Fourier Series

• Mar 15th 2009, 07:55 AM
harkapobi
Fourier Series
Hi, i really need help with this question

find the fourier series of the function f(t)= 3sin(t) + sin(4t).

I know that the function is odd (so an=0), and has period 2pi, but when i try finding bn, integrating between -pi and pi, i keep just getting 0. This cant be right, where am I going wrong?

Katy (Happy)
• Mar 15th 2009, 08:12 AM
TheEmptySet
Quote:

Originally Posted by harkapobi
Hi, i really need help with this question

find the fourier series of the function f(t)= 3sin(t) + sin(4t).

I know that the function is odd (so an=0), and has period 2pi, but when i try finding bn, integrating between -pi and pi, i keep just getting 0. This cant be right, where am I going wrong?

Katy (Happy)

It is its own fourier series on the interval $(-\pi,\pi)$(It is already a triginometric polynomial)

$\sin(x)$ is orthogonal to every $\cos(nx)$

and the only non zero coeffients of the sine sine series will be when n=1 or 4.

This is exactly like expanding the Taylor series of a polynomial centered at 0. You will get the same thing back.
• Mar 15th 2009, 08:28 AM
harkapobi
Quote:

Originally Posted by TheEmptySet
It is its own fourier series on the interval $(-\pi,\pi)$(It is already a triginometric polynomial)

$\sin(x)$ is orthogonal to every $\cos(nx)$

and the only non zero coeffients of the sine sine series will be when n=1 or 4.

This is exactly like expanding the Taylor series of a polynomial centered at 0. You will get the same thing back.

I'm sorry, i'm probably being really dumb but i dont understand. So does this mean i should integrate over a different interval? Its just i need this fourier series to work out the particular integral of a second order differential equation.
• Mar 15th 2009, 08:52 AM
TheEmptySet
You are trying to find a fourier series on the interval $(-\pi,\pi)$.

$f(t)=3\sin(t)+\sin(4t)$

To calculate the $a_n$ we use the formula

$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt$

pluggint in f(t) we get

$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\cos(nt)dt$

This will equal ZERO for all n since sine and cosine are orthongonal

Now we do the same thing for the sine terms

$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt$

$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(nt)dt$

Since sine is an orthogonal basis the only non zero terms will be when n=1 or when n=4.

$b_1=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(t)dt=3$

$b_4=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(4t)dt=1$
• Mar 15th 2009, 09:30 AM
harkapobi
Quote:

Originally Posted by TheEmptySet
You are trying to find a fourier series on the interval $(-\pi,\pi)$.

$f(t)=3\sin(t)+\sin(4t)$

To calculate the $a_n$ we use the formula

$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt$

pluggint in f(t) we get

$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\cos(nt)dt$

This will equal ZERO for all n since sine and cosine are orthongonal

Now we do the same thing for the sine terms

$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt$

$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(nt)dt$

Since sine is an orthogonal basis the only non zero terms will be when n=1 or when n=4.

$b_1=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(t)dt=3$

$b_4=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(4t)dt=1$

Ahh right, i understand, because the integral of sin(nt)sin(mt) over -pi to pi is pi if m doesn't equal n. is that right?

So now if I have the differential equation y'' + 5y = 3sin(t) + sint(4t)
and i have
$b_4= 1$ and $b_1 = 3$, using the trial particular integral:

y= $c_0$/2 + $\Sigma$( $c_n$cos(nt)+ $b_n$sin(nt))

I get that $d_1$ = 3/4 and $d_4$= -1/9 so is the particular integral just:

y = 3/4sin(t) when n=1
... -1/9sin(4t) when n=4 ? is that the answer?