# curve fitting a custom equation

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a*exp(-x/T) +c*(T*(exp(-x/T)-1)+x)

This is the equation i'm trying to fit using the custom fit tool. Since the variable x is both in exponential and linear form, the data isn't really fitting, it's just a straight line. Is there way to fit the data to this equation?

As a side note: how do I view the numerical values instead of scientific ones in curve fitting tool?

Thank you

##### 1 Comment

Matt J
on 16 Jul 2021

Edited: Matt J
on 16 Jul 2021

In the first line of your post, you show the model equation as,

a*exp(-x/T) +c*(T*(exp(-x/T)-1)+x)

However, in your screenshot of the cftool app, the model equation appears instead as,

a*exp(-x*T) +c*(T*(exp(-x/T)-1)+x)

You should edit your post so that we know which version of the equation is the intended one.

### Accepted Answer

Alex Sha
on 17 Jul 2021

Edited: Alex Sha
on 18 Jul 2021

Hi, all, as mentioned by Walter Roberson, the resule I provided above is obtained by using 1stOpt, the biggest advantage, for 1stOpt, is that the guessing of initial-start values are no longer needed for curve-fitting, equation-solving or any other optimization problem.

if the model function is: y=a*exp(-x*T) +c*(T*(exp(-x/T)-1)+x), the result is:

Root of Mean Square Error (RMSE): 15155741.0636478

Sum of Squared Residual: 4.59392974376683E15

Correlation Coef. (R): 0.996754359090101

R-Square: 0.993519252365118

Parameter Best Estimate

---------- -------------

a -33754000.9741343

t -1.00000000434611

c -33754006.8543357

Note the value of parameter t: t=-1.00000000434611, if taking t as -1.00, the chart will become:

The Sensitivity analysis of each parameter is as below:

while, if the model function is: y=a*exp(-x/T) +c*(T*(exp(-x/T)-1)+x), the result is:

Root of Mean Square Error (RMSE): 15137803.4043043

Sum of Squared Residual: 4.58306183814736E15

Correlation Coef. (R): 0.996383637725283

R-Square: 0.992780353526669

Parameter Best Estimate

---------- -------------

a -54592643.2783804

t -1.72718663083536

c -31608221.5261319

##### 2 Comments

Alex Sha
on 17 Jul 2021

The result got by Matt J (as show below) is actually a local solution, not global one.

a =

-6.2820e+07

c =

-2.9980e+07

T =

1.0481e-14

Matt J
on 17 Jul 2021

Edited: Matt J
on 17 Jul 2021

The solution I presented below constrained the search to T>=0, so it might be global subject to that constraint.

It's important to notice as well that whether T=-1 or T=0 is taken as the solution, it results in a cancelation of the exponential terms, leaving an essentially linear solution. It makes it seem like either the model was over-parametrized to begin with, or there isn't enough (x,y) data to accurately estimate the exponential terms.

### More Answers (1)

Matt J
on 15 Jul 2021

You don't need a custom type. Your model is just exp2 but with modified data (x,y-x)

##### 11 Comments

Walter Roberson
on 16 Jul 2021

Matt J
on 16 Jul 2021

Edited: Matt J
on 16 Jul 2021

Thank you so much for your input. I have opted for this method.

@Ayushi Sharma You're welcome, but please Accept-click the answer if you consider the question addressed.

Alex uses a commercial program named 1stOpt from 7d-soft. ... I would say that if you do curve fitting, that it does a very nice job.

However, the fit Alex has shown us is based on an incorrect model equation, because @Ayushi Sharma had a typo in the first line of his post. If we substitute Alex's parameter estimates into the true model, it gives strong disagreement with the data:

load doubt

a=-54592639.7090172;

c=-31608221.6797308;

T=-1.72718650949557;

fun=@(x) a*exp(-x*T) +c*(T*(exp(-x/T)-1)+x);

plot(x,y,'o'); ax=axis;

hold on

plot( x,fun(x));

hold off

axis(ax)

legend('Data','Fit')

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