Erf squared approximation I found a nice and useful approximation of squared error function 
$$
\mathrm{erf}^{2}\!\left(x\right)=1-\exp\!\left(-\frac{\pi^{2}}{8}x^{2}\right)+\varepsilon\!\left(x\right).
$$
I checked numerically that maximum error is bounded by $\left|\varepsilon\!\left(x\right)\right| < 61\cdot10^{-4}$ but I was asked if this could be somehow proved analytically. Or at least if order of the error could be determined in such manner.
Error function is defined as
$$
\mathrm{erf}\left(x\right)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}\exp\left(-t^{2}\right)\,\mathrm{d}t = 2\Phi\!\left(x\sqrt{2}\right)-1,
$$
where $\Phi\!\left(x\right)$ is normal cumulative distribution function.
And more generally: is numerical check not enough in such cases?
 A: Such kind of inequalities can be proved through Fubini's theorem, since:
$$\operatorname{erf}(x)^2 = \frac{4}{\pi}\iint_{[0,x]^2}e^{-(u^2+v^2)}\,du\,dv=\frac{1}{\pi}\iint_{[-x,x]^2}e^{-(u^2+v^2)}\,du\,dv $$
and the last integral, over a square, can be effectively approximated with an integral over a suitable circle, whose explicit form is given by:
$$ 1-\exp\left({-\rho^2 x^2}\right). $$
For instance, it is trivial that:
$$\operatorname{erf}(x)^2 \geq \frac{1}{\pi}\iint_{u^2+v^2\leq x^2}e^{-(u^2+v^2)}\,du\,dv = 1-e^{-x^2}. $$
as well as:
$$\operatorname{erf}(x)^2 \leq \frac{1}{\pi}\iint_{u^2+v^2\leq 2x^2}e^{-(u^2+v^2)}\,du\,dv = 1-e^{-2x^2}. $$
A tight approximation is given by considering the circle that has the same area of the original square of integration, $[-x,x]^2$:

$$ \operatorname{erf}(x)^2 \approx 1-e^{-4x^2/\pi}.$$

One may even try to find "the best" constant $\rho$ such that:
$$\operatorname{erf}(x)^2 \approx 1-e^{-\rho^2 x^2}$$
by solving a least-square minimization problem. Numerically, such optimal value is around:
$$ \rho = 1.1131 $$
that is extremely close to $\rho=\frac{\pi}{\sqrt{8}}=1.11072\ldots$ given by your approximation.
A: Just to be part of this interesting discussion, I fully agree that $$F(a)=\int_0^{\infty}\left(\text{erf}(x)^2-(1-e^{-a x^2})\right)^2$$ is minimum for $a\approx 1.23907$ (same value as the one given by Jack D'Aurizio) but, according to $RIES$, this number seems to be much closer to $$a=(1+\pi )^{2/3} \log ^2(2)\approx 1.23907$$ than to $\frac{\pi^2}8\approx 1.23370$ even if this does make very large difference (the maximum error is reduced from $0.006$ to $0.004$ and the value of the integral $F(a)$ changes from $0.00002769$ to $0.00002572$).
If we look for a still better approximation, we could consider $\log\big(1-\text{erf}(x)^2\big)$ (which, for sure, introduces a bias in the problem) and establish a Pade approximant and finally arrive to $$\mathrm{erf}\!\left(x\right)^2\approx1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha x^2}{1+\beta x^2}\,x^2 \Big)$$ where $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi }$$ $$\beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$ The value of the corresponding error function is $1.1568\times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation; the maximum error is $0.00035$.
