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In this excercise we show how the symmetries of a function imply certain properties of its Fourier coefficients. Let $f$ be a 2$\pi$ Riemann integrable function defined on $R$.

(a) Show that the Fourier series of the function $f$ can be written as $$f(\theta) \thicksim \hat{f}(0) + \sum_{n\ge1}[\hat{f}(n) + \hat{f}(-n)]\ cos n\theta + i[\hat{f}(n) - \hat{f}(-n)]\ sin n\theta. $$

(b)Prove that if $f$ is even, then $ \hat{f}(n)= \hat{f}(-n) $ , and we get a cosine series.

(c)Prove that if $f$ is odd, then $ \hat{f}(n)= -\hat{f}(-n) $, and we get a sine series.[This hint is helpful here Series Of Sines.

(d)Suppose that $f(\theta + \pi )$ = $f(\theta)$ for all $\theta \in R$. Show that $\hat{f}(n)$ = 0 for all odd n.(Its answer is here Effect of Symmetry of a Function on its Fourier Series)

(e)Show that $f$ is real-valued iff $\bar{\hat{f}}(n) =\hat{f}(-n)$ for all n.

For (e), we have two steps to prove 1- If $f$ is real valued, then $\bar{\hat{f}}(n) =\hat{f}(-n)$ for all n.Which is sooo easy by using the definition of Fourier Coeffient.

2- If $\bar{\hat{f}}(n) =\hat{f}(-n)$ for all n, Then $f$ is real valued.I reached this step,$$\int_{-\pi}^{\pi}\overline{f}(\theta)e^{in\theta} d\theta = \int_{-\pi}^{\pi}f(\theta)e^{in\theta} d\theta$$, Does this means that f is real valued ? If so what is the justification.

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    What about sharing some thoughts of yours? – 2017-02-27
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    b) & c) are easy for me, but I am stucked in a) – 2017-02-27
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    Can you rearrange $\sum_{n=-\infty}^{\infty} \hat{f}(n) e^{in\theta}$ accordingly? – 2017-02-27
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    accordingly to what ? – 2017-02-27

1 Answers 1

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Let's solve (a).

The Fourier series of the function $f$ is $$ \sum_{n=-\infty}^{\infty}\hat{f}(n)e^{in\theta} $$

By definition, this is $$ \sum_{n=0}^{\infty}\hat{f}(n)e^{in\theta} + \sum_{n=1}^{\infty}\hat{f}(-n)e^{i(-n)\theta} $$

Taking out the $n=0$ term in the first series, this is $$ \hat{f}(0)+\sum_{n=1}^{\infty}\hat{f}(n)e^{in\theta} + \sum_{n=1}^{\infty}\hat{f}(-n)e^{i(-n)\theta}\tag{1} $$

Applying Euler's formula we have $$ e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\\ e^{i(-n)\theta}=\cos((-n)\theta)+i\sin((-n)\theta) $$

Since $\cos$ is an even function and $\sin$ is an odd function, this is $$ e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\\ e^{i(-n)\theta}=\cos(n\theta)-i\sin(n\theta) $$ Substituting in $(1)$, we obtain $$ \hat{f}(0)+\sum_{n=1}^{\infty}\hat{f}(n)[\cos(n\theta)+i\sin(n\theta)] + \sum_{n=1}^{\infty}\hat{f}(-n)[\cos(n\theta)-i\sin(n\theta)] $$ But this equals $$ \hat{f}(0)+\sum_{n=1}^{\infty}\left\{\hat{f}(n)[\cos(n\theta)+i\sin(n\theta)] + \hat{f}(-n)[\cos(n\theta)-i\sin(n\theta)]\right\} $$ which is nothing more than $$ \hat{f}(0)+\sum_{n=1}^{\infty}\left\{[\hat{f}(n) + \hat{f}(-n)]\cos(n\theta) + i[\hat{f}(n) - \hat{f}(-n)]\sin(n\theta)\right\} $$

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    thank u very much, for (b) I understand that I will substitute in the formula that we proved in (a) by $ \hat{f}(n)= \hat{f}(-n) $ and then the last term in the summation will be cancelled and I will get a cosine series but what about the first part of (b), How could I prove? – 2017-02-28
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    @Marwa Assem What do you mean by "what about the first part of (b)"? Do you mean $\hat{f}(0)$ ? The series will reduce to $$\hat{f}(0)+\sum_{n=1}^{\infty}2\hat{f}(n)\cos(n\theta)\tag{$\star$}$$ and this is a cosine series. Indeed, if we define $$\hat{f}'(n):=\begin{cases}\hat{f}(0),&n=0,\\2\hat{f}(n),&n\geq1\end{cases}$$ then since $\cos(\color{#090}{0})=1$, we have that $(\star)$ is the same as $$\sum_{n=\color{#090}{0}}^{\infty}\hat{f}'(n)\cos(n\theta)$$ – 2017-02-28
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    I mean that he said in (b)Prove that if $f$ is even, then , $ \hat{f}(n)= \hat{f}(-n) $ How can I do this? – 2017-02-28
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    @Marwa Assem Probably checking the definition of $\hat{f}(n)$ will help... We simply have \begin{align} \hat{f}(n) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) e^{-int} \,dt \\ &=-\frac{1}{2\pi} \int_{\pi}^{-\pi} f(-u) e^{-in(-u)} \,du\tag{1} \\ &=\frac{1}{2\pi} \int_{-\pi}^{\pi} f(-u) e^{-in(-u)} \,du \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(u) e^{-i(-n)u} \,du \tag{2}\\ &= \hat{f}(-n) \end{align} where $(1)$ follows from a substitution $t=-u$ and $(2)$ from the fact that $f$ is even. – 2017-02-28
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    thank u very much. could u please see my question of Fourier Series of the characteristic function and help me in it if u do not mind. – 2017-02-28
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    @Marwa Assem There's already an answer there. Anyway you're not showing any effort so I'm not willing to help you anymore. Moreover you did not upvote nor accept my answer here. That's ungrateful behavior. Finally you asked me for help regarding another question 10 seconds after my last comment. I sense you prepared your comment and did not even check the proof I gave of $\hat{f}(n)=\hat{f}(-n)$. As if all you wanted was that the answer was there so that you could copy it. But hey, that's typical behavior. You're not the first one and certainly not the last one to do so. Have a great day. – 2017-02-28
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    Sorry, but I am new and do not know how to upvote or accept your answer . No I checked your answer and your answers are thoroughly written and neat, I am so grateful to your help, forgive me if I made an attitude that u do not like, I am so sorry. – 2017-02-28