Fourier coefficients for sine terms

Many videos ago we first looked at the idea of representing a periodic function as a set of weighted cosines and sines, as a sum, as the infinite sum of weighted cosines and sines.

Then we did some work in order to get some basics in terms of some of these integrals, which we then started to use to derive formulas for the various coefficients. We are almost there; we have figured out a formula for a sub 0. We figured out a coefficient in general for any of the coefficients on the cosines, and now let's figure out a way to figure out the coefficients on the sines.

So how can we do that? Well, we're going to use a very similar technique that we did to figure out the formula for a sub n. What we're going to do is multiply both sides of this equation right over here by sine of n t. So let's do that — sine of n t.

And so if we multiply the right-hand side by sine of n t, that means we're going to multiply every one of the infinite terms by sine of n t. So sine of n t, sine of n t. I think you—well, this is always a little time consuming, but I'll get there. Sine of n t, sine of n t, sine of n t, sine of n t, sine of n t. And of course, there's these dot dot dots, so we're just showing in general we're taking each of these terms and multiplying by sine of n t.

We'll do it for all of the terms; we're going to keep on doing this on and on forever for an infinite number of terms. Well now, let's take the integral of both sides of this equation, the definite integral from 0 to 2 pi. And once again, I'm picking that interval because 2 pi is the period of the periodic function that we care about, and everything else that I've been doing has been over that interval of 2 pi.

So I'll just keep focusing it on that from 0 to 2 pi dt. And I could say the integral of this whole thing, but we know that the integral of a sum of a bunch of things is also the same thing as the sum of the integrals. We also know that we could take these constant multipliers of expressions outside of integrals.

So this is going to be equal to—if I take the integral of the right-hand side, the definite integral is going to be from 0 to 2 pi dt, 0 to 2 pi dt, 0 to 2 pi dt, 0 to 2 pi dt. We're almost there—0 to 2 pi dt, 0 to 2 pi dt, and 0 to 2 pi dt.

And I would be doing this for every term in this Fourier expansion. Now, this is where some of that integration work is going to be valuable. We've already shown that the definite integral from zero to two pi of sine of nt dt is going to be equal to zero for n being any integer. So we saw that, we saw that right over there.

So that tells us that that term is going to be 0; so that's going to be equal to 0. And we also know that when we take a cosine of something times t, times the sine of something times t, where these are integers, and you take that definite integral, we know that's going to be zero. So all of those are going to be a zero.

And we also know that when you take the sine of two different non—two different coefficients right over, or sine of some integer coefficient times t times sine of some other integer coefficient times t, that these are also going to be equal to zero. We saw that; we saw that right over here—that is this property.

And so really, all you're left with, the only one that does not become 0 is this term right over here where it's sine of nt times sine of nt. And what is that going to be equal to? Well, this inside of the integral—that's the same thing. Let me do this in a different color since I'm using orange to cross out everything that ends up becoming 0.

So that is the same thing as the definite integral from 0 to 2 pi of sine squared of nt dt. And I'll do the—or the dt is still there, so I'm just replacing the blue stuff with the sine squared of nt. And we know what that's going to be; we know that when the coefficient on the t is a non-zero integer, that this is going to result in pi.

So we know that all of this stuff right over here—let me do that in a different color—all of that stuff is going to evaluate—or actually, maybe all of this stuff is going to become pi. So we know that b sub n times pi is going to be equal to this definite integral. Or we could write—let me write it this way: b sub n times pi is equal to, because everything else ends up becoming 0, is going to be equal to 0, the definite integral from 0 to 2 pi.

I'll put my dt out here of f of t, f of t times sine of nt, sine of n t dt. And so we could then divide both sides by—divide both sides by pi, and we get a little bit of a drum roll. We get b sub n; actually, I could write that in that same blue color—we get b sub n; I'll do it actually right here.

b sub n is going to be equal to 1 over pi times the definite integral from 0 to 2 pi, 0 to 2 pi. Remember dt here, you have your f of t here, and then you have your sine, sine n t. So once again, that was pretty straightforward now that we knew these properties of integrals, so we knew how to evaluate these integrals that are definite integrals where we're taking of cosine or sine or products of cosines and sines.

Using these three formulas, we can now attempt to find the Fourier expansion, the Fourier series, find the coefficients for our square wave.