Question Number 146548 by mathmax by abdo last updated on 13/Jul/21
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cos}\left(\alpha\mathrm{x}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie}\:\:\left(\alpha\:\mathrm{real}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 14/Jul/21
![a_0 = (1/T)∫_(−(T/2)) ^(+(T/2)) f(x)dx a_0 = (1/(2π))∫_(−π) ^(+π) cos(αx)dx a_0 = (1/(2πα))[sin(αx)]_(−π) ^(+π) a_0 = ((sin(πα))/(πα)) a_n = (2/T)∫_(−(T/2)) ^(+(T/2)) f(x)cos(((2πnx)/T))dx a_n = (1/π)∫_(−π) ^(+π) cos(αx)cos(nx)dx a_n = (1/π)∫_(−π) ^(+π) (1/2)[cos((α−n)x)−cos((α+n)x)]dx a_n = (1/(2π))[((sin((α−n)x))/(α−n))−((sin((α+n)x))/(α+n))]_(−π) ^(+π) a_n = ((sin(π(α−n)))/(π(α−n)))−((sin(π(α+n)))/(π(α+n))) b_n = 0 (f is even)](https://www.tinkutara.com/question/Q146557.png)
$${a}_{\mathrm{0}} \:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right){dx} \\ $$$${a}_{\mathrm{0}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{+\pi} \mathrm{cos}\left(\alpha{x}\right){dx} \\ $$$${a}_{\mathrm{0}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\pi\alpha}\left[\mathrm{sin}\left(\alpha{x}\right)\right]_{−\pi} ^{+\pi} \\ $$$${a}_{\mathrm{0}} \:=\:\frac{\mathrm{sin}\left(\pi\alpha\right)}{\pi\alpha} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right)\mathrm{cos}\left(\frac{\mathrm{2}\pi{nx}}{\mathrm{T}}\right){dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{+\pi} \mathrm{cos}\left(\alpha{x}\right)\mathrm{cos}\left({nx}\right){dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{+\pi} \frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\left(\left(\alpha−{n}\right){x}\right)−\mathrm{cos}\left(\left(\alpha+{n}\right){x}\right)\right]{dx} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\left[\frac{\mathrm{sin}\left(\left(\alpha−{n}\right){x}\right)}{\alpha−{n}}−\frac{\mathrm{sin}\left(\left(\alpha+{n}\right){x}\right)}{\alpha+{n}}\right]_{−\pi} ^{+\pi} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{sin}\left(\pi\left(\alpha−{n}\right)\right)}{\pi\left(\alpha−{n}\right)}−\frac{\mathrm{sin}\left(\pi\left(\alpha+{n}\right)\right)}{\pi\left(\alpha+{n}\right)} \\ $$$${b}_{{n}} \:=\:\mathrm{0}\:\left({f}\:\mathrm{is}\:\mathrm{even}\right) \\ $$
Answered by mathmax by abdo last updated on 14/Jul/21
![f(x)=(a_0 /2)+Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_(−(T/2)) ^(T/2) f(x)cos(nx)dx =(2/π)∫_0 ^π cos(αx)cos(nx)dx =(1/π)∫_0 ^π (cos(n+α)x+cos(n−α)x)dx ⇒a_n =[((sin(n+α)x)/(n+α)) +((sin(n−α)x)/(n−α))]_0 ^π =((sin(nπ+απ))/(n+α)) +((sin(nπ−απ))/(n−α)) =(((−1)^n )/(n+α))sin(πα)−(((−1)^n )/(n−α))sin(πα) =(−1)^n sin(πα){(1/(n+α))−(1/(n−α))} =−2α(−1)^n sin(απ)/n^2 −α^2 ⇒a_n =(1/π)(−2α(−1)^n sin(απ)/_(n^2 −α^2 ) =−((2α)/π)(−1)^n sin(απ)/n^2 −α^2 a_o =(2/π)∫_0 ^π cos(αx)dx =(2/(πα))sin(απ) ⇒ cos(αx)=((sin(πα))/(πα)) −((2α)/π)sin(πα)Σ_(n=1) ^∞ (((−1)^n )/(n^2 −α^2 ))cos(nx) (α ∈R−Z) let use this remark...x=π ⇒cos(απ)=((sin(απ))/(απ))−((2α)/π)sin(απ)Σ_(n=1) ^∞ (1/(n^2 −α^2 )) ⇒ cotan(απ) =(1/(απ))−((2α)/π)Σ_(n=1) ^∞ (1/(n^2 −α^2 )) απ =t ⇒cotan(t)=(1/t)−(2/π).