exponential fitting to limited data

Question

Mos_bad 2020-10-8

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https://ww2.mathworks.cn/matlabcentral/answers/608759-exponential-fitting-to-limited-data

评论： Star Strider 2020-10-9

expFit.jpg

Here is my data :

y=[2.5648    1.2495    1.1663    1.0652    2.1537    1.3080    1.2624    1.2949]

I want to fit an exponential function to them. attached is the figure that I came up and doesn't make any sense to me.

f = @(b,x) b(1).*exp(b(2).*x)+b(3);                                    
B = fminsearch(@(b) norm(y - f(b,x)), [-2; -1; 1])  ;                
figure
plot(x, y, 'ro','MarkerFaceColor','r')
hold on
plot(x, f(B,x), '-b','linewidth', 3)
hold off

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Answer 1

Star Strider 2020-10-8

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https://ww2.mathworks.cn/matlabcentral/answers/608759-exponential-fitting-to-limited-data#answer_509057

在 MATLAB Online 中打开

It needs to be changed slightly, adding an additional parameter:

x = 35:5:70;
y = [2.5648    1.2495    1.1663    1.0652    2.1537    1.3080    1.2624    1.2949];
f = @(b,x) b(1).*exp(b(2).*(x-b(4)))+b(3);                                    
B = fminsearch(@(b) norm(y - f(b,x)), [1; -1; 1; 35])  ;                
figure
plot(x, y, 'ro','MarkerFaceColor','r')
hold on
plot(x, f(B,x), '-b','linewidth', 3)
hold off

Then, it is a bit more understandable, althoug still not an excellent fit.

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Mos_bad 2020-10-9

Pol2.jpg

Long story short, each data point comes from 1000 Monte carlo simulations and based on the logic behind the concept, it should look like as attached. I guesses exponential may be the best function describing the data decaying over time. BTW, thanks for asking fundamental questions, it had me rethought about it.

Star Strider 2020-10-9

That is about the best that function can do with those data.

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Answer 2

Walter Roberson 2020-10-9

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https://ww2.mathworks.cn/matlabcentral/answers/608759-exponential-fitting-to-limited-data#answer_509089

Your proposed function is a quite bad fit for your data.

With your function, the optimal fit drives b(2) to roughly -4 so you are dealing with an exp(-4*(35:5:70)) which gives you values that are non-zero for the first result and effectively zero by comparison for the others results. Then b(1) is driven to about 1e100 to balance out the exp(-4*35) to give a notable value for the first entry and effective zeros for the rest. Your fitting then becomes "non-trivial value for x = 35, zeros for the rest" plus b(3) -- so all of the fitting is going into matching the first datapoint, and the exp() makes the rest of them into noise.

I said above roughly -4 and about 1e100 but you can get slightly better fits by driving b(1) to 1e200 and b(2) about -6.4 which has the effect of lowering the contributions of the other entries even further towards 0.

So.. you get a pulse and a straight line.

Star Strider's idea of subtracting something from x is a decent one.