Fitting an exponential function: Setting options?

2021 2 15
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更新时间：2021 2 16
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Hello,
I would like to fit a custom specific equation and get an error message. My code is as follows:
T_K = [250.150000000000 253.150000000000 263.150000000000 273.150000000000 283.150000000000 293.150000000000 303.150000000000 313.150000000000 323.150000000000 333.150000000000 343.150000000000 353.150000000000 363.150000000000 373.150000000000 383.150000000000 393.150000000000 403.150000000000 413.150000000000 423.150000000000 433.150000000000 443.150000000000 453.150000000000 463.150000000000 473.150000000000 483.150000000000 493.150000000000 503.150000000000 513.150000000000 523.150000000000 533.150000000000 543.150000000000 553.150000000000 563.150000000000 573.150000000000 583.150000000000 593.150000000000 603.150000000000 613.150000000000]'
d = [985 983 976 969 962 955 948 940 933 926 919 912 904 897 890 882 875 867 859 852 844 836 828 820 812 803 795 786 777 768 759 750 740 730 720 710 699 688]'
f = fittype('A/(B * exp(1+ (1 - T_K/C)^D))',...
'dependent',{'d'},'independent',{'T_K'},...
'coefficients',{'A','B','C','D'});
fit2 = fit(T_K, d, f);
The error I get is in the fit function: [fitobj, goodness, output, convmsg] = iFit( xdatain, ydatain, fittypeobj, ...
My idea was that it could be a mathematical problem with the exponential function. After having read this I tried to set some options (lower bounds, where I was not sure about the order:
options = fitoptions('A/(B * exp(1+ (1 - T_K/C)^D))', 'Lower', [0 Inf Inf 1]);
), but it was without success. Does anybody has an idea about what is wring or the problem could be?

4 个评论
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David Goodmanson
David Goodmanson 2021-2-15
Hi Caroline, this comment is not on the code itself but just on the fact that A and B only come into the fit as the ratio A/B. That makes the problem underdefined. So you can, e.g. drop the B factor entirely and let A stand for the previous A/B.
Caroline Otto
Caroline Otto 2021-2-15
Yes, you are right. Unfortunatley, it does not change anything. With three coefficients I still have the same error message.
Walter Roberson
Walter Roberson 2021-2-15
You did not say what the error is ?
Caroline Otto
Caroline Otto 2021-2-15
Error using fit>iFit (line 348)
Complex value computed by model function, fitting cannot continue.
Try using or tightening upper and lower bounds on coefficients.
Error in fit (line 116)
[fitobj, goodness, output, convmsg] = iFit( xdatain, ydatain, fittypeobj, ...
Error in Umrechnung_Polynom (line 36)
fit2 = fit(T_K, d, f);

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Walter Roberson
Walter Roberson 2021-2-15
编辑：Walter Roberson 2021-2-15

0 个投票

options = fitoptions('A/(B * exp(1+ (1 - T_K/C)^D))', 'Lower', [0 Inf Inf 1]);
You set the lower bounds for B and C to infinity?? You divide by C so if C is infinite T_K/C would be sign(T_K)*0
If T_K/C > 1 then 1-T_K/C would be negative, and negative ^D would be complex if D is not an integer. So you need to ensure T_K/C <= 1 so T_K <= C and C should have a lower bound of max(T_K)

