 So how can we find the coefficients of a Fourier series? So remember our starting point. Suppose f is a function with fundamental period l, it's conceivable that f can be expressed as a Fourier series. How can we find the coefficients? While we could work with any period, let's assume f has period 2π. This allows us to write a series this way, which will simplify our computations. If f has a different period, we can then use a change of variables on x to scale to the desired period. So how can we find the coefficients? To begin with, here's a useful result. For any integer n, the integral from 0 to 2π of cosine nx dx is 0, and likewise the integral from 0 to 2π of sine nx dx is equal to 0. And what this suggests is that if I integrate from 0 to 2π, that we'll get rid of most of the sine and cosine terms over on the right-hand side. Now provided we have convergence, we can switch the integral and the summation. Now since n goes from 0 to infinity, most of these terms will, when we do the integral, give us 0, and the only one that won't is when n is equal to 0. And so the only portion of this integrand that matters is the a0 cosine 0x term. But since cosine of 0 is equal to 1, that's just a0, and when I integrate, I get 2π a0. And so that means I can solve for a0. And so a0 is 1 over 2π times the integral between 0 and 2π of f of x dx. We could extend this if we use a couple more results from the integration of trigonometric functions. For n not equal to m, the integral from 0 to 2π sine nx cosine mx will be 0. And the integral from 0 to 2π of sine of nx sine of mx is equal to 0. And the integral from 0 to 2π of cosine nx cosine mx is equal to 0. Now those results held when our arguments for sine and cosine were different. What if they're the same? So let's consider the integral from 0 to 2π of sine of nx cosine of nx, and that's equal to 0. Maybe things aren't as interesting as we thought they might be. But the integral from 0 to 2π of sine of nx sine of nx is equal to π, as is the integral from 0 to 2π of cosine nx times cosine nx. The useful thing to remember here is this. The only time the integral of a product of sines and cosines over the interval 0 to 2π isn't 0. 0 is when the trigonometric functions and the arguments are the same. Either sine of nx times sine of nx, or cosine of nx times cosine of nx. And in these cases the integral is equal to π. Suppose f of x has period 2π, and so it has the form of a trigonometric series. If we multiply by sine of kx and integrate over the interval between 0 and 2π, a lot of really useful things happen. Provided we have convergence, we can switch the order of the integration and the summation. And again, provided we have convergence, we can distribute the sine of kx among the individual terms. So remember, the only time the integral of a product of sines and cosines over the interval 0 to 2π isn't 0, is when the trigonometric functions are the same. So that means all of our cosine nx sine kx terms will vanish, and most of our sine nx sine kx terms will vanish. The only one that will be left is sine kx sine kx. And we can evaluate this integral. And consequently, bk is 1 over π the integral from 0 to 2π f of x sine kx. We can find the ak's in the same way, so ak is 1 over π integral from 0 to 2π f of x cosine kx dx. And these give us the coefficients of the Fourier series. So if we put all of these results together, we have the following theorem. Let f of x be a function with period 2π, then f of x can be expressed as the Fourier series, where our coefficients a0, the odd one, is going to be 1 over 2π times the integral from 0 to 2π f of x. And the others will be 1 over π times the integral from 0 to 2π of f of x times either the cosine of nx or the sine of nx. So again, an important idea is that as with all formulas, don't memorize formulas, understand concepts. And with the Fourier series, the basic concept is the following. If f is periodic with period 2π, we assume that f of x could be written as a trigonometric series. Then, assuming convergence, we found the coefficients ak and bk by integrating the function times cosine kx or the function times sine of kx, where most of the terms of the right-hand side will vanish. Let's take a look at how that works next.