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$$ f(x) = \begin{cases} \frac{\pi}{4}-\frac{x}{2} & [0,\pi] \\ -\frac{3\pi}{4}+\frac{x}{2}, & (\pi,2\pi) \end{cases} $$enter image description here

Is it right to compute only $a_n \text{ and } a_0$ coefficient for fourier series because $f(x)$ is even for fourier? How can I proove it since $f(x)!=f(-x)$

user91500
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GorillaApe
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    How did you conclude that $f(x) \ne f(-x)$? Please show us the steps. (This is a common mistake, but if you show the steps I can point out the mistake). – M. Vinay Jun 11 '14 at 11:52
  • From visual inspection, we have $f(x)=f(2\pi-x)$. And $\int_0^{2\pi}f(x)=0$ – mike Jun 11 '14 at 11:55
  • fourier $b_n$ == 0 too? – GorillaApe Jun 11 '14 at 11:58
  • @Parhs Yes, that is true for all even functions. – M. Vinay Jun 11 '14 at 12:00
  • I guess that I did not quite understand the question yet. – mike Jun 11 '14 at 12:02
  • @mike Why? In Fourier series of even functions, $b_n = 0$, so only the $a_n$ have to be calculated. – M. Vinay Jun 11 '14 at 12:04
  • the question is if we can conclude that $b_n =0$ without calculating it. and since domain isnt at -pi,pi how can we tell – GorillaApe Jun 11 '14 at 12:05
  • @M.Vinay. I just realized the convention. By the way, I am not math-majored. – mike Jun 11 '14 at 12:07
  • @M.Vinay I know that b_n is zero for even function but because this isnt even at x=0 but $f(x)=f(2π−x)$ me an other students had trouble with that. – GorillaApe Jun 11 '14 at 12:10
  • @Parhs. Since $f(x)$ is a $2\pi$-periodic. $f(x)=f(2\pi+x)$. The definition of $f(x)$ showed the symmetry $f(x)=f(2\pi-x)$. So we have $f(-x)=f(2\pi+(-x))=f(2\pi-x)=f(x)$. – mike Jun 11 '14 at 12:10
  • @Parhs One way to see this symmetry is set $x=2\pi-y$ and $g(y)=f(x)$. Then you can prove that $g(y)=f(y)$. – mike Jun 11 '14 at 12:15
  • so is it safe to say $b_n=0$ and for $a_n$ calculate $2 \int_0^{pi}f(x)dx$ ? – GorillaApe Jun 11 '14 at 12:15
  • ok it is becoming more clear. The problem was that I heard other students saying that teacher told that it isnt even and that they have found a $b_n$ after many computations – GorillaApe Jun 11 '14 at 12:17
  • @mike You should just write an answer. – M. Vinay Jun 11 '14 at 12:20

1 Answers1

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For $2\pi$-periodic function $f(x)$ defined as: $$f(x) = \begin{cases} \frac{\pi}{4}-\frac{x}{2} & x\in [0,\pi] \\ -\frac{3\pi}{4}+\frac{x}{2}, & x\in(\pi,2\pi) \end{cases}$$

We can calculate $$f(2\pi-x) = \begin{cases} \frac{\pi}{4}-\frac{2\pi-x}{2} & x\in [\pi,2\pi] \\ -\frac{3\pi}{4}+\frac{2\pi-x}{2}, & x\in (0,\pi) \end{cases}$$

$$ = \begin{cases} \frac{-3\pi}{4}+\frac{x}{2} & x\in [\pi,2\pi] \\ +\frac{1\pi}{4}-\frac{x}{2}, & x\in (0,\pi) \end{cases}$$

Thus we proved $$f(2\pi-x)=f(x)\text{ (1)}$$.

Since for $2\pi$-periodic function $f(-x)$ we have $$f(-x)= f((-x)+2\pi)=f(2\pi-x)\text{ (2)}$$

Combining (1) and (2), we proved $$f(x)= f(-x)$$.

Thus the coefficients $b_n$ that go with $\sin(2nx)$ are zero.

mike
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