Timeline for Approximating erf by tanh
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 1, 2013 at 9:06 | comment | added | John Bentin | Thanks, Aryeh. I was confusing erf with a cumulative distribution function. | |
| Jan 1, 2013 at 8:38 | comment | added | Aryeh Kontorovich | Depends how you define erf. If you take the definition given here, which appears to be standard, then erf(0)=0. | |
| Jan 1, 2013 at 8:30 | comment | added | John Bentin | Yes. But the function $f$ here doesn't tend to zero at both ends of the interval $[0, \infty)$, because $f(0)=\mathrm{erf}\, 0-\mathrm{tanh}\,0=\frac{1}{2}-0=\frac{1}{2}$. | |
| Dec 31, 2012 at 23:36 | comment | added | Alexandre Eremenko | John, if you have a function that tends to zero from the positive side on both ends of the interval then derivative has ODD number of zeros. Only one if it positive everywhere, and at least 3 if not. | |
| Dec 31, 2012 at 21:20 | comment | added | John Bentin | Sorry, I can only see that, on your supposition, $f'$ has $2$ (or more) positive zeros. But anyway that's enough for the rest of the argument to work. | |
| Dec 31, 2012 at 17:19 | vote | accept | Aryeh Kontorovich | ||
| Dec 31, 2012 at 17:17 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |