Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
    
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.?).

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.