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Feb 26, 2018 at 12:43 review Close votes
Mar 8, 2018 at 21:41
Feb 26, 2018 at 12:25 history edited Michael E2
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Feb 26, 2018 at 12:18 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jan 27, 2018 at 12:47 review Close votes
Jan 29, 2018 at 10:45
Jan 27, 2018 at 11:43 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Dec 28, 2017 at 11:43 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Nov 28, 2017 at 11:32 history edited m0nhawk CC BY-SA 3.0
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Nov 28, 2017 at 10:31 answer added Ulrich Neumann timeline score: 1
Nov 27, 2017 at 23:39 comment added user484 Previously: Fitting a two-dimensional Gaussian to a set of 2D pixels. That question is in 2D rather than 1D but has the same zero-offset problem.
Nov 27, 2017 at 22:40 answer added Daniele timeline score: 0
Nov 27, 2017 at 20:49 comment added JimB If you have trouble with convergence following @JoséAntonioDíazNavas good suggestion, then you should include starting values (different from the default values of 1 for all parameters). You might try {{a, 100}, {b, 2500}, {c, 2675}, {d, -400}}.
Nov 27, 2017 at 20:26 comment added user53761 @JoséAntonioDíazNavas thank you, I'll try this! Not sure how to attach a file...
Nov 27, 2017 at 20:22 comment added José Antonio Díaz Navas Try Normal@NonlinearModelFit[G', a+ b*Exp[ (x-c)^2/d], {a, b, c, d}, {x, y}]. You will get the fitting equation. A better reply could be provided if you give us access to the data G'
Nov 27, 2017 at 20:09 comment added Chris K @JimB Good point -- I should avoid weighing in on statistical questions, which are outside my area of expertise!
Nov 27, 2017 at 20:07 comment added JimB @ChrisK. I disagree. This is a regression problem where the fitted curve has a similar shape to a Gaussian probability function (a + b Exp[(x-c)^2/d]) rather than fitting a probability distribution from a random sample.
Nov 27, 2017 at 20:05 review First posts
Nov 27, 2017 at 20:10
Nov 27, 2017 at 20:01 comment added Chris K Check out FindDistributionParameters
Nov 27, 2017 at 20:00 history asked user53761 CC BY-SA 3.0