I have a problem with the below fourier series

-2, -2 < t < -1

-1, -1 < t < 0

f(t) = 1, 0 < t < 1

2, 1 < t < 2

f(t + 4) = f(t)

I want to intergrate by parts to go on to solve bn but dont how what U will be.

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- Feb 6th 2011, 04:46 AMhunterage2000Problem with Fourier Series
I have a problem with the below fourier series

-2, -2 < t < -1

-1, -1 < t < 0

f(t) = 1, 0 < t < 1

2, 1 < t < 2

f(t + 4) = f(t)

I want to intergrate by parts to go on to solve bn but dont how what U will be. - Feb 6th 2011, 08:08 AMTheEmptySet
You want to find the Fourier series of

$\displaystyle f(t)=\begin{cases} -2, -2 < t < -1 \\ -1, -1 < t < 0 \\ 0, 0 < t < 1 \\ 1, 1 < t < 2 \\

\end{cases}$

You should not need integration by part. The function is constant on each interval.

Here is a PDF with the series.

Attachment 20698 - Feb 6th 2011, 08:12 AMhunterage2000
its $\displaystyle f(t)=\begin{cases} -2, -2 < t < -1 \\ -1, -1 < t < 0 \\ 1, 0 < t < 1 \\ 2, 1 < t < 2 \\

\end{cases}$ - Feb 6th 2011, 08:25 AMTheEmptySet
That make the problem easier. $\displaystyle f(x)$ is an odd function so all of the $\displaystyle a_n$ must be zero.

$\displaystyle \displaystyle b_n=\frac{1}{2}\left[ \int_{-2}^{-1}-2\sin\left(\frac{n\pi x}{2} \right)dx + \int_{-1}^{0}-\sin\left(\frac{n\pi x}{2} \right)dx+\int_{0}^{1}\sin\left(\frac{n\pi x}{2} \right)dx+\int_{1}^{2}2\sin\left(\frac{n\pi x}{2} \right)dx \right]$

$\displaystyle \displaystyle b_n=-\frac{2}{\pi n}\left[ 2\cos(n\pi)-\cos\left( \frac{n \pi}{2}\right)-1\right]$ - Feb 6th 2011, 08:28 AMhunterage2000
so what is the value of U? is it 2 + 1 or 3?

- Feb 6th 2011, 10:12 AMTheEmptySet
- Feb 6th 2011, 10:18 AMhunterage2000
U as in the intergration by parts U.

U = ?

dv/dt = cos (n*pi*t/2)

du/dt = ?

v = -2/n*pi sin(n*pi*t/2) - Feb 6th 2011, 10:23 AMTheEmptySet
As I said in my first post you do not need integration by parts.

$\displaystyle \displaystyle \int \sin(ax)dx=-\frac{1}{a}\cos(ax)$

Or in your case

$\displaystyle \displaystyle \int \sin\left(\frac{\pi n}{L}x \right)dx=-\frac{L}{\pi n}\cos\left(\frac{\pi n}{L}x \right)$ - Feb 6th 2011, 10:26 AMhunterage2000
totally new to this. can u show me from start to finish how you got the answer?