What is the formula for calculating the FWHM of a gaussian function?

AI Thread Summary
The discussion focuses on calculating the Full Width at Half Maximum (FWHM) for a Gaussian function defined as exp[-(π*x²/A²)]. The FWHM is established as 0.939A, but the proof is sought. The method involves setting the Gaussian equal to 0.5 to find the half-maximum and applying natural logarithms to simplify the equation. The variable substitution x = kA is introduced, leading to the computation of k as (-ln(0.5)/π)^(1/2). The conversation highlights the challenge in proving the FWHM formula for this specific Gaussian function.
peterjaybee
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Hi,

I have a gaussian of the form
exp[-\frac{\pi*x^{2}}{A^2}].

I know that the FWHM=0.939A, but I cannot prove it.

I Let exp[-\frac{\pi*x^{2}}{A^2}=0.5 (i.e. the half maximum part)

taking natural logs I get rid of the exponential, but then which bit represents the full width?
 
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Write x = kA and compute k=(-ln(.5)/π)1/2

Your Gaussian looks funny.
 
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