Timeline for Is $ \sum\limits_{n=0}^\infty x^n / \sqrt{n!} $ positive?
Current License: CC BY-SA 3.0
17 events
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| Jan 7, 2021 at 12:38 | review | Suggested edits | |||
| Jan 7, 2021 at 17:25 | |||||
| May 18, 2020 at 15:18 | comment | added | Hans | To be clear, I understand the integral involving $e^{-nx}$ quoted in my last comment transforms into your original integral. What I want to know is where and how this comes about or its motivation. | |
| May 18, 2020 at 3:59 | comment | added | Hans | +1. Magical feat! Could you please explain the purpose of the expression $\frac{e^{−nx}−e^{n(x+t)}}{t^{\frac12}}\frac{dt}t=n^\frac12 e^{−nx}$ in your comment revealing your motivation for the setup? I suppose the reason you consider half-derivative as opposed to other power the power on $n$ is $\frac12$? If it is $n^a$, you will use $a$-derivative, right? | |
| Nov 19, 2015 at 2:50 | comment | added | student | Very nice answer, but I have more questions... How about the power series is defined on the complex plane? Can you show there is no zero point? | |
| Jul 15, 2013 at 7:47 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Jan 8, 2012 at 15:59 | comment | added | fedja | Do you really think so? I don't. | |
| Jan 7, 2012 at 23:02 | comment | added | Noam D. Elkies | Nice!! For what its worth, $c$ is $\int_0^\infty (1-e^{-x}) \phantom. dx^{3/2} = 2 \Gamma(1/2) = 2\sqrt{\pi}$ by integration by parts. [My previous comment, now deleted, reported an incorrect answer because I left the factor $1/t$ out of the integrand in the numerical calculation.] | |
| Jan 7, 2012 at 2:38 | comment | added | GH from MO | Beautiful indeed. I had the feeling that this idea might also lead to $\forall x\in\mathbb{R}:\sum_{r=0}^{2n}x^r{2n\choose r}^{1/2}>0$ (cf. mathoverflow.net/questions/85013/…). I could not make it work, but admittedly I have not tried too hard either. | |
| Jan 6, 2012 at 22:29 | comment | added | Yemon Choi | Beautiful and educative | |
| Jan 6, 2012 at 18:57 | comment | added | J Russell | @fedja: Thank you, that's quite enlightening. | |
| Jan 6, 2012 at 16:53 | comment | added | fedja | @ J Russell. Yes, I started with the proof for the exponent you mentioned. The main difference is that you need half-derivative here, not the full one. The half-derivative (or, rather, quarter-Laplace) operator is well-known: it is just (regularized) convolution with $|t|^{-3/2}$. We needed the one-sided version here (the other side is not integrable), which works especially well on the negative exponents: $\int_0^\infty \frac{e^{-nx}-e^{n(x+t)}}{t^{1/2}}\frac{dt}{t}=n^{1/2}e^{-nx}$, so I just made the logarithmic change of variable to make it act on powers as needed. | |
| Jan 6, 2012 at 14:43 | comment | added | Liviu Nicolaescu | This is awesome! youtube.com/watch?v=b2RiI9v3io4 | |
| Jan 6, 2012 at 14:36 | comment | added | J Russell | Altogether, very nice. | |
| Jan 6, 2012 at 14:36 | vote | accept | J Russell | ||
| Jan 6, 2012 at 14:35 | comment | added | J Russell | Fedja, thank you. Because $d e^x / dx = e^x$, you can show that $e^x$ is positive by asking what would happen at a zero crossing. I've tried (unsuccessfully) to analyze $f'(x)$ with the hope of doing something similar. Your proof strikes me as having the same flavor as that sort of manipulation. Could you offer any motivation for how you came up with your integral "operator"? | |
| Jan 6, 2012 at 13:26 | comment | added | David E Speyer | Very slick! I had not seen that trick before. | |
| Jan 6, 2012 at 12:16 | history | answered | fedja | CC BY-SA 3.0 |