}{\underset{1}{\Sigma}}_{m}\left\{\varphi\left(\beta+\frac{2(m+n)\omega}{2n+1}\right)-\varphi\left(\beta+\frac{2(m-n-1)\omega}{2n+1}\right)\right\}, \end{aligned} \] but: $\varphi\left(\beta+\frac{2(m-n-1)}{2n+1}\omega\right)=\varphi\left(\beta+\frac{2(m+n)}{2n+1}\omega-2\omega\right)=\varphi\left(\beta+\frac{2(m+n)}{2n+1}\omega\right)$, therefore: \[ \varphi_{1}\left(\beta+\frac{2k\omega}{2n+1}\right)=\varphi_{1}\beta. \tag{73.} \]

On putting in the expression of $\psi\beta$, $\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}$ in place of $\beta$, we shall find \[ \psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right), \] but by virtue of (73.) we have: \[ \varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}\right), \] therefore \[ \psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}\right). \] By virtue of (72.) we have \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2(k'+\mu)\varpi i}{2n+1}\right)\\ &=\theta^{-k'}.\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right)+\overset{k'}{\underset{1}{\Sigma}}_{\mu}.\theta^{n+\mu-k'}.\varphi_{1}\left(\beta+\frac{2(\mu+n)\varpi i}{2n+1}\right)\\ &\quad-\overset{k'}{\underset{1}{\Sigma}}_{\mu}\theta^{\mu-n-1-k'}.\varphi_{1}\left(\beta+\frac{2(\mu-n-1)\varpi i}{2n+1}\right), \end{aligned} \] therefore, on remarking that $\theta^{n+\mu-k'}=\theta^{\mu-n-1-k'}$ and \[ \varphi_{1}\left(\beta+\frac{2(\mu-n-1)\varpi i}{2n+1}\right)=\varphi_{1}\left(\beta+\frac{2(\mu+n)\varpi i}{2n+1}-2\varpi i\right)=\varphi_{1}\left(\beta+\frac{2(\mu+n)\varpi i}{2n+1}\right), \] there will result \[ \psi\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\theta^{-k'}.\psi\beta. \tag{74.} \] In the same manner we shall also find \[ \psi_{1}\left(\beta+\frac{2k'\varpi i}{2n+1}+\frac{2k\omega}{2n+1}\right)=\theta^{k'}.\psi_{1}\beta. \] These two equations will give \[ \begin{aligned} &\psi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right).\psi_{1}\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)=\psi\beta.\psi_{1}\beta,\\ &\left\{\psi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right\}^{2n+1}+\left\{\psi_{1}\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right\}^{2n+1}=(\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}. \end{aligned} \]

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By virtue of these equations we shall obtain, on putting in the values of $\lambda(\Phi\beta)$ and $\lambda_{1}(\Phi\beta)$, $\beta+\frac{2k\omega+2k'\varpi i}{2n+1}$ in place of $\beta$: \[ \begin{aligned} \lambda(\Phi\beta)&=\lambda\left[\Phi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right],\\ \lambda_{1}(\Phi\beta)&=\lambda_{1}\left[\Phi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)\right]. \end{aligned} \] Now, $\Phi\left(\beta+\frac{2k\omega+2k'\varpi i}{2n+1}\right)$ expresses any root whatever of the equation \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \]

Therefore, as we have said, the functions $\lambda(x)$ and $\lambda_{1}(x)$ will have the same values, whatever root be put in place of $x$.

Let then $x_{0}$, $x_{1}$, $x_{2}....x_{2n+1}$ be these roots; we shall have \[ \begin{aligned} \lambda(x)&=\frac{1}{2n+1}\{\lambda(x_{0})+\lambda(x_{1})+....+\lambda(x_{2n})\},\\ \lambda_{1}(x)&=\frac{1}{2n+1}\{\lambda_{1}(x_{0})+\lambda_{1}(x_{1})+....+\lambda_{1}(x_{2n})\}. \end{aligned} \] Editorial note: The list of roots is printed as ending with $x_{2n+1}$, while the sums run only to $x_{2n}$. Now the second member of these equations is a rational and symmetric function of the roots of the equation $\Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}$, therefore $\lambda(x)$ and $\lambda_{1}(x)$ can be expressed rationally in $\Phi(2n+1)\beta$. On making \[ \lambda(x)=B,\ \lambda_{1}(x)=2A, \] the equations (71.) will give \[ (\psi\beta)^{2n+1}.(\psi_{1}\beta)^{2n+1}=B^{2n+1};\ (\psi\beta)^{2n+1}+(\psi_{1}\beta)^{2n+1}=2A, \] whence one derives \[ \psi\beta=\sqrt[2n+1]{[A+\sqrt{(A^{2}-B^{2n+1})}]}=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right). \tag{75.} \]

15.

The value of $\psi\beta$ having been found, one will easily deduce from it those of $\Phi_{1}\beta$.

In fact, on taking for $\theta$ successively all the imaginary roots of the equation $\theta^{2n+1}-1=0$, and denoting the corresponding values of $A$ and $B$ by $A_{1}$, $B_{1}$, $A_{2}$, $B_{2}$ etc., we shall obtain: \[ \begin{aligned} \sqrt[2n+1]{[A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}]}&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta_{1}^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\\ \sqrt[2n+1]{[A_{2}+\sqrt{(A_{2}^{2}-B_{2}^{2n+1})}]}&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta_{2}^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\\ \cdots\cdots\cdots&\cdots\cdots\cdots\\ \sqrt[2n+1]{[A_{2n}+\sqrt{(A_{2n}^{2}-B_{2n}^{2n+1})}]}&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta_{2n}^{\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right). \end{aligned} \]

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Likewise we know the sum of the roots: \[ \overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\varphi\left(\beta+\frac{2m\omega}{2n+1}+\frac{2\mu\varpi i}{2n+1}\right)=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right), \] which is equal to $(2n+1)\varphi(2n+1)\beta$, as we shall see in what follows. On adding these equations member to member, after having multiplied the first by $\theta_{0}^{-k}$, the second by $\theta_{1}^{-k}$, the third by $\theta_{2}^{-k}....$ and the $(2n)^{\text{th}}$ by $\theta_{2n}^{-k}$, there will result: \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Sigma}}_{\mu}.(\theta_{0}^{\mu-k}+\theta_{1}^{\mu-k}+\theta_{2}^{\mu-k}+....+\theta_{2n}^{\mu-k}+1).\varphi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right)\\ &=(2n+1).\varphi(2n+1)\beta+\overset{2n}{\underset{1}{\Sigma}}_{\mu}\theta_{\mu}^{-k}.\sqrt[2n+1]{[A_{\mu}+\sqrt{(A_{\mu}^{2}-B_{\mu}^{2n+1})}]}; \end{aligned} \] Editorial note: In the first sum the term $\theta_{0}^{\mu-k}$ and the final $1$ are both printed, so 1 is counted twice; the next sum is printed without $\theta_{0}^{\mu-k}$. now the sum \[ 1+\theta_{1}^{\mu-k}+\theta_{2}^{\mu-k}+....+\theta_{2n}^{\mu-k} \] reduces to zero for all values of $k$, except for $k=\mu$. In this case it becomes equal to $2n+1$. Therefore the first member of the preceding equation becomes \[ (2n+1).\varphi_{1}\left(\beta+\frac{2k\varpi i}{2n+1}\right), \] therefore, on substituting and dividing by $(2n+1)$, we have: \[ \begin{aligned} \varphi_{1}\left(\beta+\frac{2k\varpi i}{2n+1}\right)=&\\ \varphi(2n+1)\beta+\frac{1}{2n+1}.&\left\{\theta_{1}^{-k}.\sqrt[2n+1]{[A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}]}+\theta_{2}^{-k}\sqrt[2n+1]{[A_{2}+\sqrt{(A_{2}^{2}-B_{2}^{2n+1})}]}+....\right.\\ &\left.....+\theta_{2n}^{-k}\sqrt[2n+1]{[A_{2n}+\sqrt{(A_{2n}^{2}-B_{2n}^{2n+1})}]}\right\}. \end{aligned} \tag{76.} \] For $k=0$, we have: \[ \begin{aligned} \varphi_{1}\beta=\varphi(2n+1)\beta+\frac{1}{2n+1}&\left\{\sqrt[2n+1]{[A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}]}+\sqrt[2n+1]{[A_{2}+\sqrt{(A_{2}^{2}-B_{2}^{2n+1})}]}+\ldots\right.\\ &\left.....+\sqrt[2n+1]{[A_{2n}+\sqrt{(A_{2n}^{2}-B_{2n}^{2n+1})}]}\right\}. \end{aligned} \tag{77.} \]

16.

The value of $\varphi_{1}\beta$ having thus been found, the question is to derive from it that of $\varphi\beta$. Now this can easily be done as follows:

Let \[ \psi_{2}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}.\varphi\left(\beta+\frac{2m\omega}{2n+1}\right);\ \psi_{3}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}.\varphi\left(\beta-\frac{2m\omega}{2n+1}\right), \tag{78.} \] we have \[ \varphi\left(\beta\pm\frac{2m\omega}{2n+1}\right)=\frac{\varphi\beta.f\left(\frac{2m\omega}{2n+1}\right).F\left(\frac{2m\omega}{2n+1}\right)\pm f\beta.F\beta.\varphi\left(\frac{2m\omega}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{2m\omega}{2n+1}\right).\varphi^{2}\beta}. \]

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From this it follows that we may make \[ \psi_{2}\beta=r+f\beta.F\beta.s;\ \psi_{3}\beta=r-f\beta.F\beta.s, \] where $r$ and $s$ are rational functions of $\Phi\beta$.

From this we derive \[ \left\{\begin{aligned} \psi_{2}\beta.\psi_{3}\beta&=\chi(\Phi\beta),\\ (\psi_{2}\beta)^{2n+1}+(\psi_{3}\beta)^{2n+1}&=\chi_{1}(\Phi\beta), \end{aligned}\right. \tag{79.} \] $\chi(\Phi\beta)$ and $\chi_{1}(\Phi\beta)$ being two rational functions of $\Phi\beta$.

This being premised, I say that $\chi(\Phi\beta)$ and $\chi_{1}(\Phi\beta)$ can be expressed rationally in $\Phi_{1}\beta$.

We have seen that \[ \varphi_{1}\beta=\varphi\beta+\overset{n}{\underset{1}{\Sigma}}_{m}.\left\{\frac{2\varphi\beta.f\left(\frac{2m\omega}{2n+1}\right).F\left(\frac{2m\omega}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{2m\omega}{2n+1}\right)\varphi^{2}\beta}\right\}. \tag{80.} \] On making $\Phi\beta=x$, we shall have an equation in $x$ of the degree $(2n+1)$. One root of this equation is $x=\Phi\beta$; now, on putting $\beta+\frac{2k\omega}{2n+1}$ in place of $\beta$, $\Phi_{1}\beta$ does not change its value, therefore $x=\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)$ will be a root, whatever the integer $k$ may be. Now, on giving to $k$ all the integer values from $-n$ up to $+n$, $\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)$ will take $2n+1$ different values, therefore these $2n+1$ quantities will be precisely the $2n+1$ roots of the equation in $x$.

This being premised, on putting $\beta+\frac{2k\omega}{2n+1}$ in place of $\beta$ in the expression of $\psi_{2}\beta$, there will result by virtue of (72.): \[ \begin{aligned} \psi_{2}\left(\beta+\frac{2k\omega}{2n+1}\right)&=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}\Phi\left(\beta+\frac{2(k+m)\omega}{2n+1}\right)\\ &=\theta^{-k}.\psi_{2}3+\overset{k}{\underset{1}{\Sigma}}_{m}\theta^{m+n-k}.\Phi\left(\beta+\frac{2(m+n)\omega}{2n+1}\right)\\ &\quad-\overset{k}{\underset{1}{\Sigma}}_{m}\theta^{m-n-1-k}.\Phi\left(\beta+\frac{2(m-n-1)\omega}{2n+1}\right), \end{aligned} \] Editorial note: Printed $\psi_{2}3$ with a digit three instead of $\psi_{2}\beta$; an apparent misprint. therefore, on paying attention that $\theta^{m+n-k}=\theta^{m-n-1-k}$ and $\Phi\left(\beta+\frac{2(m-n-1)}{2n+1}\omega\right)=\Phi\left(\beta+\frac{2(m+n)\omega}{2n+1}\right)$, it will result \[ \psi_{2}\left(\beta+\frac{2k\omega}{2n+1}\right)=\theta^{-k}.\psi_{2}\beta. \tag{81.} \]

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Likewise we shall have \[ \psi_{3}\left(\beta+\frac{2k\omega}{2n+1}\right)=\theta^{+k}.\psi_{3}\beta. \] One sees by virtue of these relations that the equations which give the values of the functions $\chi(\Phi\beta)$ and $\chi_{1}(\Phi\beta)$Editorial note: The second printed argument looks like a damaged or misprinted $\beta$., lead to these two equalities: \[ \begin{aligned} \chi_{1}\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right]&=\chi_{1}(\Phi\beta),\\ \chi\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right]&=\chi(\Phi\beta). \end{aligned} \] From this we derive \[ \begin{aligned} \chi(\Phi\beta)&=\frac{1}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{k}.\chi\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right],\\ \chi_{1}(\Phi\beta)&=\frac{1}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{k}.\chi_{1}\left[\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)\right]. \end{aligned} \]

Now, these values of $\chi(\Phi\beta)$ and $\chi_{1}(\Phi\beta)$ are rational and symmetric functions of all the roots of the equation (80.). Therefore they can be expressed rationally by the coefficients of the same equation, that is rationally in $\Phi_{1}\beta$.

Let \[ \chi_{1}(\Phi\beta)=2C,\ \chi(\Phi\beta)=D, \] Editorial note: The subscripts of both printed $\chi$ are damaged and hard to read; they are read here as $\chi_1$ for $2C$ and $\chi$ for $D$, as (79.) requires. the equations (79.) will give \[ \psi_{2}\beta=\sqrt[2n+1]{[C+\sqrt{(C^{2}-D^{2n+1})}]}, \] whence, on restoring the value of $\psi_{2}\beta$: \[ \sqrt[2n+1]{[C+\sqrt{(C^{2}-D^{2n+1})}]}=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta^{m}.\Phi\left(\beta+\frac{2m\omega}{2n+1}\right). \tag{82.} \] From this we derive, on putting $\theta_{\mu}$ in place of $\theta$, and denoting the corresponding values of $C$ and $D$ by $C_{\mu}$ and $D_{\mu}$: \[ \theta_{\mu}^{-k}\sqrt[2n+1]{[C_{\mu}+\sqrt{(C_{\mu}^{2}-D_{\mu}^{2n+1})}]}=\overset{+n}{\underset{-n}{\Sigma}}_{m}\theta_{\mu}^{m-k}.\Phi\left(\beta+\frac{2m\omega}{2n+1}\right). \] On joining to it the equation \[ \Phi_{1}\beta=\overset{+n}{\underset{-n}{\Sigma}}_{m}\Phi\left(\beta+\frac{2m\omega}{2n+1}\right), \] one will easily derive from it: \[ (2n+1).\Phi\left(\beta+\frac{2k\omega}{2n+1}\right)=\Phi_{1}\beta+\overset{2n}{\underset{1}{\Sigma}}_{\mu}\theta_{\mu}^{-k}.\sqrt[2n+1]{[C_{\mu}+\sqrt{(C_{\mu}^{2}-D_{\mu}^{2n+1})}]}. \tag{83.} \] Supposing $k=0$, there will result: \[ \Phi\beta=\frac{1}{2n+1}\left\{\Phi_{1}\beta+\sqrt[2n+1]{[C_{1}+\sqrt{(C_{1}^{2}-D_{1}^{2n+1})}]}+\ldots+\sqrt[2n+1]{[C_{2n}+\sqrt{(C_{2n}^{2}-D_{2n}^{2n+1})}]}\right\}. \tag{84.} \]

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This equation gives $\Phi\beta$ as an algebraic function of $\Phi_{1}\beta$, and previously we found $\Phi_{1}\beta$ as an algebraic function of $\Phi(2n+1)\beta$. Therefore, on putting $\frac{\alpha}{2n+1}$ in place of $\beta$, we shall have $\Phi\left(\frac{\alpha}{2n+1}\right)$ as an algebraic function of $\Phi\alpha$.

By an entirely similar analysis one will find $f\left(\frac{\alpha}{2n+1}\right)$ in $f\alpha$ and $F\left(\frac{\alpha}{2n+1}\right)$ in $F\alpha$.

17.

The values which we have just found for the quantities $\Phi_{1}\beta$ and $\Phi\beta$, the first in $\Phi(2n+1)\beta$ and the second in $\Phi_{1}\beta$, each contain the sum of $2n$ different radicals of the $(2n+1)$ degree. There will result for $\Phi\beta$, $\Phi_{1}\beta....$ a number $(2n+1)^{2n}$ of values, whereas each of these quantities is the root of an equation of the $(2n+1)^{\text{th}}$ degree. Now one can give to the expressions of $\Phi\beta$ and $\Phi_{1}\beta$ such a form that the number of the values of these quantities shall be precisely equal to $2n+1$.

