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Can anyone see a nice way to prove the following for $0\le x \le 1$?

$$\sqrt{1-x^2}\ge \operatorname{erf}(\sqrt{-\log x})$$

$\operatorname{erf}$ is defined as

$$\operatorname{erf}(z) = \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-t^2} \, dt$$

J. M. ain't a mathematician
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Yaroslav Bulatov
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  • I am surprized since $\sqrt{1-x^2}\ge \operatorname{erf}(\sqrt{-\log x})$ is true only if $x > 0.186456$. If $x=\frac 1{e^2}$, $\sqrt{1-\frac{1}{e^4}}-\text{erf}(2)=-0.00452241$. – Claude Leibovici Mar 27 '24 at 13:18
  • @ClaudeLeibovici wow, that's a very old question of mine, I'm trying to remember what motivated this, probably something I needed when writing this paper, it appears there's a mistake, and it doesn't hold the entire range, but only for $x>0.18...$ as you point out – Yaroslav Bulatov Mar 27 '24 at 17:00
  • I am really sorry ! My comment is fully wrong : I forgot the square root in the error function. – Claude Leibovici Mar 28 '24 at 03:49

1 Answers1

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Let $y = \sqrt{-\log x}$. Then the inequality reduces to $\text{erf}(y) \leq \sqrt{(1-e^{-2y^2})}$ or equivalently $\text{erf}^2(y) + e^{-2y^2} \leq 1$. Now $\text{erf}^2(y)$ can be written as a double integral $\text{erf}^2(y) = \frac{4}{\pi} \int_{0}^y \int_{0}^{y} e^{- (a^2+b^2)} da db$ (As Qiaochu Yuan points out, the functions involved are well behaved and the double integral is well defined). Replace the area of integration from the square of side $y$ in the first quadrant to a quarter-circle of radius $y\sqrt{2}$ in the first quadrant and switch to polar co-ordinates. This would give the inequality $\text{erf}^2(y) \leq 1-e^{-2y^2}$ which is what we wanted.

Dinesh
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  • The double integral step is fine. The domain is compact and the integrand is continuous so everything is as nice as possible. Nice solution. – Qiaochu Yuan Oct 14 '10 at 23:56
  • The double integral is okay. The area of integration change should be to, instead of the quarter circle of radius $y$, the quarter circle of radius $\sqrt{2} y$ (the quarter circle has to contain the square as a subset). – Willie Wong Oct 14 '10 at 23:58
  • @Willie Wong: Typo on my part. Thanks for pointing it out. I have edited the solution. – Dinesh Oct 14 '10 at 23:59
  • Indeed. Very nice solution. – Aryabhata Oct 15 '10 at 01:39