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Question Number 98429 by mathmax by abdo last updated on 13/Jun/20
let f(x) =cos(αx) developp f at fourier serie
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{cos}\left(\alpha\mathrm{x}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$
Answered by abdomathmax last updated on 15/Jun/20
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 =(1/π)∫_(−π) ^π cos(αx)cos(nx)dx =(2/π) ∫_0 ^π cos(αx)cos(nx)dx ⇒ (π/2)a_n =(1/2) ∫_0 ^π {cos(n+α)x+cos(n−α)x}dx ⇒πa_n =[(1/(n+α))sin(n+α)x+(1/(n−α)) sin(n−α)x]_0 ^π =(1/(n+α)) sin(nπ +απ)+(1/(n−α))sin(nπ−απ) =(((−1)^n sin(απ))/(n+α))−(((−1)^n sin(απ))/(n−α)) =(−1)^n sin(απ){(1/(n+α))−(1/(n−α))} =(−1)^n sin(απ)(((−2α)/(n^2 −α^2 ))) (but α ∈ R−Z) ⇒ a_n =−2α ((sin(απ))/(π(n^2 −α^2 )))(−1)^n a_0 =((2sin(απ))/(απ)) ⇒ cos(αx) =((sin(απ))/(απ)) −((2sin(απ))/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 −α^2 ))cos(nx)
$$\mathrm{f}\:\mathrm{is}\:\mathrm{even}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \mathrm{cos}\left(\mathrm{nx}\right)\: \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}}{\mathrm{T}}\int_{\left[\mathrm{T}\right]} \mathrm{f}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\pi} \:\mathrm{cos}\left(\alpha\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:\mathrm{cos}\left(\alpha\mathrm{x}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\left\{\mathrm{cos}\left(\mathrm{n}+\alpha\right)\mathrm{x}+\mathrm{cos}\left(\mathrm{n}−\alpha\right)\mathrm{x}\right\}\mathrm{dx} \\ $$$$\Rightarrow\pi\mathrm{a}_{\mathrm{n}} =\left[\frac{\mathrm{1}}{\mathrm{n}+\alpha}\mathrm{sin}\left(\mathrm{n}+\alpha\right)\mathrm{x}+\frac{\mathrm{1}}{\mathrm{n}−\alpha}\:\mathrm{sin}\left(\mathrm{n}−\alpha\right)\mathrm{x}\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}+\alpha}\:\mathrm{sin}\left(\mathrm{n}\pi\:+\alpha\pi\right)+\frac{\mathrm{1}}{\mathrm{n}−\alpha}\mathrm{sin}\left(\mathrm{n}\pi−\alpha\pi\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\alpha\pi\right)}{\mathrm{n}+\alpha}−\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\alpha\pi\right)}{\mathrm{n}−\alpha} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\alpha\pi\right)\left\{\frac{\mathrm{1}}{\mathrm{n}+\alpha}−\frac{\mathrm{1}}{\mathrm{n}−\alpha}\right\} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\alpha\pi\right)\left(\frac{−\mathrm{2}\alpha}{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\right)\:\:\:\left(\mathrm{but}\:\alpha\:\in\:\mathrm{R}−\mathrm{Z}\right)\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =−\mathrm{2}\alpha\:\frac{\mathrm{sin}\left(\alpha\pi\right)}{\pi\left(\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} \right)}\left(−\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{0}} =\frac{\mathrm{2sin}\left(\alpha\pi\right)}{\alpha\pi}\:\Rightarrow \\ $$$$\mathrm{cos}\left(\alpha\mathrm{x}\right)\:=\frac{\mathrm{sin}\left(\alpha\pi\right)}{\alpha\pi}\:−\frac{\mathrm{2sin}\left(\alpha\pi\right)}{\pi}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right) \\ $$
Commented by abdomathmax last updated on 15/Jun/20
sorry cos(αx) =((sin(απ))/(απ)) −((2αsin(απ))/π) Σ_(n=1) ^∞ (((−1)^n )/(n^2 −α^2 ))cos(nx)
$$\mathrm{sorry}\:\: \\ $$$$\mathrm{cos}\left(\alpha\mathrm{x}\right)\:=\frac{\mathrm{sin}\left(\alpha\pi\right)}{\alpha\pi}\:−\frac{\mathrm{2}\alpha\mathrm{sin}\left(\alpha\pi\right)}{\pi}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right) \\ $$
Commented by abdomathmax last updated on 15/Jun/20
for x=π we get cos(απ) =((sin(απ))/(απ)) −((2α sin(απ))/π) Σ_(n=1) ^∞ (1/(n^2 −α^2 )) ⇒ cotan(απ) =(1/(απ)) −((2α)/π) Σ_(n=1) ^∞ (1/(n^2 −α^2 )) ⇒ πcotan(απ) =(1/α) −Σ_(n=1) ^∞ ((2α)/(n^2 −α^2 )) ⇒ πcotan(απ) −(1/α) =Σ_(n=1) ^∞ ((2α)/(α^2 −n^2 ))
$$\mathrm{for}\:\mathrm{x}=\pi\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\mathrm{cos}\left(\alpha\pi\right)\:=\frac{\mathrm{sin}\left(\alpha\pi\right)}{\alpha\pi}\:−\frac{\mathrm{2}\alpha\:\mathrm{sin}\left(\alpha\pi\right)}{\pi}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{cotan}\left(\alpha\pi\right)\:=\frac{\mathrm{1}}{\alpha\pi}\:−\frac{\mathrm{2}\alpha}{\pi}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$$\pi\mathrm{cotan}\left(\alpha\pi\right)\:=\frac{\mathrm{1}}{\alpha}\:−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}\alpha}{\mathrm{n}^{\mathrm{2}} −\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$$\pi\mathrm{cotan}\left(\alpha\pi\right)\:−\frac{\mathrm{1}}{\alpha}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}\alpha}{\alpha^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} } \\ $$

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