curve fitting exponential function with two terms

Question

Briana Canet 2023-11-11

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046025-curve-fitting-exponential-function-with-two-terms

评论： Matt J 2023-11-13

Update: I need help curve fitting this set of points with an exponential function with two terms nicely.

% Curve Fit
x = [6500 6350 6000 5400 4500];
y = [0 0.25 0.5 0.75 1.0];
theFit=fit(x' , y', 'exp2')
theFit = 
     General model Exp2:
     theFit(x) = a*exp(b*x) + c*exp(d*x)
     Coefficients (with 95% confidence bounds):
       a =         0.5
       b =           0
       c =           0
       d =           0

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Briana Canet 2023-11-11

在 MATLAB Online 中打开

It's not plotting in the following code, so I assumed it's not appropriate. See below.

clear;

clc;

close all;

% Curve Fit

x = [6500 6350 5800 4900 4500];

y = [0 0.25 0.5 0.75 1.0];

theFit=fit(x' , y', 'exp2')

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) Coefficients (with 95% confidence bounds): a = -2.905e-15 (-8.856e-13, 8.798e-13) b = 0.004984 (-0.04157, 0.05153) c = 9.409 (-64.2, 83.02) d = -0.0005043 (-0.002143, 0.001135)

plot(theFit , x , y)

% Monthly Expense Cost

exp = [6500 6350 5800 4900 4500];

expUtility = [0 0.25 0.5 0.75 1.0];

% Utility Points

figure;

h = plot(exp,expUtility,'o');

set(h(1),'MarkerFaceColor','b')

xlim([4500 6500]);ylim([0 1.25]);

yticks([expUtility 1.25]);

grid on;

xlabel('Monthly Expense Cost ($)');

ylabel('Utility');

legend('Utility Points');

% Utility Curve Fitting

a = -2.905e-15;

b = 0.004984;

c = 9.409;

d = -0.0005043;

curveX = linspace(4500,6500);

curveY = a*exp(b*curveX) + c*exp(d*curveX);

Array indices must be positive integers or logical values.

hold on;

plot(curveX,curveY,'Color','b');

legend('Utility Points','Utility Curve Fit');

% Consistency Check

% New Lottery

costCheck = 4650;

costUtilityCheck = 0.875;

plot(costCheck,costUtilityCheck,'*','Color','r');

legend('Utility Lottery Points','Utility Curve Fit','Consistency Check Point');

% Local Risk Aversion

x_q = 4500:1:6500;

q = (a*b^2*exp(b.*x_q) + c*d^2*exp(d.*x_q)) ./ (a*b*exp(b.*x_q) + c*d*exp(d.*x_q));

Briana Canet 2023-11-11

移动：Matt J 2023-11-11

x and y are as shown, but the plot just comes out like a line.

Image Analyst 2023-11-12

在 MATLAB Online 中打开

Sorry, but I don't believe you. When I swap x and y, the fit looks great.

% Curve Fit

y = [6500 6350 6000 5400 4500];

x = [0 0.25 0.5 0.75 1.0];

theFit=fit(x' , y', 'exp2')

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) Coefficients (with 95% confidence bounds): a = -5.893e+07 (-5.463e+16, 5.463e+16) b = 0.5577 (-3.089e+04, 3.089e+04) c = 5.893e+07 (-5.463e+16, 5.463e+16) d = 0.5576 (-3.089e+04, 3.089e+04)

plot(theFit , x , y)

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Answer 1

Matt J 2023-11-11

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046025-curve-fitting-exponential-function-with-two-terms#answer_1351145

编辑：Matt J 2023-11-11

在 MATLAB Online 中打开

You should normalize your x data

% Curve Fit
x = [6500 6350 6000 5400 4500]; 
      x=(x-mean(x))/std(x);
      
y = [0 0.25 0.5 0.75 1.0];

Also, I would recommend downloading fminspleas from the File Exchange

https://www.mathworks.com/matlabcentral/fileexchange/10093-fminspleas

and using it to generate an initial guess for fit():

e=@(a,xd)exp(a*xd);

flist={@(p,xd) e(p(1),xd) , @(p,xd) e(p(2),xd)};

[bd,ac]=fminspleas(flist,[-1,1],x, y);

theFit=fit(x',y','exp2','StartPoint',[ac(1),bd(1),ac(2), bd(2) ])

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) Coefficients (with 95% confidence bounds): a = 0.6747 (0.0581, 1.291) b = -0.2594 (-0.9322, 0.4135) c = -0.04679 (-0.4881, 0.3946) d = 2.634 (-6.423, 11.69)

plot(theFit,x,y)

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Torsten 2023-11-12

编辑：Torsten 2023-11-12

在 MATLAB Online 中打开

x = [6500 6350 6000 5400 4500];

xt = (x-mean(x))/std(x);

y = [0 0.25 0.5 0.75 1.0];

theFit=fit(xt',y','exp2')

