Curve fitting with integral involved

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lvenG
lvenG 2021-9-19
编辑： lvenG 2021-9-20
Hello. I am having trouble generating an equation to estimate parameters given a set of data for B and T. I am about to use lsqcurve fitting but for some reason i could not make the code run.
The equation should be in the form of :
See below:
T = [120,210,400,540,650];
B = [-344,-157,-17.5,-0.85,10.5];
fun = @(constant,T)(int(((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./T)*((1./y.^4)-(1./y.^2)))),y,0,10));
T0 = [110,200]; %I just entered a random number, but don't know what this is for
T = lsqcurvefit(fun,T0,T,B);
The error I am getting is:
Unrecognized function or variable 'y'.
Error in
samplesept19>@(constant,T)(int(((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./T)*((1./y.^4)-(1./y.^2)))),y,0,10))
(line 9)
fun =
@(constant,T)(int(((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./T)*((1./y.^4)-(1./y.^2)))),y,0,10));
Error in lsqcurvefit (line 225)
initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});
Error in samplesept19 (line 12)
T = lsqcurvefit(fun,T0,T,B);
Caused by:
Failure in initial objective function evaluation. LSQCURVEFIT cannot continue.
Hoping for answers and help. :) Thank you so much!

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采纳的回答

Walter Roberson
Walter Roberson 2021-9-19
Ran in:
format long g
T = [120,210,400,540,650];
B = [-344,-157,-17.5,-0.85,10.5];
syms t y
syms constant [1 2]
expr = ((2.*pi.*(constant(1)).^3)./3).*(1-exp((-4.*constant(2)./t)*((1./y.^4)-(1./y.^2))))
expr = 
intexpr = int(expr, y, 0, 10)
intexpr = 
RHS = 2/3.*pi.*(constant(1)).^3 .* intexpr;
fun = matlabFunction(RHS, 'vars', {constant, t})
fun = function_handle with value:
@(in1,t)in1(:,1).^3.*pi.*integral(@(y)in1(:,1).^3.*pi.*(exp((in1(:,2).*(1.0./y.^2-1.0./y.^4).*4.0)./t)-1.0).*(-2.0./3.0),0.0,1.0e+1).*(2.0./3.0)
obj = @(P,T) arrayfun(@(t) fun(P,t), T)
obj = function_handle with value:
@(P,T)arrayfun(@(t)fun(P,t),T)
P0 = [110,200]; %I just entered a random number, but don't know what this is for
P = lsqcurvefit(obj, P0, T(:), B(:))
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
P = 1×2
1.6038792022093 184.707909161448
B2_predicted = obj(P,T(:));
plot(T, B, 'k*', T, B2_predicted, 'b-+')
legend({'original B', 'predicted B'}, 'location', 'best')
  3 个评论
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Walter Roberson
Walter Roberson 2021-9-20
P(1) is constant(1) and P(2) is constant(2)
lvenG
lvenG 2021-9-20
@Walter Roberson, thank you so much Sir. This worked.... Really, thanks for the help and enlightenment... Hope you are having a great day! Stay safe.

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