# Finding Fourier Series

Question:
Given that $f( x) = ( x − 4)^2$ for all $x \in [0, 4]$. For each of the following questions, define a periodic extension function of $f(x)$ and sketch its graph on the interval $[−8, 8]$.

$$a_0=\frac{1}{L}\int_{-L}^{L}f(x)d(x)=\frac{1}{2}\int_{0}^{4}[x^2-8x+16]d(x)=\frac13[x^3/3-4x^2+16x]|^4_0$$ $$a_0=32/3$$
Full range series: $p=4,l=2$ $$a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos(\frac{n\pi x}L)d(x)$$
$$a_n=\frac{1}{2}\int_{0}^{4}[x^2-8x+16]\cos(\frac{n\pi x}L)d(x)=\frac12(0)=0$$ $$a_n=0$$

But the $a_n$ I got $0$. Is it correct? BTW could you check my overall work done to see whether I do it right or not. Thanks in advance. And also could you please show me the work done for $b_n$ only? I would like to see on how to do it.Please only $b_n$ only.

Determine the full-range Fourier series expansion corresponding to f(x).

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I'm feeling lazy, could you at least include the question in latex instead of an external link? – Matt N. Sep 5 '12 at 14:06
@David: as a question and answers forum, we desire that questions asked also be useful to future users, that is why we ask users, when posting questions, to make their best effort in making the question self-contained. In our FAQ (a link to which you can find on the top right) we provide some help to how to use the Math capabilities of our site. Please feel free to look at it and edit your question accordingly. – Willie Wong Sep 5 '12 at 14:14
Now to your actual question: you´ve not shown in your work how you get from the Fourier integral expression for $a_n$ to that the integral is 0. How were you able to conclude that? If $L= 2$ is the length of the half-period, is the graph you sketched actually correct? (What is the value of your function between $(-4,0)$?) – Willie Wong Sep 5 '12 at 14:21
I've edited your post...Just for practicing some latex$\bigstar$ – PooyaM Sep 5 '12 at 14:26
– PooyaM Sep 5 '12 at 14:29