Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.