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# 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.

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.

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 bn = 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 an, 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).

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 ao = 0 and an = 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.

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 2π outside of this range as shown by the dotted lines.

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

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