Given the square-wave function (later used to illustrate Gibbs phenomenon) $$f(x) = \left\{\begin{array}{c} \frac{h}{2} & 0 < x < \pi \\ -\frac{h}{2} & -\pi < x < 0 \end{array}\right.$$ I'm asked to show that the first $r$ terms of its Fourier series $$S_r (x) = \frac{a_0}{2} + \sum_{n=1}^{r} \left (a_n \cos nx + b_n \sin nx \right)$$ can be written as

$$S_r (x) = \frac{1}{2 \pi}\displaystyle\int_{-\pi}^{\pi} f(t) \frac{\sin \left( (r+1/2)(t-x)\right)}{\sin \left( (t-x)/2\right)} dt$$

Now I realize that since $f$ is odd we expect to have only $b_n$ terms in our sum, whence the sines. However I do not quite get the same result when using the standard definition of $b_n$. Is there some kind of change of variable I should perform to get it to the above form? And why is there a $(t-x)$ term in the arguments of the sines?

Any help would be appreciated.


Hint: $$b_n = 2\frac{h \sin ^2\left(\frac{n\pi}{2}\right)}{n\pi}=\frac{h(1-(-1)^n)}{n\pi}$$ So: $$S_{2r-1}=\sum_{n=1}^r \frac{2h}{(2n-1)\pi}\sin((2n-1)x)$$

Now, write out the Riemann sum for the function $\sin(x)/x$ on the interval $[0,\pi]$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.