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Question Number 204469 by DeArtist last updated on 18/Feb/24
Given the function f(x) =  { ((2, 0< x <2)),((−2, −2 <x < 0)) :}  of period 4  (a)  sketch the graph of y = f(x) , for −6 < x < 6  (b) Find the Fourier coefficient a_0 , a_n , and b_n   (c) write down the Fourier series.   (d) hence show that  Σ_(n=1) ^∞ (((−1)^n )/(2n−1)) = (π/4)
$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2},\:\mathrm{0}<\:{x}\:<\mathrm{2}}\\{−\mathrm{2},\:−\mathrm{2}\:<{x}\:<\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{of}\:\mathrm{period}\:\mathrm{4} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:{y}\:=\:{f}\left({x}\right)\:,\:\mathrm{for}\:−\mathrm{6}\:<\:{x}\:<\:\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{coefficient}\:{a}_{\mathrm{0}} ,\:{a}_{{n}} ,\:\mathrm{and}\:{b}_{{n}} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{series}.\: \\ $$$$\left(\mathrm{d}\right)\:\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by witcher3 last updated on 19/Feb/24
a_0 =(1/4)∫_(−2) ^2 f(x)dx=(1/4)(∫_(−2) ^0 −2dx+∫_0 ^2 2dx)=0  a_n =(1/2)∫_(−2) ^2 f(x)cos(((πnx)/2))dx  =(1/2){∫_(−2) ^0 −2cos(((πnx)/2))xx+∫_0 ^2 2cos(((πnx)/2))dx}=0  b_n =(1/2){∫_(−2) ^0 −2sin(((πnx)/2))+∫_0 ^2 2sin(((πnx)/2))dx}  =(1/2)[(4/(πn))cos(((πnx)/2))]_(−2) ^0 +(1/2)[−(4/(πn))cos(((πnx)/2))]_0 ^2   =(2/(πn))[(1−(−1)^n −(−1)^n +1)= { ((0,n=2k)),((4 ;n=2k−1)) :}  f(x)=Σ_(n≥1) (8/(π(2n−1)))sin(((πx)/2)(2n−1));for x=1  2=Σ_(n≥1) (8/(π(2n−1)))sin(ππ−(π/2))=(8/π)Σ(((−1)^(n−1) )/((2n−1)))  ⇔Σ_(n≥1) (((−1)^(n−1) )/((2n−1)))=((2π)/8)=(π/4)
$$\mathrm{a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{4}}\int_{−\mathrm{2}} ^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\left(\int_{−\mathrm{2}} ^{\mathrm{0}} −\mathrm{2dx}+\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{2dx}\right)=\mathrm{0} \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{cos}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\int_{−\mathrm{2}} ^{\mathrm{0}} −\mathrm{2cos}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)\mathrm{xx}+\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{2cos}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)\mathrm{dx}\right\}=\mathrm{0} \\ $$$$\mathrm{b}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\int_{−\mathrm{2}} ^{\mathrm{0}} −\mathrm{2sin}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)+\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{2sin}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)\mathrm{dx}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{4}}{\pi\mathrm{n}}\mathrm{cos}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)\right]_{−\mathrm{2}} ^{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{\mathrm{4}}{\pi\mathrm{n}}\mathrm{cos}\left(\frac{\pi\mathrm{nx}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}}{\pi\mathrm{n}}\left[\left(\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} −\left(−\mathrm{1}\right)^{\mathrm{n}} +\mathrm{1}\right)=\begin{cases}{\mathrm{0},\mathrm{n}=\mathrm{2k}}\\{\mathrm{4}\:;\mathrm{n}=\mathrm{2k}−\mathrm{1}}\end{cases}\right. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{8}}{\pi\left(\mathrm{2n}−\mathrm{1}\right)}\mathrm{sin}\left(\frac{\pi\mathrm{x}}{\mathrm{2}}\left(\mathrm{2n}−\mathrm{1}\right)\right);\mathrm{for}\:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{2}=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{8}}{\pi\left(\mathrm{2n}−\mathrm{1}\right)}\mathrm{sin}\left(\pi\pi−\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{8}}{\pi}\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{2n}−\mathrm{1}\right)} \\ $$$$\Leftrightarrow\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\left(\mathrm{2n}−\mathrm{1}\right)}=\frac{\mathrm{2}\pi}{\mathrm{8}}=\frac{\pi}{\mathrm{4}} \\ $$
Commented by DeArtist last updated on 19/Feb/24
I appreciate your efforts
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{efforts} \\ $$

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