Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 146758 by tabata last updated on 15/Jul/21

$$findforierseriestohalfrangof$$ $$f\left(x\right)=sinx,0<x<\pi andprovethat$$ $$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4n^2-1}=\frac{1}{2}$$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$${\mathrm{S}}_{\mathrm{N}}=\underset{n=1}{\sum ^{\mathrm{N}}}\frac{1}{4n^2-1}=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{N}}}(\frac{1}{2n-1}-\frac{1}{2n+1})$$ $${\mathrm{S}}_{\mathrm{N}}=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{N}}}(1-\frac{1}{2\mathrm{N}+1})$$ $$\Rightarrow {\mathrm{S}}_{\mathrm{\infty }}=\frac{1}{2}$$

Commented bytabata last updated on 15/Jul/21

$$andforiersir?$$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$$$$ $$f\left(x\right)=\sin x,0<x<\pi$$ $$a_0=\frac{2}{\pi }{\int }_0^{\pi }f\left(x\right)dx=\frac{2}{\pi }{\int }_0^{\pi }\sin xdx$$ $$a_0=\frac{2}{\pi }{[-\cos x]}_0^{\pi }=\frac{4}{\pi }$$ $$a_1=\frac{2}{\pi }{\int }_0^{\pi }\sin x\cos dx=\frac{1}{\pi }{\int }_0^{\pi }\sin \left(2x\right)dx$$ $$a_1=0$$ $$n>1:$$ $$a_n=\frac{2}{\pi }{\int }_0^{\pi }f\left(x\right)\cos \left(\frac{n\pi x}{\pi }\right)dx$$ $$a_n=\frac{2}{\pi }{\int }_0^{\pi }\sin x\cos \left(nx\right)dx$$ $$a_n=\frac{2}{\pi }{\int }_0^{\pi }\frac{1}{2}[\sin \left((n+1)x\right)-\sin \left((n-1)x\right)]dx$$ $$a_n=\frac{2}{\pi }{\int }_0^{\pi }\frac{1}{2}[\sin \left((n+1)x\right)-\sin \left((n-1)x\right)]dx$$ $$a_n=\frac{1}{\pi }{[-\frac{\cos \left((n+1)x\right)}{n+1}+\frac{\cos \left((n-1)x\right)}{n-1}]}_0^{\pi }$$ $$a_n=\frac{1}{\pi }[-\frac{{(-1)}^{n+1}}{n+1}+\frac{{(-1)}^{n-1}}{n-1}+\frac{1}{n+1}-\frac{1}{n-1}]$$ $$a_n=-\frac{2}{\pi (n^2-1)}[{(-1)}^n+1]$$ $$\sin x=\frac{a_0}{2}+a_1\cos x+\underset{n=2}{\sum ^{\mathrm{\infty }}}{a}_n\cos \left(nx\right)$$ $$\sin x=\frac{2}{\pi }-\frac{2}{\pi }\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n+1}{n^2-1}\cos \left(nx\right)$$ $$n=2m$$ $$\sin x=\frac{2}{\pi }-\frac{4}{\pi }\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4m^2-1}\cos \left(2mx\right)$$ $$x=0:$$ $$0=\frac{2}{\pi }-\frac{4}{\pi }\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4m^2-1}$$ $$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4m^2-1}=\frac{1}{2}$$ $$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com