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


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Question Number 98429 by mathmax by abdo last updated on 13/Jun/20

$let f (x) = \cos (α x) developp f at fourier serie$
	Answered by abdomathmax last updated on 15/Jun/20
	$f is even ⇒f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T)∫_([T]) f(x)cos(nx)dx =(1/π)∫_(−π) ^π cos(αx)cos(nx)dx =(2/π) ∫_0 ^π cos(αx)cos(nx)dx ⇒ (π/2)a_n =(1/2) ∫_0 ^π {cos(n+α)x+cos(n−α)x}dx ⇒πa_n =[(1/(n+α))sin(n+α)x+(1/(n−α)) sin(n−α)x]_0 ^π =(1/(n+α)) sin(nπ +απ)+(1/(n−α))sin(nπ−απ) =(((−1)^n sin(απ))/(n+α))−(((−1)^n sin(απ))/(n−α)) =(−1)^n sin(απ){(1/(n+α))−(1/(n−α))} =(−1)^n sin(απ)(((−2α)/(n^2 −α^2 ))) (but α ∈ R−Z) ⇒ a_n =−2α ((sin(απ))/(π(n^2 −α^2 )))(−1)^n a_0 =((2sin(απ))/(απ)) ⇒ cos(αx) =((sin(απ))/(απ)) −((2sin(απ))/π)Σ_(n=1) ^∞ (((−1)^n )/(n^2 −α^2 ))cos(nx)$
	$f is even \Rightarrow f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx)$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) \cos (nx) dx = \frac{1}{π} \int_{- π}^{π} \cos (α x) \cos (nx) dx$ $= \frac{2}{π} \int_{0}^{π} \cos (α x) \cos (nx) dx \Rightarrow$ $\frac{π}{2} a_{n} = \frac{1}{2} \int_{0}^{π} {\cos (n + α) x + \cos (n - α) x} dx$ $\Rightarrow π a_{n} = {[\frac{1}{n + α} \sin (n + α) x + \frac{1}{n - α} \sin (n - α) x]}_{0}^{π}$ $= \frac{1}{n + α} \sin (n π + α π) + \frac{1}{n - α} \sin (n π - α π)$ $= \frac{{(- 1)}^{n} \sin (α π)}{n + α} - \frac{{(- 1)}^{n} \sin (α π)}{n - α}$ $= {(- 1)}^{n} \sin (α π) {\frac{1}{n + α} - \frac{1}{n - α}}$ $= {(- 1)}^{n} \sin (α π) (\frac{- 2 α}{n^{2} - α^{2}}) (but α \in R - Z) \Rightarrow$ $a_{n} = - 2 α \frac{\sin (α π)}{π (n^{2} - α^{2})} {(- 1)}^{n}$ $a_{0} = \frac{2 \sin (α π)}{α π} \Rightarrow$ $\cos (α x) = \frac{\sin (α π)}{α π} - \frac{2 \sin (α π)}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - α^{2}} \cos (nx)$
	Commented by abdomathmax last updated on 15/Jun/20

	$sorry$ $\cos (α x) = \frac{\sin (α π)}{α π} - \frac{2 α \sin (α π)}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - α^{2}} \cos (nx)$
	Commented by abdomathmax last updated on 15/Jun/20

	$for x = π we get$ $\cos (α π) = \frac{\sin (α π)}{α π} - \frac{2 α \sin (α π)}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - α^{2}} \Rightarrow$ $cotan (α π) = \frac{1}{α π} - \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - α^{2}} \Rightarrow$ $π cotan (α π) = \frac{1}{α} - \sum_{n = 1}^{\infty} \frac{2 α}{n^{2} - α^{2}} \Rightarrow$ $π cotan (α π) - \frac{1}{α} = \sum_{n = 1}^{\infty} \frac{2 α}{α^{2} - n^{2}}$
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