For this let \[ \theta=\cos\frac{2\pi}{2n+1}+i\sin\frac{2\pi}{2n+1}, \] we may make \[ \theta_{1}=\theta,\ \theta_{2}=\theta^{2},\ \theta_{3}=\theta^{3}....\theta_{2n}=\theta^{2n}. \]

Let likewise \[ \left\{\begin{aligned} \psi^{k}(\beta)&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{k\mu}.\Phi_{1}\left(\beta+\frac{2\mu\varpi i}{2n+1}\right),\\ \psi_{1}^{k}(\beta)&=\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta^{k\mu}.\Phi_{1}\left(\beta-\frac{2\mu\varpi i}{2n+1}\right), \end{aligned}\right. \tag{85.} \] we shall have by virtue of (74.) \[ \begin{aligned} \psi^{k}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{-k\nu}.\psi^{k}(\beta),\\ \psi_{1}^{k}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{+k\nu}.\psi_{1}^{k}(\beta),\\ \psi^{1}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{-\nu}.\psi^{1}(\beta),\\ \psi_{1}^{1}\left(\beta+\frac{2\nu\varpi i}{2n+1}\right)&=\theta^{+\nu}.\psi_{1}^{1}(\beta). \end{aligned} \]

Let now \[ \left\{\begin{aligned} \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k}}+\frac{\psi_{1}^{k}(\beta)}{(\psi_{1}^{1}\beta)^{k}}&=P(\Phi\beta),\\ \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k-2n-1}}+\frac{\psi_{1}^{k}(\beta)}{\psi_{1}^{1}(\beta)^{k-2n-1}}&=Q(\Phi\beta), \end{aligned}\right. \tag{86.} \]

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$P(\Phi\beta)$ and $Q(\Phi\beta)$ will be rational functions of $\Phi\beta$; now, on putting $\beta+\frac{2m\omega+2\mu\varpi i}{2n+1}$ in place of $\beta$, it is clear by virtue of the formulas that $P$ and $Q$ do not change their values; therefore we shall have \[ P(\Phi\beta)=\frac{1}{(2n+1)^{2}}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}.P\left[\Phi\left(\beta+\frac{2m\omega+2\mu\varpi i}{2n+1}\right)\right]; \] now, the second member being a symmetric and rational function of the roots of the equation $\Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}$, $P(\Phi\beta)$ can be expressed rationally in $\Phi(2n+1)\beta$. The same holds of $Q(\Phi\beta)$. These two quantities being known, the equations (86.) will give \[ \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k}}\left\{1-\left(\frac{\psi_{1}^{1}\beta}{\psi_{1}^{1}\beta}\right)^{2n+1}\right\}=P(\Phi\beta)-\frac{Q(\varphi\beta)}{(\psi_{1}^{1}\beta)^{2n+1}}, \] now \[ \begin{aligned} \psi_{1}^{1}\beta&=A_{1}+\sqrt{(A_{1}^{2}-B_{1}^{2n+1})},\\ \psi_{1}^{1}\beta&=A_{1}-\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}, \end{aligned} \] Editorial note: In the first of these two equations the subscript of the printed $\psi$ is damaged; it should read $\psi^{1}\beta$ without subscript, and the quotient in the preceding display should likewise have $\psi^{1}\beta$ in the numerator. therefore \[ \frac{\psi^{k}(\beta)}{(\psi^{1}\beta)^{k}}.2\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}=Q(\Phi\beta)-[A_{1}-\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}].P(\Phi\beta). \] Therefore we shall have \[ \psi^{k}(\beta)=(\psi^{1}\beta)^{k}.\{F_{k}+H_{k}.\sqrt{(A_{1}^{2}-B_{1}^{2n+1})}\}, \] where $F_{k}$ and $H_{k}$ are rational functions of $\Phi(2n+1)\beta$. On replacing $A_{1}$ and $B_{1}$ and substituting the values of $\psi^{k}(\beta)$ and $(\psi^{1}\beta)^{k}$, there will result: \[ \sqrt[2n+1]{[A_{k}+\sqrt{(A_{k}^{2}-B_{k}^{2n+1})}]}=[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{k}{2n+1}}[F_{k}+H_{k}.\sqrt{(A^{2}-B^{2n+1})}], \] therefore the value of $\Phi_{1}\beta$ will become: \[ \begin{aligned} \Phi_{1}\beta=\Phi(2n+1)\beta+\frac{1}{2n+1}.&\left\{[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{1}{2n+1}}\right.\\ &+[F_{2}+H_{2}\sqrt{(A^{2}-B_{2n}^{2n+1})}]\,[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{2}{2n+1}}\\ &\left.+....[F_{2n}+H_{2n}\sqrt{(A^{2}-B^{2n+1})}]\,[A+\sqrt{(A^{2}-B^{2n+1})}]^{\frac{2n}{2n+1}}\right\}. \end{aligned} \tag{87.} \] Editorial note: In the second line the printed $B$ under the radical carries the subscript $2n$, which the first and third lines lack; an apparent misprint. By an entirely similar procedure we shall find \[ \begin{aligned} \Phi\beta=\frac{1}{2n+1}&\left\{\Phi_{1}\beta+[C+\sqrt{(C^{2}-D^{2n+1})}]^{\frac{1}{2n+1}}\right.\\ &+[K_{2}+L_{2}\sqrt{(C^{2}-D^{2n+1})}]\,[C+\sqrt{(C^{2}-D^{2n+1})}]^{\frac{2}{2n+1}}\\ &\left.+....[K_{2n}+L_{2n}\sqrt{(C^{2}-D^{2n+1})}]\,[C+\sqrt{(C^{2}-D^{2n+1})}]^{\frac{2n}{2n+1}}\right\}, \end{aligned} \tag{88.} \] where $K_{2}$, $L_{2}$, $K_{3}$, $L_{3}....K_{2n}$, $L_{2n}$ are rational functions of $\Phi_{1}\beta$.

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These expressions of $\Phi_{1}\beta$ and $\Phi\beta$ have only $2n+1$ different values, which one will obtain by attributing to the radicals their $2n+1$ values. — It follows from our analysis that one may take $\sqrt{(A^{2}-B^{2n+1})}$ and $\sqrt{(C^{2}-D^{2n+1})}$ with whatever sign one pleases.

18.

The value which we have found for $\Phi(\beta)$ or $\Phi\left(\frac{\alpha}{2n+1}\right)$ contains, besides the function $\Phi\alpha$, the following: \[ \begin{gathered} e,\ c,\ \theta,\\ \Phi\left(\frac{m\omega}{2n+1}\right),\ \Phi\left(\frac{m\varpi i}{2n+1}\right),\ f\left(\frac{m\omega}{2n+1}\right),\\ f\left(\frac{m\varpi i}{2n+1}\right),\ F\left(\frac{m\omega}{2n+1}\right),\ F\left(\frac{m\varpi i}{2n+1}\right), \end{gathered} \] for any values whatever of $m$ from $1$ to $2n$. Now whatever the value of $m$ may be, one can always express algebraically $\Phi\left(\frac{m\omega}{2n+1}\right)$, $f\left(\frac{m\omega}{2n+1}\right)$, $F\left(\frac{m\omega}{2n+1}\right)$ in $\Phi\left(\frac{\omega}{2n+1}\right)$, and $\Phi\left(\frac{\mu\varpi i}{2n+1}\right)$, $f\left(\frac{\mu\varpi i}{2n+1}\right)$, $F\left(\frac{\mu\varpi i}{2n+1}\right)$ in $\Phi\left(\frac{\varpi i}{2n+1}\right)$. Everything is therefore known in the expression of $\Phi\left(\frac{\alpha}{2n+1}\right)$, except the two quantities independent of $\alpha$; $\Phi\left(\frac{\omega}{2n+1}\right)$, $\Phi\left(\frac{\varpi i}{2n+1}\right)$. — These quantities depend only on $c$ and $e$, and they can be found by the resolution of an equation of the degree $(2n+1)^{2}-1$, namely of the equation $\frac{P_{2n+1}}{x}=0$. — We are going to see in the following paragraph how one can reduce its resolution to that of lower equations.

§. V.

C. On the equation $P_{2n+1}=0$.

19.

The expression which we have just found for $\Phi\left(\frac{\alpha}{2n+1}\right)$ will contain, as we have seen, the two constant quantities $\Phi\left(\frac{\omega}{2n+1}\right)$ and $\Phi\left(\frac{\varpi i}{2n+1}\right)$. We shall find these quantities by resolving the equation \[ P_{2n+1}=0, \] whose roots will be represented by \[ x=\Phi\left(\frac{m\omega+\mu\varpi i}{2n+1}\right), \]

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where $m$ and $\mu$ may be all the integers from $-n$ up to $+n$. One of these roots, which answers to $m=0$, $\mu=0$, is equal to zero. Therefore $P_{2n+1}$ is divisible by $x$. On removing this factor, we shall have an equation \[ R=0,\ \text{of the degree }(2n+1)^{2}-1. \] On making $x^{2}=r$, the equation (90.) $R=0$, in $r$, will be of the degree $\frac{(2n+1)^{2}-1}{2}=2.n(n+1)$, and the roots of this equation will be \[ r=\Phi^{2}\left(\frac{m\omega\pm\mu\varpi i}{2n+1}\right), \tag{91.} \] $\mu$ and $m$ having all positive values below $n$, setting aside the root zero.

We are now going to see how one can reduce the resolution of the equation $R=0$ to that of two equations, the one of the degree $n$ and the other of the degree $2n+2$.

First, I say that one can represent all the values of $r$ by \[ \Phi^{2}\left(\frac{m\omega}{2n+1}\right)\ \text{and}\ \Phi^{2}\left(m\frac{\mu\omega+\varpi i}{2n+1}\right)\ldots\ldots, \tag{92.} \] on giving to $\mu$ all the integer values from zero up to $2n$, and to $m$ all those from $1$ to $n$.

In fact $\Phi^{2}\left(\frac{m\omega}{2n+1}\right)$ represents first a number $n$ of values of $r$; now the others can be represented by $\Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)$. Let, to demonstrate it, $m\mu=(2n+1)k+m'$, where $m'$ is an integer lying between the limits $-n$ and $+n$. On substituting, we shall have \[ \begin{aligned} \Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)&=\Phi^{2}\left(k.\omega+\frac{m'\omega+\mu\varpi i}{2n+1}\right)\\ &=\Phi^{2}\left(\frac{m'\omega+\mu\varpi i}{2n+1}\right)=\Phi^{2}\left(\frac{-m'\omega-\mu\varpi i}{2n+1}\right). \end{aligned} \] $\Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)$ is therefore a value of $r$; now to each value of $m$ there answers a different value of $m'$. For if we had \[ m_{1}\mu=(2n+1)k_{1}+m', \] it would follow \[ (m_{1}-m)\mu=(2n+1).(k-k_{1}), \] which is impossible, on remarking that $2n+1$ is a prime number. Therefore $\Phi^{2}\left(m.\frac{\mu\omega+\varpi i}{2n+1}\right)$ combined with $\Phi^{2}\left(\frac{m\omega}{2n+1}\right)$ represents all the values of $r$.

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This being premised, let \[ \begin{aligned} &\left[r-\Phi^{2}\left(\frac{\omega}{2n+1}\right)\right]\left[r-\Phi^{2}\left(\frac{2\omega}{2n+1}\right)\right]\ldots\left[r-\Phi^{2}\left(\frac{n\omega}{2n+1}\right)\right]\\ &=r^{n}+p_{n-1}.r^{n-1}+p_{n-2}.r^{n-2}+\ldots\ldots+p_{1}.r+p_{0}. \end{aligned} \tag{93.} \] The quantities $p_{0}$, $p_{1}$, $\ldots p_{n-1}$, will be rational and symmetric functions of $\Phi^{2}\left(\frac{\omega}{2n+1}\right)$, $\Phi^{2}\left(\frac{2\omega}{2n+1}\right)\ldots\Phi^{2}\left(\frac{n\omega}{2n+1}\right)$; now these functions can be found by means of an equation of degree $2n+2$.

Let $p$ be any rational and symmetric function whatever of $\Phi^{2}\left(\frac{\omega'}{2n+1}\right)$, $\Phi^{2}\left(\frac{2\omega'}{2n+1}\right)\ldots\Phi^{2}\left(\frac{n\omega'}{2n+1}\right)$, where $\omega'$ denotes the quantity $m\omega+\mu\varpi i$.

By the formulas which we have given above for expressing $\Phi(n\beta)$ in $\Phi\beta$, it is clear that one can express $\Phi^{2}\left(m'.\frac{\omega'}{2n+1}\right)$ as a rational function of $\Phi^{2}\left(\frac{\omega'}{2n+1}\right)$. Therefore we may make \[ p=\psi.\left[\Phi^{2}\left(\frac{\omega'}{2n+1}\right)\right]=\theta\left[\Phi^{2}\left(\frac{\omega'}{2n+1}\right),\ \Phi^{2}\left(\frac{2\omega'}{2n+1}\right),\ \ldots\ \Phi^{2}\left(\frac{n\omega'}{2n+1}\right)\right], \tag{94.} \] $\theta$ denoting a symmetric and rational function. On putting $\nu\omega'$ in place of $\omega'$, there will result \[ \psi\left[\Phi^{2}\left(\frac{\nu\omega'}{2n+1}\right)\right]=\theta\left[\Phi^{2}\left(\frac{\nu\omega'}{2n+1}\right),\ \Phi^{2}\left(\frac{2\nu\omega'}{2n+1}\right),\ \ldots\ \Phi^{2}\left(\frac{n\nu\omega'}{2n+1}\right)\right], \tag{95.} \] now, on making: \[ a.\nu=(2n+1).k_{a}'+k_{a}, \] where $k_{a}$ is an integer and lies between $-n$ and $+n$, the series \[ k_{1},\ k_{2}\ \ldots\ldots\ k_{n} \] will have, apart from the sign, the same terms as this one: \[ 1,\ 2,\ 3\ \ldots\ldots\ n; \] therefore it is clear that the second member of the equation (95.) will have the same value as $p$. Therefore: \[ \psi\left[\Phi^{2}\left(\frac{\nu\omega'}{2n+1}\right)\right]=\psi\left[\Phi^{2}\left(\frac{\omega'}{2n+1}\right)\right], \tag{96.} \] Editorial note: The exponent of the second Phi in (96.) is damaged in the print and appears as a raised dot; read as the exponent 2 of the first member. an equation which, on making $\omega'=\omega$ and $\omega'=m\omega+\varpi i$, will give these two: \[ \left\{\begin{aligned} &\psi\left[\Phi^{2}\left(\frac{\nu\omega}{2n+1}\right)\right]=\psi.\left[\Phi^{2}\left(\frac{\omega}{2n+1}\right)\right],\\ &\psi\left[\Phi^{2}\left(\nu.\frac{m\omega+\varpi i}{2n+1}\right)\right]=\psi\left[\Phi^{2}\left(\frac{m\omega+\varpi i}{2n+1}\right)\right], \end{aligned}\right. \tag{97.} \] or else, on making, for brevity, \[ \Phi^{2}\left(\frac{\nu\omega}{2n+1}\right)=r_{\nu},\ \Phi^{2}\left(\nu.\frac{m\omega+\varpi i}{2n+1}\right)=r_{\nu,m}, \tag{98.} \]

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there will result: \[ \psi.r_{\nu}=\psi r_{1};\ \psi r_{\nu,m}=\psi r_{1,m}. \tag{99.} \]

This being premised, let \[ \left\{\begin{aligned} &(p-\psi r_{1})(p-\psi r_{1,0})(p-\psi r_{1,1})(p-\psi r_{1,2})\ldots\ (p-\psi r_{1,2n})\\ &=q_{0}+q_{1}.p+q_{2}.p^{2}+\ldots q_{2n+1}.p^{2n+1}+p^{2n}. \end{aligned}\right. \tag{100.} \] Editorial note: The exponent of the last term is partly obscured by a smudge and reads $2n$; the degree $2n+2$ is expected from (104.).

I say that the coefficients $q_{0}$, $q_{1}$ etc. can be expressed rationally in terms of $e$ and $c$.

First, by virtue of the known formulas, these coefficients can be expressed rationally in terms of $t_{1}$, $t_{2}\ldots t_{2n}$, if one makes, for brevity, \[ t_{k}=(\psi r_{1})^{k}+(\psi r_{1,0})^{k}+(\psi r_{1,1})^{k}+\ldots+(\psi r_{1,2n})^{k}. \tag{101.} \]

It is therefore a question of finding the quantities $t_{1}$, $t_{2}\ldots$, now this can easily be done by means of the relations (99.). Indeed, on making therein successively $\nu=1$, $2\ldots n$, after having raised both members to the $k^{\text{th}}$ power, one will deduce at once: \[ \left\{\begin{aligned} (\psi r_{1})^{k}&=\frac{1}{n}\left[(\psi r_{1})^{k}+(\psi r_{2})^{k}+\ldots+(\psi r_{n})^{k}\right],\\ (\psi r_{1,m})^{k}&=\frac{1}{n}\left[(\psi r_{1,m})^{k}+(\psi r_{2,m})^{k}+\ldots+(\psi r_{n,m})^{k}\right]. \end{aligned}\right. \tag{102.} \] Therefore, putting for $m$ all the integral numbers $0$, $1$, $\ldots$ $2n$, and then substituting in the expression for $t_{k}$, there will result: \[ \left\{\begin{aligned} n.t_{k}&=(\psi r_{1})^{k}+(\psi r_{2})^{k}+\ldots\ldots+(\psi r_{n})^{k}\\ &+(\psi r_{1,0})^{k}+(\psi r_{2,0})^{k}+\ldots\ldots+(\psi r_{n,0})^{k}\\ &+(\psi r_{1,1})^{k}+(\psi r_{2,1})^{k}+\ldots\ldots+(\psi r_{n,1})^{k}\\ &+\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &+(\psi r_{1,2n})^{k}+(\psi r_{2,2n})^{k}+\ldots\ldots+(\psi r_{n,2n})^{k}. \end{aligned}\right. \tag{103.} \]

This value of $t_{k}$ is, as one sees, a rational and symmetric function of the $2n+2$ quantities $r_{1}$, $r_{2}$, $\ldots r_{n}$; $r_{1,0}$, $r_{2,0}$, $\ldots r_{n,0}\ldots$ $r_{1,2n}$, $r_{2,2n}\ldots r_{n,2n}$, which are the $2n+2$ roots of the equation $R=0$. Therefore, as is known, $t_{k}$ can be expressed rationally by the coefficients of this equation, and consequently as a rational function of $e$ and $c$. Having thus found the quantities $t_{k}$, one deduces from them the values of $q_{0}$, $q_{1}$, $\ldots$ $\ldots q_{2n+1}$; which will likewise be rational functions of $e$ and $c$.

20.

This being premised, on supposing \[ 0=q_{0}+q_{1}.p+q_{2}.p^{2}+\ldots\ldots+q_{2n+1}.p^{2n+1}+p^{2n+2}, \tag{104.} \] one will have an equation of the $(2n+2)^{\text{th}}$ degree, whose roots will be \[ \psi r_{1};\ \psi r_{1,0};\ \psi r_{1,1};\ \psi r_{1,2}\ \ldots\ldots\ \psi r_{1,2n}. \]

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The function $\psi r_{1}$, that is, any rational and symmetric function whatever of the roots $r_{1}$, $r_{2}$, $r_{3}$, $\ldots\ldots r_{n}$, can therefore be found by means of an equation of the degree $2n+2$.

Therefore one will in this manner have the coefficients $p_{0}$, $p_{1}$, $\ldots\ldots p_{n-1}$, by solving a number $n$ of equations, each of the $(2n+2)^{\text{th}}$ degree.

Having determined $p_{0}$, $p_{1}$, $\ldots\ldots$, one will have, by solving the equation \[ 0=p_{0}+p_{1}r+\ldots\ldots+p_{n-1}.r^{n-1}+r^{n}, \tag{105.} \] the value of the quantities \[ r_{1},\ r_{2},\ \ldots\ldots r_{n};\ r_{1,0},\ r_{2,0},\ \ldots\ldots r_{n,0};\ r_{1,1},\ r_{2,1},\ \ldots\ldots r_{n,1}\ \text{etc. etc.,} \] of which the first is equal to $\Phi^{2}\left(\frac{\omega}{2n+1}\right)$. Therefore the determination of this quantity, or rather the solution of the equation $R=0$, which is of the degree $(2n+2).n$, is reduced to that of equations of the degrees $(2n+2)$ and $n$.

But the preceding procedure can still be simplified. Indeed, as we shall see, in order to have the quantities $p_{0}$, $p_{1}$, $\ldots\ldots$, it suffices to know any one of them, and then the others can be expressed rationally by that one.

