1. ## Fourier Series Question

Think this is a pretty simple question, just getting stuck on the last section, here's what I've got:

Let $f$ be a periodic function with period 6, such that $f(x) = x \: \text{for} \: -3 < x \leq 3$. Calculate the Fourier series of $f$.

Odd function, so we know $a_0 = a_n = 0$.

$b_n = \frac{2}{3} \int_0^3 x \cos{\frac{\pi n}{3} x} \: dx$

...

$= \frac{6}{\pi^2n^2}(\cos{\pi n} - \cos{0})$

$= \frac{6}{\pi^2n^2}((-1)^n - 1)$.

However in the answer they've got:

$\frac{6}{\pi} \displaystyle\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin{\frac{\pi n}{3} x}$.

Just wondering how they get to the $\frac{6}{\pi} \frac{(-1)^{n+1}}{n}$, as far as I can see my integration's correct?

Thanks in advance for the help

2. Originally Posted by craig
Think this is a pretty simple question, just getting stuck on the last section, here's what I've got:

Let $f$ be a periodic function with period 6, such that $f(x) = x \: \text{for} \: -3 < x \leq 3$. Calculate the Fourier series of $f$.

Odd function, so we know $a_0 = a_n = 0$.

$b_n = \frac{2}{3} \int_0^3 x \cos{\frac{\pi n}{3} x} \: dx$

...

$= \frac{6}{\pi^2n^2}(\cos{\pi n} - \cos{0})$

$= \frac{6}{\pi^2n^2}((-1)^n - 1)$.

However in the answer they've got:

$\frac{6}{\pi} \displaystyle\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin{\frac{\pi n}{3} x}$.

Just wondering how they get to the $\frac{6}{\pi} \frac{(-1)^{n+1}}{n}$, as far as I can see my integration's correct?

Thanks in advance for the help

As far as I can see your answer's correct, and I wonder what happened to the squared pi and n in the denominator, not to mention

the numerator.

Is this from some book? Perhaps you misread the number of the question...?

Tonio

3. I don't think so. I'll try find the questions and post them here.

4. Here's the question:

And here's the answer. Sorry about the quality, this is how I received it. There's no indication as to how they've come by the answer as far as I can tell.

It's solution number 2 by the way.

Thanks for the relpy