Okay, in the previous two sections we’ve looked at Fourier
sine and Fourier cosine series. It is
now time to look at a Fourier series.
With a Fourier series we are going to try to write a series
representation for on in the form,
So, a Fourier series is, in some way a combination of the
Fourier sine and Fourier cosine series.
Also, like the Fourier sine/cosine series we’ll not worry about whether
or not the series will actually converge to or not at this point. For now we’ll just assume that it will
converge and we’ll discuss the convergence of the Fourier series in a later section.
Determining formulas for the coefficients, and ,
will be done in exactly the same manner as we did in the previous two
sections. We will take advantage of the
fact that and are mutually orthogonal on as we proved earlier. We’ll also need the following formulas that
we derived when we proved the two sets were mutually orthogonal.
So, let’s start off by multiplying both sides of the series
above by and integrating from L to L. Doing this gives,
Now, just as we’ve been able to do in the last two sections
we can interchange the integral and the summation. Doing this gives,
We can now take advantage of the fact that the sines and
cosines are mutually orthogonal. The
integral in the second series will always be zero and in the first series the
integral will be zero if and so this reduces to,
Solving for gives,
Now, do it all over again only this time multiply both sides
by ,
integrate both sides from L to L
and interchange the integral and summation to get,
In this case the integral in the first series will always be
zero and the second will be zero if and so we get,
Finally, solving for gives,
In the previous two sections we also took advantage of the
fact that the integrand was even to give a second form of the coefficients in
terms of an integral from 0 to L. However, in this case we don’t know anything
about whether will be even, odd, or more likely neither even
nor odd. Therefore, this is the only
form of the coefficients for the Fourier series.
Before we start examples let’s remind ourselves of a couple
of formulas that we’ll make heavy use of here in this section, as we’ve done in
the previous two sections as well.
Provided n in an integer then,
In all of the work that we’ll be doing here n will be an integer and so we’ll use
these without comment in the problems so be prepared for them.
Also don’t forget that sine is an odd function, i.e. and that cosine is an even function, i.e. . We’ll also be making heavy use of these ideas
without comment in many of the integral evaluations so be ready for these as
well.
Now let’s take a look at an example.
Example 1 Find
the Fourier series for on .
Solution
So, let’s go ahead and just run through formulas for the
coefficients.
Note that in
this case we had and This will happen on occasion so don’t get
excited about this kind of thing when it happens.
The Fourier
series is then,

As we saw in the previous example sometimes we’ll get and Whether or not this will happen will depend
upon the function and often won’t happen, but when it does don’t
get excited about it.
Let’s take a look at another problem.
Example 2 Find
the Fourier series for on .
Solution
Because of the piecewise nature of the function the work
for the coefficients is going to be a little unpleasant but let’s get on with
it.
At this point
it will probably be easier to do each of these individually.
So, if we put
all of this together we have,
So, we’ve
gotten the coefficients for the cosines taken care of and now we need to take
care of the coefficients for the sines.
As with the
coefficients for the cosines will probably be easier to do each of these
individually.
So, if we put
all of this together we have,
So, after all
that work the Fourier series is,

As we saw in the previous example there is often quite a bit
of work involved in computing the integrals involved here.
The next couple of examples are here so we can make a nice
observation about some Fourier series and their relation to Fourier sine/cosine
series
Example 3 Find
the Fourier series for on .
Solution
Let’s start with the integrals for .
In both cases
note that we are integrating an odd function (x is odd and cosine is even so the product is odd) over the
interval and so we know that both of these integrals
will be zero.
Next here is
the integral for
In this case we’re integrating an even function (x and sine are both odd so the product
is even) on the interval and so we can “simplify” the integral as
shown above. The reason for doing this
here is not actually to simplify the integral however. It is instead done so that we can note that
we did this integral back in the
Fourier sine series section and so don’t need to redo it in this
section. Using the previous result we
get,
In this case
the Fourier series is,

If you go back and take a look at Example 1 in the
Fourier sine series section, the same example we used to get the integral out
of, you will see that in that example we were finding the Fourier sine series
for on . The important thing to note here is that the
answer that we got in that example is identical to the answer we got here.
If you think about it however, this should not be too
surprising. In both cases we were using an odd function on and because we know that we had an odd
function the coefficients of the cosines in the Fourier series, ,
will involve integrating and odd function over a symmetric interval, ,
and so will be zero. So, in these cases
the Fourier sine series of an odd function on is really just a special case of a Fourier
series.
Note however that when we moved over to doing the Fourier
sine series of any function on we should no longer expect to get the same
results. You can see this by comparing
Example 1 above with Example
3 in the Fourier sine series section.
In both examples we are finding the series for and yet got very different answers.
So, why did we get different answers in this case? Recall that when we find the Fourier sine
series of a function on we are really finding the Fourier sine series
of the odd extension of the function on and then just restricting the result down to . For a Fourier series we are actually using
the whole function on instead of its odd extension. We should therefore not expect to get the
same results since we are really using different functions (at least on part of
the interval) in each case.
So, if the Fourier sine series of an odd function is just a
special case of a Fourier series it makes some sense that the Fourier cosine
series of an even function should also be a special case of a Fourier
series. Let’s do a quick example to
verify this.
Example 4 Find
the Fourier series for on .
Solution
Here are the integrals for the and in this case because both the function
and cosine are even we’ll be integrating an even function and so can
“simplify” the integral.
As with the previous example both of these integrals were
done in Example
1 in the Fourier cosine series section and so we’ll not bother redoing
them here. The coefficients are,
Next here is
the integral for the
In this case
the function is even and sine is odd so the product is odd and we’re
integrating over and so the integral is zero.
The Fourier
series is then,

As suggested before we started this example the result here
is identical to the result from Example 1 in
the Fourier cosine series section and so we can see that the Fourier cosine
series of an even function is just a special case a Fourier series.