Let $p$, $q$ be, in general, two rational and symmetric functions of the quantities $r_{1}$, $r_{2}$, $\ldots\ldots r_{n}$; one can make, as we have seen: \[ p=\psi r_{1};\ q=\theta r_{1}, \] $\psi r_{1}$ and $\theta r_{1}$ denote two rational functions of $r_{1}$, which have this property: of remaining the same, if one changes $r_{1}$ into any other of the quantities $r_{1}$, $r_{2}$, $\ldots\ldots r_{n}$.

Suppose now: \[ s_{k}=(\psi r_{1})^{k}.\theta r_{1}+(\psi r_{1,0})^{k}.\theta r_{1,0}+(\psi r_{1,1})^{k}.\theta r_{1,1}+\ldots\ldots+(\psi r_{1,2n})^{k}.\theta r_{1,2n}, \] I say that $s_{k}$ can be expressed rationally in terms of $e$ and $c$.

Indeed, one has \[ (\psi r_{1})^{k}.\theta r_{1}=(\psi r_{\nu})^{k}.\theta r_{\nu}=\frac{1}{n}\left\{(\psi r_{1})^{k}.\theta r_{1}+(\psi r_{2})^{k}.\theta r_{2}+\ldots\ldots+(\psi r_{n})^{k}.\theta r_{n}\right\}, \] \[ (\psi r_{1,m})^{k}.\theta r_{1,m}=(\psi r_{\nu,m})^{k}.\theta r_{\nu,m}=\frac{1}{n}\left\{(\psi r_{1,m})^{k}.\theta r_{1,m}+(\psi r_{2,m})^{k}.\theta r_{2,m}+\ldots+(\psi r_{n,m})^{k}.\theta r_{n,m}\right\}. \] On making $m=0$, $1$, $2$, $\ldots\ldots 2n$, and substituting in the expression for $s_{k}$, it will be seen that $s_{k}$ will be a rational and symmetric function of the roots $r_{0}$, $r_{1}$, $\ldots\ldots r_{1,0}$ etc., etc. of the equation $R=0$; therefore $s_{k}$ can be expressed rationally in terms of $e$ and $c$.

Knowing $s_{k}$, one will obtain, on making $k=0$, $1$, $2$, $\ldots\ldots 2n$, $2n+1$ equations, from which one will easily deduce the value of $\theta r_{1}$ as a function

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rational of $\psi_{1}r$.Editorial note: Printed $\psi_{1}r$ with the subscript on the psi; $\psi r_{1}$ is expected, as on the previous page. Therefore, a function of the form $p$ being given, one can express any other function whatever of the same form as a rational function of $p$. Therefore, as we have said, one can express the coefficients $p_{0}$, $p_{1}$, $\ldots\ldots p_{n-1}$ rationally by any one of them whatever. Therefore, finally, in order to have their values, it suffices to solve a single equation of the degree $2n+2$, and consequently, in order to have the roots of the equation $R=0$, it suffices to solve one equation of the degree $2n+2$, and $2n+2$ equations of the degree $n$.

21.

Now, among the equations on which the determination of the quantities $\varphi\left(\frac{\omega}{2m+1}\right)$; $\varphi\left(\frac{\varpi i}{2n+1}\right)$,Editorial note: Printed $2m+1$ in the first denominator; $2n+1$ is apparently expected. those of the degree $n$ can be solved algebraically. The procedure by which we shall effect this solution is entirely similar to that which is due to Mr. Gauss for the solution of the equation \[ \theta^{2n+1}-1=0. \] Let the equation be proposed \[ 0=p_{1}+p_{1}.r+p_{2}.r^{2}+\ldots\ldots+p_{n-1}.r^{n-1}+r^{n}, \tag{106.} \] Editorial note: The first coefficient is printed $p_{1}$; $p_{0}$ is expected. whose roots are: \[ \Phi^{2}\left(\frac{\omega'}{2n+1}\right);\ \Phi^{2}\left(\frac{2\omega'}{2n+1}\right);\ \ldots\ldots\ \Phi^{2}\left(\frac{n\omega'}{2n+1}\right), \] where $\omega'$ has one of the values of $\omega$, $m\omega+\varpi i$. Let us denote by $\alpha$ one of the primitive roots of the number $2n+1$, that is, an integer such that $\mu=2n$$\,+1$Editorial note: The characters after $2n$ are overstruck by a heavy bar and hard to read; $\mu=2n$ is expected. is the smallest number which renders $\alpha^{\mu}-1$ divisible by $2n+1$, I say that the roots of the equation (106.) can also be represented by \[ \Phi^{2}(\varepsilon),\ \Phi^{2}(\alpha\varepsilon),\ \Phi^{2}(\alpha^{2}\varepsilon),\ \Phi^{2}(\alpha^{3}\varepsilon)\ldots\ \Phi^{2}(\alpha^{n-1}.\varepsilon), \tag{107.} \] where $\varepsilon=\frac{\omega'}{2n+1}$.

Let \[ \alpha^{m}=(2n+1)k_{m}\pm a_{m}, \] where $k$ is an integer and $a_{m}$ an integer, positive and less than $n+1$, I say that the terms of the series \[ 1,\ a_{1},\ a_{2},\ \ldots\ldots\ a_{n-1} \] will all be different from one another.

Indeed, if one has \[ a_{m}=a_{\mu}, \] it follows,

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\[ \begin{aligned} \text{or }\ \alpha^{m}-\alpha^{\mu}&=(2n+1)(k_{m}-k_{\mu}),\\ \text{or }\ \alpha^{m}+\alpha^{\mu}&=(2n+1)(k_{m}+k_{\mu}). \end{aligned} \] It is therefore necessary that one of the quantities $\alpha^{m}-\alpha^{\mu}$, $\alpha^{m}+\alpha^{\mu}$ be divisible by $2n+1$; now let $m>\mu$ be posited, which is permitted, it is necessary that $\alpha^{m-\mu}-1$ or $\alpha^{m-\mu}+1$ be divisible by $n$,Editorial note: Printed $n$ here; $2n+1$ is apparently expected. now this is impossible, for $m-\mu$ is less than $n$.

Therefore the quantities $1$, $a_{1}$, $a_{2}\ldots a_{n-1}$ are different from one another and consequently they coincide, but in a different order, with the numbers $1$, $2$, $3$, $4\ldots n$.

Therefore, remarking that \[ \varphi^{2}\{[(2n+1)k_{m}\pm a_{m}]\varepsilon\}=\varphi^{2}(a_{m}\varepsilon), \] one sees that the quantities (107.) are the same as these: \[ \Phi^{2}(\varepsilon),\ \Phi^{2}(2\varepsilon)\ \ldots\ \Phi^{2}(n\varepsilon), \] that is, the roots of the equation (106.) Q. E. D.

There is still to be remarked, that having \[ \alpha^{n}=(2n+1)k_{n}-1, \] one will have \[ \alpha^{n+m}=(2n+1)k_{n}\alpha^{m}-\alpha^{m}, \] therefore \[ a_{n+m}=\pm a_{m} \] and \[ \Phi^{2}(\alpha^{n+m}\varepsilon)=\Phi^{2}(\alpha^{m}\varepsilon). \] This being premised, let $\theta$ be any imaginary root whatever of the equation \[ \theta^{n}-1=0 \] and \[ \psi(\varepsilon)=\Phi^{2}(\varepsilon)+\Phi^{2}(\alpha\varepsilon)\theta+\Phi^{2}(\alpha^{2}\varepsilon)\theta^{2}+\ldots+\Phi^{2}(\alpha^{n-1}\varepsilon)\theta^{n-1}. \tag{108.} \] By virtue of what we have seen previously, the second member of this equation can be transformed into a rational function of $\Phi^{2}(\varepsilon)$. Let us make \[ \psi(\varepsilon)=\chi[\Phi^{2}(\varepsilon)]. \tag{109.} \] On putting in the first expression for $\psi(\varepsilon)$, $\alpha^{m}\varepsilon$ in place of $\varepsilon$, there will result: \[ \begin{aligned} \psi(\alpha^{m}\varepsilon)&=\varphi^{2}(\alpha^{m}\varepsilon)+\varphi^{2}(\alpha^{m+1}\varepsilon).\theta+\varphi^{2}(\alpha^{m+2}\varepsilon).\theta^{2}+\ldots\\ &\ldots+\Phi^{2}(\alpha^{n-1}\varepsilon).\theta^{n-m-1}+\Phi^{2}(\alpha^{n}\varepsilon).\theta^{n-m}+\ldots+\Phi^{2}(\alpha^{n+m-1}\varepsilon)\theta^{n-1}, \end{aligned} \] but we have seen that $\Phi^{2}(\alpha^{n+m}\varepsilon)=\Phi^{2}(\alpha^{m}\varepsilon)$; therefore: \[ \begin{aligned} \psi(\alpha^{m}\varepsilon)&=\theta^{n-m}.\Phi^{2}(\varepsilon)+\theta^{n-m+1}.\Phi^{2}(\alpha\varepsilon)+\theta^{n-m+2}.\Phi^{2}(\alpha^{2}\varepsilon)+\ldots\\ &\ldots+\theta^{n-1}.\Phi^{2}(\alpha^{m-1}\varepsilon)+\Phi^{2}(\alpha^{m}\varepsilon)+\theta.\Phi^{2}(\alpha^{m+1}\varepsilon)+\ldots+\theta^{n-m-1}.\Phi^{2}(\alpha^{n-1}\varepsilon). \end{aligned} \]

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On multiplying by $\theta^{m}$, the second member will become equal to $\psi(\varepsilon)$, therefore: \[ \psi(\alpha^{m}\varepsilon)=\theta^{-m}.\psi(\varepsilon), \tag{110.} \] or rather: \[ \psi(\varepsilon)=\theta^{m}.\chi[\Phi^{2}(\alpha^{m}\varepsilon)], \] whence, on raising both to the $n^{\text{th}}$ power, one deduces by virtue of the relation $\theta^{+mn}=1$: \[ [\psi(\varepsilon)]^{n}=\{\chi[\Phi^{2}(\alpha^{m}\varepsilon)]\}^{n}. \tag{111.} \]

This formula gives, on making successively $m=0$, $1$, $2$, $3\ldots\ldots n-1$, $n$ equations, which, added member to member, will give the following: \[ n[\psi(\varepsilon)]^{n}=\{\chi[\Phi^{2}(\varepsilon)]\}^{n}+\{\chi[\Phi^{2}(\alpha\varepsilon)]\}^{n}+\{\chi[\Phi^{2}(\alpha^{2}\varepsilon)]\}^{n}+\ldots+\{\chi[\Phi^{2}(\alpha^{n-1}\varepsilon)]\}^{n}; \tag{112.} \] now, the second member of this equation is a rational and symmetric function of the quantities $\Phi^{2}(\varepsilon)$, $\Phi^{2}(\alpha\varepsilon)\ldots\Phi^{2}(\alpha^{n-1}\varepsilon)$, that is, of the roots of the equation (106.); therefore $(\psi\varepsilon)^{n}$ can be expressed as a rational function of $p_{0}$, $p_{1}\ldots p_{n-1}$, consequently as a rational function of any one whatever of these quantities. Let $v$ be the value of $\psi(\varepsilon)$, one will have \[ \sqrt[n]{v}=\Phi^{2}(\varepsilon)+\theta.\Phi^{2}(\alpha\varepsilon)+\theta^{2}.\Phi^{2}(\alpha^{2}\varepsilon)+\ldots+\theta^{n-1}.\Phi^{2}(\alpha^{n-1}\varepsilon). \tag{113.} \] This being premised, let $\theta=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. The imaginary roots of the equation $\theta^{n}-1$ can be represented by \[ \theta^{1},\ \theta^{2}\ \ldots\ldots\ \theta^{n-1}. \] Therefore, on making successively $\theta$ equal to each of these roots and denoting the corresponding values of $v$ by $v_{1}$, $v_{2}\ldots v_{n-1}$, there will result \[ \begin{aligned} \sqrt[n]{v_{1}}&=\Phi^{2}(\varepsilon)+\theta.\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\theta^{n-1}.\Phi^{2}(\alpha^{n-1}\varepsilon),\\ \sqrt[n]{v_{2}}&=\Phi^{2}(\varepsilon)+\theta^{2}.\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\theta^{2n-2}.\Phi^{2}(\alpha^{n-1}\varepsilon)\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ \sqrt[n]{v_{n-1}}&=\Phi^{2}(\varepsilon)+\theta^{n-1}.\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\theta^{(n-1)^{2}}.\Phi^{2}(\alpha^{n-1}\varepsilon), \end{aligned} \] On combining these equations with the following: \[ -p_{n-1}=\Phi^{2}(\varepsilon)+\Phi^{2}(\alpha\varepsilon)+\ldots\ldots+\Phi^{2}(\alpha^{n-1}\varepsilon), \] one easily deduces from them: \[ \begin{aligned} \Phi^{2}(\alpha^{m}\varepsilon)=+\frac{1}{n}&\left\{-p_{n-1}+\theta^{-m}.\sqrt[n]{v_{1}}+\theta^{-2m}.\sqrt[n]{v_{2}}+\theta^{-3m}.\sqrt[n]{v_{3}}+\ldots\right.\\ &\left.\ldots+\theta^{-(n-1)m}\sqrt[n]{v_{n-1}}\right\}, \end{aligned} \tag{114.} \] and for $m=0$: \[ \Phi^{2}(\varepsilon)=\frac{1}{n}\left\{-p_{n-1}+\sqrt[n]{v_{1}}+\sqrt[n]{v_{2}}+\ldots+\sqrt[n]{v_{n-1}}\right\}. \tag{115.} \]

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22.

All the roots of the equation (106.) are contained in the formula (115.), but since their number is only $n$, it still remains to give to $\Phi^{2}(\varepsilon)$ a form which does not contain roots foreign to the question. Now this is easily done, as follows:

Let \[ s_{k}=\frac{\sqrt[n]{v_{k}}}{(\sqrt[n]{v_{1}})^{k}}. \] On putting here $\varepsilon$ in place of $\alpha^{m}\varepsilon$, $\sqrt[n]{v_{k}}$ will be changed into $\theta^{-km}.\sqrt[n]{v_{k}}$, and $v_{1}$ into $\theta^{-m}.v_{1}$, therefore $s_{k}$ will be changed into: \[ \frac{\theta^{-km}.\sqrt[n]{v_{k}}}{(\theta^{-m}\sqrt[n]{v_{1}})^{k}}=\frac{\sqrt[n]{v_{k}}}{(\sqrt[n]{v_{1}})^{k}}. \]

The function $s_{k}$, as one sees, does not change in value on putting $\alpha^{m}\varepsilon$ in place of $\varepsilon$. Now $s_{k}$ is a rational function of $\Phi^{2}(\varepsilon)$. Therefore, denoting $s_{k}$ by $\lambda[\Phi^{2}(\varepsilon)]$, one will have \[ s_{k}=\lambda[\Phi^{2}(\alpha^{m}\varepsilon)], \] whatever the integer $m$ may be. From this one will deduce in the same manner, and as we have found $(\psi\varepsilon)^{n}$, the value of $s_{k}$ as a rational function of one of the quantities $p_{0}$, $p_{1}$, $....p_{n-1}$. Knowing $s_{k}$, one has: \[ \sqrt[n]{v_{k}}=s_{k}.(\sqrt[n]{v_{1}})^{k}. \] Therefore, putting $v$ in place of $v_{1}$, the expression for $\Phi^{2}(\alpha^{m}\varepsilon)$ will become: \[ \Phi^{2}(\alpha^{m}\varepsilon)=\frac{1}{n}\left\{-p_{n-1}+\theta^{-m}.v^{\frac{1}{n}}+s_{2}\theta^{-2m}.v^{\frac{2}{n}}+....+s_{n-1}\theta^{-(n-1)m}.v^{\frac{n-1}{n}}\right\}, \tag{116.} \] for $m=0$: \[ \Phi^{2}(\varepsilon)=\frac{1}{n}\left\{-p_{n-1}+v^{\frac{1}{n}}+s_{2}.v^{\frac{2}{n}}+s_{3}.v^{\frac{3}{n}}+....+s_{n-1}.v^{\frac{n-1}{n}}\right\}. \tag{117.} \] This value has only $n$ different values, which correspond to the $n$ values of $v^{\frac{1}{n}}$. Therefore, in the last place, the solution of the equation $P_{2n+1}=0$ is reduced to that of a single equation of the degree $2n+2$; but this equation does not appear to be algebraically soluble in general. Nevertheless it can be completely solved in several particular cases, e.g., when $e=c$, $e=c\sqrt{3}$, $e=c(2\pm\sqrt{3})$ etc. In the course of this memoir I shall occupy myself with these cases, of which the first especially is remarkable, as much by the simplicity of the solution as by its fine application in geometry.

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Indeed, among other things, I have arrived at this theorem:

“The entire circumference of the lemniscate can be divided into $m$ equal parts by the ruler and compass alone, if $m$ is of the form $2^{n}$ or $2^{n}+1$, the latter number being at the same time prime; or else if $m$ is a product of several numbers of these two forms.”

This theorem is, as one sees, precisely the same as that of Mr. Gauss, relative to the circle.

§. VI.

Various expressions of the functions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$.

23.

On making use of the known formulas which give the values of the coefficients of an algebraic equation as functions of the roots, one can derive several expressions of the functions $\varphi(n\beta)$, $f(n\beta)$, $F(n\beta)$ from the formulas of the preceding section.

I shall consider the most remarkable.

To abridge the formulas, I shall make use of the following notations: I shall denote:

1) By $\overset{k'}{\underset{k}{\Sigma}}_{m}\psi(m)$ the sum, and by $\overset{k'}{\underset{k}{\Pi}}_{m}\psi(m)$ the product of all the quantities of the form $\psi(m)$, which one will obtain by giving to $m$ all the integral values from $k$ to $k'$, the limits $k$ and $k'$ being included.

2) By $\overset{k'}{\underset{k}{\Sigma}}_{m}\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(m,\mu)$ the sum, and by $\overset{k'}{\underset{k}{\Pi}}_{m}\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(m,\mu)$ the product of all the quantities of the form $\psi(m,\mu)$, which one will obtain by giving to $m$ all the integral values from $k$ to $k'$, and to $\mu$ the integral values from $\nu$ to $\nu'$, always including the limits therein.