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) Coefficients (with 95% confidence bounds): a = -0.04696 (-0.4897, 0.3958) b = 2.631 (-6.421, 11.68) c = 0.6749 (0.05714, 1.293) d = -0.2592 (-0.9328, 0.4145)

theFit.b = theFit.b/std(x);

Warning: Setting coefficient values clears confidence bounds information.

theFit.a = theFit.a*exp(-theFit.b*mean(x));

theFit.d = theFit.d/std(x);

theFit.c = theFit.c*exp(-theFit.d*mean(x));

theFit

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) Coefficients: a = -4.278e-10 b = 0.00322 c = 4.181 d = -0.0003172

plot(theFit,x,y)

Matt J 2023-11-12

I mentioned in the comments that I needed to change my points (noticed an error in my work).

All answers in this thread have been demonstrated using your new points.

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Answer 2

Matt J 2023-11-12

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046025-curve-fitting-exponential-function-with-two-terms#answer_1351155

编辑：Matt J 2023-11-12

在 MATLAB Online 中打开

You can also use fit()'s normalizer,

x = [6500 6350 6000 5400 4500];

y = [0 0.25 0.5 0.75 1.0];

theFit=fit(x',y','exp2','Normalize','on')

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) where x is normalized by mean 5750 and std 817 Coefficients (with 95% confidence bounds): a = -0.04696 (-0.4897, 0.3958) b = 2.631 (-6.421, 11.68) c = 0.6749 (0.05714, 1.293) d = -0.2592 (-0.9328, 0.4145)

plot(theFit,x,y)

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Briana Canet 2023-11-12

编辑：Briana Canet 2023-11-12

在 MATLAB Online 中打开

Thanks. Also I know, I can plot the 'theFit' directly. But, why isn't line 32 plotting CurveY (see last plot)? I need those a, b, c, and d values/the equation to work for future steps in the code I am developing.

clear;

clc;

close all;

% Curve Fit

x = [6500 6350 6000 5400 4500];

y = [0 0.25 0.5 0.75 1.0];

theFit=fit(x',y','exp2','Normalize','on')

theFit =

General model Exp2: theFit(x) = a*exp(b*x) + c*exp(d*x) where x is normalized by mean 5750 and std 817 Coefficients (with 95% confidence bounds): a = -0.04696 (-0.4897, 0.3958) b = 2.631 (-6.421, 11.68) c = 0.6749 (0.05714, 1.293) d = -0.2592 (-0.9328, 0.4145)

plot(theFit,x,y)

% Monthly Cost

cost = [6500 6350 6000 5400 4500];

costUtility = [0 0.25 0.5 0.75 1.0];

% Plot Utility Points

figure;

plot(cost,costUtility,'*');

xlim([4500 6500]);ylim([0 1.25]);

yticks([costUtility 1.25]);

grid on;

xlabel('Monthly Cost ($)');

ylabel('Utility');

legend('Utility Points');

% Add utility curve fit

a = -0.04696;

b = 2.631;

c = 0.6749;

d = -0.2592;

curveX = linspace(4500,6500);

curveY = a*exp(b*curveX) + c*exp(d*curveX);

hold on;

plot(curveX,curveY,'Color','b');

legend('Utility Points','Utility Curve Fit');

Matt J 2023-11-12

编辑：Matt J 2023-11-12

在 MATLAB Online 中打开

If you need to explicitly manipulate the coefficients and fit function, you'll have to do the normalization manually:

% Curve Fit

x = [6500 6350 6000 5400 4500];

xmu=mean(x);

xstd=std(x);

y = [0 0.25 0.5 0.75 1.0];

theFit=fit((x-xmu)'/xstd,y','exp2');

% Monthly Cost

cost = x;

costUtility = y;

% Plot Utility Points

figure;

plot(cost,costUtility,'*');

xlim([4500 6500]);ylim([0 1.25]);

yticks([costUtility 1.25]);

grid on;

xlabel('Monthly Cost ($)');

ylabel('Utility');

legend('Utility Points');

% Add utility curve fit

coeff=num2cell(coeffvalues(theFit));

[a,b,c,d]=deal(coeff{:});

curveX = linspace(4500,6500);

X=(curveX-xmu)/xstd;

curveY = a*exp(b*X) + c*exp(d*X);

hold on;

plot(curveX,curveY,'Color','b');

legend('Utility Points','Utility Curve Fit');

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Answer 3

Alex Sha 2023-11-12

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046025-curve-fitting-exponential-function-with-two-terms#answer_1351160

编辑：Alex Sha 2023-11-12

在 MATLAB Online 中打开

If taking the fitting function as: y=a*exp(b*x) + c*exp(d*x);

and also taking the data like below directly;

x = [6500 6350 5800 4900 4500];

y = [0 0.25 0.5 0.75 1.0];

