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This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0\le x\le 5$. I think it holds for all positive $x$, can anyone see a proof?

$$1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$$

Note: using analysis for previous question you can show that $1-\exp(-k x^2)$ is an upper bound on $\text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/\pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $\text{erf}(x)^2$ very closely.

Dashed graph below is $\text{erf}(x)^2$, red is $1-\exp(-k x^2)$ for $k=4/\pi$, other two graphs are for $k=1$ and $k=2$

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The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about $x=0$ also tells us why $k=4/\pi$ is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one. –  Dinesh Oct 16 '10 at 0:30
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up vote 10 down vote accepted

As before we consider

$$erf(x)^2={4\over \pi}\int_0^x\int_0^x \exp{-(s^2+t^2)}\ ds \ dt$$

Now compare this with the same over the area which is given by the quarter of a circle of radius $\displaystyle \frac{2x}{\sqrt{\pi}}$. The area of this is same as the area of the square of side $x$.

Since $\displaystyle e^{-(s^2 + t^2)}$ decreases as $\displaystyle s^2 + t^2$ increases, we are done!

The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $\displaystyle s^2 + t^2$ is higher in that region).

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Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) \geq 0$, such that $\partial_r f \leq 0$, we can ask: of all the sets $\Omega$ of area 1, when is the integral $\int_\Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1. –  Willie Wong Oct 16 '10 at 2:26
    
Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy! –  Yaroslav Bulatov Oct 16 '10 at 3:32
    
Ah!!! This is such a lovely proof. –  Dinesh Oct 16 '10 at 3:43
    
@Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state? –  Aryabhata Oct 16 '10 at 15:03
    
@Moron: Beautifully done! I will delete my answer. –  Byron Schmuland Oct 16 '10 at 15:31
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