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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.2010-10-16

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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 1.2010-10-16
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    Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy!2010-10-16
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    Ah!!! This is such a lovely proof.2010-10-16
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    @Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: http://en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state?2010-10-16
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    @Moron: Beautifully done! I will delete my answer.2010-10-16
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    @Moron: it is the same thing. Let $x_1\ldots x_k$ be 1, and $x_{k+1} \ldots x_n$ be 0. Let $y_1\ldots y_n$ be a bunch of positive numbers. Then one arrangement that gives the maximum $x_1 y_{\sigma(1)} + \cdots + x_n y_{\sigma(n)}$ is if you arrange the $y$ in decreasing order. Now you move to the integral case and consider $\int \chi_{\Omega} f dx$ where $\chi_{\Omega}$ is a characteristic function. The rearrangement inequality then says the the norm is the maximum if you put the support of $\chi_{\Omega}$ where $f$ is big...2010-10-16
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    In general, for arbitrary integrable non-negative function $f$, without radial symmetry, the analogous problem can be solved by first defining $A_\lambda$ as the set $\{ x : f(x) > \lamba \}$. Let $\lambda_0$ be the inf of all $\lambda$ with $|A_\lambda| < 1$. Then any $\Omega\subset $A_{\lambda_0}$ such that $\Omega\supset A_\lambda$ for all $\lambda > \lambda_0$ is a solution.2010-10-16
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    @Willie: Oh I see. What you meant(and said!) regarding the ball at origin was a _consequence_ of rearrangement inequality. I read that as being same as rearrangement inequality and hence the confusion. Thanks for clarifying.2010-10-16
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    (In other words, think of it as rearrangement where the index set is $\mathbb{R}^2$, instead of a finite set of points.) Yes, sorry about the confusion. I meant the fact that followed the colon mark in my first sentence to be the "illustration".2010-10-16