The unique stable result should be:

Sum Squared Error (SSE): 0.00221359211819696
Root of Mean Square Error (RMSE): 0.0210408750682901
Correlation Coef. (R): 0.998229714370904
R-Square: 0.996462562653016
Parameter	Best Estimate        
---------	-------------        
a        	11.4185972844776     
b        	-0.000545792212445247
c        	-1.22298024843855E-22
d        	0.00759142468815435 

If add one more parameter, that is the fitting function become: y=a*exp(b*x)+c*exp(d*x)+e; the result will be perfect:

Sum Squared Error (SSE): 4.41584921368883E-29
Root of Mean Square Error (RMSE): 2.97181736103982E-15
Correlation Coef. (R): 1
R-Square: 1
Parameter	Best Estimate        
---------	-------------        
a        	-2.47788945923639E-15
b        	0.00505297753885221  
c        	332.002937639918     
d        	-0.00141137023194644 
e        	0.420769231934917  

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Alex Sha 2023-11-12

在 MATLAB Online 中打开

For this problem, the objective value of the least squares fit (SSE) is 0.00221359211819697, corresponded parameters are:

1:

Sum Squared Error (SSE): 0.00221359211819697
Root of Mean Square Error (RMSE): 0.0210408750682901
Correlation Coef. (R): 0.998229714367005
R-Square: 0.996462562645233
Parameter	Best Estimate        
---------	-------------        
a        	11.4185970800621     
b        	-0.000545792208439445
c        	-1.22298133472063E-22
d        	0.00759142455222768

or 2: (actually same as 1, only the order of parameter is different)

Sum Squared Error (SSE): 0.00221359211819697
Root of Mean Square Error (RMSE): 0.0210408750682901
Correlation Coef. (R): 0.9982297143666
R-Square: 0.996462562644425
Parameter	Best Estimate        
---------	-------------        
a        	-1.22297963729857E-22
b        	0.007591424764864    
c        	11.4185972678769     
d        	-0.000545792211987959

There are maybe other solutions, but should be local solutions, not global ones, or in other word, If there are any other solutions with the SSE values less than 0.00221359211819697, give it out please!

Alex Sha 2023-11-13

在 MATLAB Online 中打开

@Matt J If you have tried out the 1stOpt software, maybe you'll change your views and opinions!

The code looks like below, very simple, end-users do not need to do anything else, such as data-prenormalization you said

Function y=a*exp(b*x) + c*exp(d*x);
Data;
x = [6500 6350 5800 4900 4500];
y = [0 0.25 0.5 0.75 1.0];

Less than half of second, the output will be:

Sum Squared Error (SSE): 0.00221359211819697
Root of Mean Square Error (RMSE): 0.0210408750682901
Correlation Coef. (R): 0.998229714368578
R-Square: 0.996462562648373
Parameter	Best Estimate        
---------	-------------        
a        	-1.22298041267836E-22
b        	0.00759142466699956  
c        	11.4185972934291     
d        	-0.000545792212517027

No need for users to care about anything else, such as the initial start-values of parameters, every run will output same results.

Taking one more example, if the fitting function become: y=a*exp(b*x) + c*exp(d*x) + e/x;

Function y=a*exp(b*x) + c*exp(d*x) + e/x;
Data;
x = [6500 6350 5800 4900 4500];
y = [0 0.25 0.5 0.75 1.0];

Multi-run will produce two solutions:

1:

Sum Squared Error (SSE): 2.37582717939609E-30
Root of Mean Square Error (RMSE): 6.89322446957314E-16
Correlation Coef. (R): 1
R-Square: 1
Parameter	Best Estimate        
---------	-------------        
a        	-5.46715403508488E-16
b        	0.00527771888944181  
c        	1272.58844177386     
d        	-0.00179769933992128 
e        	2743.69929384194 

2:

Sum Squared Error (SSE): 0
Root of Mean Square Error (RMSE): 0
Correlation Coef. (R): 1
R-Square: 1
Parameter	Best Estimate        
---------	-------------        
a        	-25.5545324787999    
b        	-0.000174029056569443
c        	-1.10229181033535E-13
d        	0.00449303003927482  
e        	57050.1019982362  

What it means is that the start-values of each run are not fixed, but highly likely to be random values.

Alex Sha 2023-11-13

It would be a good suggestion for Mathwork, although not claer how 1stOpt process such problem internally.

Matt J 2023-11-13

MathWorks' fit() routine does have an internal normalization step which can be enabled,

https://www.mathworks.com/matlabcentral/answers/2046025-curve-fitting-exponential-function-with-two-terms#answer_1351155

However, if 1stOpt does something similar, it appears to be smart enough to post-transform the parameters and undo the effect of the data normalization. fit() does not do that.

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curve fitting exponential function with two terms

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curve fitting exponential function with two terms

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