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We have the $n$'th function defined as

\begin{equation} f_{n}(r,\theta)=\int_{0}^{2\pi}w(r,\theta,R,\theta')cos(n\theta')d\theta' \end{equation}

where $R$ is just some constant. I know that $w$ is $2\pi$ periodic in $\theta$ (and in $\theta'$ I think).

I can't work out how we know $f_{n}$ is $2\pi n$ periodic in $\theta$?

Any help would be appreciated. Thanks!

1 Answers 1

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Note that $w$ is $2\pi$ periodic in $\theta$ and thus $2n\pi$ periodic in $\theta$, which implies that $$ \int_{0}^{2\pi}w(r,\theta+2n\pi,R,\theta')\cos(n\theta')\ d\theta'=\int_{0}^{2\pi}w(r,\theta,R,\theta')\cos(n\theta')\ d\theta' $$

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    Oh yes, ofcourse! Do you know why this then implies that the $n$'th function is symmetric with respect to rotation of $\frac{2\pi}{n}$ about the origin? – 2017-01-10
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    Would you like to post a follow-up question? – 2017-01-10
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    Yes I will do that. Thanks for the answer. – 2017-01-10