Timeline for On Mehler's formula for Hermite polynomials
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2020 at 20:52 | comment | added | Bazin | @Aleksei Kulikov You are right for my post since $x^2-\frac{(x-ry)^2}{1-r^2}$ is indeed symmetric. Sorry for the absurd question. | |
| Jun 10, 2020 at 20:51 | comment | added | Bazin | @Carlo Beenakker Take $r=0$ in your equality, you get $x^2-y^2=y^2-x^2$. Nevertheless you are right for my post since $x^2-\frac{(x-ry)^2}{1-r^2}$ is indeed symmetric. Sorry for the absurd question. | |
| Jun 10, 2020 at 19:25 | comment | added | Carlo Beenakker | $x^2-\frac{(y-r x)^2}{1-r^2}=y^2-\frac{(x-r y)^2}{1-r^2}$ | |
| Jun 10, 2020 at 19:09 | comment | added | Aleksei Kulikov | @Bazin I'm a bit lost. If you literally expand everything in the exponent in the rhs from your post the resulting expression will be symmetric in $x$ and $y$ (and would symbol-by-symbol coincide with the one from the answer by user69642). | |
| Jun 10, 2020 at 18:43 | comment | added | Bazin | Thanks for your answer and for the reference, which I need to study. However, I am still puzzled, since if user69642 is correct, the logarithms of the two different rhs must be the same. | |
| Jun 10, 2020 at 14:14 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
deleted 37 characters in body
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| Jun 10, 2020 at 14:03 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |