let f(x)=cos(αx) developp f at fourier serie (α real)


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Question Number 146548 by mathmax by abdo last updated on 13/Jul/21

$let f (x) = \cos (α x) developp f at fourier serie (α real)$
	Answered by Olaf_Thorendsen last updated on 14/Jul/21

	$a_{0} = \frac{1}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) d x$ $a_{0} = \frac{1}{2 π} \int_{- π}^{+ π} \cos (α x) d x$ $a_{0} = \frac{1}{2 π α} {[\sin (α x)]}_{- π}^{+ π}$ $a_{0} = \frac{\sin (π α)}{π α}$ $a_{n} = \frac{2}{T} \int_{- \frac{T}{2}}^{+ \frac{T}{2}} f (x) \cos (\frac{2 π n x}{T}) d x$ $a_{n} = \frac{1}{π} \int_{- π}^{+ π} \cos (α x) \cos (n x) d x$ $a_{n} = \frac{1}{π} \int_{- π}^{+ π} \frac{1}{2} [\cos ((α - n) x) - \cos ((α + n) x)] d x$ $a_{n} = \frac{1}{2 π} {[\frac{\sin ((α - n) x)}{α - n} - \frac{\sin ((α + n) x)}{α + n}]}_{- π}^{+ π}$ $a_{n} = \frac{\sin (π (α - n))}{π (α - n)} - \frac{\sin (π (α + n))}{π (α + n)}$ $b_{n} = 0 (f is even)$
	Answered by mathmax by abdo last updated on 14/Jul/21
	$f(x)=(a_0 /2)+Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_(−(T/2)) ^(T/2) f(x)cos(nx)dx =(2/π)∫_0 ^π cos(αx)cos(nx)dx =(1/π)∫_0 ^π (cos(n+α)x+cos(n−α)x)dx ⇒a_n =[((sin(n+α)x)/(n+α)) +((sin(n−α)x)/(n−α))]_0 ^π =((sin(nπ+απ))/(n+α)) +((sin(nπ−απ))/(n−α)) =(((−1)^n )/(n+α))sin(πα)−(((−1)^n )/(n−α))sin(πα) =(−1)^n sin(πα){(1/(n+α))−(1/(n−α))} =−2α(−1)^n sin(απ)/n^2 −α^2 ⇒a_n =(1/π)(−2α(−1)^n sin(απ)/_(n^2 −α^2 ) =−((2α)/π)(−1)^n sin(απ)/n^2 −α^2 a_o =(2/π)∫_0 ^π cos(αx)dx =(2/(πα))sin(απ) ⇒ cos(αx)=((sin(πα))/(πα)) −((2α)/π)sin(πα)Σ_(n=1) ^∞ (((−1)^n )/(n^2 −α^2 ))cos(nx) (α ∈R−Z) let use this remark...x=π ⇒cos(απ)=((sin(απ))/(απ))−((2α)/π)sin(απ)Σ_(n=1) ^∞ (1/(n^2 −α^2 )) ⇒ cotan(απ) =(1/(απ))−((2α)/π)Σ_(n=1) ^∞ (1/(n^2 −α^2 )) απ =t ⇒cotan(t)=(1/t)−(2/π).(t/π)Σ_(n=1) ^∞ (1/(n^2 −(t^2 /π^2 ))) ⇒cotant =(1/t)−Σ_(n=1) ^∞ ((2t)/(n^2 π^2 −t^2 )) ⇒ cotant =(1/t)+Σ_(n=1) ^∞ ((2t)/(t^2 −π^2 n^2 ))$
	$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx) with a_{n} = \frac{2}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (x) \cos (nx) dx$ $= \frac{2}{π} \int_{0}^{π} \cos (α x) \cos (nx) dx = \frac{1}{π} \int_{0}^{π} (\cos (n + α) x + \cos (n - α) x) dx$ $\Rightarrow a_{n} = {[\frac{\sin (n + α) x}{n + α} + \frac{\sin (n - α) x}{n - α}]}_{0}^{π}$ $= \frac{\sin (n π + α π)}{n + α} + \frac{\sin (n π - α π)}{n - α}$ $= \frac{{(- 1)}^{n}}{n + α} \sin (π α) - \frac{{(- 1)}^{n}}{n - α} \sin (π α) = {(- 1)}^{n} \sin (π α) {\frac{1}{n + α} - \frac{1}{n - α}}$ $= - 2 α {(- 1)}^{n} \sin (α π) / n^{2} - α^{2} \Rightarrow a_{n} = \frac{1}{π} (- 2 α {(- 1)}^{n} \sin (α π) /_{n^{2} - α^{2}}$ $= - \frac{2 α}{π} {(- 1)}^{n} \sin (α π) / n^{2} - α^{2}$ $a_{o} = \frac{2}{π} \int_{0}^{π} \cos (α x) dx = \frac{2}{π α} \sin (α π) \Rightarrow$ $\cos (α x) = \frac{\sin (π α)}{π α} - \frac{2 α}{π} \sin (π α) \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - α^{2}} \cos (nx) (α \in R - Z)$ $let use this remark . . . x = π \Rightarrow \cos (α π) = \frac{\sin (α π)}{α π} - \frac{2 α}{π} \sin (α π) \sum_{n = 1}^{\infty} \frac{1}{n^{2} - α^{2}} \Rightarrow$ $cotan (α π) = \frac{1}{α π} - \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - α^{2}}$ $α π = t \Rightarrow cotan (t) = \frac{1}{t} - \frac{2}{π} . \frac{t}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - \frac{t^{2}}{π^{2}}}$ $\Rightarrow cotant = \frac{1}{t} - \sum_{n = 1}^{\infty} \frac{2 t}{n^{2} π^{2} - t^{2}} \Rightarrow$ $cotant = \frac{1}{t} + \sum_{n = 1}^{\infty} \frac{2 t}{t^{2} - π^{2} n^{2}}$
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