Fourier series coefficients for cosine terms

So we've been spending some time now thinking about the idea of a Fourier series, taking a periodic function and representing it as the sum of weighted cosines and sines. Some of you might say, "Well, how is this constant weighted cosine or sine?" Well, you could view a sub 0 as a sub 0 times cosine of 0t, and of course, cosine of 0t is just 1. So, it'll just end up being a sub 0. But that could be a weighted cosine if you want to view it that way.

In the last video, we started leveraging some of the integrals, definite integrals of sines and cosines to establish a formula for a sub 0, which intuitively ended up being the average value of our function over one of its periods, or over 0 to 2π, which is the interval that we are caring about. What I want to do now is find a general expression for a sub n where n is not equal to 0; so for n is greater than 0, for n being an integer greater than 0.

I'm going to use a very similar technique to figure out a sub 0. I just multiplied both sides by cosine of n t. Let me do that. On the left-hand side, I'm going to multiply by cosine of nt, and on the right-hand side, by multiplying it by cosine nt. If I distribute that cosine of nt, I'm going to multiply every term by cosine nt, so it's going to be cosine of nt times cosine nt. You might see where this is going, especially because we took all that trouble to figure out the properties of definite integrals of products of cosines and sines.

This is going to be true for all of the terms; there's an infinite number of terms here. That's what that dot dot dot is representing. We're just going to keep on going on and on and on. Now, let's take the definite integral of both sides from 0 to 2π. The definite integral from 0 to 2π dt is going to be the integral of this from 0 to 2π. And actually, let me just take that coefficient out of the integral.

So, 0 to 2π dt, 0 to 2π dt. A little bit monotonous, but it's worth it. Zero to two pi dt, zero to two pi dt. We might as well have fun while we do it. Zero to two pi dt. All right, my hand is sorting. Zero to two pi dt. Once again, just integration property. I'm taking the integral from zero, the definite go from zero to two pi of both sides, and I'm just saying, "Hey, the integral of this infinite sum is equal to the infinite sum of the different integrals."

Now, what is this going to be equal to? Well, we know from before that if I just take the definite integral from 0 to 2π of cosine nt, where n is some non-zero integer, this is just going to be zero. So, this whole thing is going to be zero. We also know, and we've established it before, if we take cosine nt times sine of mt for any integers m and t over the interval from 0 to 2π, that's going to be 0.

And we also know that if we take the integral of cosine mt and cosine nt, where m does not equal n, that's going to be zero. So, we know here we are assuming that n is not equal to 1, the coefficient here, so that is going to be equal to zero. This is sines and cosines with different coefficients. Frankly, even if they had the same coefficient, that's going to be zero; we established that in the last few videos.

This is going to be zero; same argument, this is going to be zero. This is going to be zero; everything's going to be zero except for this thing right over here. But what is this thing? This is the definite integral of cosine squared nt dt, the definite integral from 0 to 2π of cosine squared nt dt. Well, what is that? Well, we established that for any m, not non-zero m, this is going to be equal to π.

So, all our work is paying off. This whole integral, this whole integral right over here, is going to evaluate to π. We did that in another video, and so now we can start solving for a sub n. We know that a sub n times π is equal to is equal to this because everything else ended up being zero, which is very nice. It's equal to the definite integral from 0 to 2π of f of t, f of t times cosine nt dt.

So, we can now solve for a sub n. We just divide both sides by π, and we get a sub n is equal to, I think we deserve a little bit of a drum roll, a sub n is equal to 1 over π times all of this business times the definite integral from 0 to 2π f of t, f of t cosine nt dt. So, if you want to find the nth coefficient for one of our cosines a sub n, well, you take your function, multiply it times cosine of nt, and then take that definite integral from 0 to 2π, and then divide by π.

So, pretty intuitive, and what's cool is that the math kind of works out this way that you can actually do this. So, hopefully, you enjoyed that as much as I did.