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Fourier coefficients

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Problem 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

Gold Member
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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

Gold Member
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Usually there are integration tables for different functions that help.
 

LCKurtz

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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.
 
1,688
50
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!
 
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50
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

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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.
 
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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

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

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