Formelsammlung Mathematik: Identitäten: Partialbruchzerlegungen

Zurück zu Identitäten

1

${\frac {\alpha ^{n}\,z^{m-1}}{\alpha ^{n}-z^{n}}}={\frac {1}{n}}\sum _{k=0}^{n-1}{\frac {(\alpha \xi ^{k})^{m}}{\alpha \xi ^{k}-z}}\qquad \xi =\exp \left({\frac {2\pi i}{n}}\right)$

ohne Beweis

2

${\frac {1}{(x-\alpha )^{n+1}\,(x-\beta )^{m+1}}}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{(\alpha -\beta )^{m+1+k}}}\,{\frac {(m+k)!}{m!\,k!}}\,{\frac {1}{(x-\alpha )^{n+1-k}}}+\sum _{k=0}^{m}{\frac {(-1)^{k}}{(\beta -\alpha )^{n+1+k}}}\,{\frac {(n+k)!}{n!\,k!}}\,{\frac {1}{(x-\beta )^{m+1-k}}}$

Beweis

${\frac {1}{x-\alpha }}\cdot {\frac {1}{x-\beta }}=-{\frac {1}{\beta -\alpha }}\cdot {\frac {1}{x-\alpha }}+{\frac {1}{\beta -\alpha }}\cdot {\frac {1}{x-\beta }}\qquad \left|\left({\frac {d}{d\alpha }}\right)^{n}\right.$

${\frac {n!}{(x-\alpha )^{n+1}}}\cdot {\frac {1}{x-\beta }}=-\sum _{k=0}^{n}{n \choose k}\cdot {\frac {k!}{(\beta -\alpha )^{k+1}}}\cdot {\frac {(n-k)!}{(x-\alpha )^{n-k+1}}}+{\frac {n!}{(\beta -\alpha )^{n+1}}}\cdot {\frac {1}{x-\beta }}$

${\frac {1}{(x-\alpha )^{n+1}}}\cdot {\frac {1}{x-\beta }}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{(\alpha -\beta )^{k+1}}}\cdot {\frac {1}{(x-\alpha )^{n+1+k}}}+{\frac {(-1)^{n+1}}{(\alpha -\beta )^{n+1}}}\cdot {\frac {1}{x-\beta }}\qquad \left|\left({\frac {d}{d\beta }}\right)^{m}\right.$

${\frac {1}{(x-\alpha )^{n+1}}}\cdot {\frac {m!}{(x-\beta )^{m+1}}}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{(\alpha -\beta )^{m+k+1}}}\cdot {\frac {(m+k)!}{k!}}\cdot {\frac {1}{(x-\alpha )^{n+1+k}}}+(-1)^{n+1}\cdot \sum _{k=0}^{m}{m \choose k}\cdot {\frac {1}{(\alpha -\beta )^{n+k+1}}}\cdot {\frac {(n+k)!}{n!}}\cdot {\frac {(m-k)!}{(x-\beta )^{m+1-k}}}$

${\frac {1}{(x-\alpha )^{n+1}\,(x-\beta )^{m+1}}}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{(\alpha -\beta )^{m+1+k}}}\,{\frac {(m+k)!}{m!\,k!}}\,{\frac {1}{(x-\alpha )^{n+1-k}}}+\sum _{k=0}^{m}{\frac {(-1)^{k}}{(\beta -\alpha )^{n+1+k}}}\,{\frac {(n+k)!}{n!\,k!}}\,{\frac {1}{(x-\beta )^{m+1-k}}}$

3

${\frac {t^{m}}{\prod \limits _{k=1}^{n}(k^{2}+t)}}=2\cdot \sum _{k=1}^{n}{\frac {(-1)^{m+k-1}\,k^{2m+2}}{(n+k)!\,(n-k)!}}\cdot {\frac {1}{k^{2}+t}}\qquad m<n$

Beweis

Ansatz: ${\frac {t^{m}}{(1+t)(2^{2}+t)\cdots (n^{2}+t)}}={\frac {C_{1}}{1+t}}+{\frac {C_{2}}{2^{2}+t}}+...+{\frac {C_{n}}{n^{2}+t}}$

Den Koeffizienten $C_{k}\,(1\leq k\leq n)$ erhält man, wenn man beide Seiten mit $(k^{2}+t)$ durchmultipliziert und $t\,$ gegen $-k^{2}$ gehen lässt.

$C_{k}={\frac {(-k^{2})^{m}}{\prod \limits _{\ell =1}^{k-1}(\ell ^{2}-k^{2})\cdot \prod \limits _{\ell =k+1}^{n}(\ell ^{2}-k^{2})}}={\frac {(-1)^{m}\cdot k^{2m}}{\prod \limits _{\ell =1}^{k-1}(\ell +k)\cdot \prod \limits _{\ell =1}^{k-1}(\ell -k)\cdot \prod \limits _{\ell =k+1}^{n}(\ell +k)\cdot \prod \limits _{\ell =k+1}^{n}(\ell -k)}}$

$={\frac {(-1)^{m}\cdot k^{2m}}{{\frac {(2k-1)!}{k!}}\cdot (-1)^{k-1}\,(k-1)!\cdot {\frac {(n+k)!}{(2k)!}}\cdot (n-k)!}}=2\cdot {\frac {(-1)^{m+k-1}\cdot k^{2m+2}}{(n+k)!\,(n-k)!}}$

4

${\frac {t^{m}}{\prod \limits _{k=0}^{n}{\Big (}(2k+1)^{2}+t{\Big )}}}=\sum _{k=0}^{n}{\frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}\cdot {\frac {1}{(2k+1)^{2}+t}}\qquad m\leq n$

Beweis

Ansatz: ${\frac {t^{m}}{(1+t)(3^{2}+t)\cdots {\Big (}(2n+1)^{2}+t{\Big )}}}={\frac {C_{1}}{1+t}}+{\frac {C_{3}}{3^{2}+t}}+...+{\frac {C_{2n+1}}{(2n+1)^{2}+t}}$

Den Koeffizienten $C_{2k+1}\,(0\leq k\leq n)$ erhält man, wenn man beide Seiten mit ${\Big (}(2k+1)^{2}+t{\Big )}$ durchmultipliziert und $t\,$ gegen $-(2k+1)^{2}$ gehen lässt.

$C_{2k+1}={\frac {{\Big (}-(2k+1)^{2}{\Big )}^{m}}{\prod \limits _{\ell =0}^{k-1}{\Big (}(2\ell +1)^{2}-(2k+1)^{2}{\Big )}\cdot \prod \limits _{\ell =k+1}^{n}{\Big (}(2\ell +1)^{2}-(2k+1)^{2}{\Big )}}}$

$={\frac {(-1)^{m}\cdot (2k+1)^{2m}}{\prod \limits _{\ell =0}^{k-1}(2\ell +2k+2)\cdot \prod \limits _{\ell =0}^{k-1}(2\ell -2k)\cdot \prod \limits _{\ell =k+1}^{n}(2\ell +2k+2)\cdot \prod \limits _{\ell =k+1}^{n}(2\ell -2k)}}$

$={\frac {(-1)^{m}\cdot (2k+1)^{2m}}{2^{k}\cdot {\frac {(2k)!}{k!}}\cdot 2^{k}\cdot (-1)^{k}\cdot k!\cdot 2^{n-k}\cdot {\frac {(n+k+1)!}{(2k+1)!}}\cdot 2^{n-k}\cdot (n-k)!}}={\frac {(-1)^{m+k}\,(2k+1)^{2m+1}}{2^{2n}\,(n+k+1)!\,(n-k)!}}$