Timeline for How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2011 at 15:20 | comment | added | B R | GarouDan, the inverse Fourier transform (maybe with a shifted contour) defines a solution to your problem. As Will Sawin points out, this usually won't have a closed-form expression. | |
| Nov 1, 2011 at 14:07 | comment | added | GarouDan | @BR , Are you saying it's possible find $f(x)$ in my exampling using Inverse Fourier Transform, but Mathematica, don't do it? | |
| Nov 1, 2011 at 14:05 | comment | added | GarouDan | @AliBleybel I can't embrace you answer as solution how it presents now. When I put the FourierTransform link and that formulas, I tried it before. I tried isolate $f(x)$ using that. And tried using ZTransform too. It's not a easy question. Maybe we can use ZTransform to several things and FourierTransform to many others and other transform and find all solutions. Don't know, but if $a=1$ or $a\implies 0$, I think there limitated solutions. And I'm searching a way to find them. | |
| Nov 1, 2011 at 5:00 | comment | added | Will Sawin | 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. | |
| Nov 1, 2011 at 3:01 | comment | added | user16974 | @BR: Thank you for the comments and the link! | |
| Nov 1, 2011 at 2:49 | comment | added | B R | 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 | |
| Nov 1, 2011 at 2:49 | comment | added | B R | I don't think so. You just need to take some care with the integral, say by shifting the contour. I can't guarantee that this works out, but this sort of thing happens when you want to solve differential equations via Fourier inversion: you end up wanting to take the inverse transform of $1/P(x)$, where $P(x)$ is some polynomial that may have poles on the real line. | |
| Nov 1, 2011 at 1:09 | comment | added | user16974 | Do the singularities at $a\xi$ mean that the solution is wrong? | |
| Nov 1, 2011 at 0:55 | comment | added | B R | Mathematica may not like the fact that the function you are taking the inverse transform of has singularities at $a\xi\in\mathbb Z$. | |
| Oct 31, 2011 at 23:23 | comment | added | GarouDan | If you have Mathematica you can try: a=1 G[x]=1/(x*(x-1)) F[y_]:=InverseFourierTransform[aFourierTransform[G[x],x,k]/(E^(2*PiIka)-1),k,y] and finally F[y] to see the results...but your kernel probably will works forever> So my new question is, Fourier and InverseFourier transforms can't solve this simple question? | |
| Oct 31, 2011 at 23:21 | comment | added | user16974 | I don't know what's wrong with mathematica, my answer is clear. | |
| Oct 31, 2011 at 23:18 | comment | added | user16974 | @GarouDan: Shall I write more details? | |
| Oct 31, 2011 at 23:17 | comment | added | GarouDan | 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, $f(x+1)=f(x)+1\times g(x) \iff \frac{1}{x+1}-\frac{1}{x}=g(x) \iff \frac{-1}{x(x+1)}=g(x)$ | |
| Oct 31, 2011 at 22:42 | history | answered | user16974 | CC BY-SA 3.0 |