According to this it is clear that one will have: \[ \overset{k'}{\underset{k}{\Sigma}}_{m}\psi(m)=\psi(k)+\psi(k+1)+.....\psi(k'), \tag{119.} \] Editorial note: No equation number 118. is printed; the numbering passes from 117. to 119. \[ \overset{k'}{\underset{k}{\Pi}}_{m}\psi(m)=\psi(k).\psi(k+1)+.....\psi(k'), \tag{120.} \] Editorial note: A plus sign is printed before the row of dots in the product; a multiplication dot is expected. \[ \overset{k'}{\underset{k}{\Sigma}}_{m}\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(k,\mu)+\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(k+1,\mu)+.....+\overset{\nu'}{\underset{\nu}{\Sigma}}_{\mu}\psi(k',\mu), \tag{121.} \] \[ \overset{k'}{\underset{k}{\Pi}}_{m}\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(m,\mu)=\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(k,\mu).\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(k+1,\mu).........\overset{\nu'}{\underset{\nu}{\Pi}}_{\mu}\psi(k',\mu). \tag{122.} \]

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This being premised, let us consider the equations \[ \left\{\begin{aligned} \Phi(2n+1)\beta&=\frac{P_{2n+1}}{Q_{2n+1}},\\ f(2n+1)\beta&=\frac{P'_{2n+1}}{Q_{2n+1}},\\ F(2n+1)\beta&=\frac{P''_{2n+1}}{Q_{2n+1}}. \end{aligned}\right. \tag{123.} \]

We have seen that $P_{2n+1}$ is a rational function of $x$ of the degree $(2n+1)^{2}$ and of the form $x.\psi(x^{2})$. Likewise $P'_{2n+1}$ and $F''_{2n+1}$ are functions of this same form, the first of $y$ and the second of $z$.Editorial note: Printed $F''_{2n+1}$ where $P''_{2n+1}$ is expected. Finally $Q_{2n+1}$ is a function which, expressed indifferently in $x$, $y$ or $z$, will be of the degree $(2n+1)^{2}-1$, and will contain only even powers. Therefore one will have \[ \begin{aligned} P_{2n+1}&=A.x^{(2n+1)^{2}}+......+B.x;&\quad P'_{2n+1}&=A'.y^{(2n+1)^{2}}+......+B'.y;\\ P''_{2n+1}&=A''.z^{(2n+1)^{2}}+......+B''.z;&\quad Q_{2n+1}&=C.x^{(2n+1)^{2}-1}+....+D;\\ Q_{2n+1}&=C'.y^{(2n+1)^{2}-1}+....+D';&\quad Q_{2n+1}&=C''.z^{(2n+1)^{2}-1}+....+D''. \end{aligned} \]

On substituting these values in (123.), there will result \[ \begin{aligned} &\{A.x^{(2n+1)^{2}}+.....+B.x\}-\varphi(2n+1)\beta.\{C.x^{(2n+1)^{2}-1}+....+D\},\\ &\{A'.y^{(2n+1)^{2}}+.....+B'.y\}-f(2n+1)\beta.\{C'.y^{(2n+1)^{2}-1}+....+D'\},\\ &\{A''.z^{(2n+1)^{2}}+.....+B''.z\}-F(2n+1)\beta.\{C''.z^{(2n+1)^{2}-1}+....+D''\}. \end{aligned} \]

In the first of these equations $A$ is the coefficient of the first term, $-\varphi(2n+1)\beta.C$ that of the second and $-\varphi(2n+1)\beta.D$ the last term. Therefore $\frac{C}{A}\varphi(2n+1)\beta$ is equal to the sum and $\pm\frac{D}{A}\varphi(2n+1)\beta$ equal to the product of the roots of the equation in question, which equation is the same as this: \[ \Phi(2n+1)\beta=\frac{P_{2n+1}}{Q_{2n+1}}. \tag{124.} \]

Therefore, remarking that $A$, $C$ and $D$ (and in general all the coefficients) are independent of $\beta$, one sees that $\Phi(2n+1)\beta$ is (to within a constant coefficient) equal to the sum and to the product of all the roots of the equation (124.).

In the same manner one sees that $f(2n+1)\beta$ and $F(2n+1)\beta$ are respectively equal to the product or to the sum of the roots of the equations \[ f(2n+1)\beta=\frac{P'_{2n+1}}{Q_{2n+1}},\ F(2n+1)\beta=\frac{P''_{2n+1}}{Q_{2n+1}}, \]

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taking care to multiply the result by a constant coefficient, suitably chosen.

Now the roots of the equations (123.), according to No. 11., are respectively: \[ \begin{aligned} x&=(-1)^{m+\mu}.\Phi\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right),\\ y&=(-1)^{m}.f\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right),\\ z&=(-1)^{\mu}.F\left(\beta+\frac{m}{2n+1}\omega+\frac{\mu}{2n+1}\varpi i\right), \end{aligned} \] where the limits of $m$ and $\mu$ are $-n$ and $+n$.

Therefore, by virtue of what has just been seen, and making use of the notations adopted, one will have the following formulas: \[ \left\{\begin{aligned} \Phi(2n+1)\beta&=A.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m+\mu}.\Phi\left(\beta+\frac{m.\omega+\mu\varpi i}{2n+1}\right),\\ f(2n+1)\beta&=A'.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m}.f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ F(2n+1)\beta&=A''.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{\mu}.F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right);\\ \Phi(2n+1)\beta&=B.\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}.\Phi\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ f(2n+1)\beta&=B'.\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}.f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ F(2n+1)\beta&=B''.\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}.F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right). \end{aligned}\right. \tag{125.} \]

To determine the constant quantities $A$, $A'$, $A''$, $B$, $B'$, $B''$, it will be necessary to give $\beta$ a particular value. Thus on making in the first three formulas $\beta=\frac{\omega}{2}+\frac{\varpi}{2}i$, after having divided both members by $\Phi\beta$, there will result, remarking that $\Phi\left(\frac{\omega}{2}+\frac{\varpi}{2}i\right)=\frac{1}{0}$: \[ \left.\begin{aligned} A&=\frac{\varphi(2n+1)\beta}{\varphi\beta}\\ A'&=\frac{f(2n+1)\beta}{f\beta}\\ A''&=\frac{F(2n+1)\beta}{f\beta} \end{aligned}\right\}\ \text{for }\beta=\frac{\omega}{2}+\frac{\varpi}{2}i. \] Editorial note: The denominator of the third fraction is printed $f\beta$; $F\beta$ is apparently expected.

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Let $\beta=\frac{\omega}{2}+\frac{\varpi}{2}i+\alpha$, one has: \[ \left.\begin{aligned} A&=\frac{\varphi\left((2n+1)\alpha+n\omega+n\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{\varphi\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}\\ &=\frac{\varphi\left((2n+1)\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{\varphi\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}=\frac{\varphi\alpha}{\varphi(2n+1)\alpha},\\ A'&=\frac{f\left((2n+1)\alpha+n\omega+n\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{f\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}\\ &=(-1)^{n}.\frac{f\left((2n+1)\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{f\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}=(-1)^{n}.\frac{f\left(\alpha+\frac{\omega}{2}\right)}{f\left((2n+1)\alpha+\frac{\omega}{2}\right)},\\ A''&=\frac{F\left((2n+1)\alpha+n\omega+n\varpi i+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{F\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}\\ &=(-1)^{n}.\frac{F\left((2n+1)\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}{F\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}=(-1)^{n}.\frac{F\left(\alpha+\frac{\varpi}{2}i\right)}{F\left((2n+1)\alpha+\frac{\varpi}{2}i\right)}. \end{aligned}\right\}\ \text{for }\alpha=0. \]

These expressions for $A$, $A'$, $A''$ will become of the form $\frac{0}{0}$, on making $\alpha=0$, therefore one will find according to the known rules: \[ A=\frac{1}{2n+1},\quad A'=A''=\frac{(-1)^{n}}{2n+1}. \] According to this the first three formulas will become: \[ \left\{\begin{aligned} \Phi(2n+1)\beta&=\frac{1}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m+\mu}.\varphi\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ f(2n+1)\beta&=\frac{(-1)^{n}}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{m}.f\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right),\\ F(2n+1)\beta&=\frac{(-1)^{n}}{2n+1}.\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}(-1)^{\mu}.F\left(\beta+\frac{m\omega+\mu\varpi i}{2n+1}\right). \end{aligned}\right. \tag{126.} \]

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To obtain the value of the constants $B$, $B'$, $B''$, I remark that there will be: \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Pi}}_{m}\overset{+n}{\underset{-n}{\Pi}}_{\mu}\psi(m,\mu)=\psi(0,0).\overset{n}{\underset{1}{\Pi}}_{m}\psi(m,0).\overset{n}{\underset{1}{\Pi}}_{\mu}\psi(0,\mu)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\psi(m,\mu).\psi(-m,-\mu).\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\psi(m,-\mu).\psi(-m,\mu). \end{aligned} \tag{127.} \]

Applying this transformation to the formulas (126.), dividing the first by $\Phi\beta$, the second by $f\beta$ and the third by $F\beta$, then making in the first $\beta=0$, in the second $\beta=\frac{\omega}{2}$ and in the third $\beta=\frac{\varpi}{2}i$, and remarking that $\frac{\varphi(2n+1)\beta}{\varphi\beta}=2n+1$, for $\beta=0$, that $\frac{f(2n+1)\beta}{f\beta}=2n+1$, for $\beta=\frac{\omega}{2}$, and that $\frac{F(2n+1)\beta}{F\beta}=2n+1$, for $\beta=\frac{\varpi}{2}i$: we shall find: \[ \left\{\begin{aligned} (2n+1)&=B.\overset{n}{\underset{1}{\Pi}}_{m}\varphi^{2}\left(\frac{m\omega}{2n+1}\right).\overset{n}{\underset{1}{\Pi}}_{\mu}\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right).\varphi^{2}\left(\frac{m\omega-\mu\varpi i}{2n+1}\right),\\ (2n+1)&=B'.\overset{n}{\underset{1}{\Pi}}_{m}f^{2}\left(\frac{\omega}{2}+\frac{m\omega}{2n+1}\right).\overset{n}{\underset{1}{\Pi}}_{\mu}f^{2}\left(\frac{\omega}{2}+\frac{\mu\varpi i}{2n+1}\right)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}f^{2}\left(\frac{\omega}{2}+\frac{m\omega+\mu\varpi i}{2n+1}\right).f^{2}\left(\frac{\omega}{2}+\frac{m\omega-\mu\varpi i}{2n+1}\right),\\ (2n+1)&=B''.\overset{n}{\underset{1}{\Pi}}_{m}F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right).\overset{n}{\underset{1}{\Pi}}_{\mu}F^{2}\left(\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right).F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right). \end{aligned}\right. \tag{128.} \]

Drawing from these equations the values of $B$, $B'$, $B''$, and then substituting them in the transformed formulas, there will result:

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These formulas may be given simpler forms, by making use of the following formulas: \[ \frac{\varphi(\beta+\alpha).\varphi(\beta-\alpha)}{\varphi^{2}\alpha}=-\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\alpha}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}}, \] \[ \frac{f(\beta+\alpha).f(\beta-\alpha)}{f^{2}\left(\frac{\omega}{2}+\alpha\right)}=-\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\alpha\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\alpha+\frac{\omega}{2}+\frac{\varpi}{2}i\right)}}, \] \[ \frac{F^{2}(\beta+\alpha).F(\beta-\alpha)}{F^{2}\left(\frac{\varpi}{2}i+\alpha\right)}=-\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\alpha\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\alpha\right)}}, \] Editorial note: The numerator of the third left-hand fraction is printed $F^{2}(\beta+\alpha).F(\beta-\alpha)$; $F(\beta+\alpha).F(\beta-\alpha)$ is apparently expected.

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which one will easily verify by means of the formulas (13.), (16.), (18.).

By virtue of these formulas it is clear that the equations (129.) may be put under the form: \[ \left\{\begin{aligned} &\Phi(2n+1)\beta=\\ &(2n+1)\Phi\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}},\\ &f(2n+1)\beta=\\ &(2n+1)f\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{m\omega}{2n+1}+\frac{\omega}{2}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}},\\ &F(2n+1)\beta=\\ &(2n+1)F\beta.\overset{n}{\underset{1}{\Pi}}_{m}\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega}{2n+1}\right)}}.\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{\mu\varpi i}{2n+1}\right)}}\\ &\quad\times\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega+\mu\varpi i}{2n+1}\right)}}.\frac{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\frac{\varpi}{2}i+\frac{m\omega-\mu\varpi i}{2n+1}\right)}}. \end{aligned}\right. \tag{130.} \]

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These formulas give, as one sees, the values of $\Phi(2n+1)\beta$, $f(2n+1)\beta$ and $F(2n+1)\beta$, expressed respectively as rational functions of $\Phi\beta$, $f\beta$ and $F\beta$, in the form of products.

We shall further give the values of $f(2n+1)\beta$, $F(2n+1)\beta$ under another form, which will be useful in what follows.

We have $f^{2}\beta=1-c^{2}\varphi^{2}\beta$, hence \[ 1-\frac{f^{2}\beta}{f^{2}\alpha}=\frac{c^{2}(\varphi^{2}\beta-\varphi^{2}\alpha)}{f^{2}\alpha}=\frac{c^{2}}{f^{2}\alpha}\Phi^{2}\beta-\frac{c^{2}\varphi^{2}\alpha}{f^{2}\alpha} \] and \[ 1-\frac{f^{2}\beta}{f^{2}\left(\frac{\varpi}{2}i+\alpha\right)}=\frac{c^{2}\left[\varphi^{2}\beta-\varphi^{2}\left(\frac{\varpi}{2}i+\alpha\right)\right]}{f^{2}\left(\frac{\varpi}{2}i+\alpha\right)}, \] now by virtue of (18.) we have: \[ f^{2}\left(\frac{\varpi}{2}i+\alpha\right)=\frac{e^{2}+c^{2}}{e^{2}}.\frac{1}{f^{2}\alpha}; \] hence: \[ \frac{1-\frac{f^{2}\beta}{f^{2}\alpha}}{1-\frac{f^{2}\beta}{f^{2}\left(\frac{\varpi}{2}i+\alpha\right)}}=\frac{1}{f^{4}\alpha}.\frac{e^{2}+c^{2}}{e^{2}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\alpha}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\varpi}{2}i+\alpha\right)}}. \] We shall find in the same way: \[ \frac{1-\frac{F^{2}\beta}{F^{2}\alpha}}{1-\frac{F^{2}\beta}{F^{2}\left(\frac{\omega}{2}+\alpha\right)}}=\frac{1}{F^{4}\alpha}.\frac{e^{2}+c^{2}}{e^{2}}.\frac{1-\frac{\varphi^{2}\beta}{\varphi^{2}\alpha}}{1-\frac{\varphi^{2}\beta}{\varphi^{2}\left(\frac{\omega}{2}+\alpha\right)}}. \]

By virtue of these formulas, and making $\beta=0$ in order to determine the constant factor, it is clear that the expressions of $f(2n+1)\beta$, $F(2n+1)\beta$ may be written as follows:

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In this paragraph we have considered the functions $\Phi(n\beta)$, $f(n\beta)$, $F(n\beta)$ only in the case of odd values of $n$. One might find analogous expressions of these functions for even values of $n$; but as there is no difficulty in this, and as moreover the formulas to which we have arrived are those which will be most useful to us in what follows, I shall not occupy myself with it.

§. VII.

Development of the functions $\Phi\alpha$, $f\alpha$, $F\alpha$ in series and in infinite products.

24.

Making in the formulas of the preceding paragraph $\beta=\frac{\alpha}{2n+1}$, we shall obtain expressions of the functions $\Phi\alpha$, $f\alpha$, $F\alpha$ which, on account of the indeterminate number $n$, may be varied in an infinity of ways.

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Among all the formulas that one will obtain in this way, those which result from the supposition of $n$ infinite are the most remarkable. Then the functions $\Phi$, $f$, $F$ will disappear from the values of $\Phi\alpha$, $f\alpha$, $F\alpha$, and one will obtain for these functions algebraic expressions, but composed of an infinity of terms. To have these expressions, one must make in the formulas (126.), (130.) $\beta=\frac{\alpha}{2n+1}$, and then seek the limit of the second member of these equations for ever increasing values of $n$. For brevity, let $v$ be a quantity whose limit is zero for ever increasing values of $n$. This being premised, let us consider successively the three formulas (126.).

Making in the first of the formulas (126.) $\beta=\frac{\alpha}{2n+1}$, and remarking that \[ \begin{aligned} &\overset{+n}{\underset{-n}{\Sigma}}_{m}\overset{+n}{\underset{-n}{\Sigma}}_{\mu}\theta(m,\mu)=\theta(0,0)+\overset{n}{\underset{1}{\Sigma}}_{m}\theta(m,0)+\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(0,\mu)\\ &\quad+\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}\{\theta(m,\mu)+\theta(-m,-\mu)+\theta(m,-\mu)+\theta(-m,\mu)\}, \end{aligned} \tag{131.} \] it is clear that one may put the formula in question under the form: \[ \begin{aligned} \Phi\alpha={}&\frac{1}{2n+1}.\varphi\left(\frac{\alpha}{2n+1}\right)+\frac{1}{2n+1}\overset{n}{\underset{1}{\Sigma}}_{m}(-1)^{m}\left\{\Phi\left(\frac{\alpha+m\omega}{2n+1}\right)+\Phi\left(\frac{\alpha-m\omega}{2n+1}\right)\right\}\\ &+\frac{1}{2n+1}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{\mu}\left\{\Phi\left(\frac{\alpha+\mu\varpi i}{2n+1}\right)+\Phi\left(\frac{\alpha-\mu\varpi i}{2n+1}\right)\right\}-\frac{i}{ec}.\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(n-m,n-\mu)\\ &+\frac{i}{ec}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(n-m,n-\mu), \end{aligned} \tag{132.} \] where one has made, for brevity, \[ \left\{\begin{aligned} \psi(m,\mu)&=\frac{1}{2n+1}.\left\{\frac{1}{\varphi\left(\frac{\alpha+(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}+\frac{1}{\varphi\left(\frac{\alpha-(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}\right\},\\ \psi_{1}(m,\mu)&=\frac{1}{2n+1}.\left\{\frac{1}{\varphi\left(\frac{\alpha+(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}+\frac{1}{\varphi\left(\frac{\alpha-(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}\right\}. \end{aligned}\right. \tag{133.} \]

Now, remarking that \[ \Phi\left(\frac{\alpha+m\omega}{2n+1}\right)+\Phi\left(\frac{\alpha-m\omega}{2n+1}\right)=\frac{2\varphi\left(\frac{\alpha}{2n+1}\right).f\left(\frac{m\omega}{2n+1}\right).F\left(\frac{m\omega}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{m\omega}{2n+1}\right).\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}=\frac{A_{m}}{2n+1}, \] \[ \Phi\left(\frac{\alpha+\mu\varpi i}{2n+1}\right)+\Phi\left(\frac{\alpha-\mu\varpi i}{2n+1}\right)=\frac{2\varphi\left(\frac{\alpha}{2n+1}\right).f\left(\frac{\mu\varpi i}{2n+1}\right).F\left(\frac{\mu\varpi i}{2n+1}\right)}{1+e^{2}c^{2}.\varphi^{2}\left(\frac{\mu\varpi i}{2n+1}\right).\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}=\frac{B_{\mu}}{2n+1}, \]

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where $A_{m}$ and $B_{\mu}$ are finite quantities, the part of the equation (132.) as far as the member which has the sign $-$ will take the form: \[ \frac{1}{2n+1}.\varphi\left(\frac{\alpha}{2n+1}\right)+\frac{1}{(2n+1)^{2}}.\overset{n}{\underset{1}{\Sigma}}_{m}(-1)^{m}(A_{m}+B_{\mu}); \] now the limit of this quantity is evidently zero; hence, taking the limit of the formula (132.), we shall have: \[ \begin{aligned} \Phi\alpha&=-\frac{i}{ec}\ \text{lim.}\ \overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(n-m,n-\mu)\\ &\quad+\frac{i}{ec}\ \text{lim.}\ \overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(n-m,n-\mu), \end{aligned} \] or else: \[ \begin{aligned} \Phi\alpha&=-\frac{i}{ec}\ \text{lim.}\ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu)\\ &\quad+\frac{i}{ec}\ \text{lim.}\ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(m,\mu). \end{aligned} \tag{134.} \]

It suffices to know one of these limits, for we shall have the other by merely changing the sign of $i$.

