# 3. Fourier Series of Even and Odd Functions

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

### Revision

Go back to Even and Odd Functions for more information.

In some of the problems that we encounter, the Fourier coefficients *a*_{o}, *a*_{n} or *b*_{n} 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.

## 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 *b*_{n } has zero value:

`b_n= 0`

So we only need to calculate *a*_{0} and *a _{n}* 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.

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.

**Interactive:** You can explore this example using the interactive Fourier Series graph.

### Online Algebra Solver

This algebra solver can solve a wide range of math problems. (Please be patient while it loads.)

Go to: Online algebra solver

### Calculus Lessons on DVD

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!