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Error fitting function to data using NIntegrate and ' (differentiation).

Christoph Bernkopf
Posted 6 years ago

I've been trying to fit a function to a data set. IT works with functions without integrals (like NIntegrate) and differentiations ( ' ). I'm trying to get it to work using both integrals and differentiations:

deltaData = {{1.00, 0}, {0.96, 0.3416}, {0.92, 0.4749}, {0.88, 0.5715}, {0.84, 0.648}, {0.80, 0.711}, {0.76, 0.764}, {0.72, 0.8089}, {0.66, 0.864}, {0.62, 0.8939}, {0.58, 0.919}, {0.54, 0.9399}, {0.50, 0.9569}, {0.46, 0.9704}, {0.42, 0.9809}, {0.38, 0.9885}, {0.34, 0.9938}, {0.30, 0.9971}, {0.26, 0.9989}, {0.22, 0.9997}, {0.16, 1}, {0.14, 1}};

deltaFit = 
  NonlinearModelFit[
   deltaData, (1 + a*T + c*T^2 + e*T^3 + g*T^4 + i*T^5 + k*T^6)/(1 + 
      b*T + d*T^2 + f*T^3 + h*T^4 + j*T^5 + l*T^6), {a, b, c, d, e, f,
     g, h, i, j, k, l}, T];
Plot[deltaFit[T], {T, 0, 1}]

data = {{0.203, 0.031}, {0.203, 0.030}, {0.246, 0.055}, {0.267, 0.072}, {0.300, 0.105}, {0.300, 0.105}, {0.330, 0.147}, {0.373, 0.214}, {0.397, 0.255}, {0.415, 0.293}, {0.445, 0.351}, {0.477, 0.430}, {0.493, 0.463}, {0.520, 0.538}, {0.541, 0.590}, {0.582, 0.717}, {0.589, 0.733}, {0.625, 0.847}, {0.645, 0.911}, {0.685, 1.029}, {0.688, 1.043}, {0.730, 1.181}, {0.751, 1.247}, {0.782, 1.338}, {0.793, 1.375}, {0.830, 1.472}, {0.856, 1.531}, {0.878, 1.585}};
AlphaBCS = 1.746;

delta[T_, p_, alpha_] := p*deltaFit[T] + (1 - p)*deltaFit[T]/alpha;
fx[T_, x_, p_, alpha_] := (Exp[AlphaBCS/T*(x^2 + delta[T, p, alpha]^2)^(1/2)] + 1)^(-1);
Ses[T_, p_, alpha_] := -6*AlphaBCS/Pi^2* NIntegrate[fx[T, x, p, alpha]*Log[fx[T, x, p, alpha]] + (1 - fx[T, x, p, alpha])*Log[1 - fx[T, x, p, alpha]], {x, 0,100}];
Ces[T_, p_, alpha_] := T*Ses'[T, p, alpha];

cesFit = NonlinearModelFit[data, Ces[T, p, alpha], {{p, 0.5}, {alpha, 1.764}}, T];

I don't know why I'm getting an Error as the code is working in the first fit I'm performing. This is the error message:

The function value ... a bunch of numbers ... Ses' ... more numbers ... is not a list of real numbers with dimensions {28} at {p,alpha} = {0.5`,1.764`}.

Does anybody know what I'm doing wrong? Is there anything I can replace the fit or integral with to make this work?
POSTED BY: Christoph Bernkopf
7 Replies
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The function Ses[T, p, alpha] has three arguments, it is necessary to define Ses'[T, p, alpha]. What is the argument for the differentiation?
POSTED BY: Alexander Trounev

Thank you for pointing that out! Here is my corrected code:

dSes[T_, p_, alpha_] = D[Ses[T, p, alpha], T];
Ces[T_, p_, alpha_] := T*dSes[T, p, alpha];

now the differentiation works. But when I run the fit, I get the following error: a long list of numbers, brackets, Compile'$... is not a list of real numbers with dimensions {28} at {p,alpha} = {0.5,1.764}.
POSTED BY: Christoph Bernkopf
Anonymous User
Anonymous User
Posted 6 years ago

did you try InterpolatingFunction[points]?

it seems to integrate differentiate your data. i can't say if it "can represent your curve" as needed though.
POSTED BY: Anonymous User

