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Could somebody please tell me if I've gotten this question correct? I'm unsure about my answer.

Consider the periodic function:

$$f(x)= \begin{cases} 0,\ -\pi \lt x \le 0\\ 1,\ 0\lt x\le \pi \end{cases}$$ with $f(x)=f(x+2\pi )$.

Find the Fourier series for $f(x)$.

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Here's my working:

$a_0=\frac {1}{2\pi }\int_{-\pi} ^\pi f(x) \,dx$

$a_0=\frac {1}{2\pi }(\pi )$

$a_0=\frac {1}{2}$

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$a_n=\frac {1}{\pi }\int_{-\pi }^\pi f(x)\cos nx\,dx$

$a_n=\frac {1}{\pi }[\int_{-\pi }^0\require{cancel}\cancelto{0}{0.\cos nx\,dx} +\int_0^{\pi }1.\cos nx\,dx]$

$a_n=\frac{1}{\pi }[\frac{1}{n}\sin nx]_0^{\pi }$

$a_n=\frac{1}{n\pi }[\cancelto{0}{\sin nx]_0^{\pi }}$

$a_n=0$

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$b_n=\frac {1}{\pi }\int_{-\pi }^\pi f(x)\sin nx\,dx$

$b_n=\frac {1}{\pi }[\int_{-\pi }^0\cancelto{0}{0.\sin nx\,dx} +\int_0^{\pi }1.\sin nx\,dx]$

$b_n=\frac{1}{\pi }[-\frac{1}{n}\cos nx]_0^{\pi }$

$b_n=-\frac{1}{n\pi }(\cos n\pi -{\cos n0})$

$b_n=-\frac{1}{n\pi }[(-1)^n-1]$

$b_n= \begin{cases} 0,\ \text{if n is even}\\ -\frac{2}{n\pi },\ \text{if n is odd} \end{cases}$

$\therefore Sf(x)=\frac {1}{2}-\frac{2}{\pi }\mathop{\LARGE \Sigma} _{n=odd}^\infty \frac{1}{n}\sin n x$

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If anyone could tell me either if I got this right or where I went wrong it'd be greatly appreciated.

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  • $\begingroup$ you just have a sign mistake for $b_n$ ... $\endgroup$ – Math-fun Jul 30 '15 at 7:36
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Some notes:

i) $$b_n = \begin{cases} 0, &\text{$n$ is even} \\[2ex] \dfrac 2{n\pi}, & \text{$n$ is odd} \end{cases}$$

ii) The Fourier series of $f$ is: $$Sf(x) \begin{array}[t]{l} = a_0 + \displaystyle \sum_{n=1}^{\infty} \left[a_n \cdot \cos \left(\dfrac{2 n \pi x}{2\pi}\right) + b_n \cdot \sin \left( \dfrac{2 n \pi x}{2\pi}\right)\right]\\[2ex] = \dfrac 12 + \dfrac 2\pi\cdot \displaystyle \sum_{n = 1,3,5,\ldots}^\infty \frac 1n\cdot sin (nx) \end{array}$$

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