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Question Number 98943 by mathmax by abdo last updated on 17/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{(1+{\mathrm{x}}^2)}^3}\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{integr}\mathrm{serie}$$

Answered by maths mind last updated on 17/Jun/20

$$f\left(x\right)=\frac{1}{(1+x^2)}$$$$f^{\prime }\left(x\right)=\frac{-2x}{{(1+x^2)}^2}$$$$f^{″}\left(x\right)=\frac{-2}{{(1+x^2)}^2}+\frac{8x^2}{{(1+x^2)}^3}$$$$f^{″}\left(x\right)=\frac{f^{\prime }\left(x\right)}{x}+8x^2.\frac{1}{{(1+x^2)}^3}$$$$f\left(x\right)=\mathrm{\Sigma }{(-1)}^n{x}^{2n}$$$$f^{\prime }\left(x\right)=\sum _{n⩾1}2n{(-1)}^n{x}^{2n-1}$$$$f^{″}\left(x\right)=\sum _{n⩾1}.2n(2n-1){(-1)}^n{x}^{2n-2}$$$$\frac{1}{{(1+x^2)}^3}=\frac{1}{8x^2}\{\sum _{n⩾0}2(n+1)(2n+1){(-1)}^{n+1}{x}^{2n}-\sum _{n⩾0}(2n+2){(-1)}^{n+1}{x}^{2n}\}$$$$=\frac{1}{2}.\sum _{n⩾1}{(-1)}^{n+1}(n^2+n)x^{2n-2}$$$$=\frac{1}{2}\sum _{n⩾0}{(-1)}^n(n^2+3n+2)x^{2n}$$

Commented by mathmax by abdo last updated on 17/Jun/20

$$\mathrm{thanks}\mathrm{sir}.$$

Answered by mathmax by abdo last updated on 17/Jun/20

$$\mathrm{we}\mathrm{have}\frac{1}{1+{\mathrm{x}}^2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\mathrm{by}\mathrm{derivation}\mathrm{we}\mathrm{get}$$$$-\frac{2\mathrm{x}}{{(1+{\mathrm{x}}^2)}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}2\mathrm{n}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-1}\Rightarrow \frac{-2}{{(1+{\mathrm{x}}^2)}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}2\mathrm{n}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-2}$$$$\Rightarrow 2\times \frac{2\left(2\mathrm{x}\right)(1+{\mathrm{x}}^2)}{{(1+{\mathrm{x}}^2)}^4}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}2\mathrm{n}(2\mathrm{n}-2){(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-3}\Rightarrow$$$$\frac{8\mathrm{x}}{{(1+{\mathrm{x}}^2)}^3}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}2\mathrm{n}(2\mathrm{n}-2){(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-3}\Rightarrow$$$$\frac{4}{{(1+{\mathrm{x}}^2)}^3}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(2\mathrm{n}-2){(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}-4}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+2)(2\mathrm{n}+2){(-1)}^{\mathrm{n}+2}{\mathrm{x}}^{2\mathrm{n}}=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1)(\mathrm{n}+2){(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\Rightarrow$$$$\frac{1}{{(1+{\mathrm{x}}^2)}^3}=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}({\mathrm{n}}^2+3\mathrm{n}+2){(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}.$$

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