 Before plunging into finding Fourier series for other functions, let's introduce the following useful concept. A function f is odd if f of x equals minus f of minus x for all x, and even if f of x equals f of minus x. The term even and odd come from our fundamental even and odd functions, which are f of x equals x squared, which has f of x equals f of minus x, and f of x equals x to the third, which has f of x equal minus f of minus x. And so our fundamental even function has an even exponent, and our fundamental odd function has an odd exponent. So let's determine whether f of x equals cosine of 5x is even or odd, and we observe that f of x equals cosine of 5x. f of minus x is, which simplifies to, and so f of x equals f of minus x for all x, and so f of x equals cosine of 5x is an even function. It can be difficult to identify odd and even functions algebraically, but if we graph y equals f of x, then, if f is odd, then f of minus x equals minus f of x, but if x, y is a point on the graph, we know that y equals f of x, and so f of minus x equals minus y, and so minus x minus y is also a point on the graph. And so for an odd function, if x, y is on the graph, so is minus x minus y, and this means the graph is symmetric about the origin. On the other hand, if f is even, then f of minus x equals f of x, and again, if x, y is a point on the graph, we know that y equals f of x, and so f of minus x equals y, so minus x, y is also a point on the graph. And so for an even function, if x, y is on the graph, so is minus x, y, and this means the graph is symmetric about the y-axis. So what's the big deal about even and odd functions? Since the sum of odd functions is odd, and the sum of even functions is even, we have the following theorem. If f is an odd function, the Fourier series will consist of sine terms only, and if f is an even function, the Fourier series will consist of cosine terms only. So let's try to find the Fourier series for this function, which might seem a little daunting in its description, but note that this function is equivalent to the following. If k is zero, then we're one between zero and pi, and again, if k is zero, then we're minus one between pi and two pi. If k is one, then we're one between two pi and three pi, and again, if k is equal to one, then we'll be minus one between three pi and four pi. And this behavior continues that infinitum, and so our graph is going to look like this, and if we look at the graph, we see that our function is an odd function with a fundamental period of two pi, and so we know the Fourier series consists of sine terms only. So assuming our function is a trigonometric series in sine only, we make the following observations. If n is equal to zero, sine of zero x is zero, so there is no b zero term, and a series can actually begin at one. And remember, if it's not written down, it didn't happen, so we'll make a note of that. And so we say that we find f of x equals a series in sine from one to infinity, where we don't yet know what the coefficients are. So let's figure those out. So remember, don't memorize formulas, understand concepts. Once we've assumed that our function has a form of a trigonometric series, remember that we can find the coefficients through integration. So we'll multiply our series by sine of kx and integrate. Assuming convergence allows us to interchange the summation and the integration, and the important idea here is that for n not equal to k, the integrals will evaluate to zero, leaving as our sole survivor the integral with sine kx. And this will be bk times pi. And so for k greater than zero, bk will be one over pi times the integral from zero to two pi f of x sine of kx. Now to evaluate this integral, we'll need to remember what our function is. And the important thing to recognize here is that the function value changes at pi. And so we split this integral to the part between zero and pi, and the part between pi and two pi. Between zero and pi, the function value is one. Between pi and two pi, our function value is minus one. So we'll evaluate each of these integrals. And it's worth noting that for any value of k, cosine of two k pi is going to equal one. And so we can simplify our expression a little bit. Again, if it's not written down, it didn't happen. So we make a note that our coefficients of the Fourier series are going to have the form one over pi times two over k minus two cosine k pi over k. Now while we could leave this in this form, we're not going to because we can do a little bit more simplification. And that emerges as follows. The value of cosine k pi depends on k. If k is even, cosine of k pi is equal to one. If k is odd, cosine of k pi is equal to negative one. So if k is even, our expression for bk will become zero. But if k is odd, our expression for k pi will be four over k pi. And so our series will look like, and if you want to show off a bit, you can rewrite our series in this form. And if possible, it's always a good idea to graph your series just to make sure it looks something like the function. So we have our original step function. If we graph the function equal to just the first term of the series, it looks like this. If we include the first two terms of the series we get. If we include the first three terms of the series we get. And it's certainly looking like our trigonometric series is going to converge to the step function.