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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

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

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1 Answer 1

up vote 20 down vote accepted

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.

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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 1:39
    
Thanks, this seems to work –  Yaroslav Bulatov Oct 15 '10 at 21:12

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