# Talk:Mexican hat wavelet

## Renaming to Ricker Wavelet

Is it necessary to name this as the Mexican Hat Wavelet? It's more recognizable name is the Ricker Wavelet so perhaps this should be the official name. Chris Engelsma 18:04, 20 October 2011 (UTC)

I agree with this. Ricker Wavelet is a much more common name. I have not heard Mexican Hat Wavelet in any place other than here. All publications I have read call it Ricker. 129.7.231.237 (talk) 06:52, 12 April 2012 (UTC)
This function is referred to as the Mexican hat function in the image processing and computer vision communities. In these areas, this is an established terminology. I have also seen a number of mathematicians using this terminology. Therefore, a redirect from Richer Wavelet may be more appropriate. I have never heard of Ricker wavelet before this occasion now, however. Tpl (talk) 09:38, 14 April 2012 (UTC)

## normalizing term

Is the normalizing term correct there? i have coded this in matlab: excit_function=sprintf('(1-((t-%e).^2/%e.^2)).*exp(-(t-%e).^2/(2*%e.^2))',5*ex_sigma,ex_sigma,5*ex_sigma,ex_sigma); t=0:dt:sim_time; excit=eval(excit_function); which is without the normalizing term and it correctly evaluates so that peak is +1 and not something like +10^27 —Preceding unsigned comment added by 188.220.32.144 (talk) 20:59, 25 February 2010 (UTC)

## Lanczos resampling

This looks a lot like the kernel of Lanczos resampling. Is this just a superficial relationship? —Ben FrantzDale (talk) 21:32, 31 December 2007 (UTC) Yes, it's just resemblance, there's no relationship. crisluengo (talk) 07:50, 15 September 2009 (UTC)

## normalizing term, second post

> I've changed the coefficient in front of the equation to be the mathematical result of what it says it should be, ie. the second derivative of a normalized Gaussian distribution.

It shouldn't be the second derivative of a normalized Gaussian distribution. It should be a wavelet, normalized by ordinary wavelet normalization rules:

${\displaystyle \int _{-\infty }^{\infty }|\psi (t)|^{2}\,dt=1}$

(see Wavelet)

## Is second image 3D or 2D?

This is 1-dimensional wavelet. It depends on one variable ${\displaystyle t}$:

${\displaystyle \psi (t)={2 \over {{\sqrt {3\sigma }}\pi ^{1 \over 4}}}\left(1-{t^{2} \over \sigma ^{2}}\right)e^{-t^{2} \over 2\sigma ^{2}}}$

This is 2-dimensional wavelet. It depends on two variables: ${\displaystyle x,y}$:

${\displaystyle \psi (x,y)=-{\frac {1}{\pi \sigma ^{4}}}\left(1-{\frac {x^{2}+y^{2}}{2\sigma ^{2}}}\right)\mathrm {e} ^{-(x^{2}+y^{2})/2\sigma ^{2}}.}$

The image itself is 3d, but what it depicts is a 2d wavelet. — Preceding unsigned comment added by Eloquent2013 (talkcontribs) 09:06, 24 December 2015 (UTC)