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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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    (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