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March 9th, 2018, 02:25 PM   #1
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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 $\displaystyle 2 \pi$.

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March 9th, 2018, 04:07 PM   #2
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$f(x+2\pi)=-f(x+\pi)=f(x)$. Therefore f(x) is periodic with period $2\pi$.
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March 9th, 2018, 04:22 PM   #3
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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.
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March 10th, 2018, 06:31 AM   #4
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Quote:
Originally Posted by v8archie View Post
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.
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?
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March 10th, 2018, 02:07 PM   #5
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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?
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March 10th, 2018, 02:09 PM   #6
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Quote:
Originally Posted by SDK View Post
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?
I double checked the question. The question is correct.
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March 10th, 2018, 02:30 PM   #7
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Quote:
Originally Posted by SDK View Post
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?
No, $a_1=1$. All other coefficients, including $a_{2k}\,(k \in \mathbb N)$ are zero.
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March 10th, 2018, 03:03 PM   #8
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Originally Posted by zollen View Post
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?
It might involve the linear independence of the trigonometric functions. I would look for a proof of that the Fourier series of an odd function is a Sine series and try to adapt it.

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.
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Last edited by v8archie; March 10th, 2018 at 04:02 PM.
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March 10th, 2018, 04:08 PM   #9
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$\displaystyle \begin{align*}\int_{-\pi}^{\pi}f(x)\cos {(2kx)}\,\mathrm dx &= \int_{-\pi}^0f(x)\cos {(2kx)}\,\mathrm dx + \int_0^{\pi}f(x)\cos {(2kx)}\,\mathrm dx \\
&= \int_{-\pi}^0f(x)\cos {(2kx)}\,\mathrm dx + \int_{-\pi}^0f(u + \pi)\cos {(2ku + 2k\pi)}\,\mathrm du \\
&= \int_{-\pi}^0f(x)\cos {(2kx)}\,\mathrm dx - \int_{-\pi}^0f(u)\cos {(2ku)}\,\mathrm du \\
&= 0 \end{align*}$

Similarly, $\displaystyle \int_{-\pi}^{\pi}f(x)\sin {(2kx)}\,\mathrm dx = 0$.
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March 10th, 2018, 04:17 PM   #10
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\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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