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Results tagged with Search options answers only user 2079
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Questions on the use of Mathematica to construct models for approximating empirical data.

If I understand the question, I would think you want to minimize the error: h[t_] := Exp[-t] f[t_] := Exp[-t/2] g[t_] := Exp[-3 t] Last@Minimize[Integrate[(h[t] - ((1 - x) f[t] + x g[t]))^2, {t, 0 …
answered Jan 4 '16 by george2079
If you fit the real part of the data directly: FindFit[ Re[data] , Re[b + a/((-0.977727 + 0.0601085 I) + s)] , {{a, -.1}, {b, -.1}}, s] {a -> -0.0791299, b -> -0.121803} Gives a prett …
answered Jul 14 '14 by george2079
It helps to look at the data..your points fall very nearly on a plane, which can be found readily: plane = Last@Minimize[Total[(a X + b Y - Z + c)^2], {a, b, c}] {a -> -0.00357726, b -> 0.00799 …
answered Dec 30 '15 by george2079
Here is a solution -- I added a constraint and manually found some good initial values. g[z_] = Simplify[ 5 Log[10,(1 + z ) Integrate[1/(a + b x + c x^2), {x, 0, z}] ] + 25 , …
answered Mar 27 '14 by george2079
Here is a stab at the discrete Zipf problem: v = {3617764, 1044065, 633141, 526307, 90386, 6060, 3038, 3044, 1533, 904, 1934, 975, 981}; bins = Partition[ {1, 5, 10, 20, 100, 500, 750, 100 …
answered Jan 2 '14 by george2079
Took another look at this. I don't really have in answer but this may help if anyone want to play with it: data = #[[{2, 3}]] & /@ Import["Cat3_0mod.txt", "Table"] Break the integral into parts …
answered Apr 2 '14 by george2079
you can remove the large value like this: ListPlot[ Select[ data , Abs[#[]]<10^4& ] , Joined->True]
answered Mar 9 '15 by george2079
Here it is with the constraint. I changed the symbols to a,x just for readability (There was nothing wrong with the y,F except that single Caps are good to avoid ) nlm = NonlinearModelFit[ test, …
answered Nov 18 '15 by george2079
If I understand correctly you have basically independent data per channel but the fit is "coupled" by having the same parameter b. Here is one approach. sol = Last@ NMinimize[ Total[(a[#[]] E …
answered Jun 28 '16 by george2079
something like this? truth[x_] = 2 x + 1; cluster1 = {#, truth[#] + RandomVariate[NormalDistribution[0, 1]]} & /@ RandomReal[{0, 5}, 20]; cluster2 = {#, truth[#] + RandomVariate[NormalDistributio …
answered Oct 28 '16 by george2079
Reduced major axis regression: Example data: SeedRandom data = Table[ {# + RandomVariate[NormalDistribution[0, .1]], 1 + # + RandomVariate[NormalDistribution[0, .1]]} &@ RandomReal[{ …
answered Apr 17 '18 by george2079
The approach mentioned in my comment using a polar representation: r[a_, b_, theta_, t_] := a Sqrt/Sqrt[ (1 + (a/b)^2) + (1-(a/b)^2) Cos[2 (t - theta)] ]; some random sample data: data …
answered Jun 25 '14 by george2079
You can pass your data to the NormFunction: myNorm[residuals_, data_] := Total[(Log[data[[All, 2]] + 1] - Log[data[[All, 2]] + residuals + 1])^2]; data = {#, 2 # + 10 + # RandomReal[{0 …
answered Dec 21 '15 by george2079
This may be useful.. sampledata = {#, Piecewise[ {{ 1 + # , # < 1} , { 6 (# - 1) + 2 , # > 1} }] (1 + RandomVariate[NormalDistribution[0, .2]])} & /@ RandomReal[{0, 2}, …
answered May 8 '15 by george2079
, but there really is a bit of an art to data fitting that goes beyond what mathematica can do in a fully automatic way. …
answered Sep 11 '13 by george2079

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