Pagina:Werkecarlf03gausrich.djvu/53

quorum altero saltem ad propositum nostrum non possumus carere, solidiori fundamento superstruemus: a secundo, tamquam faciliori, initium faciemus.

8.

Denotemus per $\rho$ functionem $${\begin{array}{rl}&{\frac {\pi (x-b)(x-c)(x-d)\dots }{(a-b)^{2}(a-c)^{2}(a-d)^{2}\dots }}\\+&{\frac {\pi (x-a)(x-c)(x-d)\dots }{(b-a)^{2}(b-c)^{2}(b-d)^{2}\dots }}\\+&{\frac {\pi (x-a)(x-b)(x-d)\dots }{(c-a)^{2}(c-b)^{2}(c-d)^{2}\dots }}\\+&{\frac {\pi (x-a)(x-b)(x-c)\dots }{(d-a)^{2}(d-b)^{2}(d-c)^{2}\dots }}\\+&{\text{ etc. }}\end{array}}$$ quae, quoniam $\pi$ per singulos denominatores est divisibilis, fit functio integra indeterminatarum $x,$ $a,$ $b,$ $c$ etc. Statuamus porro ${\frac {d{\mathfrak {u}}}{dx}}={\mathfrak {u}}',$ ita ut habeatur $${\begin{array}{rl}{\mathfrak {u}}'=&(x-b)(x-c)(x-d)\dots \\&+(x-a)(x-c)(x-d)\\&+(x-a)(x-b)(x-c)\\&+(x-a)(x-c)(x-d)\\&+{\text{ etc.}}\end{array}}$$ Manifesto pro $x=a$, fit $\rho {\mathfrak {u}}'=\pi ,$ unde concludimus, functionem $\pi -\rho {\mathfrak {u}}'$ indefinite divisibilem esse per $x-a,$ et perinde per $x-b,$ $x-c$ etc., nec non per productum ${\mathfrak {u}}$. Statuendo itaque $${\frac {\pi -\rho {\mathfrak {u}}'}{\mathfrak {u}}}=\sigma$$ erit $\sigma$ functio integra indeterminatarum $x,$ $a,$ $b,$ $c$ etc., et quidem, perinde ut $\rho ,$ symmetrica ratione indeterminatarum $a,$ $b,$ $c$ etc. Erui poterunt itaque functiones duae integrae $r,$ $s,$ indeterminatarum $x,$ $l',$ $l'',$ $l'''$ etc., tales quae per substitutiones $l'=\lambda ',$ $l''=\lambda '',$ $l'''=\lambda '''$ etc. transeant in $\rho ,$ $\sigma$ resp. Quodsi itaque analogiam sequentes, functionem $$mx^{m-1}-(m-1)l'x^{m-2}+(m-2)l''x^{m-3}-(m-3)l'''x^{m-4}+{\text{ etc. }}$$ i.e. quotientem differentialem ${\frac {dy}{dx}}$ per $y'$ denotemus, ita ut $y'$ per easdem illas