In this section we are going to start taking a look at
Fourier series. We should point out that
this is a subject that can span a whole class and what we’ll be doing in this
section (as well as the next couple of sections) is intended to be nothing more
than a very brief look at the subject.
The point here is to do just enough to allow us to do some basic
solutions to partial differential equations in the next chapter. There are many topics in the study of Fourier
series that we’ll not even touch upon here.
So, with that out of the way let’s get started, although
we’re not going to start off with Fourier series. Let’s instead think back to our Calculus
class where we looked at Taylor Series.
With Taylor Series we wrote a series
representation of a function, ,
as a series whose terms were powers of for some . With some conditions we were able to show
that,
and that the series will converge to on for some R
that will be dependent upon the function itself.
There is nothing wrong with this, but it does require that
derivatives of all orders exist at . Or in other words exists for Also for some functions the value of R may end up being quite small.
These two issues (along with a couple of others) mean that
this is not always the best way of writing a series representation for a
function. In many cases it works fine
and there will be no reason to need a different kind of series. There are times however where another type of
series is either preferable or required.
We’re going to build up an alternative series representation
for a function over the course of the next couple of sections. The ultimate goal for the rest of this chapter
will be to write down a series representation for a function in terms of sines
and cosines.
We’ll start things off by assuming that the function, ,
we want to write a series representation for is an odd function (i.e. ).
Because is odd it makes some sense that we should be able
to write a series representation for this in terms of sines only (since they
are also odd functions).
What we’ll try to do here is write as the following series representation, called
a Fourier sine series, on .
There are a couple of issues to note here. First, at this point, we are going to assume
that the series representation will converge to on . We will be looking into whether or not it
will actually converge in a later section. However, assuming that the series does
converge to it is interesting to note that, unlike Taylor
Series, this representation will always converge on the same interval and that
the interval does not depend upon the function.
Second, the series representation will not involve powers of
sine (again contrasting this with Taylor Series) but instead will involve sines
with different arguments.
Finally, the argument of the sines, ,
may seem like an odd choice that was arbitrarily chosen and in some ways it
was. For Fourier sine series the
argument doesn’t have to necessarily be this but there are several reasons for
the choice here. First, this is the
argument that will naturally arise in the next chapter when we use Fourier
series (in general and not necessarily Fourier sine series) to help us solve
some basic partial differential equations.
The next reason for using this argument is the fact that
the set of functions that we chose to work with, in this case, need to be orthogonal on
the given interval, in this case, and note that in the last
section we showed
that in fact they are. In other words,
the choice of functions we’re going to be working with and the interval we’re
working on will be tied together in some way.
We can use a different argument, but will need to also choose an
interval on which we can prove that the sines (with the different argument) are
orthogonal.
So, let’s start off by assuming that given an odd function, ,
we can in fact find a Fourier sine series, of the form given above, to
represent the function on . This means we will have,
As noted above we’ll discuss whether or not this even can be
done and if the series representation does in fact converge to the function in
later section. At this point we’re
simply going to assume that it can be done.
The question now is how to determine the coefficients, ,
in the series.
Let’s start with the series above and multiply both sides by
where m
is a fixed integer in the range . In other words we multiply both sides by any
of the sines in the set of sines that we’re working with here. Doing this gives,
Now, let’s integrate both sides of this from to .
At this point we’ve got a small issue to deal with. We know from Calculus that an integral of a
finite series (more commonly called a finite sum….) is nothing more than the
(finite) sum of the integrals of the pieces.
In other words for finite series we can interchange an integral and a
series. For infinite series however, we
cannot always do this. For some
integrals of infinite series we cannot interchange an integral and a series. Luckily enough for us we actually can
interchange the integral and the series in this case. Doing this and factoring the constant, ,
out of the integral gives,
Now, recall from the last section we proved that is orthogonal on and that,
So, what does this mean for us. As we work through the various values of n in the series and compute the value of
the integrals all but one of the integrals will be zero. The only nonzero integral will come when we
have ,
in which case the integral has the value of L. Therefore, the only nonzero term in the
series will come when we have and our equation becomes,
Finally all we need to do is divide by L and we now have an equation for each of the coefficients.
Next, note that because we’re integrating two odd functions
the integrand of this integral is even and so we also know that,
Summarizing all this work up the Fourier sine series of an
odd function on is given by,
Let’s take a quick look at an example.
Example 1 Find
the Fourier sine series for on .
Solution
First note that the function we’re working with is in fact
an odd function and so this is something we can do. There really isn’t much to do here other
than to compute the coefficients for .
Here is that work and note that we’re going to leave the
integration by parts details to you to verify. Don’t forget that n, L, and are constants!
These integrals
can, on occasion, be somewhat messy especially when we use a general L for the endpoints of the interval
instead of a specific number.
Now, taking
advantage of the fact that n is an
integer we know that and that . We therefore have,
The Fourier
sine series is then,

