Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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)}

$${S}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left({x}^{\mathrm{2}} +\mathrm{1}\right){arctan}\left({x}\right)\:−{x}−{x}^{\mathrm{3}} \:+\frac{{x}^{\mathrm{5}} }{\mathrm{3}}\right\} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com