# REALLY fun problem

1. Feb 4, 2013

### iRaid

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Honestly no idea where to start, I don't even know what this means.

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2. Feb 4, 2013

### cepheid

Staff Emeritus
What part of it, exactly, is confusing you? I mean, you say "I don't know what 'this' means," but what does "this" refer to?

Have you considered that you are given an expression for f(x) in the first equation, and then f(x) also appears in the second equation? Seems to naturally suggest a way in which the two equations can be combined, doesn't it?

3. Feb 4, 2013

### iRaid

$$a_{m}=\frac{1}{\pi} \int_{-\pi}^\pi \displaystyle\sum_{n=1}^{N} a_{n}\:sin\,nx\:sin\,mx\:dx$$

Is that what you mean? I still have no idea what I would do with that

4. Feb 4, 2013

### cepheid

Staff Emeritus
Yeah, that's what I meant.

Well, for one thing, you can pull the summation sign outside of the integral sign, because there is a property of integration that says the integral of a sum of two or more functions is equal to the sum of the integrals of the individual functions.

So, now you can just focus your attention on evaluating the integral on the inside. Hint: consider two separate cases n = m, and n ≠ m, and try to evaluate the integral for each of those cases.

5. Feb 4, 2013

### iRaid

So you're saying:
$$a_{m}=\frac{1}{\pi} \displaystyle\sum_{n=1}^{N} a_{n}\:sin\,nx\: \int_{-\pi}^\pi sin\,mx\:dx$$

6. Feb 4, 2013

### cepheid

Staff Emeritus
NO. You can't take sin(nx) outside the integral, because it depends on x.

It seems like the summation notation is tripping you up, so let's consider a stupid trivial example. What if N = 2? Then:

$$a_{m}=\frac{1}{\pi} \int_{-\pi}^\pi \sum_{n=1}^{2} a_{n}\sin(nx)\sin(mx)\,dx$$

This sum only has two terms. One for n = 1 and one for n = 2. So it's not too tedious to expand the sum in full:

$$a_m = \frac{1}{\pi}\int_{-\pi}^\pi \left[ a_1 \sin(x)\sin(mx) + a_2\sin(2x)\sin(mx)\right]\,dx$$

Now, recall what I already said before (and something you probably already know):

The integral of a sum of functions is equal to the sum of the integrals of those functions

So our expression above becomes:

$$a_m = \frac{1}{\pi} \left[ a_1 \int_{-\pi}^\pi\sin(x)\sin(mx)\,dx + a_2\int_{-\pi}^\pi\sin(2x)\sin(mx)\,dx\right]$$

Now, if we were to put this BACK in summation notation, we would write it as:

$$a_m = \frac{1}{\pi}\sum_{n=1}^2 a_n\int_{-\pi}^{\pi} \sin(nx)\sin(mx)\,dx$$

Does that makes sense? That's why you can switch the Ʃ and the ∫ around. Because of this property of integration. The same thing holds for "N" terms in the summation.

7. Feb 5, 2013

### iRaid

Sorry it was really late and I wasn't thinking. Yes I see what you're saying.

So: 0 if m≠n and ∏ if m=n. Do I plug these in to prove it? I think my problem now is how to "prove" this.

8. Feb 5, 2013

### cepheid

Staff Emeritus
Yeah, you can now just focus on evaluating the integral that is inside the sum for those two cases. For n = m, you're integrating sine squared, which should be doable. For n ≠ m, you have something of the form sin(a)*sin(b), for which trig identities should help.

9. Feb 5, 2013

### iRaid

I did solve for both of those cases already. What do I do with this information now?

10. Feb 5, 2013

### cepheid

Staff Emeritus
Oh, well that part is obvious. What happens to all the terms in your sum if the the integral is 0 when n is not equal to m, and non-zero when n = m?

11. Feb 5, 2013

### Ray Vickson

The standard way is to use trigonometric addition formulas to replace $\sin (mx) \sin (nx)$ by sines/cosines of sums and differences. Alternatively, try using repeated integration by parts.

12. Feb 6, 2013

### iRaid

When the integral is 0, then am would be 0 right?

13. Feb 6, 2013

### cepheid

Staff Emeritus
THINK. The integral is not always 0. The integral is equal to 0 if n is NOT equal to m, and the integral is equal to pi if n = m. We *just* established that. So, which terms in the sum survive and which ones vanish?