Timeline for Bounds on the inverse multivariate beta function
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 1, 2020 at 18:04 | comment | added | lrnv | Yes i agree with you, sorry. | |
| Dec 1, 2020 at 17:34 | comment | added | Iosif Pinelis | I am afraid that your question cannot be satisfactorily answered unless you specify the bounds you need. | |
| Dec 1, 2020 at 12:11 | comment | added | lrnv | I did try to bound the IB function wiht the stirling gbounds, but this does not help me in the surrounding problem. Bounding hte IB function might not be exactly what i need. Thanks anyway ! | |
| Nov 29, 2020 at 21:34 | comment | added | Iosif Pinelis | If $x_1\downarrow 0$, then $\Gamma(x_1)=\Gamma(x_1+1)/x_1\sim1//x_1\to\infty$, whence $IB(x)\to0$. Also, if the coordinates of $\mathbf b$ are allowed to depend on $d$, just choose $\mathbf b=(d,\dots,d)$. Also, how about using Stirling bounds on the gamma function? | |
| Nov 29, 2020 at 17:25 | comment | added | lrnv | $\mathbf a,\mathbf b$ are allowed to depend on $d$. I don get why you first point holds ? | |
| Nov 28, 2020 at 22:53 | comment | added | Iosif Pinelis | Lettiing $x_1\downarrow0$, we see that any lower bound of the form $\mathbf a^{\mathbf x} $ with $\mathbf a\in(0,\infty)^n$ will not hold. Also, if $\mathbf b$ is not allowed to depend on $d$, then any upper bound of the form $\mathbf b^{\mathbf x}$ will not hold either. | |
| Nov 27, 2020 at 13:30 | history | edited | lrnv | CC BY-SA 4.0 |
added 16 characters in body
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| Nov 27, 2020 at 13:11 | history | asked | lrnv | CC BY-SA 4.0 |