Let us seek the limit of \[ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu). \] For this, one must try to put the preceding quantity under the form \[ P+v, \] where $P$ is independent of $n$, and $v$ is a quantity which has zero for its limit; for then the quantity $P$ will be precisely the limit in question.

25.

Let us consider first the expression: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\psi(m,\mu). \]

Let \[ \theta(m,\mu)=\frac{2\alpha}{\alpha^{2}-(m\omega+\mu\varpi i)^{2}}, \tag{135.} \] and let us make \[ \psi(m,\mu)-\theta(m,\mu)=\frac{2\alpha}{(2n+1)^{2}}.R_{\mu}, \tag{136.} \] we shall have: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\psi(m,\mu)-\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)=2\alpha.\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{n}.\frac{R_{\mu}}{(2n+1)^{2}}. \tag{137.} \] Editorial note: The exponent of the factor $-1$ in the last sum is apparently printed $n$; $\mu$ is expected, as in the two sums on the left.

This being premised, I say that the second member of this equation is a quantity of the form $\frac{v}{2n+1}$.

page 157

According to (12.), (13.) we shall have \[ \frac{1}{\varphi(\beta+\varepsilon)}+\frac{1}{\varphi(\beta-\varepsilon)}=\frac{\varphi(\beta+\varepsilon)+\varphi(\beta-\varepsilon)}{\varphi(\beta+\varepsilon).\varphi(\beta-\varepsilon)}=\frac{2\varphi\beta.f\varepsilon.F\varepsilon}{\varphi^{2}\beta-\varphi^{2}\varepsilon}, \] hence, making $\beta=\frac{\alpha}{2n+1}$ and $\varepsilon=\frac{m\omega+\mu\varpi i}{2n+1}=\frac{\varepsilon_{\mu}}{2n+1}$ and $f\varepsilon.F\varepsilon=\theta\varepsilon$, we have: \[ \psi(m,\mu)=\frac{1}{2n+1}.\frac{2\varphi\left(\frac{\alpha}{2n+1}\right).\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}. \] Now we have: \[ \varphi\left(\frac{\alpha}{2n+1}\right)=\frac{\alpha}{2n+1}+\frac{A\alpha^{3}}{(2n+1)^{3}}, \] hence: \[ \psi(m,\mu)=\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}.\left\{\frac{2A\alpha^{3}}{(2n+1)^{4}}+\frac{2\alpha}{(2n+1)^{2}}\right\}, \] and consequently: \[ \begin{aligned} \psi(m,\mu)-\theta(m,\mu)&=\frac{2\alpha}{(2n+1)^{2}}\left\{\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}-\frac{1}{\left(\frac{\alpha}{2n+1}\right)^{2}-\left(\frac{\varepsilon_{\mu}}{2n+1}\right)^{2}}\right\}\\ &\quad+\frac{2A\alpha^{3}}{(2n+1)^{4}}.\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}. \end{aligned} \] Hence the value of $R_{\mu}$ will become \[ R_{\mu}=\frac{\theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)}.\left\{1+\frac{A\alpha^{2}}{(2n+1)^{2}}\right\}-\frac{1}{\left(\frac{\alpha}{2n+1}\right)^{2}-\left(\frac{\varepsilon_{\mu}}{2n+1}\right)^{2}}. \tag{138.} \]

This being premised, there are two cases to consider, namely whether $\frac{\varepsilon_{\mu}}{2n+1}$ has zero for its limit or not.

$a)$ If $\frac{\varepsilon_{\mu}}{2n+1}$ has zero for its limit, we shall have: \[ \begin{aligned} \varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)&=\frac{\varepsilon_{\mu}^{2}}{(2n+1)^{2}}+\frac{B_{\mu}.\varepsilon_{\mu}^{4}}{(2n+1)^{4}},\\ \theta\left(\frac{\varepsilon_{\mu}}{2n+1}\right)&=\sqrt{\left[1-c^{2}\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)\right]}.\sqrt{\left[1+e^{2}\varphi^{2}\left(\frac{\varepsilon_{\mu}}{2n+1}\right)\right]}=1+\frac{C_{\mu}.\varepsilon_{\mu}^{2}}{(2n+1)^{2}},\\ \varphi^{2}\left(\frac{\alpha}{2n+1}\right)&=\frac{\alpha^{2}}{(2n+1)^{2}}+\frac{D.\alpha^{4}}{(2n+1)^{4}}, \end{aligned} \] where $B_{\mu}$, $C_{\mu}$, $D$ have finite limits; hence, substituting:

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\[ \begin{aligned} R_{\mu}&=A\alpha^{2}.\frac{\frac{1}{\varepsilon_{\mu}^{2}}+\frac{C_{\mu}}{(2n+1)^{2}}}{\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}-1+\frac{D\alpha^{4}}{(2n+1)^{2}.\varepsilon_{\mu}^{2}}-B_{\mu}\frac{\varepsilon_{\mu}^{2}}{(2n+1)^{2}}}\\ &\quad+\frac{C_{\mu}.\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}-\frac{D\alpha^{4}}{\varepsilon_{\mu}^{4}}-C_{\mu}+B_{\mu}}{\left(1-\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}\right)^{2}-\left(1-\frac{\alpha^{2}}{\varepsilon_{\mu}^{2}}\right)\left\{\frac{D\alpha^{4}}{(2n+1)^{2}.\varepsilon_{\mu}^{2}}-B_{\mu}.\frac{\varepsilon_{\mu}^{2}}{(2n+1)^{2}}\right\}}, \end{aligned} \tag{139.} \] now whether $\varepsilon_{\mu}$ be finite or infinite, it is clear that this quantity will always converge towards a finite quantity for ever increasing values of $n$. Hence we shall have \[ R_{\mu}=r_{\mu}+v_{\mu}, \tag{140.} \] where $r_{\mu}$ is a finite quantity independent of $n$.

$b)$ If $\frac{\varepsilon_{\mu}}{2n+1}$ has for its limit a finite quantity, it is clear that, naming this limit $\delta_{\mu}$, we shall have: \[ R_{\mu}=-\frac{\theta(\delta_{\mu})}{\varphi^{2}(\delta_{\mu})}+\frac{1}{\delta_{\mu}^{2}}+v_{\mu}^{1}. \tag{141.} \] This being premised, let us consider the expression $\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}$. We have \[ \begin{aligned} \overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}&=\frac{1}{(2n+1)^{2}}.\{R_{0}-R_{1}+R_{2}-R_{3}+....\\ &\quad....+(-1)^{\nu-1}.R_{\nu-1}+(-1)^{\nu+1}[R_{\nu+1}-R_{\nu+1}+R_{\nu+2}-R_{\nu+3}...+(-1)^{n-\nu-1}.R_{n-1}]\}. \end{aligned} \tag{142.} \] Editorial note: The first term inside the square bracket is printed $R_{\nu+1}$; $R_{\nu}$ is apparently expected. Let us suppose first that $\frac{\varepsilon_{\mu}}{2n+1}$ has for its limit a finite quantity, whatever be the value of $\mu$. Then, remarking that \[ \delta_{\mu+1}=\delta_{\mu}, \] we shall have \[ R_{\mu}-R_{\mu+1}=v_{\mu}^{1}-v_{\mu+1}^{1}, \] hence: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{1}{(2n+1)^{2}}(v_{0}^{1}-v_{1}^{1}+v_{2}^{1}-v_{3}^{1}+....+v_{k-2}^{1}-v_{k-1}^{1})+\frac{B}{(2n+1)^{2}}, \] where $k=n$ or $n-1$, according as $n$ is even or odd. The quantity $B$ always has for its limit a finite quantity, namely $B=0$ if $n$ is even, and $B=R_{n-1}$ if $n$ is odd.

Now it is known that a sum such as \[ v_{0}^{1}-v_{1}^{1}+v_{2}^{1}-....+v_{k-2}^{1}-v_{k-1}^{1}, \] may be put under the form $k.v$, $v$ having zero for its limit. Hence, substituting:

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\[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{k.v+B}{(2n+1)^{2}}; \] now, $k$ being equal to $n$ or to $n-1$, and $B$ finite, the limit of $\frac{k.v+B}{2n+1}$ will be zero, hence: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{v}{2n+1}. \tag{143.} \]

Let us now suppose that $\frac{m}{2n+1}$ has zero for its limit. Then $\frac{\varepsilon_{\mu}}{2n+1}$ likewise has zero for its limit, unless at the same time $\frac{\mu}{2n+1}$ has for its limit a finite quantity. Let $\nu$ in this case be the whole number immediately below $\sqrt{n}$, and let us consider the sum \[ R_{0}-R_{1}+R_{2}-R_{3}+....+(-1)^{\nu-1}R_{\nu-1}. \] Supposing that $\mu$ is one of the numbers $0$, $1$, $....\nu$, it is clear that $\frac{\varepsilon_{\mu}}{(2n+1)}=\frac{m\omega+\mu\varpi i}{2n+1}$ has zero for its limit; hence, according to what we have seen, $R_{\mu}$ will be a finite quantity, and consequently \[ R_{0}-R_{1}+R_{2}-....+(-1)^{\nu-1}R_{\nu-1}=\nu.R, \] where $R$ is likewise a finite quantity.

Let us now consider the sum \[ (-1)^{\nu}\{R_{\nu}-R_{\nu+1}+R_{\nu+2}-....+(-1)^{n-\nu-1}.R_{n-1}\}. \]

If $\frac{\varepsilon_{\mu}}{2n+1}$ has for its limit a quantity different from zero, we have, as will be seen: \[ R_{\mu}-R_{\mu+1}=v_{\mu}^{1}-v_{\mu+1}^{1}; \] if on the contrary $\frac{\varepsilon_{\mu}}{2n+1}$ has zero for its limit, we have: \[ R_{\mu}=v_{\mu}+v_{\mu}^{1}; \] now, if at the same time $\mu>\sqrt{n}$, it is clear that, by virtue of the value of $R_{\mu}$, \[ v_{\mu}=B_{\mu}-C_{\mu}; \] now it is clear that $B_{\mu}$ and $C_{\mu}$ both have for limits quantities independent of $\mu$, hence, naming these limits $B$ and $C$, we shall have: \[ R_{\mu}=B-C+v_{\mu}, \] and consequently, in this case also, \[ R_{\mu}-R_{\mu+1}=v_{\mu}-v_{\mu+1}. \] Editorial note: The last subscript is printed indistinctly, apparently $v_{\mu+1}$ with the $\mu$ damaged in print.

Hence, as in the case where $\frac{\varepsilon_{\mu}}{2n+1}$ would have a limit different from zero for all values of $\mu$, we shall demonstrate that \[ \frac{(-1)^{\nu}}{(2n+1)^{2}}\{R_{\nu}-R_{\nu+1}+....+(-1)^{n-\nu-1}R_{n-1}\}=\frac{v}{(2n+1)}. \]

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Now, combining the equations above, we shall derive from them \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{1}{(2n+1)^{2}}.(\nu R)+\frac{v}{2n+1}; \] but $\frac{v}{2n+1}$ has zero for its limit, therefore \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{R_{\mu}}{(2n+1)^{2}}=\frac{v}{2n+1}. \] Therefore this formula always holds, and consequently the formula (137.) will become: \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\psi(m,\mu)-\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)=\frac{v}{2n+1}. \tag{144.} \] This being premised, the point is to put $\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)$ under the form $P+\frac{v}{2n+1}$. Now this can be done as follows. We have: \[ \left\{\begin{aligned} &\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)=\overset{\frac{1}{0}}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)-\overset{\frac{1}{0}}{\underset{n}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)\ \text{and}\\ &\overset{\frac{1}{0}}{\underset{n}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)=(-1)^{n}\{\theta(m,n)-\theta(m,n+1)+\theta(m,n+2)-....\text{etc.}\}. \end{aligned}\right. \tag{145.} \] Now, by a known formula, we have: \[ \begin{aligned} &\theta(m,n)-\theta(m,n+1)+\theta(m,n+2)-......\\ &=\tfrac{1}{2}\theta(m,n)+A\frac{\partial\theta(m,n)}{\partial n}+B.\frac{\partial^{3}\theta(m,n)}{\partial^{3}n}+......, \end{aligned} \] where $A$, $B$$....$ are numbers; now \[ \theta(m,n)=\frac{2\alpha}{\alpha^{2}-(m\omega+n\varpi i)^{2}}, \] therefore, on substituting: \[ \theta(m,n)-\theta(m,n+1)+....=\frac{\alpha}{\alpha^{2}-(m\omega+n\varpi i)^{2}}+\frac{4A\alpha\varpi i.n}{\{\alpha^{2}-(m\omega+n\varpi i)^{2}\}^{2}}+.... \] Editorial note: The exponent outside the braces in the denominator of the second term is printed indistinctly; $2$ is read here. From this it follows that \[ \theta(m,n)-\theta(m,n+1)+....=\frac{\alpha}{\varpi^{2}.n^{2}}+\frac{v}{n^{2}}=\frac{v}{2n+1}. \] Therefore, by virtue of the equations (145.) \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\theta(m,\mu)=\overset{\frac{1}{0}}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)+\frac{v}{2n+1}, \] and consequently \[ \overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}\psi(m,\mu)=\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu)+\frac{v}{2n+1}. \tag{146.} \]

26.

Having transformed in this way the quantity $\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\psi(m,\mu)$, we derive from the equation (146.):

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\[ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}\psi(m,\mu)=\overset{n-1}{\underset{0}{\Sigma}}(-1)^{m}.\varrho_{m}+\overset{n-1}{\underset{0}{\Sigma}}.\frac{v_{m}}{2n+1}, \tag{147.} \] on putting \[ \varrho_{m}=\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta(m,\mu); \tag{148.} \] now \[ \overset{n-1}{\underset{0}{\Sigma}}.\frac{v_{m}}{2n+1}=\frac{v_{0}+v_{1}+v_{2}\ldots v_{n-1}}{2n+1}=\frac{nv}{2n+1}=\frac{v}{2}, \] $v$ having zero for its limit. Therefore the equation (147.) will give, on making $n$ infinite: \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu)=\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho_{m}. \tag{149.} \] In the same way, if one puts, for brevity: \[ \left\{\begin{aligned}\theta_{1}(m,\mu)&=\frac{2\alpha}{\alpha^{2}-(m\omega-\mu\varpi i)^{2}},\\ \varrho'_{m}&=\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\theta_{1}(m,\mu),\end{aligned}\right. \tag{150.} \] we shall have: \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi_{1}(m,\mu)=\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho'_{m}. \tag{151.} \] Having found these two quantities, of which the expression of $\Phi\alpha$ is composed, we shall have, on substituting: \[ \Phi\alpha=-\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho_{m}+\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}.\varrho'_{m}=\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}(-1)^{m}\{\varrho'_{m}-\varrho_{m}\}, \] or else, on restoring the values of $\varrho'_{m}$ and $\varrho_{m}$, \[ \Phi\alpha=\frac{i}{ec}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}-\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}}\right\}\right\}. \tag{152.} \] Now \[ \frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega\pm(\mu+\tfrac{1}{2})\varpi i]^{2}}=\frac{1}{\alpha-(m+\tfrac{1}{2})\omega\mp(\mu+\tfrac{1}{2})\varpi i}+\frac{1}{\alpha+(m+\tfrac{1}{2})\omega\pm(\mu+\tfrac{1}{2})\varpi i}, \] therefore: \[ \begin{aligned}&\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}-\frac{2\alpha}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}}\\ &=+\frac{(2\mu+1)\varpi i}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{(2\mu+1)\varpi i}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}},\end{aligned} \] therefore the expression of $\Phi\alpha$ will become, under a real form: \[ \Phi\alpha=\frac{1}{ec}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\left\{\frac{(2\mu+1)\varpi}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{(2\mu+1)\varpi}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right\}, \tag{153.} \] that is, we shall have:

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\[ \begin{aligned}\Phi\alpha&=\frac{\varpi}{ec}(\delta_{0}-\delta_{1}+\delta_{2}-\delta_{3}+....+(-1)^{m}\delta_{m}-....)\\ &-\frac{\varpi}{ec}(\delta'_{0}-\delta'_{1}+\delta'_{2}-\delta'_{3}+....+(-1)^{m}\delta'_{m}-....),\end{aligned} \tag{154.} \] or else: \[ \left\{\begin{aligned}\delta_{m}&=\frac{1}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+\frac{\varpi^{2}}{4}}-\frac{3}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+\frac{9\varpi^{2}}{4}}+\frac{5}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+\frac{25\varpi^{2}}{4}}-\ldots\\ \delta'_{\mu}&=\frac{1}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+\frac{\varpi^{2}}{4}}-\frac{3}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+\frac{9\varpi^{2}}{4}}+\frac{5}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+\frac{25\varpi^{2}}{4}}-\ldots\end{aligned}\right. \tag{155.} \]

If one begins the search for the limit of the function $\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m+\mu}.\psi(m,\mu)$ with that of $\overset{n-1}{\underset{0}{\Sigma}}(-1)^{m}\psi(m,\mu)$ instead of that of $\overset{n-1}{\underset{0}{\Sigma}}(-1)^{\mu}\psi(m,\mu)$, as we have done, one will find in place of the formula (53.)Editorial note: Printed (53.); the formula meant is presumably (153.), an apparent misprint. the following: \[ \Phi\alpha=\frac{1}{ec}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{(2\mu+1)\varpi}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{(2\mu+1)\varpi}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right\}, \tag{156.} \] that is: \[ \begin{aligned}\Phi\alpha&=\frac{\varpi}{ec}(\varepsilon_{0}-3\varepsilon_{1}+5\varepsilon_{2}-7\varepsilon_{3}+....+(-1)^{\mu}.(2\mu+1)\varepsilon_{\mu}-....)\\ &-\frac{\varpi}{ec}(\varepsilon'_{0}-3\varepsilon'_{1}+5\varepsilon'_{2}-7\varepsilon'_{3}+....+(-1)^{\mu}.(2\mu+1)\varepsilon_{\mu}-....),\end{aligned} \tag{157.} \] Editorial note: In the second row the last term is printed with $\varepsilon_{\mu}$ without the prime; $\varepsilon'_{\mu}$ is expected, an apparent misprint. where \[ \left\{\begin{aligned}\varepsilon_{\mu}&=\frac{1}{\left(\alpha-\frac{\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{1}{\left(\alpha-\frac{3\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}+\frac{1}{\left(\alpha-\frac{5\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-....\\ \varepsilon'_{\mu}&=\frac{1}{\left(\alpha+\frac{\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{1}{\left(\alpha+\frac{3\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}+\frac{1}{\left(\alpha+\frac{5\omega}{2}\right)^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-....\end{aligned}\right. \tag{158.} \]

27.

