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Half range Fourier series

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Fourier series defined on an interval
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In mathematics, a half range Fourier series is a Fourier series defined on an interval [ 0 , L ] {\displaystyle } instead of the more common [ L , L ] {\displaystyle } , with the implication that the analyzed function f ( x ) , x [ 0 , L ] {\displaystyle f(x),x\in } should be extended to [ L , 0 ] {\displaystyle } as either an even (f(-x)=f(x)) or odd function (f(-x)=-f(x)). This allows the expansion of the function in a series solely of sines (odd) or cosines (even). The choice between odd and even is typically motivated by boundary conditions associated with a differential equation satisfied by f ( x ) {\displaystyle f(x)} .

Example

Calculate the half range Fourier sine series for the function f ( x ) = cos ( x ) {\displaystyle f(x)=\cos(x)} where 0 < x < π {\displaystyle 0<x<\pi } .

Since we are calculating a sine series, a n = 0   n {\displaystyle a_{n}=0\ \quad \forall n} Now, b n = 2 π 0 π cos ( x ) sin ( n x ) d x = 2 n ( ( 1 ) n + 1 ) π ( n 2 1 ) n 2 {\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }\cos(x)\sin(nx)\,\mathrm {d} x={\frac {2n((-1)^{n}+1)}{\pi (n^{2}-1)}}\quad \forall n\geq 2}

When n is odd, b n = 0 {\displaystyle b_{n}=0} When n is even, b n = 4 n π ( n 2 1 ) {\displaystyle b_{n}={4n \over \pi (n^{2}-1)}} thus b 2 k = 8 k π ( 4 k 2 1 ) {\displaystyle b_{2k}={8k \over \pi (4k^{2}-1)}}

With the special case b 1 = 0 {\displaystyle b_{1}=0} , hence the required Fourier sine series is

cos ( x ) = 8 π n = 1 n ( 4 n 2 1 ) sin ( 2 n x ) {\displaystyle \cos(x)={{8 \over \pi }\sum _{n=1}^{\infty }{n \over (4n^{2}-1)}\sin(2nx)}}


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