Solve Fun Problem: Honestly No Idea Where to Start

[iRaid]

Homework Statement

Homework Equations

The Attempt at a Solution

Honestly no idea where to start, I don't even know what this means.

 

[cepheid]

Homework Statement

Homework Equations

The Attempt at a Solution

Honestly no idea where to start, I don't even know what this means.

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?

 

[iRaid]

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?

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

 

[cepheid]

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

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.

 

[iRaid]

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.

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

 

[cepheid]

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

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.

 

[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 into prove it? I think my problem now is how to "prove" this.

 

[cepheid]

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 into prove it? I think my problem now is how to "prove" this.

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.

 

[iRaid]

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.

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

 

[cepheid]

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

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?

 

[Ray Vickson]

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 into prove it? I think my problem now is how to "prove" this.

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.

 

[iRaid]

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?

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

 

[cepheid]

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

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?