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Question Number 98943 by mathmax by abdo last updated on 17/Jun/20
let f(x) =(1/((1+x^2 )^3 )) developp f at integr serie
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$
Answered by maths mind last updated on 17/Jun/20
f(x)=(1/((1+x^2 ))) f′(x)=((−2x)/((1+x^2 )^2 )) f′′(x)=((−2)/((1+x^2 )^2 ))+((8x^2 )/((1+x^2 )^3 )) f′′(x)=((f′(x))/x)+8x^2 .(1/((1+x^2 )^3 )) f(x)=Σ(−1)^n x^(2n) f′(x)=Σ_(n≥1) 2n(−1)^n x^(2n−1) f′′(x)=Σ_(n≥1) .2n(2n−1)(−1)^n x^(2n−2) (1/((1+x^2 )^3 ))=(1/(8x^2 )){Σ_(n≥0) 2(n+1)(2n+1)(−1)^(n+1) x^(2n) −Σ_(n≥0) (2n+2)(−1)^(n+1) x^(2n) } =(1/2).Σ_(n≥1) (−1)^(n+1) (n^2 +n)x^(2n−2) =(1/2)Σ_(n≥0) (−1)^n (n^2 +3n+2)x^(2n)
$${f}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${f}'\left({x}\right)=\frac{−\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${f}''\left({x}\right)=\frac{−\mathrm{2}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{8}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$${f}''\left({x}\right)=\frac{{f}'\left({x}\right)}{{x}}+\mathrm{8}{x}^{\mathrm{2}} .\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$${f}\left({x}\right)=\Sigma\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} \\ $$$${f}'\left({x}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$${f}''\left({x}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}.\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}−\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }\left\{\underset{{n}\geqslant\mathrm{0}} {\sum}\mathrm{2}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{2}{n}} −\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\mathrm{2}{n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{2}{n}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({n}^{\mathrm{2}} +{n}\right){x}^{\mathrm{2}{n}−\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right){x}^{\mathrm{2}{n}} \\ $$
Commented by mathmax by abdo last updated on 17/Jun/20
thanks sir.
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$
Answered by mathmax by abdo last updated on 17/Jun/20
we have (1/(1+x^2 )) =Σ_(n=0) ^∞ (−1)^n x^(2n) by derivation we get −((2x)/((1+x^2 )^2 )) =Σ_(n=1) ^∞ 2n(−1)^n x^(2n−1) ⇒((−2)/((1+x^2 )^2 )) =Σ_(n=1) ^∞ 2n(−1)^n x^(2n−2) ⇒2 ×((2(2x)(1+x^2 ))/((1+x^2 )^4 )) =Σ_(n=2) ^∞ 2n(2n−2)(−1)^n x^(2n−3) ⇒ ((8x)/((1+x^2 )^3 )) =Σ_(n=2) ^∞ 2n(2n−2)(−1)^n x^(2n−3) ⇒ (4/((1+x^2 )^3 )) =Σ_(n=2) ^∞ n(2n−2)(−1)^n x^(2n−4) =Σ_(n=0) ^∞ (n+2)(2n+2)(−1)^(n+2) x^(2n) =2 Σ_(n=0) ^∞ (n+1)(n+2)(−1)^n x^(2n) ⇒ (1/((1+x^2 )^3 )) =(1/2)Σ_(n=0) ^∞ (n^2 +3n+2)(−1)^n x^(2n) .
$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} \:\mathrm{by}\:\mathrm{derivation}\:\mathrm{we}\:\mathrm{get} \\ $$$$−\frac{\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{2n}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}−\mathrm{1}} \:\Rightarrow\frac{−\mathrm{2}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{2n}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}−\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\:×\frac{\mathrm{2}\left(\mathrm{2x}\right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:=\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\mathrm{2n}\left(\mathrm{2n}−\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}−\mathrm{3}} \:\Rightarrow \\ $$$$\frac{\mathrm{8x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\mathrm{2n}\left(\mathrm{2n}−\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}−\mathrm{3}} \:\Rightarrow \\ $$$$\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\mathrm{n}\left(\mathrm{2n}−\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}−\mathrm{4}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{2n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{2}} \:\mathrm{x}^{\mathrm{2n}} \:=\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(\mathrm{n}^{\mathrm{2}} \:+\mathrm{3n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} . \\ $$

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