Timeline for Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?
Current License: CC BY-SA 2.5
7 events
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| Aug 1, 2022 at 21:18 | comment | added | Pietro Majer | if $f$ and $g$ are two generic such solutions (i.e. those with 0 formal Taylor series at non-negative integer points), I'd say that $f(x)+g\big(x+\frac12\big)$ represents all solutions. | |
| Jan 14, 2010 at 14:56 | comment | added | Douglas Zare | Ok, now I understand. Thanks. I misread the integral, and incorrectly assumed that these functions would be 0 at -1. | |
| Jan 14, 2010 at 14:27 | comment | added | David E Speyer | No. For z a a very small positive number, g(-z) is $- e^{-1} \int_{1-z}^1 g(u) du$, which approaches zero as z goes to 0. (I've just rewritten Bjorn's formula to not have any hidden negatives.) | |
| Jan 14, 2010 at 9:39 | comment | added | Douglas Zare | The limit from the left is the average value of g on $[0,1]$ divided by $e$, no? | |
| Jan 14, 2010 at 9:36 | comment | added | Jonas Meyer | Douglas Zare, how is that? The limit from the left at 0 is clearly 0=g(0), whether or not g is nonnegative on [0,1] | |
| Jan 14, 2010 at 9:08 | comment | added | Douglas Zare | Perhaps I'm confused, but if g is all nonnegative on [0,1], it seems to me that you are getting a discontinuity at 0. | |
| Jan 14, 2010 at 7:41 | history | answered | Bjorn Poonen | CC BY-SA 2.5 |