Pagina:Werkecarlf03gausrich.djvu/185

Si igitur per evolutionem ipsius ${\textstyle y}$ in seriem secundum potestates ipsius ${\textstyle t-{\frac {1}{2}}}$ progredientem prodit $$y=L+L^{\prime }(t-{\frac {1}{2}})+L^{\prime \prime }(t-{\frac {1}{2}})^{2}+L^{\prime \prime \prime }(t-{\frac {1}{2}})^{3}+{\text{etc. }}$$ correctio valori approximato integralis ${\textstyle \int y\operatorname {d} t}$ a ${\textstyle t=0}$ usque ad ${\textstyle t=1}$ applicanda erit $$Ll+L^{\prime \prime }l^{\prime \prime }+L^{\prime \prime \prime }l^{\prime \prime \prime }+L^{\scriptscriptstyle VI}l^{\scriptscriptstyle VI}+{\text{etc. }}$$ aut potius, quum ${\textstyle l^{(m)}}$ necessario evanescat pro valore quovis integro ipsius ${\textstyle m}$ haud maiori quam ${\textstyle n,}$ correctio erit $$L^{(n+2)}l^{(n+2)}+L^{(n+4)}l^{(n+4)}+L^{(n+6)}l^{(n+6)}+{\text{etc. }}$$ pro ${\textstyle n}$ pari, vel $$L^{(n+1)}l^{(n+1)}+L^{(n+3)}l^{(n+3)}+L^{(n+5)}l^{(n+5)}+{\text{etc. }}$$ pro ${\textstyle n}$ impari.

Facillime iam correctiones ${\textstyle l^{(m)}}$ ad ${\textstyle k^{(m)}}$ reducuntur et vice versa. Quum enim habeatur $$(t-{\frac {1}{2}})^{m}=t^{m}-{\frac {1}{2}}m.t^{m-1}+{\frac {1}{4}}\cdot {\frac {m(m-1)}{1.2}}t^{m-2}+{\text{etc. }}$$ erit $$l^{(m)}=k^{(m)}-{\frac {1}{2}}mk^{(m-1)}+{\frac {1}{4}}\cdot {\frac {m(m-1)}{1.2}}k^{(m-2)}+{\text{etc. }}$$

Et perinde fit $$k^{(m)}=l^{(m)}+{\frac {1}{2}}ml^{(m-1)}+{\frac {1}{4}}\cdot {\frac {m(m-1)}{1.2}}l^{(m-1)}+{\text{etc. }}$$ Ex posteriori formula eiicientur termini, ubi ${\textstyle l}$ afficitur indice impari: utraque autem continuanda est tantummodo usque ad indicem ${\textstyle n+1}$ (inclus.). Manifesto itaque habebimus

pro ${\textstyle n}$ pari${\textstyle {\phantom {k^{(n+3)}={\frac {n+3}{2}}.l^{(n+2)}}}}$
${\textstyle {\begin{aligned}&k^{(n+1)}=0\\&k^{(n+2)}=l^{(n+2)}\\&k^{(n+3)}={\frac {n+3}{2}}.l^{(n+2)}\end{aligned}}}$
pro ${\textstyle n}$ impari${\textstyle {\phantom {k^{(n+3)}={\frac {n+3}{2}}.l^{(n+2)}}}}$

${\textstyle {\begin{aligned}&k^{(n+1)}=l^{(n+1)}\\&k^{(n+2)}={\frac {n+2}{2}}.l^{(n+1)}\end{aligned}}}$

unde demanant observationes supra indicatae.