Steps / Historical AI

Leonhard Euler · 1748 · Internet Archive

Introductio in analysin infinitorum

All 320 leaves · leaf 65

Transcription

PER SUBSTITUTIONEM. 37 CAP. III. 48. Si fuerit $y=(a+bz)^{m:n}$; nova variabilis $x$, per quam tam $y$ quam $z$ rationaliter exprimatur, sic reperietur. Ponatur $y=x^m$, fietque $(a+bz)^{m:n}=x^m$ ideoque $(a+bz)^{1:n}=x$: ergo $a+bz=x^n$ & $z=\frac{x^n-a}{b}$. Sic ergo utraque quantitas $y$ & $z$ rationaliter per $x$ definietur, ope scilicet substitutionis $z=\frac{x^n-a}{b}$, quae praebet $y=x^m$. Quamvis igitur neque $y$ per $z$, neque vicissim $z$ per $y$ rationaliter exprimi possit; tamen utraque reddita est Functio rationalis novae quantitatis variabilis $x$ per substitutionem introducta, scopo substitutionis omnino convenienter. 49. Si fuerit $y=\left(\frac{a+bz}{f+gz}\right)^{m:n}$; requiritur nova quantitas variabilis $x$ per quam utraque $y$ & $z$ rationaliter exprimatur. Manifestum primo est si ponatur $y=x^m$, quaesito satisfieri; erit enim $\left(\frac{a+bz}{f+gz}\right)^{m:n}=x^m$, ideoque $\frac{a+bz}{f+gz}=x^n$; ex qua aequatione elicitur $z=\frac{a-fx^n}{gx^n-b}$; quae substitutio praebet $y=x^m$. Hinc quoque intelligitur si fuerit $\left(\frac{\alpha+\beta y}{\gamma+\delta y}\right)^n = \left(\frac{a+bz}{f+gz}\right)^m$, tam $y$ quam $z$ rationaliter per $x$ expressum iri, si utraque formula ponatur $=x^{mn}$; reperietur enim $y=\frac{\alpha-\gamma x^m}{\delta x^m-\beta}$ & $z=\frac{a-fx^n}{gx^n-b}$: qui casus nil habent difficultatis. 50. Si fuerit $y=\sqrt{((a+bz)(c+dz))}$; substitutio idonea invenietur, qua $y$ & $z$ rationaliter exprimuntur, hoc modo. Ponatur $\sqrt{((a+bz)(c+dz))}=(a+bz)x$, facile enim perspicitur hinc valorem rationalem pro $z$ esse proditurum; quia valor ipsius $z$ per aequationem simplicem determinatur. Erit ergo E 3
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