# 3. Fourier Series of Even and Odd Functions

This section can make our lives a lot easier because it reduces the work required.

### Revision

In some of the problems that we encounter, the Fourier coefficients ao, an or bn become zero after integration.

Finding zero coefficients in such problems is time consuming and can be avoided. With knowledge of even and odd functions, a zero coefficient may be predicted without performing the integration.

## Even Functions

Recall: A function y = f(t) is said to be even if f(-t) = f(t) for all values of t. The graph of an even function is always symmetrical about the y-axis (i.e. it is a mirror image).

### Example 1 - Even Function

f(t) = 2 cos πt

The graph of f(t) = 2 cos πt, which has amplitude 2 and period 2.

Continues below

## Fourier Series for Even Functions

For an even function f(t), defined over the range -L to L (i.e. period = 2L), we have the following handy short cut.

Since

b_n=1/Lint_(-L)^Lf(t)\ sin {:(npit)/L:}dt

and

f(t) is even,

it means the integral will have value 0. (See Properties of Sine and Cosine Graphs.)

So for the Fourier Series for an even function, the coefficient bn has zero value:

b_n= 0

So we only need to calculate a0 and an when finding the Fourier Series expansion for an even function f(t):

a_0=1/Lint_(-L)^Lf(t)dt

a_n=1/Lint_(-L)^Lf(t)cos{:(n pi t)/L:}dt

An even function has only cosine terms in its Fourier expansion:

f(t)=a_0/2+sum_(n=1)^oo\ a_n\ cos (n pi t)/L

## Fourier Series for Odd Functions

Recall: A function y = f(t) is said to be odd if f(-t) = - f(t) for all values of t. The graph of an odd function is always symmetrical about the origin.

### Example 2 - Odd Function

f(t) = sin t

The graph of f(t) = sin t, which has amplitude 1 and period 2pi.

## Fourier Series for Odd Functions

For an odd function f(t) defined over the range -L to L (i.e. period = 2L), we find that a_n= 0 for all n.

We have:

a_n=1/Lint_(-L)^Lf(t)\ cos{:(n pi t)/L:}dt

So the zero coefficients in this case are: a_0= 0 and a_n= 0. The coefficients b_n are given by:

b_n=1/Lint_(-L)^L\ f(t)\ sin{:(n pi t)/L:}dt

The Fourier Series for an odd function is:

f(t)=sum_(n=1)^oo\ b_n\ sin{:(n pi t)/L:}

An odd function has only sine terms in its Fourier expansion.

## Exercises

1. Find the Fourier Series for the function for which the graph is given by:

Graph of an odd periodic square wave function.

First, we need to define the function after observing the graph:

f(t)={(-3, ,-pi {:<=:}t <0),(3, ,0 {:<=:}t <pi) :}

We can see from the graph that it is periodic, with period 2pi.

So f(t) = f(t + 2π).

Also, L=pi.

We can also see that it is an odd function, so we know a_0= 0 and a_n= 0. So we will only need to find bn.

Since L=pi, the necessary formulae become:

b_n=1/piint_(-pi)^pif(t) sin nt\ dt

f(t)=sum_(n=1)^oob_n sin nt

Now

b_n=1/pi int_(-pi)^pif(t)sin nt\ dt

=1/pi(int_(-pi)^0 -3\ sin nt\ dt {:+int_0^pi 3\ sin nt\ dt)

=3/pi([(cos nt)/n]_(-pi)^0+[-(cos nt)/n]_0^pi)

=3/(pi n)(cos 0-cos(-pi n) -cos n pi {:+cos 0)

=3/(pi n)(2-cos pi n-cos pi n)

=6/(pi n)(1-cos pi n)

=12/(pi n)\ (n\ "odd") or

=0\ (n\ "even")

We could write this as: b_n=12/(pi(2n-1))\ \ n=1,2,3... (Substitute n=1,2,3... to see how this works.)

So the Fourier series for our odd function is given by:

f(t)=sum_(n=1)^oo b_n\ sin nt

=sum_(n=1)^oo 12/(pi(2n-1))sin(2n-1)t

=12/pisum_(n=1)^oo (sin(2n-1)t)/((2n-1))

NOTE: Since bn is non-zero for n odd, we must also have odd multiples of t within the sine expression (the even ones are multiplied by 0, so will be 0).

Checking, we graph the first 5 terms. It should closely resemble the square wave we started with.

12/pi sum_(n=1)^5 (sin(2n-1)t)/((2n-1))

=12/pi(sin t +1/3 sin 3t +1/5 sin 5t +1/7 sin 7t {:+1/9 sin 9t)

Graph of f(t), the Fourier series approximation of a square wave.

We see that the graph of the first 5 terms is certainly approaching the shape of the graph that was in the question. We can be confident we have the correct answer.