(t/π)Σ_(n=1) ^∞ (1/(n^2 −(t^2 /π^2 ))) ⇒cotant =(1/t)−Σ_(n=1) ^∞ ((2t)/(n^2 π^2 −t^2 )) ⇒ cotant =(1/t)+Σ_(n=1) ^∞ ((2t)/(t^2 −π^2 n^2 ))](https://www.tinkutara.com/question/Q146567.png)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \mathrm{cos}\left(\mathrm{nx}\right)\:\:\mathrm{with}\:\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{\frac{\mathrm{T}}{\mathrm{2}}} \:\mathrm{f}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\mathrm{cos}\left(\alpha\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{cos}\left(\mathrm{n}+\alpha\right)\mathrm{x}+\mathrm{cos}\left(\mathrm{n}−\alpha\right)\mathrm{x}\right)\mathrm{dx} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} =\left[\frac{\mathrm{sin}\left(\mathrm{n}+\alpha\right)\mathrm{x}}{\mathrm{n}+\alpha}\:+\frac{\mathrm{sin}\left(\mathrm{n}−\alpha\right)\mathrm{x}}{\mathrm{n}−\alpha}\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{n}\pi+\alpha\pi\right)}{\mathrm{n}+\alpha}\:+\frac{\mathrm{sin}\left(\mathrm{n}\pi−\alpha\pi\right)}{\mathrm{n}−\alpha} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}+\alpha}\mathrm{sin}\left(\pi\alpha\right)−\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}−\alpha}\mathrm{sin}\left(\pi\alpha\right)\:=\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{sin}\left(\pi\alpha\right)\left\{\frac{\mathrm{1}}{\mathrm{n}+\alpha}−\frac{\mathrm{1}}{\mathrm{n}−\alpha}\right\} \\ $$$$=−\mathrm{2}\alpha\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{sin}\left(\alpha\pi\right)/\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} \:\Rightarrow\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\pi}\left(−\mathrm{2}\alpha\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{sin}\left(\alpha\pi\right)/_{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \right. \\ $$$$=−\frac{\mathrm{2}\alpha}{\pi}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\alpha\pi\right)/\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{o}} =\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\mathrm{cos}\left(\alpha\mathrm{x}\right)\mathrm{dx}\:=\frac{\mathrm{2}}{\pi\alpha}\mathrm{sin}\left(\alpha\pi\right)\:\Rightarrow \\ $$$$\mathrm{cos}\left(\alpha\mathrm{x}\right)=\frac{\mathrm{sin}\left(\pi\alpha\right)}{\pi\alpha}\:−\frac{\mathrm{2}\alpha}{\pi}\mathrm{sin}\left(\pi\alpha\right)\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right)\:\:\:\left(\alpha\:\in\mathrm{R}−\mathrm{Z}\right) \\ $$$$\mathrm{let}\:\mathrm{use}\:\mathrm{this}\:\mathrm{remark}…\mathrm{x}=\pi\:\Rightarrow\mathrm{cos}\left(\alpha\pi\right)=\frac{\mathrm{sin}\left(\alpha\pi\right)}{\alpha\pi}−\frac{\mathrm{2}\alpha}{\pi}\mathrm{sin}\left(\alpha\pi\right)\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{cotan}\left(\alpha\pi\right)\:=\frac{\mathrm{1}}{\alpha\pi}−\frac{\mathrm{2}\alpha}{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$$$\alpha\pi\:=\mathrm{t}\:\Rightarrow\mathrm{cotan}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{2}}{\pi}.\frac{\mathrm{t}}{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\frac{\mathrm{t}^{\mathrm{2}} }{\pi^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{cotant}\:=\frac{\mathrm{1}}{\mathrm{t}}−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2t}}{\mathrm{n}^{\mathrm{2}} \pi^{\mathrm{2}} −\mathrm{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{cotant}\:=\frac{\mathrm{1}}{\mathrm{t}}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} −\pi^{\mathrm{2}} \mathrm{n}^{\mathrm{2}} } \\ $$