At this point we should probably point out that we’ll be
doing most, if not all, of our work here on a general interval (
or ) instead of intervals with specific
numbers for the endpoints. There are a
couple of reasons for this. First, it
gives a much more general formula that will work for any interval of that form
which is always nice. Secondly, when we
run into this kind of work in the next chapter it will also be on general
intervals so we may as well get used to them now.
Now, finding the Fourier sine series of an odd function is
fine and good but what if, for some reason, we wanted to find the Fourier sine
series for a function that is not odd?
To see how to do this we’re going to have to make a change. The above work was done on the interval . In the case of a function that is not odd
we’ll be working on the interval . The reason for this will be made apparent in
a bit.
So, we are now going to do is to try to find a series
representation for on the interval that is in the form,
or in other words,
As we did with the Fourier sine series on we are going to assume that the series will in
fact converge to and we’ll hold off discussing the convergence
of the series for a later section.
There are two methods of generating formulas for the
coefficients, ,
although we’ll see in a bit that they really the same way, just looked at from
different perspectives.
The first method is to just ignore the fact that is not odd and proceed in the same manner that we
did above only this time we’ll take advantage of the fact that we proved in the previous
section that also forms an orthogonal set on and that,
So, if we do this then all we need to do is multiply both
sides of our series by ,
integrate from 0 to L and interchange
the integral and series to get,
Now, plugging in for the integral we arrive at,
Upon solving for the coefficient we arrive at,
Note that this is identical to the second form of the
coefficients that we arrived at above by assuming was odd and working on the interval . The fact that we arrived at essentially the
same coefficients is not actually all the surprising as we’ll see once we’ve
looked the second method of generating the coefficients.
Before we look at the second method of generating the
coefficients we need to take a brief look at another concept. Given a function, ,
we define the odd extension of to be the new function,
It’s pretty easy to see that this is an odd function.
and we can also know that on we have that . Also note that if is already an odd function then we in fact get
on .
Let’s take a quick look at a couple of odd extensions before
we proceed any further.
Example 2 Sketch
the odd extension of each of the given functions.
(a) on [Solution]
(b) on [Solution]
(c) [Solution]
Solution
Not much to do with these other than to define the odd
extension and then sketch it.
(a) on
Here is the odd extension of this function.
Below is the graph of both the function and its odd
extension. Note that we’ve put the
“extension” in with a dashed line to make it clear the portion of the
function that is being added to allow us to get the odd extension.
[Return to Problems]
(b) on
First note that this is clearly an even function. That does not however mean that we can’t
define the odd extension for it. The
odd extension for this function is,
The sketch of the original function and its odd extension
are ,
[Return to Problems]
(c)
Let’s first write down the odd extension for this
function.
The sketch of
the original function and its odd extension are,
[Return to Problems]

With the definition of the odd extension (and a couple of
examples) out of the way we can now take a look at the second method for
getting formulas for the coefficients of the Fourier sine series for a function
on .
First, given such a function define its odd extension as above. At this point,
because is an odd function, we know that on the Fourier sine series for (and NOT yet)
is,
However, because we know that on we can also see that as long as we are on we have,
So, exactly the same formula for the coefficients regardless
of how we arrived at the formula and the second method justifies why they are
the same here as they were when we derived them for the Fourier sine series for
an odd function.
Now, let’s find the Fourier sine series for each of the
functions that we looked at in Example 2.
Note that again we are working on general intervals here
instead of specific numbers for the right endpoint to get a more general
formula for any interval of this form and because again this is the kind of
work we’ll be doing in the next chapter.
Also, we’ll again be leaving the actually integration
details up to you to verify. In most
cases it will involve some fairly simple integration by parts complicated by
all the constants (n, L, ,
etc.) that show up in the integral.
Example 3 Find
the Fourier sine series for on .
Solution
There really isn’t much to do here other than computing
the coefficients so here they are,
In the
simplification process don’t forget that n
is an integer.
So, with the
coefficients we get the following Fourier sine series for this function.

In the next example it is interesting to note that while we
started out this section looking only at odd functions we’re now going to be
finding the Fourier sine series of an even function on . Recall however that we’re really finding the
Fourier sine series of the odd extension of this function and so we’re
okay.
Example 4 Find
the Fourier sine series for on .
Solution
In this case the coefficients are liable to be somewhat messy
given the fact that the integrals will involve integration by parts
twice. Here is the work for the
coefficients.
As noted above
the coefficients are not the most pleasant ones, but there they are. The Fourier sine series for this function
is then,

In the last example of this section we’ll be finding the
Fourier sine series of a piecewise function and can definitely complicate the
integrals a little but they do show up on occasion and so we need to be able to
deal with them.
Example 5 Find
the Fourier sine series for on .
Solution
Here is the integral for the coefficients.
Note that we
need to split the integral up because of the piecewise nature of the original
function. Let’s do the two integrals
separately
Putting all of
this together gives,
So, the Fourier
sine series for this function is,

As the previous two examples has shown the coefficients for
these can be quite messy but that will often be the case and so we shouldn’t
let that get us too excited.