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

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

My answer :

Full range series: $p=4,l=2$

$\begin{align*} a_0&=\frac1L\int\limits_{-L}^Lf(x)\mathrm dx\\ &=\frac22\int\limits_0^4\left[x^2-8x+16\right]\mathrm dx\\ &=\left[\frac13x^3-4x^2+16x\right]\Big|^4_0\\ a_0&=64/3 \end{align*}$

$\begin{align*} a_n&=\frac1L\int\limits_{-L}^Lf(x)\cos\left(\frac{n\pi x}L\right)\mathrm dx\\ &=\int\limits_0^4x^2\cos\left(\frac{n\pi x}2\right)\mathrm dx-8\int\limits_{0}^{4}x\cos\left(\frac{n\pi x}2\right)\mathrm dx+16\int\limits_{0}^{4}\cos\left(\frac{n\pi x}2\right)\mathrm dx\\ &=\frac22\left[\frac{32}{n^2\pi^2}-8(0)+16\left(\frac{2}{n\pi}\sin\left(\frac{nx\pi}{2}\right)\right)\Big|^4_0\right]\\ a_n&=\frac{16}{n^2\pi^2} \end{align*}$

Is my fourier series right ?

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Where are the "following questions"? – Christian Blatter Sep 5 '12 at 19:33
A quick calculation gives me $a_n = \frac{64}{n^2 \pi^2}$. – copper.hat Sep 5 '12 at 23:05
sorry.Now i already edited. Could yo check now Christian blatter – David Sep 6 '12 at 6:00
no, you still have $L = 2$ in the denominator of $\cos$. Those need to be "4"'s instead. – BaronVT Sep 6 '12 at 8:10
why l need to be 4. The period is 4 right. Therefore p=2L and L=2. Because the interval is [0,4] – David Sep 6 '12 at 10:54
up vote 1 down vote accepted

For this, you need $L = 4$ (the length of the original interval), not 2 (as it appears in the denominator of $\cos$). Also, the new function will (probably) have a period of 8, though it seems like there are some specifics of the problem missing (are you supposed to define even periodic extensions, odd periodic extensions, etc.?).

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