 So finding the coefficients of a Fourier series has a lot of work. Is there a way we can make the process more efficient? And we can do that by considering some properties of functions. And the following three theorems are particularly useful. The first theorem involves the algebra of even and odd functions. If f and g are odd functions, their product is even. On the other hand, if they're even functions, their product is even. And finally, if one function is even and the other function is odd, their product is odd. Now, by itself this isn't very useful, but we'll bring in another theorem, which reminds us that the coefficients of the Fourier series can be found by integrating over any interval that spans one full period. The significance of this theorem is that it means we can move the interval of integration to anywhere we want. Now, since even and odd functions are reflected... No pun intended. Well, maybe. Yeah, yeah, enough comments from the Penet Gallery. What was I saying? Oh yeah, odd and even functions are reflected in the graphs of the function and correspond to certain symmetries, which means that they also correspond to certain simplifications on the definite integral. And in particular, if f is odd and g is even, then the integral of an odd function over an interval symmetric about the origin is guaranteed to be zero, and the integral of an even function over an interval symmetric about the origin is twice the integral from zero to the endpoint. So, for example, let's find a Fourier series for the zigzag, which is just the absolute value of x with fundamental interval minus pi to pi. So, we'll graph this function. It requires us taking our absolute value of x graph and repeating it over and over again. And we see that this graph is symmetric about the y-axis, so the function is even, and that means the Fourier series includes cosine terms only. So, again, it's important to keep in mind don't memorize formulas, understand concepts. And in this case, the thing to remember is that when we integrate a trigonometric function of one complete period, then most of the time the value is going to be zero. And since the interval from minus pi to pi includes one complete period of our function, we'll take a look at the integral from minus pi to pi of our function. We've determined that our function is a Fourier series in cosine integrating from minus pi to pi covers one complete period of cosine. Since f is an even function, this integral from minus pi to pi is the same as two times the integral from zero to pi. Meanwhile, on the right-hand side, if n is not equal to zero, every one of these integrals will vanish, and that leaves just the a zero value. Now, on the right-hand side, we can find the value of that integral easily. Over on the left-hand side, remember our function over the interval between minus pi and pi is just the absolute value of x, and on the interval between zero and pi, the absolute value of x is just x itself, and so our integral becomes which we evaluate, and find a zero is pi over two. What if k is greater than zero? So we'll take our function and multiply by cosine of kx, and it will integrate from minus pi to pi. Assuming convergence, we're allowed to switch the integration and the summation, and remember the reason that we're able to find the Fourier coefficients is that for almost every value of n, this integral will be equal to zero. The only value of n that doesn't cancel out is going to be for n equals k, and so this infinite series collapses down to just the integral from minus pi to pi, ak cosine kx, cosine kx, and that's just equal to pi times ak. And so pi times ak is the integral from minus pi to pi of f of x cosine kx, or ak is one of our pi times this integral. Since f and cosine kx are both even functions, we know that their product is also going to be an even function. And so f of x cosine kx is even, which means the integral from minus pi to pi is going to be the same as two times the integral from zero to pi. And again, the function is the absolute value function, but on the interval between zero and pi, that absolute value function is just x. So to evaluate the integral from zero to pi of x cosine kx, we'll use integration by parts. We'll let u equals x, dv equals cosine kx, and that gives us completing the integration by parts and evaluating. And so we'll drop this into our formula for ak. And again, if it's not written down, it didn't happen. And again, while we could leave our coefficients in this form, we should show a little class and sophistication and try and evaluate them a little bit more specifically. If n is odd, then cosine of n pi is going to be negative one. And so an will be, on the other hand, if n is even, cosine of n pi is going to be one, and so an is going to be... And so our actual series is going to start out this way. And again, it's always nice to see how this series develops. So remember, we wanted to find the Fourier series for the zigzag. If I graph just the first term of the series, I get... If I graph the first two terms, I get... If we graph the first three terms, or the first four terms, and it appears that our Fourier series is in fact converging to this zigzag function.