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 $[−8, 8]$.
David's Answer:
$$a_0=\frac{1}{L}\int_{-L}^{L}f(x)d(x)=\frac{1}{2}\int_{0}^{4}[x^2-8x+16]d(x)=\frac13[x^3/3-4x^2+16x]|^4_0$$
$$a_0=32/3$$
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).
