exp(-b/x) fit, inf problem when fitting

I am trying to fit some data with the model: exp(-b/x); When x goes to zero, y should go to zero as well since anything power negative infinity is zero. However Matlab sees the infinity and terminates everything. Here is my code:

vv=data(:,1);
ii=data(:,2);

g = fittype('exp(-b/x)');
f0 = fit(vv,ii,g);
xx = linspace(-1,1);
plot(vv,ii,'o',xx,f0(xx),'r-');
grid('on')

[Mathieu NOE]

hello

could you share the data as well ?

tx

[Basil Eldeeb]

it's unable to upload, the data is visibly linear, However my model is exponential:

0.000000 -4.333102E-5

0.050000 0.030123

0.100000 0.061024

0.150000 0.092435

0.200000 0.123947

0.250000 0.155877

0.300000 0.187916

0.350000 0.219968

0.400000 0.251753

0.450000 0.283991

0.500000 0.316017

0.550000 0.347719

0.600000 0.379196

0.650000 0.410459

0.700000 0.441492

0.750000 0.472394

0.800000 0.503174

0.850000 0.533827

0.900000 0.564569

0.950000 0.595374

1.000000 0.626224

[Mathieu NOE]

yes , your data show a very linear trend

how can you expect to fit an exponential model to these data ? it will never work

[Basil Eldeeb]

supposedly when the exponent is quite small it will behave linearly to a first order. The model is more complex, actually, I am just facing a problem with the exp(-b/x) term. I want matlab to evaluate exp(-Inf) without giving error. The other answer shows promise. However it gave an error, you can see it in my response

[Mathieu NOE]

not sure it's really a good model...

data = readmatrix('data.txt');
x  = data(:,1);
y = data(:,2);
% exponential fit method 
% model : y = exp(-b/x)

f = @(b,x) exp(b./x);
obj_fun = @(params) norm(f(params(1), x)-y);
sol = fminsearch(obj_fun, -0.1);
b_sol = sol(1)
y_fit = f(b_sol, x);

figure
plot(x,y,'r',x,y_fit,'-.k');
legend('data','exp fit');

Answer 1

[Matt J]

When x goes to zero, y should go to zero as well since anything power negative infinity is zero.

Only if b>=0.

g = fittype('exp(-b/x)',  'options', fitoptions('Lower',0)  );

[Basil Eldeeb]

Hello, Thanks for the response. I take it matlab applies limit evaluation on its own. However, the code is giving me this error:

Error using curvefit.basefitoptions/set

The name 'Lower' is not an accessible

property for an instance of class

'basefitoptions'.

now I have two questions:

Why am I receiving this error?

How do I specify which parameter is to be limited if I have more than one?

[Matt J]

I guess they wanted a different syntax...

data=[0.000000	-4.333102E-5
0.050000	0.030123
0.100000	0.061024
0.150000	0.092435
0.200000	0.123947
0.250000	0.155877
0.300000	0.187916
0.350000	0.219968
0.400000	0.251753
0.450000	0.283991
0.500000	0.316017
0.550000	0.347719
0.600000	0.379196
0.650000	0.410459
0.700000	0.441492
0.750000	0.472394
0.800000	0.503174
0.850000	0.533827
0.900000	0.564569
0.950000	0.595374
1.000000	0.626224];

vv=data(:,1);
ii=data(:,2);

g = fittype('exp(-b/x)');
options=fitoptions('exp(-b/x)','Lower',0);

f0 = fit(vv,ii,g,options);

Warning: Start point not provided, choosing random start point.

plot(f0,vv,ii)
grid('on')

[Basil Eldeeb]

Thank you for the response. I wanted to ask how to specify multiple options for multiple parameters, for example:

g = fittype('a*x^2*exp(-b/x)');

One more question; For my model the exp(-b/x) is only valid for positive x. So is there a way where I can apply two fittings simultaneously to the same data? i.e. one including exp(-b/x) as a multiple for +ve x and excludes it for -ve x. Thanks in advance!

[Matt J]

You should divide the data into two sets and fit each one separately, e.g.,

pos=vv>0; neg=~pos;

fpos = fit(vv(pos),ii(pos),gpos,options);
fneg = fit(vv(neg),ii(neg),gneg,options);

[Basil Eldeeb]

I appreciate it, thank you for the help!.