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Question Number 74499 by mathmax by abdo last updated on 25/Nov/19

decompose inside C(x) the fraction f(x)=(1/((x^2 +1)^n ))

$${decompose}\:{inside}\:{C}\left({x}\right)\:{the}\:{fraction} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$

Commented by mathmax by abdo last updated on 28/Nov/19

f(x)=(1/((x^2 +1)^n )) =(1/((x−i)^n (x+i)^n )) =Σ_(k=1) ^n (a_k /((x−i)^k )) +Σ_(k=1) ^n (b_k /((x+i)^k )) ch. x−i =t give f(x)=g(t)=(1/(t^n (t+2i)^n )) let find D_(n−1) (o) for h(t)=(1/((t+2i)^n )) =(t+2i)^(−n) we have h(t)=Σ_(k=0) ^(n−1) ((h^((k)) (0))/(k!)) t^k +(t^n /(n!))ξ(t) we have {(t+2i)^(−n) }^((1)) =(−n) (t+2i)^(−n−1) {(t+2i)^(−n) }^((2)) =(−n)(−n−1)(t+2i)^(−n−2) =(−1)^2 n(n+1) (t+2i)^(−n−2) { (t+2i)^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(t+2i)^(−n−k) ⇒ h^((k)) (0) =(−1)^k n(n+1)....(n+k−1)×(2i)^(−n−k) ⇒ h(t)=Σ_(k=0) ^(n−1) (((−1)^k n(n+1)...(n+k−1)(2i)^(−n−k) )/(k!)) t^k +(t^n /(n!))ξ(t) ⇒ g(t)=((h(t))/t^n ) =Σ_(k=0) ^(n−1) (((−1)^k n(n+1)....(n+k−1)(2i)^(−n−k) )/(k! t^(n−k) )) +(1/(n!))ξ(t) =_(n−k =j) Σ_(j=1) ^n (((−1)^(n+j) n(n+1)....(2n−j−1)(2i)^(−2n+j) )/((n−j)! t^j )) +(1/(n!))ξ(t) t=x−i ⇒ a_j =(((−1)^(n+j) n(n+1)....(2n−j−1)(2i)^(−2n+j) )/((n−j)!)) we have f^− =f ⇒Σ_(k=1) ^n (a_k ^− /((x+i)^k )) +Σ_(k=1) ^n (b_k ^− /((x−i)^k )) =f ⇒ b_k =a_k ^− ⇒ b_j =(((−1)^(n+j) n(n+1)....(2n−j−1)(−2i)^(−2n+j) )/((n−j)!))

$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:=\frac{\mathrm{1}}{\left({x}−{i}\right)^{{n}} \left({x}+{i}\right)^{{n}} }\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{a}_{{k}} }{\left({x}−{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{b}_{{k}} }{\left({x}+{i}\right)^{{k}} } \\ $$$${ch}.\:{x}−{i}\:={t}\:{give}\:{f}\left({x}\right)={g}\left({t}\right)=\frac{\mathrm{1}}{{t}^{{n}} \left({t}+\mathrm{2}{i}\right)^{{n}} }\:\:{let}\:{find}\:{D}_{{n}−\mathrm{1}} \left({o}\right)\:{for} \\ $$$${h}\left({t}\right)=\frac{\mathrm{1}}{\left({t}+\mathrm{2}{i}\right)^{{n}} }\:=\left({t}+\mathrm{2}{i}\right)^{−{n}} \:\:\:{we}\:{have} \\ $$$${h}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{{h}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}\right) \\ $$$${we}\:{have}\:\left\{\left({t}+\mathrm{2}{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} =\left(−{n}\right)\:\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{1}} \\ $$$$\left\{\left({t}+\mathrm{2}{i}\right)^{−{n}} \right\}^{\left(\mathrm{2}\right)} =\left(−{n}\right)\left(−{n}−\mathrm{1}\right)\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{2}} \:=\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\:\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{2}} \\ $$$$\left\{\:\left({t}+\mathrm{2}{i}\right)^{−{n}} \right\}^{\left({k}\right)} \:=\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)...\left({n}+{k}−\mathrm{1}\right)\left({t}+\mathrm{2}{i}\right)^{−{n}−{k}} \:\Rightarrow \\ $$$${h}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{k}} \:{n}\left({n}+\mathrm{1}\right)....\left({n}+{k}−\mathrm{1}\right)×\left(\mathrm{2}{i}\right)^{−{n}−{k}} \:\Rightarrow \\ $$$${h}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)...\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} }{{k}!}\:{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}\right)\:\Rightarrow \\ $$$${g}\left({t}\right)=\frac{{h}\left({t}\right)}{{t}^{{n}} }\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)....\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} }{{k}!\:{t}^{{n}−{k}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left({t}\right) \\ $$$$=_{{n}−{k}\:={j}} \:\:\sum_{{j}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{n}+{j}} {n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}−{j}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{2}{n}+{j}} }{\left({n}−{j}\right)!\:{t}^{{j}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left({t}\right) \\ $$$${t}={x}−{i}\:\Rightarrow\:{a}_{{j}} =\frac{\left(−\mathrm{1}\right)^{{n}+{j}} {n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}−{j}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{2}{n}+{j}} }{\left({n}−{j}\right)!} \\ $$$${we}\:{have}\:\:\overset{−} {{f}}={f}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\overset{−} {{a}}_{{k}} }{\left({x}+{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\overset{−} {{b}}_{{k}} }{\left({x}−{i}\right)^{{k}} }\:={f}\:\Rightarrow \\ $$$${b}_{{k}} =\overset{−} {{a}}_{{k}} \:\Rightarrow \\ $$$${b}_{{j}} =\frac{\left(−\mathrm{1}\right)^{{n}+{j}} {n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}−{j}−\mathrm{1}\right)\left(−\mathrm{2}{i}\right)^{−\mathrm{2}{n}+{j}} }{\left({n}−{j}\right)!} \\ $$

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