Talk:Difference of Gaussians

Concerning the new addition, does the author behind it have a reference on the new statement "with K~1.6 for approximating a Laplacian Of Gaussian and K~5 in the retina"? According to my humble opinion, the Laplacian can be well approximated also for other values of K. Tpl 13:51, 19 November 2006 (UTC)

Since no reply has been received concerning the previous question and this statement is neither correct and nor supported by any references I have removed this statement. Tpl 08:34, 3 December 2006 (UTC)

I've not been fast enough! 1.6 is the best fit using MSE. 5 is the best fit using physiological data (though there exist a wide heteregenoity) see http://retina.anatomy.upenn.edu/~lance/modelmath/enroth_freeman.html. main fact is that it's wider than a LOG. Meduz 13:22, 27 December 2006 (UTC)

Please clarify one point. I'm almost sure, but not positive, that K is the constant scaling factor between the standard deviations of the two Gaussians? As in sigma2 = K*sigma1? Also any comments of the practical effects of different K values? As in: larger K values would result in wider but shallower lobes on the function so larger K values will respond less strongly if 'blobs' of become too close (spatially). —Preceding unsigned comment added by Craigyk (talk • contribs) 21:34, 31 March 2008 (UTC)

zero crossing

Just a note: the zero crossing, r, of the DoG function is given by

$r=\sigma _{1}{\sqrt {\frac {2k^{2}\ln {k}}{k^{2}-1}}}$

where $\sigma _{2}=k\sigma _{1}$. This can easily be derived by the setting function:

${\frac {1}{\sigma _{1}{\sqrt {2\pi }}}}\,\exp \left(-{\frac {x^{2}}{2\sigma _{1}^{2}}}\right)-{\frac {1}{\sigma _{2}{\sqrt {2\pi }}}}\,\exp \left(-{\frac {x^{2}}{2\sigma _{2}^{2}}}\right)=0$

equal to zero and substituting $\sigma _{2}=k\sigma _{1}$.

I came here looking for exactly this (saved me a little work). Seems like this should be included in the article. Thanks! 24.91.117.221 (talk) 00:05, 10 August 2008 (UTC)