Question Number 74499 by mathmax by abdo last updated on 25/Nov/19
$${decompose}\:{inside}\:{C}\left({x}\right)\:{the}\:{fraction} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$
Commented by mathmax by abdo last updated on 28/Nov/19
$${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:=\frac{\mathrm{1}}{\left({x}−{i}\right)^{{n}} \left({x}+{i}\right)^{{n}} }\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{a}_{{k}} }{\left({x}−{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{b}_{{k}} }{\left({x}+{i}\right)^{{k}} } \\ $$$${ch}.\:{x}−{i}\:={t}\:{give}\:{f}\left({x}\right)={g}\left({t}\right)=\frac{\mathrm{1}}{{t}^{{n}} \left({t}+\mathrm{2}{i}\right)^{{n}} }\:\:{let}\:{find}\:{D}_{{n}−\mathrm{1}} \left({o}\right)\:{for} \\ $$$${h}\left({t}\right)=\frac{\mathrm{1}}{\left({t}+\mathrm{2}{i}\right)^{{n}} }\:=\left({t}+\mathrm{2}{i}\right)^{−{n}} \:\:\:{we}\:{have} \\ $$$${h}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{{h}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}\right) \\ $$$${we}\:{have}\:\left\{\left({t}+\mathrm{2}{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} =\left(−{n}\right)\:\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{1}} \\ $$$$\left\{\left({t}+\mathrm{2}{i}\right)^{−{n}} \right\}^{\left(\mathrm{2}\right)} =\left(−{n}\right)\left(−{n}−\mathrm{1}\right)\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{2}} \:=\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\:\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{2}} \\ $$$$\left\{\:\left({t}+\mathrm{2}{i}\right)^{−{n}} \right\}^{\left({k}\right)} \:=\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left({t}+\mathrm{2}{i}\right)^{−{n}−{k}} \:\Rightarrow \\ $$$${h}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{k}} \:{n}\left({n}+\mathrm{1}\right)….\left({n}+{k}−\mathrm{1}\right)×\left(\mathrm{2}{i}\right)^{−{n}−{k}} \:\Rightarrow \\ $$$${h}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} }{{k}!}\:{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}\right)\:\Rightarrow \\ $$$${g}\left({t}\right)=\frac{{h}\left({t}\right)}{{t}^{{n}} }\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)….\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} }{{k}!\:{t}^{{n}−{k}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left({t}\right) \\ $$$$=_{{n}−{k}\:={j}} \:\:\sum_{{j}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{n}+{j}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{j}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{2}{n}+{j}} }{\left({n}−{j}\right)!\:{t}^{{j}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left({t}\right) \\ $$$${t}={x}−{i}\:\Rightarrow\:{a}_{{j}} =\frac{\left(−\mathrm{1}\right)^{{n}+{j}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{j}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{2}{n}+{j}} }{\left({n}−{j}\right)!} \\ $$$${we}\:{have}\:\:\overset{−} {{f}}={f}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\overset{−} {{a}}_{{k}} }{\left({x}+{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\overset{−} {{b}}_{{k}} }{\left({x}−{i}\right)^{{k}} }\:={f}\:\Rightarrow \\ $$$${b}_{{k}} =\overset{−} {{a}}_{{k}} \:\Rightarrow \\ $$$${b}_{{j}} =\frac{\left(−\mathrm{1}\right)^{{n}+{j}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{j}−\mathrm{1}\right)\left(−\mathrm{2}{i}\right)^{−\mathrm{2}{n}+{j}} }{\left({n}−{j}\right)!} \\ $$