This is going to be a short section. We just need to have a brief discussion about
a couple of ideas that we’ll be dealing with on occasion as we move into the
next topic of this chapter.
Periodic Function
The first topic we need to discuss is that of a periodic
function. A function is said to be periodic with period T if the following
is true,
The following is a nice little fact about periodic
functions.
Fact 1
If f and g
are both periodic functions with period T
then so is and fg.

This is easy enough to prove so let’s do that.
The two periodic functions that most of us are familiar are
sine and cosine and in fact we’ll be using these two functions regularly in the
remaining sections of this chapter. So,
having said that let’s close off this discussion of periodic functions with the
following fact,
Fact 2
Even and Odd
Functions
The next quick idea that we need to discuss is that of even
and odd functions.
Recall that a function is said to be even if,
and a function is said to be odd if,
The standard examples of even functions are and while the standard examples of odd functions
are and . The following fact about certain integrals of
even/odd functions will be useful in some of our work.
Fact 3
Note that this fact is only valid on a “symmetric” interval,
i.e. an interval in the form . If we aren’t integrating on a “symmetric”
interval then the fact may or may not be true.
Orthogonal Functions
The final topic that we need to discuss here is that of
orthogonal functions. This idea will be
integral to what we’ll be doing in the remainder of this chapter and in the
next chapter as we discuss one of the basic solution methods for partial
differential equations.
Let’s first get the definition of orthogonal functions out
of the way.
Definition
Note that in the case of for the second definition we know that we’ll
get a positive value from the integral because,
Recall that when we integrate a positive function we know
the result will be positive as well.
Also note that the nonzero requirement is important because
otherwise the integral would be trivially zero regardless of the other function
we were using.
Before we work some examples there are a nice set of trig
formulas that we’ll need to help us with some of the integrals.
Now let’s work some examples that we’ll need over the course
of the next couple of sections.
Example 1 Show
that is mutually orthogonal on .
Solution
This is not too difficult to do. All we really need to do is evaluate the
following integral.
Before we start evaluating this integral let’s notice that
the integrand is the product of two even functions and so must also be
even. This means that we can use Fact
3 above to write the integral as,
There are two
reasons for doing this. First having a
limit of zero will often make the evaluation step a little easier and that
will be the case here. We’ll discuss
the second reason after we’re done with the example.
Now, in order to do this integral we’ll actually need to
consider three cases.
In this case the integral is very easy and is,
This integral
is a little harder than the first case, but not by much (provided we recall a
simple trig formula). The integral for
this case is,
Now, at the point we need to recall that n is an integer and so and our final value for the is,
The first two
cases are really just showing that if the integral is not zero (as it shouldn’t
be) and depending upon the value of n
(and hence m) we get different values
of the integral. Now we need to do the
third case and this, in some ways, is the important case since we must get
zero out of this integral in order to know that the set is an orthogonal set. So, let’s take care of the final case.
This integral is the “messiest” of the three that we’ve
had to do here. Let’s just start off
by writing the integral down.
In this case we
can’t combine/simplify as we did in the previous two cases. We can however, acknowledge that we’ve got
a product of two cosines with different arguments and so we can use one of
the trig formulas above to break up the product as follows,
Now, we know
that n and m are both integers and so and are also integers and so both of the sines
above must be zero and all together we get,
So, we’ve shown
that if the integral is zero and if the value of the integral is a positive
constant and so the set is mutually orthogonal.

In all of the work above we kept both forms of the integral
at every step. Let’s discuss why we did
this a little bit. By keeping both forms
of the integral around we were able to show that not only is mutually orthogonal on but it is also mutually orthogonal on . The only difference is the value of the
integral when and we can get those values from the
work above.
Let’s take a look at another example.
Example 2 Show
that is mutually orthogonal on and on .
Solution
First we’ll acknowledge from the start this time that
we’ll be showing orthogonality on both of the intervals. Second, we need to start this set at because if we used the first function would be zero and we
don’t want the zero function to show up on our list.
As with the first example all we really need to do is
evaluate the following integral.
In this case integrand is the product of two odd functions
and so must be even. This means that
we can again use Fact 3 above to write the integral as,
We only have
two cases to do for the integral here.
Not much to
this integral. It’s pretty similar to
the previous examples second case.
Summarizing up we get,
As with the previous example this can be a little messier
but it is also nearly identical to the third case from the previous example
so we’ll not show a lot of the work.
As with the
previous example we know that n and
m are both integers a and so both
of the sines above must be zero and all together we get,
So, we’ve shown
that if the integral is zero and if the value of the integral is a positive
constant and so the set is mutually orthogonal.

We’ve now shown that mutually orthogonal on and on .
We need to work one more example in this section.
Example 3 Show
that and are mutually orthogonal on .
Solution
This example is a little different from the previous two
examples. Here we want to show that
together both sets are mutually orthogonal on . To show this we need to show three
things. First (and second actually) we
need to show that individually each set is mutually orthogonal and we’ve
already done that in the previous two examples. The third (and only) thing we need to show
here is that if we take one function from one set and another function from
the other set and we integrate them we’ll get zero.
Also, note that this time we really do only want to do the
one interval as the two sets, taken together, are not mutually orthogonal on . You might want to do the integral on this
interval to verify that it won’t always be zero.
So, let’s take care of the one integral that we need to do
here and there isn’t a lot to do. Here
is the integral.
The integrand in this case is the product of an odd
function (the sine) and an even function (the cosine) and so the integrand is
an odd function. Therefore, since the
integral is on a symmetric interval, i.e.
,
and so by Fact 3 above we know the integral must be zero or,
So, in previous
examples we’ve shown that on the interval the two sets are mutually orthogonal
individually and here we’ve shown that integrating a product of a sine and a
cosine gives zero. Therefore, as a
combined set they are also mutually orthogonal.

We’ve now worked three examples here dealing with
orthogonality and we should note that these were not just pulled out of the air
as random examples to work. In the
following sections (and following chapter) we’ll need the results from these
examples. So, let’s summarize those
results up here.
We will also be needing the results of the integrals
themselves, both on and on so let’s also summarize those up here as well
so we can refer to them when we need to.
With this summary we’ll leave this section and move off into
the second major topic of this chapter : Fourier Series.