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Caroline Otto
Caroline Otto 2021-2-15
Ok, I just focussed on not dividing by zero. I think for D the lower bound better might be 0 than 1. I tried it now like this:
f = fittype('A/(B * exp(1+ (1 - T_K/C)^D))',...
'dependent',{'d'},'independent',{'T_K'},...
'coefficients',{'A', 'B', 'C','D'});
options = fitoptions('A/(B * exp(1+ (1 - T_K/C)^D))', 'Lower', [0 1 T_K 0]);
fit2 = fit(T_K, d, f, options);
ant the new error (I unfortunatley cannot handle) is the following one:
Error using horzcat
Dimensions of arrays being concatenated are not consistent.
Error in Umrechnung_Polynom (line 36)
options = fitoptions('A/(B * exp(1+ (1 - T_K/C)^D))', 'Lower', [0 1 T_K 0]);
Walter Roberson
Walter Roberson 2021-2-15
options = fitoptions('A/(B * exp(1+ (1 - T_K/C)^D))', 'Lower', [0 1 max(T_K) 0]);
Caroline Otto
Caroline Otto 2021-2-15
编辑：Caroline Otto 2021-2-15
Ok, but does it work when you execute the code? ...I get a new bunch of errors:
Error using fittype>iDeduceCoefficients (line 621)
The independent variable x does not appear in the equation expression.
Use x in the expression or indicate another variable as the independent variable.
Error in fittype>iCreateCustomFittype (line 477)
obj = iDeduceCoefficients(obj);
Error in fittype>iCreateFittype (line 353)
obj = iCreateCustomFittype( obj, varargin{:} );
Error in fittype (line 330)
obj = iCreateFittype( obj, varargin{:} );
Error in fitoptions (line 147)
obj = fittype(varargin{1});
Error in Umrechnung_Polynom (line 36)
options = fitoptions('A/(B * exp(1+ (1 - T_K/C)^D))', 'Lower', [0 1 max(T_K) 0]);
I tried to set T_K as the independant varibale as you can see in the code above. But are there syntax problems additionally?
Walter Roberson
Walter Roberson 2021-2-15
Ran in:
Ran in:
T_K = [250.150000000000 253.150000000000 263.150000000000 273.150000000000 283.150000000000 293.150000000000 303.150000000000 313.150000000000 323.150000000000 333.150000000000 343.150000000000 353.150000000000 363.150000000000 373.150000000000 383.150000000000 393.150000000000 403.150000000000 413.150000000000 423.150000000000 433.150000000000 443.150000000000 453.150000000000 463.150000000000 473.150000000000 483.150000000000 493.150000000000 503.150000000000 513.150000000000 523.150000000000 533.150000000000 543.150000000000 553.150000000000 563.150000000000 573.150000000000 583.150000000000 593.150000000000 603.150000000000 613.150000000000].';
d = [985 983 976 969 962 955 948 940 933 926 919 912 904 897 890 882 875 867 859 852 844 836 828 820 812 803 795 786 777 768 759 750 740 730 720 710 699 688].';
f = fittype('A/(B * exp(1+ (1 - T_K/C)^D))',...
'dependent',{'d'},'independent',{'T_K'},...
'coefficients',{'A', 'B', 'C','D'});
options = fitoptions(f);
options = fitoptions(options, 'Lower', [0 1 max(T_K) 0], 'startpoint', [1 1 1000 1]);
fit2 = fit(T_K, d, f, options);
fit2
fit2 =
General model: fit2(T_K) = A/(B * exp(1+ (1 - T_K/C)^D)) Coefficients (with 95% confidence bounds): A = 2310 B = 1 C = 952 D = 184
You will get notably different C and D results with different start points.
Alex Sha
Alex Sha 2021-2-16
编辑：Alex Sha 2021-2-16
As commented by David, if you don't drop the B or A, there will be multi-solutions, with the different combinations of A and B, but unique of C and D:
1:
Root of Mean Square Error (RMSE): 1.0064732367416
Sum of Squared Residual: 38.4935582985301
Correlation Coef. (R): 0.999934086786684
R-Square: 0.99986817791792
Parameter Best Estimate
---------- -------------
a -46890.8524606817
b -5.63376277792465
c 1128.65575737085
d -0.508557122720511
2:
Root of Mean Square Error (RMSE): 1.0064732367416
Sum of Squared Residual: 38.4935582985303
Correlation Coef. (R): 0.999934086786694
R-Square: 0.99986817791794
Parameter Best Estimate
---------- -------------
a 103527.516316089
b 12.4384487994102
c 1128.6557688586
d -0.508557130714601
3:
Root of Mean Square Error (RMSE): 1.0064732367416
Sum of Squared Residual: 38.4935582985303
Correlation Coef. (R): 0.999934086786667
R-Square: 0.999868177917885
Parameter Best Estimate
---------- -------------
a 17590.6485257921
b 2.11345146634875
c 1128.65575372825
d -0.508557120581391
...
If drob B, the fitting equation becomes: y = A/(exp(1+ (1 - x/C)^D));
There will be an unique result:
Root of Mean Square Error (RMSE): 1.00647323674159
Sum of Squared Residual: 38.4935582985297
Correlation Coef. (R): 0.999934086786666
R-Square: 0.999868177917884
Parameter Best Estimate
---------- -------------
a 8323.18547243371
c 1128.6557657083
d -0.508557129233705
While If drob A, the fitting equation becomes: y = 1/(B * exp(1+ (1 - x/C)^D));
There will also be an unique result:
Root of Mean Square Error (RMSE): 1.0064732367416
Sum of Squared Residual: 38.4935582985304
Correlation Coef. (R): 0.999934086786664
R-Square: 0.999868177917879
Parameter Best Estimate
---------- -------------
b 0.000120146307262581
c 1128.65577814606
d -0.508557138276278
Caroline Otto
Caroline Otto 2021-2-16
Yes, that's what I wanted to get. Thank you so much for this helpful answer. To be honest, I had to adapt the bound for D to get this solution, but now it works.

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