This is the best I could get using this model

deltaData = {{1.00, 0}, {0.96, 0.3416}, {0.92, 0.4749}, {0.88, 
    0.5715}, {0.84, 0.648}, {0.80, 0.711}, {0.76, 0.764}, {0.72, 
    0.8089}, {0.66, 0.864}, {0.62, 0.8939}, {0.58, 0.919}, {0.54, 
    0.9399}, {0.50, 0.9569}, {0.46, 0.9704}, {0.42, 0.9809}, {0.38, 
    0.9885}, {0.34, 0.9938}, {0.30, 0.9971}, {0.26, 0.9989}, {0.22, 
    0.9997}, {0.16, 1}, {0.14, 1}};

deltaFit = 
  NonlinearModelFit[
   deltaData, (1 + a*T + c*T^2 + e*T^3 + g*T^4 + i*T^5 + k*T^6)/(1 + 
      b*T + d*T^2 + f*T^3 + h*T^4 + j*T^5 + l*T^6), {a, b, c, d, e, f,
     g, h, i, j, k, l}, T];
Plot[deltaFit[T], {T, 0, 1}]

data = {{0.203, 0.031}, {0.203, 0.030}, {0.246, 0.055}, {0.267, 
    0.072}, {0.300, 0.105}, {0.300, 0.105}, {0.330, 0.147}, {0.373, 
    0.214}, {0.397, 0.255}, {0.415, 0.293}, {0.445, 0.351}, {0.477, 
    0.430}, {0.493, 0.463}, {0.520, 0.538}, {0.541, 0.590}, {0.582, 
    0.717}, {0.589, 0.733}, {0.625, 0.847}, {0.645, 0.911}, {0.685, 
    1.029}, {0.688, 1.043}, {0.730, 1.181}, {0.751, 1.247}, {0.782, 
    1.338}, {0.793, 1.375}, {0.830, 1.472}, {0.856, 1.531}, {0.878, 
    1.585}};
AlphaBCS = 1.746;

delta[T_, p_, alpha_] := p*deltaFit[T] + (1 - p)*deltaFit[T]/alpha;
fx[T_, x_, p_, 
   alpha_] := (Exp[AlphaBCS/T*(x^2 + delta[T, p, alpha]^2)^(1/2)] + 
     1)^(-1);
Ses[T_, p_, alpha_] := -6*AlphaBCS/Pi^2*
   NIntegrate[
    fx[T, x, p, alpha]*
      Log[fx[T, x, p, alpha]] + (1 - fx[T, x, p, alpha])*
      Log[1 - fx[T, x, p, alpha]], {x, 0, 10}, AccuracyGoal -> 5];
ces = D[fx[T, x, p, alpha]*
     Log[fx[T, x, p, alpha]] + (1 - fx[T, x, p, alpha])*
     Log[1 - fx[T, x, p, alpha]], T];
Ces[T0_, p0_, alpha0_] := -6*AlphaBCS/Pi^2*T0*
  NIntegrate[ces /. {T -> T0, p -> p0, alpha -> alpha0}, {x, 0, 10}, 
   AccuracyGoal -> 5]
ff = Interpolation[Table[{t, Ces[t, .5, 1.764]}, {t, 0.01, .95, .01}]];nlm = NonlinearModelFit[data, a*ff[T*b], {{a, 1}, {b, 1}}, T];
 Normal[nlm]
Show[ListPlot[data], Plot[nlm[T], {T, 0.015, .95}], Frame -> True]

fig1
POSTED BY: Alexander Trounev

Thank you for the model! I have one question:

When I use this model I get a fit for the parameters a & b. Are these p & alpha? I would like to be able to cross-validate the results with this model: (at the moment this is not the same plot as when I use your method + plot)

p = 0.787; (* a *)
alpha = 1.5092; (* b *)

delta[T_] := p*deltaFit[T] + (1 - p)*deltaFit[T]/alpha;
fx[T_, x_] := (Exp[AlphaBCS/T*(x^2 + delta[T]^2)^(1/2)] + 1)^(-1);
Ses[T_] := -6*AlphaBCS/Pi^2*
   NIntegrate[
    fx[T, x]*Log[fx[T, x]] + (1 - fx[T, x])*Log[1 - fx[T, x]], {x, 0, 
     100}];
Ces[T_] := T*Ses'[T];
Show[ListPlot[expHCap], Plot[Ces[T], {T, 0.01, 0.879}]]
Attachment

Attachments:
cross-validate-plot.png
POSTED BY: Christoph Bernkopf

For each pair of p-alpha, it is possible to compute a pair a-b, but this is not the same.
POSTED BY: Alexander Trounev

Okay, how can I fit for p-alpha? I need these parameters so I can put them back into my first function (delta). Is there a possibility to perform the NonlinearModelFit for p-alpha?
POSTED BY: Christoph Bernkopf
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