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Question Number 147971 by mathmax by abdo last updated on 24/Jul/21

$$\mathrm{find}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^3\mathrm{dx}}{{({\mathrm{x}}^2+\mathrm{x}+1)}^4}$$

Commented by tabata last updated on 24/Jul/21

$$msrabdocanhelpmeinquestion147915plese?$$

Commented by mathmax by abdo last updated on 24/Jul/21

$$3)\mathrm{f}\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{z}}}{1-\mathrm{z}}\mathrm{by}\mathrm{using}\mathrm{double}\mathrm{sum}$$$${\mathrm{e}}^{\mathrm{z}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{and}\frac{1}{1-\mathrm{z}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{z}}^{\mathrm{n}}\mathrm{for}\mid \mathrm{z}\mid <1\Rightarrow$$$$\mathrm{f}\left(\mathrm{z}\right)=\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}{\mathrm{z}}^{\mathrm{n}}\right).\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{z}}^{\mathrm{n}}\right)=\left(\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{n}}{\mathrm{z}}^{\mathrm{n}}\right).\left(\mathrm{\Sigma }{\mathrm{b}}_{\mathrm{n}}{\mathrm{z}}^{\mathrm{n}}\right)$$$$=\mathrm{\Sigma }{\mathrm{c}}_{\mathrm{n}}{\mathrm{z}}^{\mathrm{n}}/{\mathrm{c}}_{\mathrm{n}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{b}}_{\mathrm{j}}=\sum _{\mathrm{i}=0}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{b}}_{\mathrm{n}-\mathrm{i}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{1}{\mathrm{i}!}\times 1$$$$\Rightarrow \mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=0}^{\mathrm{n}}\frac{1}{\mathrm{i}!}\right){\mathrm{z}}^{\mathrm{n}}$$

Commented by mathmax by abdo last updated on 24/Jul/21

$$\mathrm{another}\mathrm{way}\mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}{\mathrm{z}}^{\mathrm{n}}\mathrm{we}\mathrm{have}\mathrm{by}\mathrm{leibniz}$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{z}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\frac{1}{1-\mathrm{z}}\right)}^{\left(\mathrm{k}\right)}{\left({\mathrm{e}}^{\mathrm{z}}\right)}^{(\mathrm{n}-\mathrm{k})}=-{\mathrm{e}}^{\mathrm{z}}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}\mathrm{k}!}{{(\mathrm{z}-1)}^{\mathrm{k}+1}}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)=-\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}\mathrm{k}!}{{(-1)}^{\mathrm{k}+1}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\mathrm{k}!\Rightarrow$$$$\mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{\mathrm{k}!}{\mathrm{n}!}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\right){\mathrm{z}}^{\mathrm{n}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{\mathrm{k}!}{\mathrm{n}!}\times \frac{\mathrm{n}!}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}\right){\mathrm{z}}^{\mathrm{n}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{1}{(\mathrm{n}-\mathrm{k})!}\right){\mathrm{z}}^{\mathrm{n}}(\mathrm{n}-\mathrm{k}=\mathrm{i})$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=0}^{\mathrm{n}}\frac{1}{\mathrm{i}!}\right){\mathrm{z}}^{\mathrm{n}}$$

Commented by mathmax by abdo last updated on 24/Jul/21

$$\mathrm{f}\left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}^2}\mathrm{at}\mathrm{z}={\mathrm{z}}_0\Rightarrow \mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left({\mathrm{z}}_{\mathrm{o}}\right)}{\mathrm{n}!}{(\mathrm{z}-{\mathrm{z}}_{\mathrm{o}})}^{\mathrm{n}}({\mathrm{z}}_0\ne \mathrm{o})$$$${\mathrm{f}}^{\left(1\right)}\left(\mathrm{z}\right)=-\frac{2\mathrm{z}}{{\mathrm{z}}^4}=-\frac{2}{{\mathrm{z}}^3}$$$${\mathrm{f}}^{\left(2\right)}\left(\mathrm{z}\right)=\frac{2.3{\mathrm{z}}^2}{{\mathrm{z}}^6}=\frac{2.3}{{\mathrm{z}}^4}\mathrm{let}\left[\mathrm{suppose}{\mathrm{f}}^{(\mathrm{n}}\right)\left(\mathrm{z}\right)=\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{z}}^{\mathrm{n}+1}}\Rightarrow$$$${\mathrm{f}}^{(\mathrm{n}+1)}\left(\mathrm{z}\right)=\frac{\mathrm{d}}{\mathrm{dz}}\left\{\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{z}}^{\mathrm{n}+1}}\right\}={(-1)}^{\mathrm{n}}\mathrm{n}!\times \frac{-(\mathrm{n}+1){\mathrm{z}}^{\mathrm{n}}}{{\mathrm{z}}^{2\mathrm{n}+2}}$$$$=\frac{{(-1)}^{\mathrm{n}+1}(\mathrm{n}+1)!}{{\mathrm{z}}^{\mathrm{n}+2}}\left(\mathrm{true}\right)\Rightarrow {\mathrm{f}}^{(\mathrm{n}})\left({\mathrm{z}}_0\right)=\frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{z}}_0^{\mathrm{n}+1}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}\times \frac{{(-1)}^{\mathrm{n}}\mathrm{n}!}{{\mathrm{z}}_0^{\mathrm{n}+1}}{(\mathrm{z}-{\mathrm{z}}_0)}^{\mathrm{n}}$$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{1}{{\mathrm{z}}_0}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{(\frac{\mathrm{z}}{{\mathrm{z}}_0}-1)}^{\mathrm{n}}\mathrm{for}\mid \mathrm{z}\mid <\mid {\mathrm{z}}_0\mid$$

Commented by tabata last updated on 25/Jul/21

$$thankyoumsrmathmaxbyabdthankyou$$$$verymuch$$$$$$$$msrareyoutheretomorrowat12oclock$$$$atafternoon?$$

Commented by mathmax by abdo last updated on 25/Jul/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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