Timeline for Show that $f(t)=\sum_{i=1}^n a_i e^{-(x_i-t)^2}-c$ has at most $2n$ zeros
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2019 at 23:29 | comment | added | Gjergji Zaimi | @Boby Your problem is translation invariant. You can assume all $x_i$ are positive without loss of generality. | |
| Jan 21, 2019 at 23:15 | comment | added | Boby | Also, is it a problem if one of $x_i$'s equal to zero? We might not have a distinct number of zeros condition. Or am I wrong here | |
| Jan 21, 2019 at 23:10 | comment | added | Gjergji Zaimi | @Boby Yes, just combine the exponentials and complete the square. | |
| Jan 21, 2019 at 23:09 | comment | added | Boby | @MateuszKwaśnicki Could you put your solutions as an answer too? | |
| Jan 21, 2019 at 23:07 | comment | added | Boby | I have a quick question. Is the function $e^{(-(1+\epsilon)(x_i-t)^2} \cdot e^{-\frac{t^2}{2}}$ still treated as Gaussian? | |
| Jan 21, 2019 at 20:47 | comment | added | Mateusz Kwaśnicki | Very nice argument! A slightly more direct variant would be to write $f(t)$ as a convolution of the Gauss–Weierstrass kernel $g(t) = (\pi (1-\delta))^{-1/2} e^{-t^2/(1-\delta)}$ and $h(t) = \sum_{i=1}^n a_i \delta^{-1/2} e^{-(x_i-t)^2/\delta} - c$ for a $\delta$ small enough, so that $h$ has $2 n$ zeroes. Convolution with a Gaussian does not increase the number of zeroes (as a Pólya frequency function, it is a variation diminishing kernel; for more on this, see, for example, review sections written by Karlin here). | |
| Jan 21, 2019 at 20:00 | history | answered | Gjergji Zaimi | CC BY-SA 4.0 |