Timeline for Positivity of a finite sum involving Stirling numbers
Current License: CC BY-SA 3.0
5 events
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| Sep 17, 2016 at 22:55 | comment | added | Tomeu Fiol | Ok, it is clear now. Thanks also for mentioning the discrete Chebyshev transform. In the past, I also tried to prove $a_{n,m} \geq 0$ from the last formula in your answer, but I didn't manage to. | |
| Sep 17, 2016 at 21:34 | history | edited | Pietro Majer | CC BY-SA 3.0 |
added 452 characters in body
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| Sep 17, 2016 at 21:28 | comment | added | Pietro Majer | Sorry, maybe it is not very clear. The complete sentence is "for n+m odd the integrand is positive about $\pm 1$ . $P_nT_m$ of course changes sign several times in (-1,1), but it is positive and large close to $\pm 1$. | |
| Sep 17, 2016 at 20:10 | comment | added | Tomeu Fiol | Thanks for your detailed discussion. Indeed, this is precisely the problem that led me to these coefficients. However, I don't understand what you mean by the claim "for $n+m$ odd the integrand is positive". The polynomials that you call $P_n(x)$ have $n-1$ real roots in $(-1,1)$, so for instance, for $T_0(x)=1$, the function $P_n(x)T_0(x)=P_n(x)$ changes sign $n-1$ times in $(-1,1)$. | |
| Sep 17, 2016 at 17:23 | history | answered | Pietro Majer | CC BY-SA 3.0 |