If you parametrize the Gaussian exponential, $e^{-kx^2}$, in terms of the x value at the point where it reaches half its height, $x_h$ (y( $x_h$)=0.5), you end up changing exponent bases to get a surprisingly simple form, $2^{-(\frac{x}{x_h})^2}$. Does anyone have a reference for this derivation?

According to wikipedia, the 'Full Width at Half Max' is defined, $FWHM=2 \sigma \sqrt{2 \log (2)}$, where $\sigma$ is the standard deviation as seen in the statistics-friendly representation of the Gaussian, $e^{-\frac{x^2}{2 \sigma ^2}}$.

Solving for sigma $$\sigma = \frac{FWHM}{2 \sqrt{2 \log (2)}}$$ Which means that the exponential is $$ e^{-\frac{x^2}{2 \sigma ^2}}= e^{-4 \log (2)(\frac{x}{FWHM})^2} $$

Now the exponential and log go off together, and we are left with:

$$2^{-(\frac{2 x}{FWHM})^2} $$

Which gets even simpler using $x_h \equiv \frac{FWHM}{2}$:

$$2^{-(\frac{x}{x_h})^2} $$

Mathematica verifies the result:

\[Sigma][fwhm_] := fwhm/(2 Sqrt[2 Log[2]])
k[s_] := 1/(2 s^2)
G[x_, fwhm_] := Exp[-k[\[Sigma][fwhm]]  x*x]
G[X, fwh]

Output:

2^(-((4 X^2)/fwh^2))