Timeline for Do roots of these polynomials approach the negative of the Euler-Mascheroni constant?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 6, 2023 at 17:00 | comment | added | David E Speyer | @StevenStadnicki Mathematica is very cautious about assuming that two square roots have the same choice of sign; you usually have to twist its arm a bit. | |
| Mar 3, 2023 at 18:13 | comment | added | AccidentalFourierTransform | @StevenStadnicki I asked Mathematica to try a little harder and it found a simpler expression. Thanks! | |
| Mar 3, 2023 at 18:11 | history | edited | AccidentalFourierTransform | CC BY-SA 4.0 |
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| Mar 3, 2023 at 17:17 | comment | added | Steven Stadnicki | I'm legitimately a bit surprised that Mathematica wasn't able to simplify e.g. $(\sqrt{8x+1}-1)^2(\sqrt{8x+1}+1)^2$ to $64x^2$. | |
| Mar 3, 2023 at 16:33 | comment | added | Peter Mueller | The first identity can be written as $(p(n+2)-p(n+1))(2n+3)(n+2)^2=(p(n+1)-p(n))(2n+1)(n^2+3n-2x+2)$, so a multiplicative telescope first yields a closed form of $p(n+1)-p(n)$, and then an additive telescope gives $p(n)$. This would give an alternative proof of David Speyer's explicit formula. | |
| Mar 3, 2023 at 15:17 | history | answered | AccidentalFourierTransform | CC BY-SA 4.0 |