Steps / Historical AI

Leonhard Euler · 1744 · Internet Archive

Methodus inveniendi lineas curvas

All 336 leaves · leaf 125

Transcription

118 DE METHODO MAX. ET MIN. $+ [L][P] - \frac{[Q]d[L]+2[L]d[Q]}{dx} + N - \frac{d[P]}{dx} + \frac{dd[Q]}{dx^2}$, seu $\mathfrak{d}=Vdx$, erit sequens: $nv. dx(L'''dx + L''''dx(1+[L''']dx) + L'''''dx(1+[L''']dx)(1+[L'''']dx) + L''''''dx(1+[L''']dx)(1+[L'''']dx)(1+[L''''']dx) + \&c.)V$. Quamobrem hujus seriei summa est indaganda, hunc in finem, scribamus $L$ loco $L'''$, & $[L]$ loco $[L''']$, sitque summa, quam quaerimus, $=S$: erit $S = Ldx + L'dx(1+[L]dx) + L''dx(1+[L]dx)(1+[L']dx) + L'''dx(1+[L]dx)(1+[L']dx)(1+[L'']dx) + \&c.$ Jam ipsius $S$ sumatur valor sequens $S' = S+dS$ erit $S+dS = L'dx + L''dx(1+[L']dx) + L'''dx(1+[L']dx)(1+[L'']dx) + \&c.$ Hincque $-dS = Ldx + L'[L]dx^2 + [L]dx \cdot L'dx(1+[L']dx) + [L]dx \cdot L''dx(1+[L']dx)(1+[L'']dx) + \&c.$ quae series cum ad priorem reduci queat, erit $-dS = Ldx+S[L]dx$, sive ob $S'=S$, $dS+S[L]dx = -Ldx$; quae integrata dat $e^{\int [L]dx} S = C - \int e^{\int [L]dx} Ldx$; quae constans $C$ ita debet accipi, ut posito $x=a$ fiat $S=0$. Hanc ob rem erit valor illius seriei $S = e^{-\int [L]dx} (C - \int e^{\int [L]dx} Ldx)$. Ex his igitur formulae propositae $\int Zdx$ orietur sequens valor differentialis: $nv. dx(N - \frac{dP}{dx} + \frac{ddQ}{dx^2} + L[L][Q] + L[P] - \frac{[Q]dL+2Ld[Q]}{dx} + S([L'][Q] + [L][P] - \frac{[Q]d[L]+2[L]d[Q]}{dx} + [N] - \frac{d[P]}{dx} + \frac{dd[Q]}{dx^2}))$, qui transmutatur in hanc formam commodiorem, $nv. dx(N - \frac{dP}{dx} + \frac{ddQ}{dx^2} + [N]S - \frac{d([P]S)}{dx} + \frac{dd([Q]S)}{dx^2})$. Hinc autem formari potest valor differentialis formulae $\int Zdx$, si tam in $Z$ quam in $[Z]$ differentialia
confidence 0.35 27 expressions source scan at the Internet Archive

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