Question Number 146758 by tabata last updated on 15/Jul/21
$${find}\:{forier}\:{series}\:{to}\:{half}\:{rang}\:{of}\: \\ $$ $${f}\left({x}\right)={sinx}\:\:,\mathrm{0}<{x}<\pi\:{and}\:{prove}\:{that} \\ $$ $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
$$\mathrm{S}_{\mathrm{N}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$ $$\mathrm{S}_{\mathrm{N}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2N}+\mathrm{1}}\right) \\ $$ $$\Rightarrow\:\mathrm{S}_{\infty} \:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented bytabata last updated on 15/Jul/21
$${and}\:{forier}\:{sir}\:? \\ $$
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)](Q146766.png)
$$ \\ $$ $${f}\left({x}\right)\:=\:\mathrm{sin}{x},\:\mathrm{0}<{x}<\pi \\ $$ $${a}_{\mathrm{0}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right)\:{dx}\:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}{x}\:{dx} \\ $$ $${a}_{\mathrm{0}} \:=\:\frac{\mathrm{2}}{\pi}\left[−\mathrm{cos}{x}\right]_{\mathrm{0}} ^{\pi} \:=\:\frac{\mathrm{4}}{\pi} \\ $$ $${a}_{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}{x}\mathrm{cos}\:{dx}\:=\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\left(\mathrm{2}{x}\right){dx} \\ $$ $${a}_{\mathrm{1}} \:=\:\mathrm{0} \\ $$ $${n}\:>\:\mathrm{1}\:: \\ $$ $${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right)\mathrm{cos}\left(\frac{{n}\pi{x}}{\pi}\right)\:{dx} \\ $$ $${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}{x}\mathrm{cos}\left({nx}\right)\:{dx} \\ $$ $${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\left(\left({n}+\mathrm{1}\right){x}\right)−\mathrm{sin}\left(\left({n}−\mathrm{1}\right){x}\right)\right]\:{dx} \\ $$ $${a}_{{n}} \:=\:\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\left(\left({n}+\mathrm{1}\right){x}\right)−\mathrm{sin}\left(\left({n}−\mathrm{1}\right){x}\right)\right]\:{dx} \\ $$ $${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\left[−\frac{\mathrm{cos}\left(\left({n}+\mathrm{1}\right){x}\right)}{{n}+\mathrm{1}}+\frac{\mathrm{cos}\left(\left({n}−\mathrm{1}\right){x}\right)}{{n}−\mathrm{1}}\right]_{\mathrm{0}} ^{\pi} \\ $$ $${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\left[−\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}−\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}−\mathrm{1}}\right] \\ $$ $${a}_{{n}} \:=\:−\frac{\mathrm{2}}{\pi\left({n}^{\mathrm{2}} −\mathrm{1}\right)}\left[\left(−\mathrm{1}\right)^{{n}} +\mathrm{1}\right] \\ $$ $$\mathrm{sin}{x}\:=\:\frac{{a}_{\mathrm{0}} }{\mathrm{2}}+{a}_{\mathrm{1}} \mathrm{cos}{x}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{n}} \mathrm{cos}\left({nx}\right) \\ $$ $$\mathrm{sin}{x}\:=\:\frac{\mathrm{2}}{\pi}−\frac{\mathrm{2}}{\pi}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} +\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\mathrm{cos}\left({nx}\right) \\ $$ $${n}\:=\:\mathrm{2}{m} \\ $$ $$\mathrm{sin}{x}\:=\:\frac{\mathrm{2}}{\pi}−\frac{\mathrm{4}}{\pi}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}}\mathrm{cos}\left(\mathrm{2}{mx}\right) \\ $$ $${x}\:=\:\mathrm{0}\:: \\ $$ $$\mathrm{0}\:=\:\frac{\mathrm{2}}{\pi}−\frac{\mathrm{4}}{\pi}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}} \\ $$ $$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$ \\ $$