how to find the Fourier series of f(x) = x , 0


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Question Number 77285 by jagoll last updated on 05/Jan/20

$how to find the$ $Fourier series of f (x) = x, 0 < x < \frac{1}{8}$
	Answered by john santu last updated on 05/Jan/20
	$because f(x) an odd function , we used the sine series. periode T = (1/8) and L = (1/(16)) f(x) = Σ_(n=1) ^∞ b_n sin (((nπx)/L)) b_n =(2/((1/16)))∫_0 ^(1/16) x sin(16nπx) dx = 32 {−(x/(16nπ)) cos (16nπx)+(1/((16nπ)^2 ))sin (16nπx)}∣_0 ^(1/16) = ((32)/(nπ))(−1)^n then we get f(x)= Σ_(n=1) ^∞ (((32)/(πn)))(−1)^n sin(16nπx) = −((32)/π)sin (16πx)+((16)/π)sin (32πx)−((32)/(3π))sin (48πx)+(8/π)sin (64πx)−...$
	$because f (x) an odd function$ $, we used the sine series .$ $periode T = \frac{1}{8} and L = \frac{1}{16}$ $f (x) = \underset{n = 1}{\sum^{\infty}} b_{n} \sin (\frac{n π x}{L})$ $b_{n} = \frac{2}{(1 / 16)} \underset{0}{\int^{1 / 16}} x \sin (16 n π x) dx$ $= 32 {- \frac{x}{16 n π} \cos (16 n π x) + \frac{1}{{(16 n π)}^{2}} \sin (16 n π x)} \underset{0}{\overset{1 / 16}{∣}}$ $= \frac{32}{n π} {(- 1)}^{n}$ $then we get$ $f (x) = \underset{n = 1}{\sum^{\infty}} (\frac{32}{π n}) {(- 1)}^{n} \sin (16 n π x)$ $= - \frac{32}{π} \sin (16 π x) + \frac{16}{π} \sin (32 π x) - \frac{32}{3 π} \sin (48 π x) + \frac{8}{π} \sin (64 π x) - . . .$
	Commented byjagoll last updated on 05/Jan/20

	$thanks you sir$
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