Question Number 68036 by mathmax by abdo last updated on 03/Sep/19
$${let}\:{f}\left({x}\right)\:={cos}\left(\alpha{x}\right)\:\:,\mathrm{2}\pi\:{periodic}\:\:\:{developp}\:{f}\:{at}\:{fourier}\:{serie}. \\ $$$$\alpha\:\in\:{R}−{Z} \\ $$
Commented by mathmax by abdo last updated on 05/Sep/19
![f is even ⇒f(x) =(a_0 /2)+Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T)∫_([T]) f(x)cos(nx)dx =(2/(2π))∫_(−π) ^π cos(αx)cos(nx)dx =(2/π)∫_0 ^π (1/2){ cos(n+α)x +cos(n−α)x}dx ⇒ π a_n =∫_0 ^π (cos(n+α)x +cos(n−α)x)dx =[(1/(n+α))sin(n+α)+(1/(n−α))sin(n−α)x]_0 ^π =(1/(n+α))sin(n+α)π +(1/(n−α))sin(n−α)π =(1/(n+α))sin(nπ +απ)+(1/(n−α))sin(nπ −απ) =(((−1)^n sin(απ))/(n+α)) −(((−1)^n )/(n−α))sin(απ) =(−1)^n sin(απ){(1/(n+α))−(1/(n−α))} =(−1)^n sin(απ)(((−2α)/(n^2 −α^2 ))) ⇒ a_n =(((−1)^n )/π)sin(απ)×(((−2α))/(n^2 −α^2 )) we have πa_0 = ∫_0 ^π 2cos(αx)dx =2 [(1/α)sin(αx)]_0 ^π =(2/α)sin(απ) ⇒ a_0 =(2/(πα))sin(απ) ⇒(a_0 /2) =((sin(απ))/(απ)) ⇒ cos(αx) =((sin(απ))/(απ)) −((2α)/π)sin(απ)Σ_(n=1) ^∞ (((−1)^n cos(nx))/(n^2 −α^2 ))](https://www.tinkutara.com/question/Q68128.png)
$${f}\:{is}\:{even}\:\Rightarrow{f}\left({x}\right)\:=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {cos}\left({nx}\right) \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{T}}\int_{\left[{T}\right]} \:\:\:{f}\left({x}\right){cos}\left({nx}\right){dx}\:=\frac{\mathrm{2}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} \:{cos}\left(\alpha{x}\right){cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left({n}+\alpha\right){x}\:+{cos}\left({n}−\alpha\right){x}\right\}{dx}\:\Rightarrow \\ $$$$\pi\:{a}_{{n}} =\int_{\mathrm{0}} ^{\pi} \left({cos}\left({n}+\alpha\right){x}\:+{cos}\left({n}−\alpha\right){x}\right){dx} \\ $$$$=\left[\frac{\mathrm{1}}{{n}+\alpha}{sin}\left({n}+\alpha\right)+\frac{\mathrm{1}}{{n}−\alpha}{sin}\left({n}−\alpha\right){x}\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{{n}+\alpha}{sin}\left({n}+\alpha\right)\pi\:+\frac{\mathrm{1}}{{n}−\alpha}{sin}\left({n}−\alpha\right)\pi \\ $$$$=\frac{\mathrm{1}}{{n}+\alpha}{sin}\left({n}\pi\:+\alpha\pi\right)+\frac{\mathrm{1}}{{n}−\alpha}{sin}\left({n}\pi\:−\alpha\pi\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} \:{sin}\left(\alpha\pi\right)}{{n}+\alpha}\:−\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\alpha}{sin}\left(\alpha\pi\right) \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {sin}\left(\alpha\pi\right)\left\{\frac{\mathrm{1}}{{n}+\alpha}−\frac{\mathrm{1}}{{n}−\alpha}\right\}\:=\left(−\mathrm{1}\right)^{{n}} {sin}\left(\alpha\pi\right)\left(\frac{−\mathrm{2}\alpha}{{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\pi}{sin}\left(\alpha\pi\right)×\frac{\left(−\mathrm{2}\alpha\right)}{{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$$\pi{a}_{\mathrm{0}} =\:\int_{\mathrm{0}} ^{\pi} \:\mathrm{2}{cos}\left(\alpha{x}\right){dx}\:=\mathrm{2}\:\left[\frac{\mathrm{1}}{\alpha}{sin}\left(\alpha{x}\right)\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{2}}{\alpha}{sin}\left(\alpha\pi\right)\:\Rightarrow \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi\alpha}{sin}\left(\alpha\pi\right)\:\Rightarrow\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\frac{{sin}\left(\alpha\pi\right)}{\alpha\pi}\:\Rightarrow \\ $$$${cos}\left(\alpha{x}\right)\:=\frac{{sin}\left(\alpha\pi\right)}{\alpha\pi}\:−\frac{\mathrm{2}\alpha}{\pi}{sin}\left(\alpha\pi\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{cos}\left({nx}\right)}{{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$