Timeline for Riemann $P$-symbol for ODEs
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2023 at 12:15 | history | edited | J. M. isn't a mathematician | CC BY-SA 4.0 |
link to the paper
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| Aug 17, 2019 at 17:07 | comment | added | Alexandre Eremenko | Yes, there is a misprint, and you wrote it correctly. Sorry: I knew about this misprint, but forgot when I was answering your question. | |
| Aug 17, 2019 at 15:05 | comment | added | Pavel Mostovykh | or there is a misprint, and it should be $$ \frac{d^2y}{dx^2} +\left[\sum \limits_{i=1}^r \frac{1-(\lambda_i'+\lambda_i'')}{x-e_i} \right]\frac{dy}{dx} +\left[\sum \limits_{i=1}^r \frac{\lambda_i'\lambda_i''}{(x-e_i)^2} +\frac{\left(\lambda_\infty'\lambda_\infty'' -\sum \limits_{i=1}^r \lambda_i'\lambda_i''\right) x^{r-2}+a_1x^{r-3}+\ldots +a_{r-2}}{\prod \limits_{i=1}^r (x-e_i)} \right]y=0. $$ (parenthesis added for the $x^{r-2}$ term)? | |
| Aug 17, 2019 at 15:04 | comment | added | Pavel Mostovykh | Excuse me, does the formula (1) in Edward Van Vleck you mentioned really read $$ \frac{d^2y}{dx^2} +\left[\sum \limits_{i=1}^r \frac{1-(\lambda_i'+\lambda_i'')}{x-e_i} \right]\frac{dy}{dx} +\left[\sum \limits_{i=1}^r \frac{\lambda_i'\lambda_i''}{(x-e_i)^2} +\frac{\lambda_\infty'\lambda_\infty'' -\sum \limits_{i=1}^r \lambda_i'\lambda_i'' x^{r-2}+a_1x^{r-3}+\ldots +a_{r-2}}{\prod \limits_{i=1}^r (x-e_i)} \right]y=0. $$ | |
| Aug 17, 2019 at 10:44 | vote | accept | Pavel Mostovykh | ||
| Aug 16, 2019 at 12:00 | history | answered | Alexandre Eremenko | CC BY-SA 4.0 |