The objective is to find the Fourier series of  on [-3, 3]. Since  is even on [-3, 3], we already know . That saves us a lot of working. We can now move on the find the other coefficients.

and for n = 1, 2, 3, …,

At this stage of the calculation, notice that the evaluated integrated on the left is equal to 0 by the identity . Be sure to make use for these ‘hidden’ identities. Now we are left with evaluating the integral on the right.

since . The Fourier series of  on [-3, 3] is

Remember way back in the earlier lessons on Fourier analysis, I emphasize the point to not be too quick in equating the function with the Fourier series. Well, let us take a look at that aspect using this function as an example.

Let x = 3. The function gives us . From the Fourier series, we know that we will have the 3 on the first term. What about the summation term? By letting x = 3, we evaluate it as

At our level, we have don’t have a way of evaluating this summation and all we can hope is that it is equal to 6. How about we try another value of x.

This time, let x = 3/2. Clearly from the function, . Again, we look at the summation term. This time, by substituting this value of x, the summation becomes,

Now things get a little more unclear. The value of  can either be -1, 0 and 1. Moreover, the  term doesn’t make things any easier for us. We have no idea at this point in confirming whether the Fourier series equals to 9/4.

Finally, let us look at another separate example. The Fourier series of  on  [-3, 3] is given by

We shall look at the endpoints of the Fourier series by substituting the value of x = 3. Our function gives us . However, putting x = 3 into the Fourier series, we get 0 because . It is now obvious at the function and Fourier series does not agree at this point. This sets the stage for a convergence theorem, a test as you may it, that tells use at what domain does the Fourier series actually converges to the function itself.