# 4. Half Range Fourier Series

If a function is defined over half the range, say `0` to
*L*, instead of the full range from `-L` to `L`,
it may be expanded in a series of sine terms only or of cosine
terms only. The series produced is then called a **half range
Fourier series**.

Conversely, the Fourier Series of an even or odd function can be analysed using the half range definition.

## Even Function and Half Range Cosine Series

An **even** function can be expanded using half its range
from

- `0` to
*L*or - `-L` to `0` or
- `L` to `2L`

That is, the range of integration is *L*. The Fourier
series of the **half range even function** is given by:

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

for `n = 1, 2, 3, ...` , where

`a_0=2/Lint_0^Lf(t)dt`

`a_n=2/Lint_0^Lf(t)\ cos {:(n pi t)/L:}dt`

and `b_n= 0`.

### Illustration

In the figure below, `f(t) = t` is sketched from `t = 0` to `t = π`.

Graph of `f(t)`.

An even function means that it must be symmetrical about the `f(t)` axis and this is shown in the following figure by the broken line between `t = -π` and `t = 0`.

Graph of `f(t)`, illustrating it's an even function.

It is then assumed that the "triangular wave form" produced is periodic with period `2pi` outside of this range as shown by the dotted lines.

Graph of `f(t)`, a triangular waveform.

### Example

We are given that

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

and `f(t)` is periodic with period `2π`.

a) Sketch the function for 3 cycles.

b) Find the Fourier trigonometric series for `f(t)`, using half-range series.

Answer

a) Sketch:

(This is the same function we saw in the above Illustration example.)

Graph of `f(t)`, a triangular waveform.

b) Since the function is even, we have
*b*_{n} = 0.

In this example, `L=pi`.

We have:

`a_0=2/Lint_0^Lf(t)dt`

`=2/piint_0^pit\ dt`

`=2/pi[t^2/2]_0^pi`

`=2/pi(pi^2)/2`

`=pi`

To find *a*_{n}, we use a result from
before (see Table of Common Integrals):

`intt\ cos nt\ dt` `=1/n^2(cos nt+nt\ sin nt)`

We have:

`a_n=2/Lint_0^Lf(t)cos{:(n pi t)/L:}dt`

`=2/piint_0^pi t\ cos nt\ dt`

`=2/pi[1/n^2(cos nt+nt\ sin nt)]_0^pi`

`=2/(pi n^2)[(cos n pi+0)-(cos 0+0)]`

`=2/(pi n^2)[(cos n pi-1)]`

`=2/(pi n^2)[(-1)^n-1]`

When *n* is odd, the last line gives us `-4/(pin^2`.

When *n* is even, the last line equals `0`.

For the series, we need to generate odd values for *n*.
We need to use `(2n - 1)` for `n = 1, 2, 3,...`.

So we have:

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

`=pi/2-4/pi sum_(n=1)^oo(cos(2n-1)t)/((2n-1)^2)`

`=pi/2` `-4/pi(cos t+1/9cos 3t` `{:+1/25cos 5t+...)`

**Check:** We graph different terms of the above expression to make sure our answer is correct. Here is the first term involving the cosine expression:

Graph of `f(t)~~pi/2-4/pi cos(t)`.

Now add another term:

Graph of `f(t)` `~~pi/2-4/pi (cos(t)` `{: + 1/9 cos(3t))`.

Graph of `f(t)` `~~pi/2-4/pi (cos(t)` ` + 1/9 cos(3t)` `{:+1/25 cos(5t))`.

We can see our series is rapidly converging to our original triangular function.

## Odd Function and Half Range Sine Series

An odd function can be expanded using half its range from `0` to
*L*, i.e. the range of integration has value *L*. The Fourier
series of the odd function is:

Since *a*_{o} = 0 and *a*_{n} = 0, we have:

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

n`= 1, 2, 3, ...`

`b_n=2/Lint_0^Lf(t)\ sin {: (n pi t)/L:}dt`

In the figure below, *f*(*t*) = *t* is sketched
from *t* = 0 to *t* = *π*, as before.

Graph of `f(t)`, for `0 < t < pi`.

An **odd** function means that it is symmetrical about the
origin and this is shown by the red broken lines between *t* = −*π* and `t = 0`.

Graph of `f(t)`, showing it's an odd function.

It is then assumed that the waveform produced is periodic of
period 2*π* outside of this range as shown by the dotted lines.

Graph of `f(t)`, a periodic odd function.

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