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Question Number 146758 by tabata last updated on 15/Jul/21
find forier series to half rang of   f(x)=sinx  ,0<x<π and prove that  Σ_(n=1) ^∞ (1/(4n^2 −1))=(1/2)
$${find}\:{forier}\:{series}\:{to}\:{half}\:{rang}\:{of}\: \\ $$$${f}\left({x}\right)={sinx}\:\:,\mathrm{0}<{x}<\pi\:{and}\:{prove}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
S_N  = Σ_(n=1) ^N (1/(4n^2 −1)) = (1/2)Σ_(n=1) ^N ((1/(2n−1))−(1/(2n+1)))  S_N  = (1/2)Σ_(n=1) ^N (1−(1/(2N+1)))  ⇒ S_∞  = (1/2)
$$\mathrm{S}_{\mathrm{N}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\mathrm{S}_{\mathrm{N}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2N}+\mathrm{1}}\right) \\ $$$$\Rightarrow\:\mathrm{S}_{\infty} \:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by tabata last updated on 15/Jul/21
and forier sir ?
$${and}\:{forier}\:{sir}\:? \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
  f(x) = sinx, 0<x<π  a_0  = (2/π)∫_0 ^π f(x) dx = (2/π)∫_0 ^π sinx dx  a_0  = (2/π)[−cosx]_0 ^π  = (4/π)  a_1  = (2/π)∫_0 ^π sinxcos dx = (1/π)∫_0 ^π sin(2x)dx  a_1  = 0  n > 1 :  a_n  = (2/π)∫_0 ^π f(x)cos(((nπx)/π)) dx  a_n  = (2/π)∫_0 ^π sinxcos(nx) dx  a_n  = (2/π)∫_0 ^π (1/2)[sin((n+1)x)−sin((n−1)x)] dx  a_n  = (2/π)∫_0 ^π (1/2)[sin((n+1)x)−sin((n−1)x)] dx  a_n  = (1/π)[−((cos((n+1)x))/(n+1))+((cos((n−1)x))/(n−1))]_0 ^π   a_n  = (1/π)[−(((−1)^(n+1) )/(n+1))+(((−1)^(n−1) )/(n−1))+(1/(n+1))−(1/(n−1))]  a_n  = −(2/(π(n^2 −1)))[(−1)^n +1]  sinx = (a_0 /2)+a_1 cosx+Σ_(n=2) ^∞ a_n cos(nx)  sinx = (2/π)−(2/π)Σ_(n=2) ^∞ (((−1)^n +1)/(n^2 −1))cos(nx)  n = 2m  sinx = (2/π)−(4/π)Σ_(m=1) ^∞ (1/(4m^2 −1))cos(2mx)  x = 0 :  0 = (2/π)−(4/π)Σ_(m=1) ^∞ (1/(4m^2 −1))  Σ_(m=1) ^∞ (1/(4m^2 −1)) = (1/2)
$$ \\ $$$${f}\left({x}\right)\:=\:\mathrm{sin}{x},\:\mathrm{0}<{x}<\pi \\ $$$${a}_{\mathrm{0}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right)\:{dx}\:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}{x}\:{dx} \\ $$$${a}_{\mathrm{0}} \:=\:\frac{\mathrm{2}}{\pi}\left[−\mathrm{cos}{x}\right]_{\mathrm{0}} ^{\pi} \:=\:\frac{\mathrm{4}}{\pi} \\ $$$${a}_{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}{x}\mathrm{cos}\:{dx}\:=\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\left(\mathrm{2}{x}\right){dx} \\ $$$${a}_{\mathrm{1}} \:=\:\mathrm{0} \\ $$$${n}\:>\:\mathrm{1}\:: \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right)\mathrm{cos}\left(\frac{{n}\pi{x}}{\pi}\right)\:{dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}{x}\mathrm{cos}\left({nx}\right)\:{dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\left(\left({n}+\mathrm{1}\right){x}\right)−\mathrm{sin}\left(\left({n}−\mathrm{1}\right){x}\right)\right]\:{dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\left(\left({n}+\mathrm{1}\right){x}\right)−\mathrm{sin}\left(\left({n}−\mathrm{1}\right){x}\right)\right]\:{dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\left[−\frac{\mathrm{cos}\left(\left({n}+\mathrm{1}\right){x}\right)}{{n}+\mathrm{1}}+\frac{\mathrm{cos}\left(\left({n}−\mathrm{1}\right){x}\right)}{{n}−\mathrm{1}}\right]_{\mathrm{0}} ^{\pi} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\left[−\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}−\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}−\mathrm{1}}\right] \\ $$$${a}_{{n}} \:=\:−\frac{\mathrm{2}}{\pi\left({n}^{\mathrm{2}} −\mathrm{1}\right)}\left[\left(−\mathrm{1}\right)^{{n}} +\mathrm{1}\right] \\ $$$$\mathrm{sin}{x}\:=\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}+{a}_{\mathrm{1}} \mathrm{cos}{x}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{n}} \mathrm{cos}\left({nx}\right) \\ $$$$\mathrm{sin}{x}\:=\:\frac{\mathrm{2}}{\pi}−\frac{\mathrm{2}}{\pi}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} +\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\mathrm{cos}\left({nx}\right) \\ $$$${n}\:=\:\mathrm{2}{m} \\ $$$$\mathrm{sin}{x}\:=\:\frac{\mathrm{2}}{\pi}−\frac{\mathrm{4}}{\pi}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}}\mathrm{cos}\left(\mathrm{2}{mx}\right) \\ $$$${x}\:=\:\mathrm{0}\:: \\ $$$$\mathrm{0}\:=\:\frac{\mathrm{2}}{\pi}−\frac{\mathrm{4}}{\pi}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

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