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Question Number 95697 by i jagooll last updated on 27/May/20

	Answered by john santu last updated on 27/May/20

	$a_{0} = \frac{1}{2 π} \underset{- π}{\int^{π}} \cos a t d t = \frac{1}{2 π a} {[\sin a t]}_{- π}^{π}$ $= \frac{\sin π a}{a π} .$ $a_{n} = \frac{1}{2 π} \underset{- π}{\int^{π}} \cos (a + n) t + \cos (a - n) t d t$ $a_{n} = \frac{2 a {(- 1)}^{n} \sin a π}{π (a^{2} - n^{2})} .; b_{n} = 0$ $\Rightarrow \frac{1}{π} \underset{- π}{\int^{π}} \cos^{2} a t d t = 2 a_{0} + \underset{n = 1}{\sum^{\infty}} (a_{n}^{2} + b_{n}^{2})$ $\Rightarrow 1 + \frac{\sin 2 π a}{2 a} = 2 {(\frac{\sin π a}{a π})}^{2} + \underset{n = 1}{\sum^{\infty}} \frac{4 a^{2} \sin^{2} a π}{π^{2} {(n^{2} - a^{2})}^{2}}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{1}{{(n^{2} - a^{2})}^{2}} = \frac{π^{2}}{4 a^{2} \sin^{2} a π} [1 + \frac{\sin 2 π a}{2 a}] - \frac{1}{2 a^{4}}$
	Answered by mathmax by abdo last updated on 27/May/20
	$let developp f at fourier serie f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T)∫_([T]) f(x)cos(nx)dx =(1/π)∫_(−π) ^π cos(αx)cos(nx)dx =(2/π) ∫_0 ^π cos(αx)cos(nx)dx ⇒(π/2)a_n =(1/2)∫_0 ^π (cos(n+α)x +cos(n−α)x)dx ⇒πa_n =[(1/(n+α)) sin(n+α)x +(1/(n−α)) sin(n−α)x]_0 ^π =(1/(n+α))sin(nπ +απ)+(1/(n−α))sin(nπ−απ) =(((−1)^n sin(απ))/(n+α)) +((−(−1)^n sin(απ))/(n−α)) =(−1)^n sin(απ){(1/(n+α))−(1/(n−α))} =(−1)^n sin(απ)(((−2α)/(n^2 −α^2 ))) a_0 =(2/π)∫_0 ^π cos(αx)dx =(2/π)((1/α)sin(απ)) ⇒(a_0 /2) =((sin(πα))/(πα)) ⇒ cos(αx) =((sin(πα))/(πα)) −((2α)/π)Σ_(n=1) ^∞ (((−1)^n sin(απ))/(n^2 −α^2 )) cos(nx) ⇒ ((cos(αx))/(sin(πα))) =(1/(πα)) −((2α)/π) Σ_(n=1) ^∞ (((−1)^n cos(nx))/(n^2 −α^2 )) x=π ⇒cota(πα) =(1/(πα))−((2α)/π) Σ_(n=1) ^∞ (1/((n^2 −α^2 ))) ⇒ cotan(πα)−(1/(πα)) =−((2α)/π) Σ_(n=1) ^∞ (1/(n^2 −α^2 )) ⇒ ((cotan(πα))/α)−(1/(πα^2 )) =−(2/π) Σ_(n=1) ^∞ (1/(n^2 −α^2 )) ⇒ (d/dα)( ((cotan(πα))/α)−(1/(πα^2 ))) =(2/π) Σ_(n=1) ^∞ ((−2α)/((n^2 −α^2 )^2 )) =((−4α)/π) Σ_(n=1) ^∞ (1/((n^2 −α^2 )^2 )) ⇒ Σ_(n=1) ^∞ (1/((n^2 −α^2 )^2 )) =−(π/(4α))×(d/dα)(((cotan(πα))/α)−(1/(πα^2 ))) rest to finish the calculus$
	$let developp f at fourier serie f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx)$ $a_{n} = \frac{2}{T} \int_{[T]} f (x) \cos (nx) dx = \frac{1}{π} \int_{- π}^{π} \cos (α x) \cos (nx) dx$ $= \frac{2}{π} \int_{0}^{π} \cos (α x) \cos (nx) dx \Rightarrow \frac{π}{2} a_{n} = \frac{1}{2} \int_{0}^{π} (\cos (n + α) x + \cos (n - α) x) dx$ $\Rightarrow π a_{n} = {[\frac{1}{n + α} \sin (n + α) x + \frac{1}{n - α} \sin (n - α) x]}_{0}^{π}$ $= \frac{1}{n + α} \sin (n π + α π) + \frac{1}{n - α} \sin (n π - α π)$ $= \frac{{(- 1)}^{n} \sin (α π)}{n + α} + \frac{- {(- 1)}^{n} \sin (α π)}{n - α} = {(- 1)}^{n} \sin (α π) {\frac{1}{n + α} - \frac{1}{n - α}}$ $= {(- 1)}^{n} \sin (α π) (\frac{- 2 α}{n^{2} - α^{2}})$ $a_{0} = \frac{2}{π} \int_{0}^{π} \cos (α x) dx = \frac{2}{π} (\frac{1}{α} \sin (α π)) \Rightarrow \frac{a_{0}}{2} = \frac{\sin (π α)}{π α} \Rightarrow$ $\cos (α x) = \frac{\sin (π α)}{π α} - \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} \sin (α π)}{n^{2} - α^{2}} \cos (nx) \Rightarrow$ $\frac{\cos (α x)}{\sin (π α)} = \frac{1}{π α} - \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} \cos (nx)}{n^{2} - α^{2}}$ $x = π \Rightarrow cota (π α) = \frac{1}{π α} - \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{1}{(n^{2} - α^{2})} \Rightarrow$ $cotan (π α) - \frac{1}{π α} = - \frac{2 α}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - α^{2}} \Rightarrow$ $\frac{cotan (π α)}{α} - \frac{1}{π α^{2}} = - \frac{2}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - α^{2}} \Rightarrow$ $\frac{d}{d α} (\frac{cotan (π α)}{α} - \frac{1}{π α^{2}}) = \frac{2}{π} \sum_{n = 1}^{\infty} \frac{- 2 α}{{(n^{2} - α^{2})}^{2}} = \frac{- 4 α}{π} \sum_{n = 1}^{\infty} \frac{1}{{(n^{2} - α^{2})}^{2}} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{1}{{(n^{2} - α^{2})}^{2}} = - \frac{π}{4 α} \times \frac{d}{d α} (\frac{cotan (π α)}{α} - \frac{1}{π α^{2}}) rest to finish the calculus$
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