Easy to understand math videos:
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2. Sketch 3 cycles of the function represented by

f(t)= {(0",", if , -1<=t<-1/2),(cos 3 pi t",",if , -1/2<=t<1/2),(0",",if , 1/2<=t<1):}

and f(t) = f(t + 2).

Find the Fourier Series for this function.

First, we sketch the curve:

This function is an even function, so b_n= 0. We only need to find a0 and an.

a_0 = 1/Lint_(-L)^Lf(t)dt

=int_(-1)^1f(t)dt

=int_(-1)^(-0.5)0\ dt +int_(-0.5)^0.5cos 3pi t\ dt +int_0.5^1 0\ dt

=[(sin 3pi t)/(3 pi)]_(-0.5)^0.5

=-2/(3pi)

Now for an. We will use Scientific Notebook to perform the integration:

a_n=1/Lint_(-L)^Lf(t) cos {:(n pi t)/L:}dt

=int_(-1)^1f(t)\ cos n pi t\ dt

=0 +int_(-0.5)^(0.5)cos 3 pi t\ cos n pi t\ dt +0

=6 (cos {:(n pi)/2 :})/(pi(-9+n^2))

(We used Scientific Notebook for the final answer.)

Recall that cos((nπ)/2) = 0 for n odd and +1 or -1 for n even. (See the Helpful Revision page.)

So we expect 0 for every odd term.

However, we cannot have n = 3 in this expression, since the denominator would be 0. In this situation, we need to integrate for n = 3 to see if there is a value. In fact, we will use SNB to find the values up to n = 5, to see what is happening:

When n=1,

int_(-1//2)^(1//2)cos 3 pi t\ cos pi t\ dt=0

When n=2,

int_(-1//2)^(1//2)cos 3 pi t\ cos 2pi t\ dt=6/(5pi)

When n=3,

int_(-1//2)^(1//2)cos 3 pi t\ cos 3pi t\ dt=1/2

(We see there is in fact a value for the integral when n=3.)

When n=4,

int_(-1//2)^(1//2)cos 3 pi t\ cos 4pi t\ dt=6/(7pi)

When n=5,

int_(-1//2)^(1//2)cos 3 pi t\ cos 5pi t\ dt=0

So we will start our series by writing out the terms for n = 2 and n = 3, then use summation notation from n = 4:

f(t)=a_0/2+sum_(n=1)^oo a_n cos {:(n pi t)/L:} +sum_(n=1)^oo b_n sin {:(n pi t)/L:}

=a_0/2+6/(5pi) cos 2 pi t+1/2 cos 3pit +sum_(n=4)^oo a_n cos {:(npit)/L:}

=-1/(3pi)+(6\ cos 2pit)/(5pi) +(cos 3pit)/2 +sum_(n=4)^oo 6 (cos{:(npi)/2:})/(pi(-9+n^2)) cos npit

As usual, we graph the first few terms and see that our series is correct:

Graph of f(t), the Fourier series approximation of our given periodic function.

#### Solution without Scientific Notebook:

The integration for an could have been performed as follows. We re-express the function using a trick based on what we learned in Sum and Difference of Two Angles.

In general, cos(A+B) =cosAcosB-sinAsinB.

If we let

A=3pit and B=npit,

then the cos expression becomes:

cos(3pit+npit) =cos3pit cos npit - sin3pit sin npit ... (1)

Taking the negative case:

cos(3pit-npit) =cos3pit cos npit + sin3pit sin npit ... (2)

Adding equations (1) and (2) gives:

cos(3pit+npit)+cos(3pit-npit) =2cos3pit cos npit

Reversing sides, dividing both sides by 2, and factoring, we have:

cos 3pi t cos n pi t =1/2[cos(3+n)pit+cos(3-n)pit]

So our integral becomes

int_(-1//2)^(1//2)cos 3pit\ cos npi t\ dt

=1/2int_(-1//2)^(1//2)[cos(3+n)pit {:+cos(3-n)pit]dt

=1/2[(sin(3+n)pit)/((3+n)pi) {:+(sin(3-n)pit)/((3-n)pi)]_(-1//2)^(1//2)

It is then necessary to substitute t = 1/2 and t = -1/2, and subtract as usual, then simplify the expression in n.

After integrating, we could have expressed an as follows:

a_n=6((-1)^(n//2))/(pi(-9+n^2)), \ \ n even (0 if n odd, n!=3), or as:

 =6((-1)^n)/(pi(-9+4n^2)),\ \ \n=1,2,3,...

Then we could have substituted this expression into the series. However, we would still need to consider separately the case when n = 3.

Easy to understand math videos:
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Interactive: You can explore this example using the interactive Fourier Series graph.