Formelsammlung Mathematik: Identitäten: Partialbruchzerlegungen

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

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

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

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

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

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

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

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

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

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

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

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

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

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