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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<(1/8)

$$\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\: \\ $$ $$\mathrm{Fourier}\:\mathrm{series}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}\:,\:\mathrm{0}\:<\:\mathrm{x}<\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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)−...

$$\mathrm{because}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function}\: \\ $$ $$,\:\mathrm{we}\:\mathrm{used}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{series}. \\ $$ $$\mathrm{periode}\:\mathrm{T}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\:\mathrm{and}\:\mathrm{L}\:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{b}_{\mathrm{n}} \mathrm{sin}\:\left(\frac{\mathrm{n}\pi\mathrm{x}}{\mathrm{L}}\right) \\ $$ $$\mathrm{b}_{\mathrm{n}} =\frac{\mathrm{2}}{\left(\mathrm{1}/\mathrm{16}\right)}\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{16}} {\int}}\mathrm{x}\:\mathrm{sin}\left(\mathrm{16n}\pi\mathrm{x}\right)\:\mathrm{dx} \\ $$ $$=\:\mathrm{32}\:\left\{−\frac{\mathrm{x}}{\mathrm{16n}\pi}\:\mathrm{cos}\:\left(\mathrm{16n}\pi\mathrm{x}\right)+\frac{\mathrm{1}}{\left(\mathrm{16n}\pi\right)^{\mathrm{2}} }\mathrm{sin}\:\left(\mathrm{16n}\pi\mathrm{x}\right)\right\}\underset{\mathrm{0}} {\overset{\mathrm{1}/\mathrm{16}} {\mid}} \\ $$ $$=\:\frac{\mathrm{32}}{\mathrm{n}\pi}\left(−\mathrm{1}\right)^{\mathrm{n}} \\ $$ $$\mathrm{then}\:\mathrm{we}\:\mathrm{get}\: \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{32}}{\pi\mathrm{n}}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}\left(\mathrm{16n}\pi\mathrm{x}\right)\: \\ $$ $$=\:−\frac{\mathrm{32}}{\pi}\mathrm{sin}\:\left(\mathrm{16}\pi\mathrm{x}\right)+\frac{\mathrm{16}}{\pi}\mathrm{sin}\:\left(\mathrm{32}\pi\mathrm{x}\right)−\frac{\mathrm{32}}{\mathrm{3}\pi}\mathrm{sin}\:\left(\mathrm{48}\pi\mathrm{x}\right)+\frac{\mathrm{8}}{\pi}\mathrm{sin}\:\left(\mathrm{64}\pi\mathrm{x}\right)−... \\ $$

Commented byjagoll last updated on 05/Jan/20

thanks you sir

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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