Determine Coefficients of Fourier Sine Series in [0,π]

[Silversonic]

Homework Statement

This is for an exam I had a couple weeks ago. It was a part of a "show that" question and the answer I got was completely wrong. Yet oddly I've been given full marks for the question, and looking at the paper now I still don't see why I have been. Full solutions haven't been given, and I don't really want to ask my tutor.

Determine the coefficients a_{n} for the Fourier sine series;

x = \sum_{n=1}^{\infty} a_{n} sin(nx)

In the interval x = [0,\pi]Hence show that \pi = \sum_{r=0}^{\infty} \frac{4}{2r+1}(-1)^{r} = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + ...

Homework Equations

Just knowledge of orthogonal sine functions.

The Attempt at a Solution

This is what I've done, and cheers for anyone that's patient enough to look through.

We multiply both sides of the equation given by sin(mx) and integrate with respect to x over pi and zero. Due to the orthogonality of the sin functions, if n =/= m then the integral is zero. So we are left with;

\int_{0}^{\pi} xsin(nx) dx = \int_{0}^{\pi} a_{n} sin^{2}(nx) dxLooking at the right hand side first, we get;

\frac{1}{2} a_{n} \int_{0}^{\pi} 1- cos(2nx) dx

=
\frac{1}{2} a_{n} [x- \frac{1}{2n}sin(2nx)]_{0}^{\pi}

= \frac{1}{2} a_{n} \piNow to the left hand side we get;

\int_{0}^{\pi} xsin(nx) dx

= [-\frac{xcos(nx)}{n}]_{0}^{\pi} + \int_{0}^{\pi} \frac{cos(nx)}{n} dx

= -\frac{\pi cos(n\pi)}{n}So if n is even, cos(n\pi) = 1

If n is odd cos(n\pi) = -1

So the LHS is equal to

-\frac {\pi (-1)^{n}}{n}Equating the LHS to the RHS, I get;

a_{n} = -\frac {2(-1)^{n}}{n} So I have an equation for the co-efficients, if I shove this in the original equation I get;

x = \sum_{n=1}^{\infty} -\frac {2(-1)^{n}}{n} sin(nx)
Now for the second part of the question, I assume I'd shove in x = \pi, but doing that I get;

\pi = \sum_{n=1}^{\infty} -\frac {2(-1)^{n}}{n} sin(n\pi)

But sin(n\pi) is always zero since n is a natural number! So I would get \pi = 0?! Lol. That's where my answer is so massively wrong, and is nothing like the "show that" part. Yet I have full marks for the question. What have I done wrong?

 

[Stimpon]

Perhaps the question itself was wrong, it seems like it wouldn't matter would you get for a_{n} as you'd always get \pi=0. I suspect it should have been x{\in}[0,\frac{\pi}{2}]. Since \sin(n0)=\sin(n\pi)

If the lecturer doesn't know your name, perhaps ask and run away if your name is asked for.

 

[Silversonic]

Perhaps the question itself was wrong, it seems like it wouldn't matter would you get for a_{n} as you'd always get \pi=0. I suspect it should have been x{\in}[0,\frac{\pi}{2}]. Since \sin(n0)=\sin(n\pi)

If the lecturer doesn't know your name, perhaps ask and run away if your name is asked for.

Quite possibly. I'll see what I get with that interval. I've tried with cos(nx) in front of the a_n instead and not gotten the answer. If he's wrong though it's kind of ridiculous, I spent quite a bit of time faffing over why I was wrong with this question rather than going ahead and answering another (I ran out of time).

Maybe I'll anonymously e-mail him.

 

[Silversonic]

Gah nope. That didn't work. Can't see what I've done wrong. I'm really doubtful that my lecturer has asked a question wrong on a test that accounts for the module percentage - but it would make sense as to why I was given full marks. I'm really perplexed on how I was meant to attain the answer given.

 

[Silversonic]

Ahhhhhhhhhhhh I realize how.

Instead of putting x = \pi for the last part, I was meant to put x = \frac {\pi}{2}. This gives me the answer.

So I was wrong, but given full marks. I'll just leave it at that.I still don't understand why you can't directly use x = \pi to attain an the answer. Is it because I can't use the boundary values as values for my initial condition?

I didn't include it in my original post, but it was actually;

u_0 (x) = x = \sum_{n=1}^{\infty} a_{n} sin(nx)

 

[Stimpon]

I still say the question was wrong, as the question says that you have x in [0,pi] which clearly doesn't work, x in [0,pi) would have been fine though. Basically that series does not distinguish between x=0 and x=pi, so that's why you either can't have x=pi.

 

[Ray Vickson]

To represent x as a sin series, don't you need it to be an odd function, so shouldn't the interval be
-pi < x < pi?

RGV