Let us now seek the expression of $f\alpha$, by means of the second of the formulas (126.). On making therein $\beta=\frac{\alpha}{2n+1}$, and remarking that the limit of the quantities contained in the first two lines then becomes equal to zero, we shall have: \[ \begin{aligned}f\alpha&=\text{lim.}(-1)^{n}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m}.\psi(n-m,n-\mu)\\ &+\text{lim.}(-1)^{n}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}(-1)^{m}.\psi_{1}(n-m,n-\mu),\end{aligned} \tag{159.} \]

page 163

where we have put, for brevity: \[ \begin{aligned}\psi(n-m,n-\mu)&=\frac{1}{2n+1}\left\{f\left(\frac{\alpha+m\omega+\mu\varpi i}{2n+1}\right)+f\left(\frac{\alpha-m\omega-\mu\varpi i}{2n+1}\right)\right\},\\ \psi_{1}(n-m,n-\mu)&=\frac{1}{2n+1}\left\{f\left(\frac{\alpha+m\omega-\mu\varpi i}{2n+1}\right)+f\left(\frac{\alpha-m\omega+\mu\varpi i}{2n+1}\right)\right\}.\end{aligned} \] Now we have: \[ f(\beta-\varepsilon)+f(\beta-\varepsilon)=\frac{2f\beta.f\varepsilon}{1+e^{2}c^{2}\varphi^{2}\varepsilon.\varphi^{2}\beta}=\frac{f\varepsilon}{e^{2}c^{2}\varphi^{2}\varepsilon}.\frac{f\beta}{\varphi^{2}\beta+\frac{1}{e^{2}c^{2}\varphi^{2}\varepsilon}}. \] Editorial note: Printed $f(\beta-\varepsilon)$ in both terms; the first term is presumably $f(\beta+\varepsilon)$, an apparent misprint.

Let \[ \varepsilon=\frac{m\omega+\mu\varpi i}{2n+1}, \] we shall have: \[ \begin{aligned}\frac{1}{\varphi\varepsilon}&=-iec.\Phi\left(\frac{\omega}{2}+\frac{\varpi}{2}i-\varepsilon\right)=-iec.\Phi\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right),\\ \frac{f\varepsilon}{\varphi\varepsilon}&=-\frac{i\sqrt{(e^{2}+c^{2})}}{F\left(\varepsilon-\frac{\varpi}{2}i\right)}=-i\sqrt{(e^{2}+c^{2})}.\frac{c}{\sqrt{(e^{2}+c^{2})}}.F\left(\varepsilon-\frac{\omega}{2}-\frac{\varpi}{2}i\right)\\ &=-ci.F.\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right).\end{aligned} \] Therefore we shall have, on substituting and putting $m$ and $\mu$ respectively in place of $n-m$ and $n-\mu$: \[ \psi(m,\mu)=\frac{1}{e}.\frac{2\varphi\left(\frac{(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right).F\left(\frac{(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)}{(2n+1)\left[\varphi^{2}\left(\frac{\alpha}{2n+1}\right)-\varphi^{2}\left(\frac{(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)\right]}. \] We shall have the value of $\psi_{1}(m,\mu)$ by changing only the sign of $i$. Making now \[ \theta(m,\mu)=\frac{(2m+1)\omega+(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}} \] and \[ \theta_{1}(m,\mu)=\frac{(2m+1)\omega-(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}, \] and then seeking the limit of the function \[ \overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m}.\psi(m,\mu), \] in the same manner as before, we shall find: \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m}.\psi(m,\mu)=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\theta(m,\mu)\right\} \] and \[ \text{lim.}\overset{n-1}{\underset{0}{\Sigma}}_{m}\overset{n-1}{\underset{0}{\Sigma}}_{\mu}(-1)^{m}.\psi_{1}(m,\mu)=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\theta_{1}(m,\mu)\right\}; \]

page 164

therefore, on substituting in (159.), and restoring the values of $\theta(m,\mu)$ and $\theta_{1}(m,\mu)$, we have: \[ f\alpha=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\left\{\frac{(2m+1)\omega+(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega+(\mu+\tfrac{1}{2})\varpi i]^{2}}+\frac{(2m+1)\omega-(2\mu+1)\varpi i}{\alpha^{2}-[(m+\tfrac{1}{2})\omega-(\mu+\tfrac{1}{2})\varpi i]^{2}}\right\}. \tag{160.} \]

The quantity enclosed between the brackets may also be put under the form: \[ \frac{2[\alpha-(m+\tfrac{1}{2})\omega]}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\frac{2[\alpha+(m+\tfrac{1}{2})\omega]}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}, \] therefore also: \[ f\alpha=\frac{1}{e}.\overset{\infty}{\underset{0}{\Sigma}}_{\mu}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\frac{2[\alpha-(m+\tfrac{1}{2})\omega]}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}-\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\frac{2[\alpha+(m+\tfrac{1}{2})\omega]}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}}\right\}. \tag{161.} \]

We shall have in the same manner: \[ F(\alpha)=\frac{1}{c}.\overset{\infty}{\underset{0}{\Sigma}}_{m}\left\{\overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{(2\mu+1)\varpi}{[\alpha-(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})\varpi^{2}}+\overset{\infty}{\underset{0}{\Sigma}}(-1)^{\mu}.\frac{(2\mu+1)\varpi}{[\alpha+(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})\varpi^{2}}\right\}. \tag{162.} \] Editorial note: Both denominators are printed with $(\mu+\tfrac{1}{2})\varpi^{2}$; the square on the bracket, $(\mu+\tfrac{1}{2})^{2}\varpi^{2}$, is expected as in (161.), an apparent misprint.

28.

Let us now come to the formulas (130.). To find the value of the second member, after having made $\beta=\frac{\alpha}{2n+1}$, and supposed $n$ infinite, we shall first seek the limit of the following expression: \[ t=\overset{n}{\underset{1}{\Pi}}_{m}\overset{n}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}, \tag{163.} \] where $k$ and $l$ are two quantities independent of $n$, $m$, $\mu$.

On taking the logarithm, and putting, for brevity: \[ \psi(m,\mu)=\log\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}, \tag{164.} \]

page 165

we shall have: \[ \log t=\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}.\psi(m,\mu). \tag{165.} \]

Let us first consider the expression: $\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)$. Let \[ \theta(m,\mu)=\log\left\{\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}}\right\}, \tag{166.} \] we shall have: \[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}.\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}}\right\}. \] This being premised, I say that the second member of this equation is, for every value of $m$ and $\mu$, of the form: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{v}{(2n+1)^{2}}. \]

To demonstrate this, it is necessary to distinguish two cases: whether the limit of $\frac{m\omega+\mu\varpi i}{2n+1}$ is a quantity different from zero, or whether it is equal to zero.

$a)$ In the first case we shall have, on naming $a$ the limit in question: \[ \begin{aligned}\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)&=\Phi^{2}a+v,\\ \varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)&=\Phi^{2}a+v',\\ \varphi^{2}\left(\frac{\alpha}{2n+1}\right)&=\frac{\alpha^{2}}{(2n+1)^{2}}+\frac{v'}{(2n+1)^{2}},\end{aligned} \] therefore: \[ \begin{aligned}1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}&=1-\frac{\alpha^{2}}{(2n+1)^{2}\varphi^{2}a}+\frac{v}{(2n+1)^{2}},\\ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+l}{2n+1}\right)}&=1-\frac{\alpha^{2}}{(2n+1)^{2}.\varphi^{2}a}+\frac{v'}{(2n+1)^{2}}.\end{aligned} \]

page 166

We have likewise \[ \begin{aligned}1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}&=1-\frac{\alpha^{2}}{(2n+1)^{2}\left\{\frac{m\omega+\mu\varpi i+k}{(2n+1)}\right\}}=1-\frac{\alpha^{2}}{(2n+1)^{2}a^{2}}+\frac{v}{(2n+1)^{2}},\\ 1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}&=1-\frac{\alpha^{2}}{(2n+1)^{2}a^{2}}+\frac{v'}{(2n+1)^{2}}.\end{aligned} \] Editorial note: In the first line the braced quotient in the middle denominator is printed without a square; a square is expected, as in the neighbouring terms.

On substituting these values, the expression of $\psi(m,\mu)-\theta(m,\mu)$ will take the form: \[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1-\frac{v}{(2n+1)^{2}}}{1-\frac{v'}{(2n+1)^{2}}}.\frac{1-\frac{v_{1}}{(2n+1)^{2}}}{1-\frac{v'_{1}}{(2n+1)^{2}}}\right\}, \] the quantities $v$, $v'$, $v_{1}$, $v'_{1}$ all having zero for their limit.

We have therefore: \[ \log\left(1-\frac{v}{(2n+1)^{2}}\right)=\frac{v}{(2n+1)^{2}}\ \text{etc.}, \] and consequently: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{v}{(2n+1)^{2}}. \]

$b)$ If the limit of the quantity $\frac{m\omega+\mu\varpi i}{2n+1}$ is equal to zero, we shall have: \[ \begin{aligned}\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)&=\frac{(m\omega+\mu\varpi i+k)^{2}}{(2n+1)^{2}}+A.\frac{(m\omega+\mu\varpi i+k)^{4}}{(2n+1)^{4}}+....,\\ \varphi^{2}\left(\frac{\alpha}{2n+1}\right)&=\frac{\alpha^{2}}{(2n+1)^{2}}+A'.\frac{\alpha^{4}}{(2n+1)^{4}}+....,\end{aligned} \] therefore: \[ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}=1-\frac{\alpha^{2}+A'.\frac{\alpha^{4}}{(2n+1)^{2}}+.....}{(m\omega+\mu\varpi i+k)^{2}+A.\frac{(m\omega+\mu\varpi i+k)^{4}}{(2n+1)^{2}}}. \]

If now $m\omega+\mu\varpi i$ does not go on increasing indefinitely with $n$, we shall have: \[ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+m\varpi i+k}{2n+1}\right)}=1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}+\frac{B}{(2n+1)^{2}}, \] Editorial note: Printed $m\varpi i$ in the argument of the denominator function instead of $\mu\varpi i$; an apparent misprint. likewise: \[ 1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}=1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+l)^{2}}+\frac{C}{(2n+1)^{2}}, \] Editorial note: The argument of the denominator function is printed with $+k$ (the glyph is blotted) where $+l$ is expected, as in the right-hand side. therefore in this case:

page 167

\[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1-\frac{B'}{(2n+1)^{2}}}{1-\frac{C'}{(2n+1)^{2}}}\right\}, \] $B'$ and $C'$ having finite limits, or else: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{D}{(2n+1)^{2}}, \] the limit of $D$ being likewise a finite quantity.

If on the contrary the quantity $m\omega+\mu\varpi i$ increases indefinitely with $n$, we have: \[ \begin{aligned}\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}&=1+\frac{\alpha^{2}}{(2n+1)^{2}}.\left\{\frac{A-A'\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}}{1+A\left(\frac{m\omega+\mu\varpi i+k}{2n+1}\right)^{2}}\right\}\\ &\times\frac{1}{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i+k)^{2}}};\end{aligned} \] now the quantities $\frac{1}{m\omega+\mu\varpi i+k}$, $\frac{m\omega+\mu\varpi i+k}{2n+1}$ have zero for their limit; therefore the preceding quantity will be of the form: \[ 1+\frac{\alpha^{2}}{(2n+1)^{2}}.A'', \] $A''$ having a finite quantity for its limit. On changing $k$ into $l$, and designating the corresponding value of $A''$ by $A''_{1}$, the value of $\psi(m,\mu)-\theta(m,\mu)$ will become: \[ \psi(m,\mu)-\theta(m,\mu)=\log\left\{\frac{1+\frac{\alpha^{2}}{(2n+1)^{2}}A''}{1+\frac{\alpha^{2}}{(2n+1)^{2}}A''_{1}}\right\}=\frac{\alpha^{2}(A''-A''_{1})}{(2n+1)^{2}}+\frac{v}{(2n+1)^{2}}. \] Now the limit of $A''$ is the same as that of $A$; but it is clear that this latter limit is independent of $k$, $m$, $\mu$ (it is in fact equal to the coefficient of $\alpha^{4}$ in the development of $\Phi^{2}\alpha$). Therefore we shall have: \[ A''=M+v, \] and on changing $k$ into $l$: \[ A''_{1}=M+v', \] whence $A''-A''_{1}=v-v'=v$. Therefore $A''-A''_{1}$ has zero for its limit, and consequently we have: \[ \psi(m,\mu)-\theta(m,\mu)=\frac{v}{(2n+1)^{2}}. \]

page 168

Therefore we have demonstrated that, on making \[ \psi(m,\mu)-\theta(m,\mu)=\frac{A_{m,\mu}}{(2n+1)^{2}}, \tag{167.} \] the limit of $A_{m,\mu}$ will be equal to zero every time that $m\omega+\mu\varpi i$ increases indefinitely with $n$, and that it will be equal to a finite quantity in the contrary case.

29.

This being premised, let us consider the quantity \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu). \] On substituting the value of $\psi(m,\mu)$, there will result: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)+\frac{1}{(2n+1)^{2}}.\overset{n}{\underset{1}{\Sigma}}_{\mu}(A_{m,\mu}). \tag{168.} \] Let $\nu$ be the greatest whole number contained in $\sqrt{n}$; we may put: \[ \begin{aligned}\overset{n}{\underset{1}{\Sigma}}_{\mu}A_{m,\mu}&=A_{m,1}+A_{m,2}+.....+A_{m,\nu}\\ &+A_{m,\nu+1}+A_{m,\nu+2}+.....+A_{m,n}.\end{aligned} \]

Now, from the nature of the quantities $A_{m,\mu}$, the sum contained in the first line will be equal to $\nu.A_{m}$, and the second equal to $A'_{m}(n-\nu-1)$, where $A_{m}$ is a finite quantity and $A'_{m}$ a quantity which has zero for its limit, therefore: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}A_{m,\mu}=\nu.A_{m}+(n-\nu-1)A'_{m}=(2n+1).B_{m}, \] where \[ B_{m}=\frac{\nu}{2n+1}A_{m}+\frac{n-\nu-1}{2n+1}A'_{m}. \] Therefore the quantity $B_{m}$ has zero for its limit, remarking that $\nu$ does not exceed $\sqrt{n}$.

Hereby the expression of $\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)$ is changed into: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)+\frac{B_{m}}{2n+1}. \tag{169.} \]

To have the limit of $\overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)$, I write \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)-\overset{\infty}{\underset{n+1}{\Sigma}}_{\mu}\theta(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)-\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n). \] Now the value of $\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)$ may be found as follows.

page 169

We have: \[ \begin{aligned}\theta(m.\,\mu+n)&=\log\left\{\frac{1-\frac{\alpha^{2}}{[m\omega+(\mu+n)\varpi i+k]^{2}}}{1-\frac{\alpha^{2}}{[m\omega+(\mu+n)\varpi i+l]^{2}}}\right\}\\ &=\alpha^{2}\left\{\frac{1}{[m\omega+(\mu+n)\varpi i+l]^{2}}-\frac{1}{[m\omega+(\mu+n)\varpi i+k]^{2}}\right\}\\ &-\tfrac{1}{2}\alpha^{4}\left\{\frac{1}{[m\omega+(\mu+n)\varpi i+l]^{2}}-\frac{1}{[m\omega+(\mu+n)\varpi i+k]^{2}}\right\}\\ &+\ \text{etc.}\end{aligned} \] Editorial note: Printed $\theta(m.\,\mu+n)$ with a full stop instead of a comma between the arguments; an apparent misprint. From this we derive: \[ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)=\frac{\alpha^{2}}{n}.\Sigma\frac{1}{n}.\theta\left(\frac{\mu}{n}\right)-\frac{\alpha^{4}}{2n^{3}}.\Sigma\frac{1}{n}.\theta_{1}\left(\frac{\mu}{n}\right)+...., \] where \[ \begin{gathered}\theta\left(\frac{\mu}{n}\right)=\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{2}}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{2}},\\ \theta_{1}\left(\frac{\mu}{n}\right)=\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{4}}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+\frac{\mu}{n}\varpi i\right)^{4}},\\ \text{etc.}\end{gathered} \] Now it is known that the limit of \[ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\frac{1}{n}\theta\left(\frac{\mu}{n}\right)\ \text{is equal to}\ \int_{0}^{x}\theta(x)\,\partial x, \] therefore: \[ \begin{gathered}\overset{\mu}{\underset{1}{\Sigma}}_{\mu}\frac{1}{n}\theta\left(\frac{\mu}{n}\right)=\int_{0}^{x}\theta(x)\,\partial x+v,\\ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\frac{1}{n}\theta_{1}\left(\frac{\mu}{n}\right)=\int_{0}^{x}\theta_{1}(x)\,\partial x+v_{1},\\ \text{etc},\end{gathered} \] and consequently, on substituting: \[ \overset{\mu}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)=\frac{\alpha^{2}}{n}.\int_{0}^{x}\theta(x)-\frac{\alpha^{4}}{2n^{3}}.\int_{0}^{x}\theta_{1}(x)+....+\frac{v\alpha^{2}}{n}-\frac{v_{1}\alpha^{3}}{2n^{3}}+.... \] Editorial note: The integrals are printed without the differential $\partial x$, and the last fraction has $\alpha^{3}$ where $\alpha^{4}$ is expected; both apparent misprints, reproduced as printed. where \[ \begin{gathered}\theta(x)=\frac{1}{\left(\frac{m\omega+l}{n}+\varpi i+x\varpi i\right)^{2}}-\frac{1}{\left(\frac{m\omega+k}{n}+\varpi i+x\varpi i\right)^{2}}\\ \text{etc.}\end{gathered} \] therefore we shall have:

page 170

The limit of this expression of $\displaystyle\int_{0}^{x}\theta(x)\,\partial x$ is zero for any value whatever of $x$. In the same way it will be found that the limit of $\displaystyle\int_{0}^{x}\theta_{1}(x)\,\partial x$ is zero, therefore: \[ \begin{aligned}\overset{\mu}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)&=\frac{\alpha^{2}}{n}.v-\frac{\alpha^{4}}{2n^{3}}.v'+\frac{\alpha^{6}}{3n^{5}}.v''-....\\ &=\frac{\alpha^{2}}{2n+1}.\left\{v-\frac{\alpha^{2}}{2n^{2}}v'+\frac{\alpha^{4}}{3n^{4}}v''-....\right\}\frac{2n+1}{n}=\frac{v}{2n+1},\end{aligned} \] therefore also, on making $\mu=\infty$: \[ \overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu+n)=\frac{v}{2n+1}, \] whence: \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu+n)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)-\frac{v}{2n+1},\ \text{and} \] \[ \overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)+\frac{v_{m}}{2n+1}, \tag{170.} \] $v_{m}$ having zero for its limit. From this we draw \[ \overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{n}{\underset{1}{\Sigma}}_{m}\left\{\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)\right\}+\overset{n}{\underset{1}{\Sigma}}\frac{v_{m}}{2n+1}. \] On taking the limit of the two members and remarking that \[ \overset{n}{\underset{1}{\Sigma}}\frac{v_{m}}{2n+1}=\frac{v_{1}+v_{2}+.....v_{m}}{2n+1}=v, \] there will be: \[ \text{lim.}\overset{n}{\underset{1}{\Sigma}}_{m}\overset{n}{\underset{1}{\Sigma}}_{\mu}\psi(m,\mu)=\overset{\infty}{\underset{1}{\Sigma}}_{m}\left\{\overset{\infty}{\underset{1}{\Sigma}}_{\mu}\theta(m,\mu)\right\}. \tag{171.} \] On restoring the values of $\psi(m,\mu)$ and $\theta(m,\mu)$, and passing from logarithms to numbers, we draw from it:

page 171

By an analysis quite similar to the preceding, but simpler, we shall find in the same way: \[ \text{lim.}\overset{n}{\underset{1}{\Pi}}_{m}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{m\omega+l}{2n+1}\right)}}\right\}=\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\frac{1-\frac{\alpha^{2}}{(m\omega+k)^{2}}}{1-\frac{\alpha^{2}}{(m\omega+l)^{2}}}\right\}, \tag{173.} \] \[ \text{lim.}\overset{n}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{\mu\varpi i+k}{2n+1}\right)}}{1-\frac{\varphi^{2}\left(\frac{\alpha}{2n+1}\right)}{\varphi^{2}\left(\frac{\mu\varpi i+k}{2n+1}\right)}}\right\}=\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left\{\frac{1-\frac{\alpha^{2}}{(\mu\varpi i+k)^{2}}}{1-\frac{\alpha^{2}}{(\mu\varpi i+l)^{2}}}\right\}. \tag{174.} \] Editorial note: In the left side of the last equation the lower fraction is printed with $k$ in the argument of the denominator function, as the upper one; presumably $l$ was meant, as in the right side. An apparent misprint.

30.

Now nothing is easier than to find the values of $\Phi\alpha$, $f\alpha$, $F\alpha$.

Let us consider first the first formula (130.). We have: \[ \begin{aligned}e^{2}c^{2}\Phi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}\right)&=-\frac{1}{\varphi^{2}\left(\frac{m\omega+\mu\varpi i}{2n+1}-\frac{\omega}{2}-\frac{\varpi}{2}i\right)}\\ &=-\frac{1}{\varphi^{2}\left(\frac{(n-m+\tfrac{1}{2})\omega+(n-\mu+\tfrac{1}{2})\varpi i}{2n+1}\right)},\end{aligned} \] therefore:

page 172

This being premised, if we make $\beta=\frac{\alpha}{2n+1}$ and suppose $n$ infinite, there will result, on making use of the formulas (172.), (173.), (174.), and remarking that $\alpha$ is the limit of $(2n+1)\varphi\left(\frac{\alpha}{2n+1}\right)$ and equal to $\alpha$: \[ \begin{aligned}\Phi\alpha&=\alpha\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{1-\frac{\alpha^{2}}{(m\omega)^{2}}\right\}.\overset{\infty}{\underset{1}{\Pi}}\left\{1+\frac{\alpha^{2}}{(\mu\varpi)^{2}}\right\}\\ &\times\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{(m\omega+\mu\varpi i)^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right).\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{(m\omega-\mu\varpi i)^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega-(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\right\}.\end{aligned} \tag{175.} \]

The two formulas (130.) will give in the same manner, on making $\beta=\frac{\alpha}{2n+1}$, and remarking that $f(0)=1$, $F(0)=1$: \[ \begin{aligned}f\alpha&=\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+\mu\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\left(\frac{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega-\mu\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega-(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\right\}\\ &\times\overset{\infty}{\underset{0}{\Pi}}_{\mu}\left\{1-\frac{\alpha^{2}}{(m+\tfrac{1}{2})^{2}\omega^{2}}\right\},\end{aligned} \tag{176.} \] \[ \begin{aligned}F\alpha&=\overset{\infty}{\underset{0}{\Pi}}_{\mu}\left\{1+\frac{\alpha^{2}}{(\mu+\tfrac{1}{2})\varpi^{2}}\right\}.\overset{\infty}{\underset{1}{\Pi}}_{m}\left\{\overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{\alpha^{2}}{[m\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\left(\frac{1-\frac{\alpha^{2}}{[m\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}{1-\frac{\alpha^{2}}{[(m-\tfrac{1}{2})\omega+(\mu-\tfrac{1}{2})\varpi i]^{2}}}\right)\right\}.\end{aligned} \tag{177.} \] Editorial note: The two factors of the last product are printed identically, where the earlier formulas have opposite signs; an apparent misprint, reproduced as printed. In the preceding formulas the factor $(m+\tfrac{1}{2})^{2}\omega^{2}$ of (176.) carries its square, and the product index there is printed $\mu$ although $m$ occurs; the factor $(\mu+\tfrac{1}{2})\varpi^{2}$ of (177.) is printed without a square on the bracket. All reproduced as printed.

We can also give a real form to the preceding expressions as follows:

page 173

These transformations are easily effected by means of the formula: \[ \begin{aligned} \left(1-\frac{\alpha^{2}}{(a+bi)^{2}}\right)\left(1-\frac{\alpha^{2}}{(a-bi)^{2}}\right) &= \left(1+\frac{\alpha}{a+bi}\right)\left(1+\frac{\alpha}{a-bi}\right)\left(1-\frac{\alpha}{a+bi}\right)\left(1-\frac{\alpha}{a-bi}\right)\\ &= \frac{(\alpha+a)^{2}+b^{2}}{a^{2}+b^{2}}.\frac{(\alpha-a)^{2}+b^{2}}{a^{2}+b^{2}} = \left(1+\frac{(\alpha+a)^{2}}{b^{2}}\right)\left(1+\frac{(\alpha-a)^{2}}{b^{2}}\right).\frac{1}{\left(1+\frac{a^{2}}{b^{2}}\right)^{2}}. \end{aligned} \]

31.

In what precedes we have arrived at two kinds of expressions of the functions $\Phi\alpha$, $f\alpha$, $F\alpha$: the one kind gives these functions decomposed into partial fractions, the totality of which forms infinite double series, the other gives these same functions decomposed into an infinite number of factors, each of which is in its turn composed of an infinity of factors.

Now the preceding formulas can be much simplified by means of the exponential and circular functions. This is what we shall see in what follows:

Let us consider first the equations (178.), (179.), (180.). By virtue of the known formulas, we have:

page 174

\[ \frac{\sin y}{y} = \overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(1-\frac{y^{2}}{\mu^{2}\pi^{2}}\right);\ \cos y = \overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(1-\frac{y^{2}}{(\mu-\tfrac{1}{2})^{2}\pi^{2}}\right); \] therefore: \[ \overset{\infty}{\underset{1}{\Pi}}_{\mu}\frac{\left(1-\frac{z^{2}}{\mu^{2}\pi^{2}}\right)}{\left(1-\frac{y^{2}}{(\mu-\tfrac{1}{2})\pi^{2}}\right)} = \frac{\sin z}{z\cos y};\ \overset{\infty}{\underset{1}{\Pi}}_{\mu}\left(\frac{1-\frac{z^{2}}{(\mu-\tfrac{1}{2})\pi^{2}}}{1-\frac{y^{2}}{(\mu-\tfrac{1}{2})\pi^{2}}}\right) = \frac{\cos z}{\cos y}. \] Editorial note: In both products the bracket $(\mu-\tfrac{1}{2})$ under $\pi^{2}$ is printed without the square (the denominator of the first product, and the numerator and denominator of the second); an apparent misprint for $(\mu-\tfrac{1}{2})^{2}\pi^{2}$, reproduced as printed.

By virtue of this formula it is clear that the expressions of $\Phi\alpha$, $f\alpha$, $F\alpha$ may be put under the form: \[ \begin{aligned} \Phi\alpha &= \frac{\varpi}{\pi}.\frac{\sin\left(\alpha\frac{\pi}{\varpi}i\right)}{i}.\overset{\infty}{\underset{1}{\Pi}}_{m}\left(1-\frac{\alpha^{2}}{m^{2}\omega^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\left(\frac{\sin(\alpha+m\omega)\frac{\pi}{\varpi}i.\sin(\alpha-m\omega)\frac{\pi}{\varpi}i.\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}{\cos[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\sin^{2}m\omega\frac{\pi}{\varpi}i}.\frac{\left(m\omega\frac{\pi}{\varpi}i\right)^{2}}{(\alpha+m\omega)(\alpha-m\omega).\frac{\pi^{2}}{\omega^{2}}i^{2}}\right), \end{aligned} \] Editorial note: In the denominator of the second fraction the factor $\frac{\pi^{2}}{\omega^{2}}$ is printed with $\omega$, where $\varpi$ is expected (compare the factor $\frac{\pi}{\varpi}$ elsewhere in the line); an apparent misprint, reproduced as printed. \[ \begin{aligned} f\alpha &= \overset{\alpha}{\underset{1}{\Pi}}_{m}\left(1-\frac{\alpha^{2}}{(m-\tfrac{1}{2})^{2}\omega^{2}}\right)\\ &\quad\times\overset{\infty}{\underset{1}{\Pi}}_{m}\left(\text{tang}[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\text{tang}[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cot^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i.\frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{\alpha^{2}-(m-\tfrac{1}{2})^{2}\omega^{2}}\right), \end{aligned} \] Editorial note: In the first line of the expression for $f\alpha$ the upper limit of the product is printed as $\alpha$, where $\infty$ is expected; an apparent misprint, reproduced as printed. \[ F\alpha = \cos\left(\alpha\frac{\pi}{\varpi}i\right).\overset{\infty}{\underset{1}{\Pi}}_{m}\left(\frac{\cos(\alpha+m\omega)\frac{\pi}{\varpi}i.\cos(\alpha-m\omega)\frac{\pi}{\varpi}i.\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}{\cos[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos^{2}m\omega.\frac{\pi}{\varpi}i}\right). \]

Real expressions will be found by substituting for the circular functions their expressions in exponential functions.

We have: \[ \begin{gathered} \sin(a-b).\sin(a+b) = \sin^{2}a-\sin^{2}b,\\ \cos(a+b).\cos(a-b) = \cos^{2}a-\sin^{2}b, \end{gathered} \] therefore: \[ \frac{\sin(\alpha+m\omega)\frac{\pi}{\varpi}i.\sin(\alpha-m\omega)\frac{\pi}{\varpi}i}{\sin^{2}m\omega\frac{\pi}{\varpi}i} = -\left\{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}m\omega\frac{\pi}{\varpi}i}\right\}, \] \[ \frac{\cos[\alpha+(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i.\cos[\alpha-(m-\tfrac{1}{2})\omega]\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i} = 1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}, \]

page 175

According to this, and remarking that \[ \frac{m^{2}\omega^{2}}{\alpha^{2}-m^{2}\omega^{2}} = -\frac{1}{1-\frac{\alpha^{2}}{m^{2}\omega^{2}}}\ \text{and} \] \[ \frac{(m-\tfrac{1}{2})^{2}\omega^{2}}{\alpha^{2}-(m-\tfrac{1}{2})^{2}\omega^{2}} = -\frac{1}{1-\frac{\alpha^{2}}{(m-\tfrac{1}{2})^{2}\omega^{2}}}, \] it is clear that we shall have: \[ \Phi\alpha = \frac{\pi}{\varpi}.\frac{\sin\left(\frac{\alpha}{\varpi}\pi i\right)}{i}.\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}m\omega\frac{\pi}{\varpi}i}}{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m-\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}, \tag{181.} \] Editorial note: The factor before the sine is printed as the fraction with $\pi$ over $\varpi$, whereas the preceding expression has $\varpi$ over $\pi$; an apparent misprint, reproduced as printed. \[ f\alpha = \overset{\infty}{\underset{0}{\Pi}}_{m}.\frac{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\sin^{2}(m+\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}{1-\frac{\sin^{2}\alpha\frac{\pi}{\varpi}i}{\cos^{2}(m+\tfrac{1}{2})\omega\frac{\pi}{\varpi}i}}, \tag{182.} \] \[ F\alpha = \cos\left(\frac{\alpha}{\varpi}\pi i\right).\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\frac{\sin^{2}\frac{\alpha}{\varpi}\pi i}{\cos^{2}m\frac{\omega}{\varpi}\pi i}}{1-\frac{\sin^{2}\frac{\alpha}{\varpi}\pi i}{\cos^{2}(m-\tfrac{1}{2})\frac{\omega}{\varpi}\pi i}}. \tag{183.} \]

page 176

On substituting for the cosines and sines of imaginary arcs their values in exponential quantities, these formulas will become: \[ \Phi\alpha = \tfrac{1}{2}.\frac{\varpi}{\pi}\left(h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}\right).\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1-\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{m\frac{\omega}{\varpi}\pi}-h^{-m\frac{\omega}{\varpi}\pi}}\right)^{2}}{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}+h^{-(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}, \tag{184.} \] \[ f\alpha = \overset{\infty}{\underset{0}{\Pi}}_{m}.\frac{1-\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}-h^{-(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}+h^{-(m+\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}, \tag{185.} \] \[ F\alpha = \tfrac{1}{2}\left(e^{\frac{\alpha}{\varpi}\pi}+e^{-\frac{\alpha}{\varpi}\pi}\right).\overset{\infty}{\underset{1}{\Pi}}_{m}\frac{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{m\frac{\omega}{\varpi}\pi}+h^{-m\frac{\omega}{\varpi}\pi}}\right)^{2}}{1+\left(\frac{h^{\frac{\alpha}{\varpi}\pi}-h^{-\frac{\alpha}{\varpi}\pi}}{h^{(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}+h^{-(m-\frac{1}{2})\frac{\omega}{\varpi}\pi}}\right)^{2}}, \tag{186.} \] where $h$ is the number 2,1718218..... Editorial note: The value of $h$ is printed as 2,1718218; the base of the natural logarithms is 2,7182818; an apparent misprint, reproduced as printed. Editorial note: In the first factor of the last formula the base is printed as $e$ in the sum, where the other exponentials of the page use $h$; an apparent misprint, reproduced as printed.

We can further transform this formula in the following manner:

If we replace $\alpha$ by $\alpha i$, we shall have the values of $\Phi(\alpha i)$, $f(\alpha i)$, $F(\alpha i)$. On now changing $c$ into $e$ and $e$ into $c$, the quantities: \[ \omega;\ \varpi;\ \Phi(\alpha i);\ f(\alpha i);\ F(\alpha i) \] will be changed respectively into: \[ \varpi;\ \omega;\ i\Phi\alpha;\ F\alpha;\ f\alpha, \] therefore the preceding formulas will give: \[ \Phi\alpha = \frac{\omega}{\pi}.\sin\frac{\alpha\pi}{\omega}.\overset{\infty}{\underset{1}{\Pi}}_{m}.\frac{1+\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{m\varpi\pi}{\omega}}-h^{-\frac{m\varpi\pi}{\omega}}\right)^{2}}}{1-\frac{4\sin^{2}\left(\frac{\alpha\pi}{\omega}\right)}{\left(h^{\frac{(2m-1)\varpi\pi}{2\omega}}+h^{-\frac{(2m-1)\varpi\pi}{2\omega}}\right)^{2}}}, \tag{187.} \]

page 177

32.

Let us now consider the formulas (160.), (161.), (162.).

