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March 9th, 2018, 02:25 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  Proving Fourier property
A function f has the property that $\displaystyle f(x+\pi) = −f(x)$ for all x. Show that all its even Fourier coefficients are zero (i.e., a0 = a2 = a4 = a6 = . . . = 0, b2 = b4 = b6 = . . . = 0). Hint: Show that f must be periodic with period $\displaystyle 2 \pi$. Any tips would be much grateful. Thanks. 
March 9th, 2018, 04:07 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
$f(x+2\pi)=f(x+\pi)=f(x)$. Therefore f(x) is periodic with period $2\pi$.

March 9th, 2018, 04:22 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 
You should also look at the properties of $\cos(nx)$ and $\sin(nx)$ for odd $n$ and for even $n$. Bear in mind the properties of a sine series and the properties of a cosine series.

March 10th, 2018, 06:31 AM  #4 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  I agree that cos(nx) and sin(nx) for old n and for even n... would be the starting place for this proof. Would you offer little more tips for me to go foward?

March 10th, 2018, 02:07 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics 
Something is wrong with this question I think. Take $f(x) = \cos x$ for example which of course satisfies $f(x + \pi) = f(X)$ but obviously its even Fourier coefficients aren't zero. Namely, the first even coefficient is 1. Is there something miss from the question? 
March 10th, 2018, 02:09 PM  #6 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  I double checked the question. The question is correct.

March 10th, 2018, 02:30 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra  No, $a_1=1$. All other coefficients, including $a_{2k}\,(k \in \mathbb N)$ are zero.

March 10th, 2018, 03:03 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra  Quote:
Although, on reflection, it would be worth looking carefully at the integral $$\int_{\pi}^{\pi}f(x)\cos {(2nx)}\,\mathrm dx$$ You should be able to show that the negative part cancels the positive part. Last edited by v8archie; March 10th, 2018 at 04:02 PM.  
March 10th, 2018, 04:08 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007 
$\displaystyle \begin{align*}\int_{\pi}^{\pi}f(x)\cos {(2kx)}\,\mathrm dx &= \int_{\pi}^0f(x)\cos {(2kx)}\,\mathrm dx + \int_0^{\pi}f(x)\cos {(2kx)}\,\mathrm dx \\ &= \int_{\pi}^0f(x)\cos {(2kx)}\,\mathrm dx + \int_{\pi}^0f(u + \pi)\cos {(2ku + 2k\pi)}\,\mathrm du \\ &= \int_{\pi}^0f(x)\cos {(2kx)}\,\mathrm dx  \int_{\pi}^0f(u)\cos {(2ku)}\,\mathrm du \\ &= 0 \end{align*}$ Similarly, $\displaystyle \int_{\pi}^{\pi}f(x)\sin {(2kx)}\,\mathrm dx = 0$. 
March 10th, 2018, 04:17 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 
\begin{align*} && \cos{\big(2n(u+\pi)\big)} &= \cos{(2nu)}\cancelto{1}{\cos{(2n\pi)}}  \sin{(2nu)}\cancelto{0}{\sin{(2n\pi)}} \\ &&& = \cos{(2nu)} \\[8pt] && \int_{\pi}^{\pi} f(x)\cos{(2nx)}\,\mathrm dx &= \int_{\pi}^0 f(x)\cos{(2nx)}\,\mathrm dx + \int_0^{\pi} f(x)\cos{(2nx)}\,\mathrm dx \\ &\text{with $u = x  \pi \implies \mathrm du = \mathrm dx$} &&= \int_{\pi}^0 f(x)\cos{(2nx)}\,\mathrm dx + \int_{\pi}^0 f(u + \pi)\cos{\big(2n(u+\pi)\big)}\,\mathrm du \\ &&&=\int_{\pi}^0 f(x)\cos{(2nx)}\,\mathrm dx + \int_{\pi}^0 f(u)\cos(2nu)\,\mathrm du \\ &&&=\int_{\pi}^0 f(x)\cos{(2nx)}\,\mathrm dx  \int_{\pi}^0f(x)\cos(2nx)\,\mathrm dx \\ &&&= 0 \end{align*} Something similar for the $\sin{(2nx)}$ terms. Gah! Skipjack beat me to it. 

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fourier, proofing, property, proving 
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