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

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