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Current time:0:00Total duration:6:08

Fourier coefficients for cosine terms

Video transcript

- [Voiceover] 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, and some of you might say, well, how is this constant a weighted cosine or sine? Well, you can view a sub zero as a sub zero times cosine of zero t, and of course cosine of zero t is just 1, so this, it'll just end up being a sub zero, but that could be a weighted cosine, if you want to view it that way, and in the last video, we started leveraging some of the integrals, definite integrals of sines and cosines to establish a formula for a sub zero which intuitively ended up being the average value of our function over, over 1 of its period, or over zero to 2 pi 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 zero. So for n is greater than zero, for n is an integer greater than zero and I'm going to use a very similar, a very similar technique. To figure out a sub zero, I just multiply, I just took the integral of both sides. What I'm going to do now is I'm going to multiply both sides of this equation by cosine of nt. So let me do that, so on the left-hand side, I'm going to multiply by cosine of nt, and on the right-hand side, multiplying it by cosine of 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, cosine of nt, you might see where this is going, especially 'cause we took all that trouble to take, to figure out the properties of, of definite integrals. Cosine of nt, cosine of nt, cosine of nt, or the properties of definite integrals of products of cosines and sines. Cosine of nt. And 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. And now, let's take the definite integral of both sides from zero to 2 pi. So the definite integral from zero to 2 pi, dt, well that's going to be the integral of this, from zero to 2 pi, and actually, let me just take that coefficient out of the integral, so zero to 2 pi, dt, so zero to 2 pi, dt, a little bit monotonous but it's worth it, zero to 2 pi, dt, zero to 2 pi, dt, we might as well have fun while we do it, zero to 2 pi, dt, all right. Zero, my hand is hurting, zero to 2 pi, dt, zero to 2 pi, dt, once again, just integration property. I'm taking the integral from zero, the definite integral from zero to 2 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, we know from before that if I just take the definite integral from zero to 2 pi 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, so I just used this property, we also know if we take cosine nt times sine of mt for any integers mt over the interval from zero to 2 pi, that's going to be zero, and we also know, we also know that if we take the integral of cosine mt, cosine nt, where m does not equal n, where m does not equal n, then that is going to be zero. And so, we know, well, here, we have, we're assuming that, we're assuming here 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 the different coefficients, and frankly, 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, what we're taking the integral of, this is the same thing as cosine-squared of nt, dt, the definite integral from zero to 2 pi of cosine-squared nt, dt. Well, what is that? Well, we established that for any m, non-zero m, this is going to be equal to pi. So all our work is paying off. So, this whole integral, this whole integral right over here is going to evaluate to pi. We did that in another video and so now we can start solving for a sub n. We know that a sub n times pi, a sub n times pi, times pi is equal to, is equal to this 'cause everything else ended up being zero which is very nice. It's equal to the definite integral from zero to 2 pi of f of t, of f of t times cosine nt, cosine of nt, dt. Cosine of nt, dt, and so we can now solve for a sub n. We just divide both sides by, we just divide both sides by pi, 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 pi, times all of this business, times the definite integral from zero to 2 pi, f of t, f of t, cosine nt, cosine nt, dt. 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 zero to 2 pi, and then divide by pi, 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.