We have: \[ \overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{(2\mu+1)\pi}{y^{2}+(\mu+\tfrac{1}{2})^{2}\pi^{2}} = \frac{2}{h^{y}+h^{-y}}, \] therefore, on making \[ y = [\alpha\pm(m+\tfrac{1}{2})\omega]\frac{\pi}{\varpi}: \] \[ \overset{\infty}{\underset{0}{\Sigma}}_{\mu}(-1)^{\mu}.\frac{(2\mu+1)\varpi}{[\alpha\pm(m+\tfrac{1}{2})\omega]^{2}+(\mu+\tfrac{1}{2})^{2}\varpi^{2}} = \frac{2\pi}{\varpi}.\frac{1}{h^{[\alpha\pm(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha\pm(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}. \] By virtue of this formula it is easy to see that the expressions (153.), (162.) of $\Phi\alpha$ and $F\alpha$ will become: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{1}{h^{[\alpha-(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha-(m-\frac{1}{2})\omega]\frac{\pi}{\varpi}}}-\frac{1}{h^{[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}\right\}, \tag{190.} \] Editorial note: In the first denominator the second exponent is printed with $m-\tfrac{1}{2}$ where the first has $m+\tfrac{1}{2}$; an apparent misprint, reproduced as printed. \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}\left\{\frac{1}{h^{[\alpha-(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha-(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}+\frac{1}{h^{[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}+h^{-[\alpha+(m+\frac{1}{2})\omega]\frac{\pi}{\varpi}}}\right\}. \tag{191.} \]

The preceding expressions of $\Phi\alpha$, $F\alpha$ may be put yet under many other forms: I shall recall the most remarkable. First, on uniting the terms of the second member, we shall find:

page 178

\[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{\left(h^{\frac{\alpha\pi}{\varpi}}-h^{-\frac{\alpha\pi}{\varpi}}\right)\left(h^{(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}-h^{-(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}\right)}{h^{\frac{2\alpha\pi}{\varpi}}+h^{-\frac{2\alpha\pi}{\varpi}}+h^{(2m+1)\frac{\omega\pi}{\varpi}}+h^{-(2m+1)\frac{\omega\pi}{\varpi}}}\right\}, \tag{192.} \] \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}.\left\{\frac{\left(h^{\frac{\alpha\pi}{\varpi}}+h^{-\frac{\alpha\pi}{\varpi}}\right)\left(h^{(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}+h^{-(m+\frac{1}{2})\frac{\omega\pi}{\varpi}}\right)}{h^{\frac{2\alpha\pi}{\varpi}}+h^{-\frac{2\alpha\pi}{\varpi}}+h^{(2m+1)\frac{\omega\pi}{\varpi}}+h^{-(2m+1)\frac{\omega\pi}{\varpi}}}\right\}. \tag{193.} \] If, for brevity, we suppose: \[ h^{\frac{\alpha\pi}{\varpi}} = \varepsilon\ \text{and}\ h^{\frac{\omega\pi}{\varpi}} = r^{2}, \tag{194.} \] these formulas, on developing the second member, will become: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}\left(\varepsilon-\frac{1}{\varepsilon}\right).\left\{\frac{r-\frac{1}{r}}{r^{2}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{2}}}-\frac{r^{3}-\frac{1}{r^{3}}}{r^{6}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{6}}}+\frac{r^{5}-\frac{1}{r^{5}}}{r^{10}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{10}}}-\ldots\right\}, \tag{195.} \] \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\left(\varepsilon+\frac{1}{\varepsilon}\right).\left\{\frac{r+\frac{1}{r}}{r^{2}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{2}}}+\frac{r^{3}+\frac{1}{r^{3}}}{r^{6}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{6}}}+\frac{r^{5}+\frac{1}{r^{5}}}{r^{10}+\varepsilon^{2}+\frac{1}{\varepsilon^{2}}+\frac{1}{r^{10}}}-\ldots\right\}. \tag{196.} \]

On putting $\alpha i$ in place of $\alpha$ in the formulas (192.), (193.), then changing $c$ into $e$ and $e$ into $c$, and remarking that the quantities \[ \omega,\ \varpi,\ \Phi(\alpha i),\ F(\alpha i),\ h^{\alpha i\frac{\pi}{\varpi}}-h^{-\alpha i\frac{\pi}{\varpi}},\ h^{\alpha i\frac{\pi}{\varpi}}+h^{\alpha i\frac{\pi}{\varpi}}, \] Editorial note: In the last quantity the sign of the second exponent is printed without a minus sign; an apparent misprint for $h^{-\alpha i\frac{\pi}{\varpi}}$, reproduced as printed. will be changed respectively into: \[ \varpi,\ \omega,\ i\Phi(\alpha),\ f\alpha,\ 2i.\sin\alpha\frac{\pi}{\omega},\ 2\cos\alpha\frac{\pi}{\omega}, \] there will result: \[ \Phi\alpha = \frac{4}{ec}.\frac{\pi}{\omega}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.\left\{\frac{\sin\frac{\alpha\pi}{\omega}.\left(h^{(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}-h^{-(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}\right)}{h^{(2m+1)\frac{\varpi\pi}{\omega}}+2\cos2\alpha\frac{\pi}{\omega}+h^{-(2m+1)\frac{\varpi\pi}{\omega}}}\right\}, \tag{197.} \] \[ f\alpha = \frac{4}{e}.\frac{\pi}{\omega}.\overset{\infty}{\underset{0}{\Sigma}}_{m}.\left\{\frac{\cos\frac{\alpha\pi}{\omega}.\left(h^{(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}+h^{-(m+\frac{1}{2})\frac{\varpi\pi}{\omega}}\right)}{h^{(2m+1)\frac{\varpi\pi}{\omega}}+2\cos2\alpha\frac{\pi}{\omega}+h^{-(2m+1)\frac{\varpi\pi}{\omega}}}\right\}. \tag{198.} \] On making for brevity \[ h^{\frac{\varpi\pi}{2\omega}} = \varrho, \tag{199.} \]

page 179

and developing, we shall obtain: \[ \varphi\left(\alpha\frac{\omega}{2}\right) = \frac{4}{ec}.\frac{\pi}{\omega}.\sin\left(\alpha\frac{\pi}{2}\right).\left\{\frac{\varrho-\frac{1}{\varrho}}{\varrho^{2}+2\cos(\alpha\pi)+\frac{1}{\varrho^{2}}}-\frac{\varrho^{3}-\frac{1}{\varrho^{3}}}{\varrho^{6}+2\cos(\alpha\pi)+\frac{1}{\varrho^{6}}}+\frac{\varrho^{5}-\frac{1}{\varrho^{5}}}{\varrho^{10}+2\cos(\alpha\pi)+\frac{1}{\varrho^{10}}}-....\right\}, \tag{200.} \] \[ f\left(\alpha\frac{\omega}{2}\right) = \frac{4}{e}.\frac{\pi}{\omega}.\cos\left(\alpha\frac{\pi}{2}\right).\left\{\frac{\varrho+\frac{1}{\varrho}}{\varrho^{2}+2\cos(\alpha\pi)+\frac{1}{\varrho^{2}}}+\frac{\varrho^{3}+\frac{1}{\varrho^{3}}}{\varrho^{6}+2\cos(\alpha\pi)+\frac{1}{\varrho^{6}}}+\frac{\varrho^{5}+\frac{1}{\varrho^{5}}}{\varrho^{10}+2\cos(\alpha\pi)+\frac{1}{\varrho^{10}}}+....\right\}. \tag{201.} \]

On substituting in the formulas (190.), (191.) for $h^{\alpha\frac{\pi}{\varpi}}$ and $h^{\frac{\omega\pi}{2\varpi}}$ their values $\varepsilon$ and $r$, there will result: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}(-m).\left\{\frac{1}{\varepsilon.r^{-(2m+1)}+\varepsilon^{-1}.r^{2m+1}}-\frac{1}{\varepsilon.r^{2m+1}+\varepsilon^{-1}.r^{-(2m+1)}}\right\}, \tag{202.} \] Editorial note: The factor is printed as $(-m)$ where $(-1)^{m}$ is expected; an apparent misprint, reproduced as printed. \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\overset{\infty}{\underset{0}{\Sigma}}_{m}\left\{\frac{1}{\varepsilon.r^{-2m-1}+\varepsilon^{-1}.r^{2m+1}}+\frac{1}{\varepsilon.r^{2m+1}+\varepsilon^{-1}.r^{-2m-1}}\right\}. \tag{203.} \]

Supposing now $\alpha<\frac{\omega}{2}$, we shall have: \[ \frac{1}{\varepsilon.r^{-2m-1}+\varepsilon^{-1}.r^{2m+1}} = \frac{\varepsilon^{-1}.r^{-2m-1}}{1+\varepsilon^{2}.r^{-4m-2}} = \varepsilon.r^{-2m-1}-\varepsilon^{3}.r^{-6m-3}+\varepsilon^{5}.r^{-10m-5}-...., \] Editorial note: In the first fraction the numerator is printed $\varepsilon^{-1}.r^{-2m-1}$ where $\varepsilon.r^{-2m-1}$ is expected (the expansion beside it begins with $\varepsilon.r^{-2m-1}$); an apparent misprint, reproduced as printed. \[ \frac{1}{\varepsilon.r^{2m+1}+\varepsilon^{-1}.r^{-2m-1}} = \frac{\varepsilon^{-1}.r^{-2m-1}}{1+\varepsilon^{-2}.r^{-4m-2}} = \varepsilon^{-1}.r^{-2m-1}-\varepsilon^{-3}.r^{-6m-3}+\varepsilon^{-5}.r^{-10m-5}-...., \] therefore: \[ \begin{aligned} \Phi\alpha &= \frac{2}{ec}.\frac{\pi}{\varpi}.\left\{\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}\left\{(\varepsilon-\varepsilon^{-1}).r^{-2m-1}-(\varepsilon^{3}-\varepsilon^{-3}).r^{-6m-3}+(\varepsilon^{5}-\varepsilon^{-5})\,r^{-10m-5}-....\right\}\right\}\\ &= \frac{2}{ec}.\frac{\pi}{\varpi}.\left\{(\varepsilon-\varepsilon^{-1}).\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{-2m-1}-(\varepsilon^{3}-\varepsilon^{-3}).\overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{-6m-3}+....\right\}. \end{aligned} \] Now: \[ \overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{-2m-1} = r^{-1}-r^{-3}+r^{-5}-.... = \frac{r^{-1}}{1+r^{-2}} = \frac{r}{r^{2}+1}, \] \[ \overset{\infty}{\underset{0}{\Sigma}}_{m}(-1)^{m}.r^{3m-3} = r^{-3}-r^{-6}+r^{-9}-.... = \frac{r^{-3}}{1+r^{-6}} = \frac{r^{3}}{r^{6}+1} \] Editorial note: The exponent in the summand is printed $3m-3$ and the series exponents as $-3$, $-6$, $-9$; the sum is evidently meant to be that of $(-1)^{m}.r^{-6m-3}$, with series $r^{-3}-r^{-9}+r^{-15}-\ldots$; apparent misprints, reproduced as printed. etc., therefore: \[ \Phi\alpha = \frac{2}{ec}.\frac{\pi}{\varpi}.\left\{\frac{\varepsilon-\varepsilon^{-1}}{r+r^{-1}}-\frac{\varepsilon^{3}-\varepsilon^{-3}}{r^{3}+r^{-3}}+\frac{\varepsilon^{5}-\varepsilon^{-5}}{r^{5}+r^{-5}}-.....\right\}. \tag{204.} \] In the same manner we shall find: \[ F\alpha = \frac{2}{c}.\frac{\pi}{\varpi}.\left\{\frac{\varepsilon+\varepsilon^{-1}}{r-r^{-1}}-\frac{\varepsilon^{3}+\varepsilon^{-3}}{r^{3}-r^{-3}}+\frac{\varepsilon^{5}+\varepsilon^{-5}}{r^{5}-r^{-5}}-.....\right\}. \tag{205.} \]

page 180

On putting $\alpha\frac{\varpi}{2}i$ in place of $\alpha$, and then changing $e$ into $c$ and $c$ into $e$, \[ \varepsilon;\ \varpi;\ \Phi\left(\alpha\frac{\varpi}{2}i\right);\ F\left(\alpha\frac{\varpi}{2}i\right);\ r;\ \varepsilon^{m}+\varepsilon^{-m};\ \varepsilon^{m}-\varepsilon^{-m} \] are changed into: \[ \varpi;\ \omega;\ i\varphi\left(\alpha\frac{\omega}{2}\right);\ f\left(\alpha\frac{\omega}{2}\right);\ \varrho;\ 2\cos\left(m\alpha\frac{\pi}{2}\right);\ 2i\sin\left(m\alpha\frac{\pi}{2}\right); \] therefore: \[ \varphi\left(\alpha\frac{\omega}{2}\right) = \frac{4}{ec}.\frac{\pi}{\omega}.\left\{\frac{\sin\left(\alpha\frac{\pi}{2}\right)}{\varrho+\frac{1}{\varrho}}-\frac{\sin\left(3\alpha\frac{\pi}{2}\right)}{\varrho^{3}+\frac{1}{\varrho^{3}}}+\frac{\sin\left(5\alpha\frac{\pi}{2}\right)}{\varrho^{5}+\frac{1}{\varrho^{5}}}-....\right\}, \tag{206.} \] \[ f\left(\alpha\frac{\omega}{2}\right) = \frac{4}{e}.\frac{\pi}{\omega}.\left\{\frac{\cos\left(\alpha\frac{\pi}{2}\right)}{\varrho-\frac{1}{\varrho}}-\frac{\cos\left(3\alpha\frac{\pi}{2}\right)}{\varrho^{3}-\frac{1}{\varrho^{3}}}+\frac{\cos\left(5\alpha\frac{\pi}{2}\right)}{\varrho^{5}-\frac{1}{\varrho^{5}}}-....\right\}. \tag{207.} \]

These last four formulas offer very simple expressions of the functions $\Phi\alpha$, $f\alpha$, $F\alpha$. By differentiations and integrations a multitude of others, more or less remarkable, may be deduced from them.

33.

In the case $e=c$, the preceding formulas take a simpler form, on account of the relation $\omega=\varpi$, which holds in this case. Let, for greater simplicity, $e=c=1$. We have: \[ r = h^{\frac{\omega\pi}{2\varpi}} = h^{\frac{\pi}{2}},\ \varrho = h^{\frac{\varpi\pi}{2\omega}} = h^{\frac{\pi}{2}}, \] therefore on substituting, and making in (204.), (205.) $\alpha=\alpha\frac{\varpi}{2}$, there results: \[ \varphi\left(\alpha\frac{\omega}{2}\right) = 2\frac{\pi}{\omega}\left\{\frac{h^{\frac{\alpha\pi}{2}}-h^{-\frac{\alpha\pi}{2}}}{h^{\frac{\pi}{2}}+h^{-\frac{\pi}{2}}}-\frac{h^{\frac{3\alpha\pi}{2}}-h^{-\frac{3\alpha\pi}{2}}}{h^{\frac{3\pi}{2}}+h^{-\frac{3\pi}{2}}}+\frac{h^{\frac{5\alpha\pi}{2}}-h^{-\frac{5\alpha\pi}{2}}}{h^{\frac{5\pi}{2}}+h^{-\frac{5\pi}{2}}}-.....\right\}, \] \[ F\left(\alpha\frac{\omega}{2}\right) = 2\frac{\pi}{\omega}\left\{\frac{h^{\frac{\alpha\pi}{2}}+h^{-\frac{\alpha\pi}{2}}}{h^{\frac{\pi}{2}}-h^{-\frac{\pi}{2}}}-\frac{h^{\frac{3\alpha\pi}{2}}+h^{-\frac{3\alpha\pi}{2}}}{h^{\frac{3\pi}{2}}-h^{-\frac{3\pi}{2}}}+\frac{h^{\frac{5\alpha\pi}{2}}+h^{-\frac{5\alpha\pi}{2}}}{h^{\frac{5\pi}{2}}-h^{-\frac{5\pi}{2}}}-.....\right\}, \] \[ \varphi\left(\alpha\frac{\omega}{2}\right) = \frac{4\pi}{\omega}\left\{\sin\left(\alpha\frac{\pi}{2}\right).\frac{h^{\frac{\pi}{2}}}{1+h^{\pi}}-\sin\left(3\alpha\frac{\pi}{2}\right).\frac{h^{\frac{3\pi}{2}}}{1+h^{3\pi}}+\sin\left(5\alpha\frac{\pi}{2}\right).\frac{h^{\frac{5\pi}{2}}}{1+h^{5\pi}}-....\right\}, \] \[ f\left(\alpha\frac{\omega}{2}\right) = \frac{4\pi}{\omega}\left\{\cos\left(\alpha\frac{\pi}{2}\right).\frac{h^{\frac{\pi}{2}}}{h^{\pi}-1}-\cos\left(3\alpha\frac{\pi}{2}\right).\frac{h^{\frac{3\pi}{2}}}{h^{3\pi}-1}+\sin\left(5\alpha\frac{\pi}{2}\right).\frac{h^{\frac{5\pi}{2}}}{h^{5\pi}-1}-....\right\}. \] Editorial note: In the last display the third term has a sine, where the pattern of the first two terms requires a cosine; an apparent misprint, reproduced as printed.

page 181

The functions $\varphi$, $f$, $F$ are determined by the equations \[ \begin{gathered} \alpha.\frac{\omega}{2} = \int_{0}^{x}\frac{\partial x}{\sqrt{(1-x^{4})}};\ \frac{\omega}{2} = \int_{0}^{1}\frac{\partial x}{\sqrt{(1-x^{4})}};\\ x = \varphi\left(\alpha\frac{\omega}{2}\right) = \sqrt{(1-x^{2})} = f\left(\alpha\frac{\omega}{2}\right) = \sqrt{(1+x^{2})} = F\left(\alpha\frac{\omega}{2}\right). \end{gathered} \]

If in the last two formulas we make $\alpha=0$, and remark that the value of $\dfrac{\varphi\left(\alpha\frac{\omega}{2}\right)}{\sin\left(\alpha\frac{\pi}{2}\right)}$ is then $=\frac{\omega}{\pi}$, and that of $\dfrac{\sin\left(m\alpha\frac{\pi}{2}\right)}{\sin\left(\alpha\frac{\pi}{2}\right)}=m$, we shall find: \[ \frac{\omega}{2} = \frac{\pi}{2}.\left\{\frac{h^{\frac{\pi}{2}}}{h^{\pi}-1}-\frac{h^{\frac{3\pi}{2}}}{h^{3\pi}-1}+\frac{h^{\frac{5\pi}{2}}}{h^{5\pi}-1}-\ldots\right\} = \int_{0}^{1}\frac{\partial x}{\sqrt{(1-x^{4})}}, \] \[ \frac{\omega^{2}}{4} = \pi^{2}.\left\{\frac{h^{\frac{\pi}{2}}}{h^{\pi}+1}-3.\frac{h^{\frac{3\pi}{2}}}{h^{3\pi}+1}+5.\frac{h^{\frac{5\pi}{2}}}{h^{5\pi}+1}-\ldots\right\} = \left(\int_{0}^{1}\frac{\partial x}{\sqrt{(1-x^{4})}}\right)^{2}. \]

(The continuation in the next number.)

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Revision history (2)

2026-10-07 — Translation (en) added (AI draft).
2026-10-07 — Transcription added (AI draft), pp. 101–181 (first part) from the 1827 Crelle's Journal scan.

@article{Abel1827Recherches,
  author = {Abel, Niels Henrik},
  title = {Recherches sur les fonctions elliptiques.},
  journal = {Journal für die reine und angewandte Mathematik (Crelle's Journal)},
  volume = {2},
  pages = {101–181},
  year = {1827},
  url = {https://gdz.sub.uni-goettingen.de/id/PPN243919689_0002}
}

@misc{Abel1827Recherches-rtm,
  author = {Abel, Niels Henrik},
  title = {Recherches sur les fonctions elliptiques. --- LaTeX transcription},
  howpublished = {ReadTheMasters, \url{https://readthemasters.org/works/abel-1827-recherches-fonctions-elliptiques/}},
  note = {AI transcription (claude-sonnet-5-5); status: ai-draft. Original: Journal für die reine und angewandte Mathematik (Crelle's Journal) 1827},
  year = {2026}
}

@misc{Abel1827Recherches-en,
  author = {Abel, Niels Henrik},
  title = {Researches on the elliptic functions. --- English translation},
  howpublished = {ReadTheMasters, \url{https://readthemasters.org/works/abel-1827-recherches-fonctions-elliptiques/}},
  note = {English translation (AI, claude-sonnet-5-5); status: ai-draft},
  language = {English},
  year = {2026}
}

Recherches sur les fonctions elliptiques.

Original (French)

Recherches sur les fonctions elliptiques.

§. I.

1.

2.

3.

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6.

7.

8.

§. II.

9.

§. III.

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11.

§. IV.

12.

13.

14.

15.

16.

17.

18.

§. V.

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20.

21.

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§. VI.

23.

§. VII.

24.

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27.

28.

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33.

English translation

Researches on elliptic functions.

§. I.

1.

2.

3.

4.

5.

6.

7.

8.

§. II.

9.

§. III.

10.

11.

§. IV.

12.

13.

14.

15.

16.

17.

18.

§. V.

19.

20.

21.

22.

§. VI.

23.

§. VII.

24.

25.

26.

27.

28.