Timeline for Limit involving regularized gamma function and its inverse
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2023 at 9:14 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
|
| Aug 25, 2013 at 5:01 | comment | added | Bullmoose | Thank you for the proof! I really like how you inverted the $Q$ function -- that's a neat trick that I'll use in the future. While the proof works as is, after a brief search I found this paper by Temme. I haven't had a chance to read it in detail, but it seems to apply to a very general range of scenarios, so I am now wondering whether your derivation of $Q^{-1}$ matches the form given in that paper (I'll try to read it carefully this coming week.) | |
| Aug 25, 2013 at 4:22 | vote | accept | Bullmoose | ||
| Aug 23, 2013 at 7:08 | comment | added | Johannes Trost | Thank you Bullmoose for going through the proof ! I had indeed to flip a lot of DLMF pages. I corrected one typo (the ln is now under the square root). Thank you for that. . $s>0$ is no typo. The proof works perfect for all $s>0$, I guess. One just needs that $g(x)$ grows infinitely as $x \rightarrow \infty$. | |
| Aug 23, 2013 at 7:02 | history | edited | Johannes Trost | CC BY-SA 3.0 |
edited body
|
| Aug 23, 2013 at 2:29 | comment | added | Bullmoose | I finished reading the proof -- I think it's correct (that was quite a tour de force through DLMF). I think there is one more typo in the final approximation of $Q^{-1}$ (after "This gives"): I think you want the $\ln x$ under the square root sign (it's there for the remainder of the proof). Regarding your last statement, my numerical evaluations weren't perfect and indeed $L(x)$ might be going to $\frac{1}{2}$ instead of 1. Still, it looks like that there is indeed a threshold on the asymptotics of $f(x)$. Thanks! | |
| Aug 23, 2013 at 2:03 | comment | added | Bullmoose | Back to going through your proof and noticed that I made a mistake in my comment: indeed $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx \frac{1}{2}\operatorname{erfc}(y)$ since Tricomi's formula in DLMF is for the complement $P(\alpha+1,\alpha+\sqrt{2\alpha}y)=1-Q(\alpha+1,\alpha+\sqrt{2\alpha}y)$... my mistake, though point #1 in my previous comment still stands. :) Continuing reading the proof... | |
| Aug 22, 2013 at 22:31 | comment | added | Bullmoose | Thanks for the derivation! I haven't gotten a chance to completely go through it yet, however, there are two things that I noticed right away in the beginning: 1) I am fine with the assumption of a $g(x)$ being a power function of $x$, but I think you meant $s>1$ (you have $s>0$ which I think is just a typo); 2) Are you sure that $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx\frac{1}{2}\operatorname{erfc}(y)$? The DLMF link you posted states that $Q(\alpha+1,\alpha+\sqrt{2\alpha}y)\approx\frac{1}{2}\operatorname{erfc}(-y)$... Does that impact the result? | |
| Aug 22, 2013 at 12:07 | history | edited | Johannes Trost | CC BY-SA 3.0 |
Reformulated the last ($\lambda=1$) steps. Improved language a bit.
|
| Aug 22, 2013 at 9:40 | history | edited | Johannes Trost | CC BY-SA 3.0 |
added 76 characters in body
|
| Aug 22, 2013 at 8:30 | history | answered | Johannes Trost | CC BY-SA 3.0 |