Understanding Fourier Coefficients in the Fourier Transform of a Function

[member 428835]

Homework Statement

Let $f:\mathbb R \to \mathbb C$ be a smooth function such that $|f| < C/(1+|x|^2)$ for some $C$. Put $\phi(x) = \sum_{n = -\infty}^\infty f(x+n)$. Find Fourier coefficients of $\phi$ on [0,1] in terms of the Fourier transform of $f$.

Relevant Equations

Fourier transform $\hat f$ of $f$ is $\hat f(k) = 1/\sqrt{2\pi}\int_\mathbb R f(x) e^{-i k x}$

Hi PF!

Unsure how to begin. Fourier transform of $f$ I've given as an equation. I'm not sure what is meant by Fourier coefficients. Fourier coefficients of what?

 

[osilmag]

It seems like it is asking you to find the Fourier coefficients of $\phi(x)$. Fourier coefficients are found by integrating the multiple of a function $f(x)$ and $sin$ or $cos( \frac {2\pi rt}{T})$ with respect to time and where $r$ is the order number of the coefficient. So $r=4$ would be the fourth harmonic.

 

[osilmag]

Usually there are integration tables for different functions that help.

 

[LCKurtz]

First, note that you don't have a formula for $f(x)$, just an inequality that probably helps convergence. So you aren't going to do any integrations. So you have $$\phi(x) = \sum_{k=-\infty}^\infty f(x+k),~x\in [0,1]$$When you are given a function defined on $[0,1]$ and asked to find its Fourier coefficients, I guess you would assume its period is $1$ and extend it periodically. So the complex form of the Fourier coefficients would be$$
c_n =\frac 1 2\int_{-\frac 1 2}^{\frac 1 2} \sum_{k=-\infty}^\infty f(x+k)e^{-i n \pi x}~dx$$Now, I haven't worked with this stuff for years, so you should check my formulas. I am guessing that if you swap the sum and integral and make appropriate variable changes you might be able to make it look like a Fourier transform. I haven't worked it but that's what I would try. Good luck with it.

 

[member 428835]

First, note that you don't have a formula for $f(x)$, just an inequality that probably helps convergence. So you aren't going to do any integrations. So you have $$\phi(x) = \sum_{k=-\infty}^\infty f(x+k),~x\in [0,1]$$When you are given a function defined on $[0,1]$ and asked to find its Fourier coefficients, I guess you would assume its period is $1$ and extend it periodically. So the complex form of the Fourier coefficients would be$$
c_n =\frac 1 2\int_{-\frac 1 2}^{\frac 1 2} \sum_{k=-\infty}^\infty f(x+k)e^{-i n \pi x}~dx$$Now, I haven't worked with this stuff for years, so you should check my formulas. I am guessing that if you swap the sum and integral and make appropriate variable changes you might be able to make it look like a Fourier transform. I haven't worked it but that's what I would try. Good luck with it.

Thanks for the responses everyone, and especially this insight!

 

[member 428835]

So to summarize; since we are given a function $[0,1]$, the period must be 1. If we extend the function periodically, the complex form of the Fourier coefficients is
$$ c_k = \frac 1 2\int_{-\frac 1 2}^{\frac 1 2} \sum_{n=-\infty}^\infty f(x+n)e^{-i k \pi x}\,dx\\ = \frac 1 2 \sum_{n=-\infty}^\infty \int_{-\frac 1 2}^{\frac 1 2} f(x+n)e^{-i k \pi x}\,dx. $$
Let $u = x + n$. Then
$$ c_k = \frac 1 2 \sum_{n=-\infty}^\infty \int_{-\frac 1 2 + n}^{\frac 1 2 + n} f(u)e^{-i k \pi (u-n)}\,du\\ = \frac 1 2 \sum_{n=-\infty}^\infty e^{i k \pi n} \hat f(k) : \hat f(k) \equiv \int_{-\frac 1 2 + n}^{\frac 1 2 + n} f(u)e^{-i k \pi u}\,du. $$
Note that $\hat f(k)$ is actually the definition of a Fourier transform of $f(k)$. But now I'm not really sure what to do. Any ideas?

 

[LCKurtz]

What happens if you write out a few terms of$$
\sum_{n=-\infty}^\infty \int_{-\frac 1 2 +n}^{\frac 1 2 + n}$$and combine the integrals? Starting at $0$ and working both ways.

 

[member 428835]

What happens if you write out a few terms of$$
\sum_{n=-\infty}^\infty \int_{-\frac 1 2 +n}^{\frac 1 2 + n}$$and combine the integrals? Starting at $0$ and working both ways.

The sum looks something like
$$
\frac 1 2\exp\left( -2 i k \pi \right) \int_{-5/2}^{-3/2} f(u)\exp(-i k \pi u)\, du + \frac 1 2\exp\left( -i k \pi \right) \int_{-3/2}^{-1/2} f(u)\exp(-i k \pi u)\, du+\\
\frac 1 2 \int_{-1/2}^{-1/2} f(u)\exp(-i k \pi u)\, du + \frac 1 2\exp\left( i k \pi \right) \int_{1/2}^{3/2} f(u)\exp(-i k \pi u)\, du + \\\frac 1 2\exp\left( i 2 k \pi \right) \int_{3/2}^{5/2} f(u)\exp(-i k \pi u)\, du =\\
\frac 1 2 \int_{-\infty}^\infty f(u) \exp(-i k \pi u)\, du
$$

since $\exp(i k \pi n) = 1$.

 

[LCKurtz]

So doesn't that pretty much solve your problem? That's a multiple of $\hat f(k)$ isn't it? Do you see why the given bound on $|f(x)|$ helps you to get convergence for $\hat f$? And you probably better check your advanced calculus book to see whether swapping the sum and integral was legit.

 

[member 428835]

Thanks a ton! Yep, everything makes good sense. I appreciate it!