Menu Close

let-S-x-n-1-1-n-1-x-2n-1-4n-2-1-1-find-the-radius-of-convergence-2-calculate-the-sum-S-x-

Tinku Tara 0 Comments
Pin




Question Number 34309 by prof Abdo imad last updated on 03/May/18
let S(x)= Σ_(n=1) ^∞ (−1)^(n−1) (x^(2n+1) /(4n^2 −1)) 1) find the radius of convergence 2) calculate the sum S(x).
$${let}\:{S}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{4}{n}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{radius}\:{of}\:{convergence} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{the}\:{sum}\:\:{S}\left({x}\right). \\ $$
Commented by math khazana by abdo last updated on 07/May/18
S(x)=(1/2)Σ_(n=1) ^∞ (−1)^(n−1) ((1/(2n−1)) −(1/(2n+1)))x^(2n+1) 2S(x)= Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n−1))x^(2n+1) +Σ_(n=1) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) =Σ_(n=2) ^∞ (((−1)^n )/(2n+1))x^(2n+3) +Σ_(n=1) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) =Σ_(n=0) ^∞ (((−1)^n x^(2n+3) )/(2n+1)) −x^3 +(x^5 /3) +Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) −x =(x^2 +1) Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) −x^3 +(x^5 /3) let put w(x)= Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/(2n+1)) w^′ (x) =Σ_(n=0) ^∞ (−1)^n x^(2n) =Σ_(n=0) ^∞ (−x^2 )^n = (1/(1+x^2 )) ⇒ w(x)= arctanx +λ but λ =w(0) =0⇒ w(x)= arctanx so S(x)=(1/2){(x^2 +1)arctanx +(x^5 /3) −x^3 } .
$${S}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right){x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{2}{S}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{3}} \:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}+\mathrm{3}} }{\mathrm{2}{n}+\mathrm{1}}\:\:\:−{x}^{\mathrm{3}} \:\:+\frac{{x}^{\mathrm{5}} }{\mathrm{3}}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$−{x} \\ $$$$=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:\:−{x}^{\mathrm{3}} \:\:+\frac{{x}^{\mathrm{5}} }{\mathrm{3}} \\ $$$${let}\:{put}\:{w}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$${w}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{2}} \right)^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{w}\left({x}\right)=\:{arctanx}\:+\lambda\:{but}\:\:\lambda\:={w}\left(\mathrm{0}\right)\:=\mathrm{0}\Rightarrow \\ $$$${w}\left({x}\right)=\:{arctanx}\:{so}\: \\ $$$${S}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({x}^{\mathrm{2}} +\mathrm{1}\right){arctanx}\:\:+\frac{{x}^{\mathrm{5}} }{\mathrm{3}}\:\:−{x}^{\mathrm{3}} \:\right\}\:. \\ $$
Commented by math khazana by abdo last updated on 07/May/18
the radius of convergence is R =1 .
$${the}\:{radius}\:{of}\:{convergence}\:{is}\:{R}\:=\mathrm{1}\:. \\ $$
Commented by math khazana by abdo last updated on 08/May/18
S(x)= (1/2){ (x^2 +1)arctan(x) −x−x^3 +(x^5 /3)}
$${S}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left({x}^{\mathrm{2}} +\mathrm{1}\right){arctan}\left({x}\right)\:−{x}−{x}^{\mathrm{3}} \:+\frac{{x}^{\mathrm{5}} }{\mathrm{3}}\right\} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *