Timeline for Bounds of periodic functions formed from infinite series of shifts
Current License: CC BY-SA 4.0
14 events
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| Aug 13, 2023 at 16:43 | comment | added | Bobby Ocean | The point here is that I successfully eliminate the $\sin(xt)$ and integral. The only question is now if the periodic function is positive on the first half of it's period. | |
| Aug 13, 2023 at 16:41 | vote | accept | Bobby Ocean | ||
| Aug 13, 2023 at 16:40 | comment | added | Bobby Ocean | Sure, I went ahead and split off the second part for you. We should note that rewriting the periodic infinite sum back as a fourier transform appears to only be useful when we actually know what the fourier transform is. In the case of $f(t)=t*\exp(-t^2)$ this is obvious, but in the case of $f(t)=\tanh(t)\exp(-\cosh(t))$ this is not obvious at all. The explicit relationship is $$\int_0^\infty f(t) \sin(x t) dt= \int_0^{\pi/x} f(t,\pi/x) \sin(x t) dt$$ In other words any fourier transform can be written as the finite fouier transform of a periodic function. | |
| Aug 13, 2023 at 9:30 | comment | added | user196574 | @BobbyOcean Whatever you decide re: the question, please check out my most recent edit - there's a nice formula for the Fourier series for $f(t,x)$ in terms of the Fourier transform of $k(t)$ in my notation (in your notation, you use $f(t)$ for what I call $k(t)$, which I feel overloads $f$ a little too much for my tastes). This explains the small $x$ behavior of $f(x,t)$. | |
| Aug 13, 2023 at 9:26 | history | edited | user196574 | CC BY-SA 4.0 |
Added a discussion of the qualitative behavior of $f(t,x)$ at "small" $x$
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| Aug 13, 2023 at 6:52 | comment | added | user196574 | @BobbyOcean Would you consider maybe changing the question to be your original question 1 of whether one can have simple zeroes within the interval $t \in (0,x)$ (and potentially accepting this answer :) ), and then asking your edited question as a separate question? | |
| Aug 12, 2023 at 20:25 | comment | added | Bobby Ocean | I have modified the question to make it more accurate and straight to the question. Moreover, it appears your observation is non-trivial. Indeed, the $f4$ function also appears to have the property you found. | |
| Aug 12, 2023 at 19:22 | comment | added | Bobby Ocean | It does appear to be the case that $f$ has simple roots for $x=1$. However, this does appear to be very very brief and does not appear to be true for $x$ outside of $[.9999,1.001]$. Likewise, at $x=1.01$, $f$ appears to have a maximum value of $\approx 0.45$ which is several hundred times larger than the function at $x=1$. There is some property here like "if $f$ has a simple zero, then $f$ is uniformly very close to zero". | |
| Aug 11, 2023 at 18:08 | comment | added | user196574 | @BobbyOcean I think I understand why $\sum_{m=-\infty}^\infty (-1)^m (t+m) e^{-(t+m)^2}$ and $\sum_{m=-\infty}^\infty (-1)^m (t+m)^3 e^{-(t+m)^2}$ look nearly proportional - it's because both are nearly proportional to $\sin(\pi t)$! That is, the Fourier series for both of these functions has coefficients that are rapidly decaying, leaving mostly just $\sin(\pi t)$. I think the rapid decay of the Fourier coefficients might be a key clue. | |
| Aug 11, 2023 at 8:47 | history | edited | user196574 | CC BY-SA 4.0 |
Clarified my final remark
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| Aug 11, 2023 at 8:44 | comment | added | user196574 | @BobbyOcean Fingers crossed, a small modification of my original proposed function indeed gets a couple simple zeros, see my edit above. However, there's a mystery remaining of why $\sum_{m=-\infty}^\infty (-1)^m (t+m) e^{-(t+m)^2}$ and $\sum_{m=-\infty}^\infty (-1)^m (t+m)^3 e^{-(t+m)^2}$ appear to be very close to proportional on the interval of $t \in (0,1)$. | |
| Aug 11, 2023 at 8:41 | history | edited | user196574 | CC BY-SA 4.0 |
Added an example that numerically answers the question in the negative
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| Aug 11, 2023 at 8:40 | comment | added | Bobby Ocean | Ah yes, this is an interesting point. I was avoiding non-simple zeros on purpose, mostly as an effort to keep things simplistic. While technically your example does have simple zeros, in the limit you are essentially creating very tiny perturbations of a function with a non-simple zero. Will have to do some investigation of this. It would be interesting if the distance to the imaginary roots plays a role in the observations. | |
| Aug 11, 2023 at 8:19 | history | answered | user196574 | CC BY-SA 4.0 |