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$a \in \mathbb{R}$

$f:\mathbb{R} \rightarrow \mathbb{R}$

$g:\mathbb{R} \rightarrow \mathbb{R}$

For generic functions $f$ and $g$, how isolate $f(x)$ in the equation below?

$f(x+a)=f(x)+a\times g(x)$

I tried to use Fourier Transform and Inverse Fourier Transform but looks like this don't work very well.

$f(x - a)=$ $e^{-2\pi i a \xi} \hat{f}(\xi)$

$\hat{f}(\xi)=$ $\int_{-\infty}^{\infty}f(x) e^{-2\pi i x\xi}\, dx \quad$ (Fourier Transform)

I tried ZTransform too, but again, didn't worked very well.

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Does your equation have to hold for every $x\in\mathbb{R}$? Most importantly, what do you mean by "isolate"? –  Qfwfq Oct 31 '11 at 22:45
    
You are actually trying to calculate the indefinite sum of terms defined by $f$ (just look at the case $a=1$), so you really should narrow down the class of functions to have summation algorithms. –  user11235 Oct 31 '11 at 23:01
    
@Qfwfq, isolate is find f or $f(x)=\cdots$, where $\cdots$ is something using $g$ and $a$. –  GarouDan Oct 31 '11 at 23:02
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This should not have the special-functions tag –  Yemon Choi Nov 1 '11 at 19:18
    
I think use DiracDelta is special functions no? And using Fourier Transforms, for example, DiracDelta appears frequently. –  GarouDan Nov 1 '11 at 21:10
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3 Answers

The answer is not unique - you can add any function of period $a$ to f. This is what the singularities are trying to tell you. In a distributional sense, the Fourier transform of a function of period a is supported exactly on the zeros of $e^{2 \pi i a \xi}-1$.

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@shrdlu Yes I agree, but if $f$ is unknown, how to find, at least, one solution to $f$? –  GarouDan Nov 1 '11 at 14:09
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If you're not concerned with continuity, integrability, or any of the niceties of real analysis (and there's nothing in the problem that says you are), then, given $a\ne 0$ and $g$, you can let $f$ be any function on the interval $[0,|a|)$ and simply extend it to all real numbers by repeated applications of the given functional equation (written in the form $f(x) = f(x+a) - ag(x)$ to extend it in the direction opposite to the sign of $a$). If $a=0$, it's clear $f$ can be anything.

Addition: Let me be somewhat more explicit. If $a=1$, we can let $f$ be identically zero on $[0,1)$ and equal to $\sum_{k=1}^{[x]} g(x-k)$ for $x\ge1$, with a similar formula for $x<0$.

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Didn't understand very well. Can you give an example? For example, using $a=1$ ang $g(x)=\frac{-1}{x(x-1)}$ who mathematica don't solves. –  GarouDan Nov 2 '11 at 0:04
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@GarouDan: Barry Cipra's point is that not every function is of the nice form that Mathematica would give you - not every function is given by a formula. If you are only interested in functions that are "physically realistic" in some sense then you should probably make this clear in the question –  Yemon Choi Nov 2 '11 at 2:38
    
Great answer! :) –  B R Nov 2 '11 at 14:15
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Just as you said:$e^{2\pi i a \xi} \hat{f}(\xi)= \hat{f}(\xi) +a \hat{g}(\xi)$. In this way you get $\hat{f}(\xi)$ and you can use it to find $f(x)$.

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In Mathematica, if we try $a=1$ and $g(x)=\frac{-1}{x(x+1)}$ $f(x)=a (\mathcal{F}_{\xi }^{-1}[\frac{\mathcal{F}_x[G(x)](\xi )}{-1+e^{2 i \pi a \xi }}])(x)$ We have no answer, but, is easy to know the answer. $f(x)=\frac{1}{x}$ because, $f(x+1)=f(x)+1\times g(x) \iff \frac{1}{x+1}-\frac{1}{x}=g(x) \iff \frac{-1}{x(x+1)}=g(x)$ –  GarouDan Oct 31 '11 at 23:17
    
@GarouDan: Shall I write more details? –  user16974 Oct 31 '11 at 23:18
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I don't know what's wrong with mathematica, my answer is clear. –  user16974 Oct 31 '11 at 23:21
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I was going to say "unfortunately I don't know a good source for this, as I learnt it from Folland's excellent Lectures on Partial Differential Equations, which is very hard to find nowadays", but I see that the Tata Institute has made it available online (and re-typeset)! math.tifr.res.in/~publ/ln/tifr70.pdf –  B R Nov 1 '11 at 2:49
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I mean, what the poles are expressing is that you only know $f$ up to a periodic function with period $a$. It's up to you to coax a reasonable closed form out of it, and Fourier shouldn't work in general. In fact, nothing should. –  Will Sawin Nov 1 '11 at 5:00
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