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Question Number 204469 by DeArtist last updated on 18/Feb/24

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} = \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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