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I believe that (i.e., it would be convenient if, and visually appears that) the product of the two complementary error functions:

$$\operatorname{erfc}\left[\frac{a-x}{b}\right]\operatorname{erfc}\left[\frac{a+x}{b}\right]$$

will have a solution, or can be approximated with a solution, of a Gaussian form (i.e., $c\operatorname{exp}\left[-\frac{x^2}{2d^2}\right]$, where $c$ and $d$ are functions of $a$ and $b$) when $a>0$ and $c>0$, however I cannot find a proof of this. Any help?

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In this paper, the authors provide the following simple approximation of the error function:

$$\operatorname{erf}(z) = 1- \exp\{-c_1z-c_2z^2\},\; z\ge 0 $$

with

$$ c_1 = 1.09500814703333,\;\; c_2 = 0.75651138383854$$

Set $\frac{a-x}{b} \equiv z_1$, and $\frac{a+x}{b} =\equiv z_2$. Given also that $\operatorname{erfc} = 1- \operatorname{erf}$, and under the implied restrictions so that $\frac{a-x}{b}\ge 0,\;\frac{a+x}{b}\ge 0$, (for example, for $b>0,\; a>0,\; x\le |a|$) we have, using the approximation,

$$\operatorname{erfc}(z_1)\operatorname{erfc}(z_2) = [1-\operatorname{erf}(z_1)][1-\operatorname{erf}(z_1)] $$

$$\approx [1-1+ \exp\{-c_1z_1-c_2z_2^2\}][1-1+ \exp\{-c_1z_2-c_2z_2^2\}]$$

$$=\exp\{-c_1(z_1+z_2)-c_2(z_1^2+z_2^2)\} \\= \exp\left\{-c_1\left(\frac{a-x}{b}+\frac{a+x}{b}\right)-c_2\left[\left(\frac{a-x}{b}\right)^2+\left(\frac{a+x}{b}\right)^2\right]\right\}$$

$$ =\exp\left\{-\frac{2a}{b}c_1-c_2\left[\left(\frac{a-x}{b}+\frac{a+x}{b}\right)^2-2\frac{a-x}{b}\frac{a+x}{b}\right]\right\} $$

$$=\exp\left\{-\frac{2a}{b}c_1-c_2\left[4\left(\frac{a}{b}\right)^2-2\frac{a^2-x^2}{b^2}\right]\right\} $$

$$=\exp\left\{-\frac{2a}{b}c_1-c_2\left[2\left(\frac{a}{b}\right)^2+2\frac{x^2}{b^2}\right]\right\} $$

$$ = \exp\left\{-\frac{2a}{b}c_1-2c_2\left(\frac{a}{b}\right)^2\right\}\cdot\exp\left\{-2c_2\frac{x^2}{b^2}\right\}$$

$$=C\cdot\exp\left\{-\frac{x^2}{2d^2}\right\} $$

with

$$ C\equiv \exp\left\{-\frac{2a}{b}c_1-2c_2\left(\frac{a}{b}\right)^2\right\},\;\; d^2 = \frac {b^2}{4c_2}$$

Let $x = a$. Then, the product becomes $\mathrm{erfc}(2a/b)$. If the product were equal to an exponential function, that would mean we could express $\mathrm{erfc}$ in terms of elementary functions, which, as far as I know, is impossible.

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