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Question Number 57922 by maxmathsup by imad last updated on 14/Apr/19

$$decomposeinsideC\left(x\right)thefractionF\left(x\right)=\frac{1}{{(x^2+1)}^n}withnintegrnatural$$$$andn⩾1$$

Commented by maxmathsup by imad last updated on 30/Apr/19

$$wehaveF\left(x\right)=\frac{1}{{(x-i)}^n{(x+i)}^n}=\sum _{k=1}^n\frac{a_k}{{(x-i)}^k}+\sum _{k=1}^n\frac{b_k}{{(x+i)}^k}$$$$fromanothersidech.x-i=tgiveF\left(x\right)=G\left(t\right)=\frac{1}{t^n{(t+2i)}^n}$$$$letdetermine{D}_{n-1}\left(0\right)foru\left(t\right)=\frac{1}{{(t+2i)}^n}\Rightarrow u\left(t\right)=\sum _{k=0}^{n-1}\frac{u^{\left(k\right)}\left(0\right)}{k!}{t}^k+\frac{t^n}{n!}\xi \left(t^n\right)$$$$wehaveu\left(t\right)={(t+2i)}^{-n}\Rightarrow u^{\left(1\right)}\left(t\right)=(-n){(t+2i)}^{-n-1}\Rightarrow$$$$u^{\left(2\right)}\left(t\right)={(-1)}^{2}n(n+1){(t+2i)}^{-n-2}\Rightarrow u^{\left(p\right)}\left(t\right)={(-1)}^{p}n(n+1)...(n+p-1){(t+2i)}^{-n-p}$$$$\Rightarrow u^{\left(k\right)}\left(0\right)={(-1)}^{k}n(n+1)...(n+k-1){\left(2i\right)}^{-n-k}\Rightarrow$$$$u\left(t\right)=\sum _{k=0}^{n-1}\frac{{(-1)}^{k}n(n+1)....(n+k-1)}{k!{\left(2i\right)}^{n+k}}{t}^k+\frac{t^n}{n!}\xi \left(t^n\right)\Rightarrow$$$$\frac{u\left(t\right)}{t^n}=\sum _{k=0}^{n-1}\frac{{(-1)}^{k}n(n+1)....(n+k-1)}{{\left(2i\right)}^{n+k}k!t^{n-k}}+\frac{1}{n!}\xi \left(1\right)(n-k=p)$$$$=\sum _{p=1}^n\frac{{(-1)}^{n-p}n(n+1)....(n+n-p)}{(n-p)!t^p{\left(2i\right)}^{2n-p}}+\frac{1}{n!}\xi \left(1\right)$$$$=\sum _{p=1}^n\frac{{(-1)}^{n-p}n(n+1)....(2n-p)}{(n-p)!{(x-i)}^p{\left(2i\right)}^{2n-p}}+\frac{1}{n!}\xi \left(1\right)\Rightarrow$$$$a_k=\frac{{(-1)}^{n-k}n(n+1)....(2n-k)}{(n-k)!{\left(2i\right)}^{2n-k}}nowletfind{b}_{k}weusech.x+i=t\Rightarrow$$$$F\left(x\right)=H\left(t\right)=\frac{1}{t^n{(t-2i)}^n}letdetermine{D}_{n-1}\left(0\right)forv\left(t\right)={(t-2i)}^{-n}\Rightarrow$$$$v\left(t\right)=\sum _{k=0}^{n-1}\frac{v^{\left(k\right)}\left(0\right)}{k!}{t}^k+t^n\xi \left(t^n\right)but{v}^{\left(k\right)}\left(0\right)=\frac{{(-1)}^{k}n(n+1)....(n+k-1)}{{(-2i)}^{n+k}}\Rightarrow$$$$\frac{v\left(t\right)}{t^n}=\sum _{p=1}^n\frac{{(-1)}^{n-p}n(n+1)...(2n-p)}{(n-p)!t^p{(-2i)}^{2n-p}}+\frac{1}{n!}\xi \left(1\right)\Rightarrow$$$$b_k=\frac{{(-1)}^{n-k}n(n+1)....(2n-k)}{(n-k)!{(-2i)}^{2n-k}}\Rightarrow$$$$F\left(x\right)=\sum _{k=1}^n\frac{{(-1)}^{n-k}n(n+1)....(2n-k)}{(n-k)!{\left(2i\right)}^{2n-k}{(x-i)}^k}+\sum _{k=1}^n\frac{{(-1)}^{n-k}n(n+1)...(2n-k)}{(n-k)!{(-2i)}^{2n-k}{(x+i)}^k}$$$$andweseethat{b}_k=conj\left(a_k\right)sowecansearchonly{a}_k...$$$$erroroftypingchange\xi \left(t^n\right)by\xi \left(t\right)...$$

Commented by maxmathsup by imad last updated on 30/Apr/19

$$F\left(x\right)=\sum _{k=1}^n\frac{{(-1)}^{n-k}n(n+1)...(2n-k-1)}{{\left(2i\right)}^{}(n-k)!{(x-i)}^k}+\sum _{k=1}^n\frac{{(-1)}^{n-k}n(n+1)...(2n-k-1)}{{(-2\mathit{i})}^{2\mathit{n}-\mathit{k}}(\mathit{n}-\mathit{k})!{(x+i)}^k}$$

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