Newsgroups: comp.ai.neural-nets
Path: cantaloupe.srv.cs.cmu.edu!bb3.andrew.cmu.edu!newsfeed.pitt.edu!portc02.blue.aol.com!howland.erols.net!torn!nott!cu23.crl.aecl.ca!usenet
From: BurtonG@wl.aecl.ca (Gordon R. Burton)
Subject: Re: Adding two Gaussians
X-Nntp-Posting-Host: wp4099.wl.aecl.ca
Message-ID: <E3sqMn.99A@cu23.crl.aecl.ca>
Sender: usenet@cu23.crl.aecl.ca (USENET News System)
Reply-To: BurtonG@wl.aecl.ca
Organization: Atomic Energy of Canada
X-Newsreader: Forte Free Agent 1.0.82
References: <82u3os4t0r.fsf@eagle.eecs.umich.edu>
Date: Fri, 10 Jan 1997 14:26:16 GMT
Lines: 23

Here is the bare bones solution.

1) Properly normalized gaussian function has the form
g(x) = 1/(sigma)/(2*Pi)^0.5*EXP(-0.5*x^2/(sigma)^2)

2) What you want to do is find out the point at which the
second derivative of h(x) = g(x) + g(x-x0) becomes
positive at x = x0/2 (the midpoint between the two gaussians centred
at x = 0 and x = x0).

3) If you do the math, and make the substitution, I find that the
second derivative of h(x) goes to zero when x0 = 2*sigma.  So, the gut
feeling of the other posters was correct, the value of the constant
multiplying sigma was between 1 and 2.  Good gut instinct!

4) If x0 < 2 * sigma then the second derivative is negative (as you
would expect, take for example the case of x0 = 0).  If x0 > 2 * sigma
then the second derivative is positive and you should "see" the dip.

	Hope this helps.

	Gordon

