Steps / Historical AI

Leonhard Euler · 1748 · Internet Archive

Introductio in analysin infinitorum

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Transcription

FRACTARUM EVOLUTIONE. 171 CAP. XII. $B = \frac{RN+rN}{N^2+n^2} + \frac{RN-rN}{N^2+n^2} \frac{\cos.\Phi}{\sin.\Phi}$ $B = -\frac{RN+rN}{N^2+n^2} \cdot \frac{q}{p\sin.\Phi}$ 209. Hinc jam generaliter concludere licet quomodo resolutio institui debeat, si denominator Functionis propositae $\frac{M}{N}$, Factorem habeat $(pp - 2pqz.\cos.\Phi + qqzz)^k$: fit enim $N=(pp-2pqz.\cos.\Phi+qqzz)^k Z$, ita ut haec resolvenda fit Functio fracta $\frac{M}{(pp-2pqz.\cos.\Phi+qqzz)^k Z}$. Praebeat ergo Factor denominatoris $(pp-2pqz.\cos.\Phi+qqzz)^k$ has partes: $\frac{A+Az}{(pp-2pqz.\cos.\Phi+qqzz)^k} + \frac{B+Bz}{(pp-2pqz.\cos.\Phi+qqzz)^{k-1}} + \frac{C+Cz}{(pp-2pqz.\cos.\Phi+qqzz)^{k-2}} + \frac{D+Dz}{(pp-2pqz.\cos.\Phi+qqzz)^{k-3}} + \&c.$ Jam, posito $z^n = \frac{p^n}{q^n} . \cos.n\Phi$, fit $M=M$, & $z=N$; &, posito $z^n = \frac{p^n}{q^n} . \sin.n\Phi$, fit $M=M$; & $z=N$; erit $A = \frac{MN+mN}{N^2+n^2} + \frac{MN-mN}{N^2+n^2} \frac{\cos.\Phi}{\sin.\Phi}$ $A = -\frac{MN+mN}{N^2+n^2} \cdot \frac{q}{p\sin.\Phi}$ Deinde vocetur $\frac{M-(A+Az)Z}{pp-2pqz.\cos.\Phi+qqzz} = P$; atque, posito
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