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 Original Message (ID '70032') By Bill Simpson: In Response To 'Re: Re: Plotting non-linear values and solving' --------- I originally looked at Fit and FindFit and couldn't see a way to apply these. I go to http://reference.wolfram.com/mathematica/ref/FindFit.html and click on Scope and then Multivariate data. That example certainly isn't the simplicity I would want when I'm confused and trying to see a brilliantly clear example, but I type it all back in and look at each of the intermediate values and try to reverse guess how to apply that. I think for your problem I want to do this In[1]:= data = {{0.0000079, 0.244},{0.0000119, 0.295}, {0.0000158, 0.428},{0.0000198,0.488}, {0.0000237, 0.703}, {0.0000277,0.711}, {0.0000316, 0.859}, {0.0000356,0.878}, {0.0000395, 0.926}, {0.0000435,0.915}, {0.0000474, 0.991}, {0.0000593, 1.04}, {0.0000632, 1.02}}; y = ((1 + (K(0.000011501)) + (K(x)/(2s))) - ((1 + (K(0.000011501)) + (K(x)/(2s)))^2 - (2K^2(0.000011501(x)))/s))^.5/((2K(0.000011501))); FindFit[data, y, {{K, 2*10^6}, {s, 1}}, x] and that gives me this FindFit::nrlnum: The function value {-0.244+0.350I, <<9>>, <<3>>} is not a list of real numbers with dimensions {13} {K, s} = {2*^6, 1.} which again is showing that for some, maybe lots, of values of K your square roots are resulting in complex values and most of the functions looking for real results give up the moment they see a complex value. All this is part of the reason I fell back on using Norm and minimize to look for the best approximation I could find. Is any of this getting you closer to your solution? Is there some way to find a domain for K and s that give real results? Perhaps that would help.