Complex Fourier Series: Uncovering the Mystery of Different Results for x^2

[TehAdzMan]

Hello,

First post. I will attempt to use latex, something that involves me jabbing my keyboard with a pen since my \ key is missing.

We have an assignment question which I have solved, but there is a deeper issue I don't understand.

We are asked to find the complex Fourier series representation of

$$f(x) = x^2, \ -\pi < x < \pi, \ f(x+2\pi) = f(x)$$

Initially I did the working, used integration by parts twice, and got the result

$$f(x) \sim \sum \sinh (in\pi)(\frac{\pi^2 n^2 + 2\pi in+2}{\pi i n^3})e^{-inx})$$
summed for n between negative infinity to infinity.

This is obviously incorrect because $\sinh (in\pi) = 0$.

I received a tip off about how to go about solving the question, and early in the working I split the complex exponential term into its cos and isin terms.
Because the region of integration is symmetric and $x^2 \times i\sin(nx)$
is odd, this term becomes 0 and the complex coefficients become real, the working proceeds and I get the correct answer which is
$$f(x) \sim \frac{\pi^2}{3} + 2\sum \frac{(-1)^n}{n^2} e^{-inx}$$

where the sum is over all infinity again, except n = 0 which was worked out seperately.

I've omitted the working because it took so long just to do that, but if I need to I can show it.

I'm pretty sure the working in both cases is right.
What I don't understand is how they come to different answers, excluding for n = 0, which we work out separately, and is the same in either case.
The first way comes up with a sum of 0s, unless I've done something wrong.

I could understand if the first method returned an undefined answer.
Then it is just undefined and we have to try something else to get a defined answer.
But in both cases (excluding n = 0 obviously), the sums are well defined as far as I can tell.

Why are they different? Have I just done something wrong in the working or is there something I don't understand happening?

Adam.

 

[chiro]

Hey TehAdzMan.

Can you show us the working for the first problem?

 

[TehAdzMan]

Thanks for replying.

Ok so here is the working for the first, incorrect part.

$$f(x) \sim \sum^{\infty}_{n = -\infty} C_n e^{-inx}, \ where \ C_n = \frac{1}{2 \pi} \int^{\pi}_{-\pi} x^2 e^{inx} dx \\$$

Proceeding to calculate Cn using integration by parts

$$C_n = \frac{1}{2 \pi} \{ \left[ \frac {x^2 e^{inx}}{in} \right]^{\pi}_{-\pi} - \int^{\pi}_{-\pi} \frac {2xe^{inx}}{in} dx \} \\$$

and a second application of integration by parts

$$= \frac{1}{2 \pi} \{ \frac{\pi^2 e^{in \pi} - \pi^2 e^{-in \pi}}{in} - \frac{2}{in} \left[ \left[ \frac {xe^{inx}}{in} \right]^{\pi}_{-\pi} - \int^{\pi}_{ -\pi} \frac{e^{inx}}{in} dx \right] \} \\ = \frac{1}{2 \pi} \{ \frac{\pi^2}{in} (e^{in \pi} - e^{-in \pi}) - \frac{2}{in} \left[ \frac{\pi e^{in \pi}}{in} + \frac{\pi e^{-in \pi}}{in} - \left[ \frac {e^{inx}}{i^2 n^2} \right]^{\pi}_{-\pi} \right] \} \\ = \frac{1}{2 \pi} \{ \frac{2 \pi^2}{in} \sinh (in \pi) - \frac{4 \pi}{i^2 n^2} \sinh (in \pi) - \frac{2}{i^3 n^3} (e^{in \pi} - e^{-in \pi}) \}$$

which, with some arrangement gives

$$C_n = \sinh (in\pi)(\frac{\pi^2 n^2 + 2\pi in+2}{\pi i n^3})$$

So basically sinh(in pi) is coming out at every step, which = 0.

The correct calculation just involves splitting the complex exponential up into cos and i sin, getting rid of the i sin for the aforementioned reason, and proceeding in a very similar fashion, using integration by parts twice.

Is something wrong in that working?

 

[TehAdzMan]

Ok there was a working error. My bad.
It turns out it is the same.

I learned latex tho so that's good.
I'm trying to work out how to delete this or mark it as DONT READ or something now.

 

[TehAdzMan]

Why can I only edit my last post? Can I change the thread name etc?
Can I delete the post to rid this forum of a useless post?
I looked for this info but couldn't find it.