# Inverting the normalized Gaussian to get a kernel window radius

I am seeking to compute a kernel radius to use with my gaussian convolution filter, and inspired by https://stackoverflow.com/a/68050503/, I came up with:

$$r=\sqrt{-2\sigma^2\ln\left( \epsilon\sigma\sqrt{2\pi} \right)}$$

This is basically inverting the normalized gaussian, setting $$g(x,\sigma)=\epsilon$$ and solving for $$x$$ given $$\sigma$$ and $$\epsilon$$.

However, I noticed that with $$\epsilon=0.01$$ at about $$\sigma>22$$ my kernel radius computation started get smaller instead of bigger, and at $$\sigma>40$$ it went completely undefined (NaN).

So I re-derived this little formula (derivation at bottom) and graphed it in desmos, and mysteriously, it indeed showed a graph that peaked at about $$20$$ and fell to $$0$$ at about $$40$$ and was thereafter was undefined!

Here is the desmos graph, with multiple steps through the derivation, all showing the same graph (only the first is activated though): https://www.desmos.com/calculator/ucrqzrahxs

Where is my math wrong?

Derivation:

$$\begin{eqnarray} g(x,\sigma)&=&\frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-x^2}{2\sigma^2}} &\hspace{5em}\text{Normalized Guassian function.}\\ g(r,\sigma)&=&\frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-r^2}{2\sigma^2}} &\hspace{5em}\text{Set radius to x.} \\ \epsilon&=&g(r,\sigma) &\hspace{5em}\text{Set }\epsilon\text{ to output} \\ \epsilon&=&\frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-r^2}{2\sigma^2}} &\hspace{5em}\text{Replace }g(r,\sigma). \\ \ln \epsilon&=& \ln \left[\frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-r^2}{2\sigma^2}}\right] &\hspace{5em}\text{Log both sides.} \\ \ln \epsilon&=& \ln \left[\frac{1}{\sigma\sqrt{2\pi}} \right]-\frac{r^2}{2\sigma^2} &\hspace{5em}\text{Get rid of }e. \\ \frac{r^2}{2\sigma^2} + \ln \epsilon&=& \ln \left[\frac{1}{\sigma\sqrt{2\pi}} \right] &\hspace{5em}\text{Move x to the left hand side.} \\ \frac{r^2}{2\sigma^2}&=& \ln \left[\frac{1}{\sigma\sqrt{2\pi}} \right] - \ln \epsilon &\hspace{5em}\text{Move }\epsilon\text{ to the right hand side.} \\ \frac{r^2}{2\sigma^2}&=& \ln \left[\frac{1}{\epsilon\sigma\sqrt{2\pi}} \right] &\hspace{5em}\text{Move }\epsilon\text{ into the log function.} \\ r^2&=& 2\sigma^2\ln \left[\frac{1}{\epsilon\sigma\sqrt{2\pi}} \right] &\hspace{5em}\text{Move the denominator to the right.} \\ r &=& \sqrt{2\sigma^2\ln \left[\frac{1}{\epsilon\sigma\sqrt{2\pi}} \right]} &\hspace{5em}\text{Square root both sides.} \\ r &=& \sqrt{-2\sigma^2\ln \left[\epsilon\sigma\sqrt{2\pi} \right]} &\hspace{5em}\text{Flip the fraction inside the logarithm.} \\ \end{eqnarray}$$

I think this should be correct, because when $$\sigma$$ increases, the peak of the Gaussian curve becomes smaller and eventually less than 0.01.
Edit: Indeed, this is the curve for $$\sigma=39$$ (red) and for $$\sigma=41$$ (blue). The latter stays below 0.01
• 3$\sigma$ is usually a good guess, but it really depends what you need it for. Otherwise you can retry your approach using $g(x,\sigma)/g(0,\sigma)=\epsilon$