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))