How to fit to an infinite series function?

Question

Qili Hu 2021-1-24

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/725382-how-to-fit-to-an-infinite-series-function

评论： Vladimir Sovkov 2024-3-19

采纳的回答： Vladimir Sovkov

在 MATLAB Online 中打开

qt is a dependent variable; t is an independent variable; qe B are undetermined parameters.

syms n t;
x=[0 5 10 15 20 30 45 60 75 90 105 120];
y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.53];
plot(x,y,'bo');
hold on
beta0=[39,0.002];
fun=@(beta,xdata) beta(1)*(1-6/(pi^2)*symsum((1/n^2)*exp(-beta(2)*(n^2)*t),n,1,inf))
betafit = nlinfit(x,y,fun,beta0);
plot(x,y,fun,beta0)

However, it does not work well. How to do it in MATLAB? Help me. Many thanks.

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Answer 1

Vladimir Sovkov 2021-1-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/725382-how-to-fit-to-an-infinite-series-function#answer_604897

在 MATLAB Online 中打开

An iterative solution instead of the symbolic one can be more productive this case, like this one

x=[0 5 10 15 20 30 45 60 75 90 105 120];
y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.53];
plot(x,y,'bo');
hold on
pause(0.1);
beta0=[39,0.002];
% syms n t
% fun=@(beta,t) beta(1)*(1-6/(pi^2)*symsum((1./n.^2).*exp(-beta(2)*(n.^2).*t),n,1,Inf));
% betafit = nlinfit(x,y,fun,beta0);
beta1=beta0;
delta = 1e-8; % desired objective accuracy
R0=Inf; % initial objective function
for K=1:10000
    fun=@(beta,t) beta(1)*(1-6/(pi^2)*sum((1./(1:K)'.^2).*exp(-beta(2)*((1:K)'.^2).*t),1));
    [betafit,R] = nlinfit(x,y,fun,beta1);
    R = sum(R.^2);
    if abs(R0-R)<delta
        break;
    end
    beta1=betafit;
    R0 = R;
end
plot(x,fun(betafit,x),'.-r');
xlabel('x');
ylabel('y');
legend('experiment','model');
title(strcat('\beta=[',num2str(betafit),'];----stopped at--','K=',num2str(K)));

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Qili Hu 2024-3-19

Dear Sovkov

Excuse me. Thank you for answering how to solve the an infinite series function for me earlier. I have another question about how to obtain the standard deviation of the fitting parameters. I haven't learned how to obtain it yet. I hope you can help me continue to answer it. Thank you very much!

These are the program codes you helped me answer earlier.

x=[0 5 10 15 20 30 45 60 75 90 105 120];

y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.53];

plot(x,y,'bo');

hold on

pause(0.1);

beta0=[39,0.002];

% syms n t

% fun=@(beta,t) beta(1)*(1-6/(pi^2)*symsum((1./n.^2).*exp(-beta(2)*(n.^2).*t),n,1,Inf));

% betafit = nlinfit(x,y,fun,beta0);

beta1=beta0;

delta = 1e-8; % desired objective accuracy

R0=Inf; % initial objective function

for K=1:10000

fun=@(beta,t) beta(1)*(1-6/(pi^2)*sum((1./(1:K)'.^2).*exp(-beta(2)*((1:K)'.^2).*t),1));

[betafit,R] = nlinfit(x,y,fun,beta1);

R = sum(R.^2);

if abs(R0-R)<delta

break;

end

beta1=betafit;

R0 = R;

end

plot(x,fun(betafit,x),'.-r');

xlabel('x');

ylabel('y');

legend('experiment','model');

title(strcat('\beta=[',num2str(betafit),'];----stopped at--','K=',num2str(K)));

Vladimir Sovkov 2024-3-19

Hi! The mathematics of how to estimate the standard deviations in LSF using the local linear approximation is described in many textbooks. Briefly, it requires estimating derivatives of your computed function over its parameters, construction from them the design matrix (Jacoby matrix), and applying some standard equations to find the covariance matrix of the parameters; its diagonal elements are the squares of the sought standard deviation. Programming of all this "from scratch" is somewhat complucated, and I do not have time to do it for you now. However, you can download my package Optimizer from https://sourceforge.net/projects/optimizer-sovkov/, adopt the interface of your program to its regulations, and this would privide you with those estimates automatuclly. If you believe that the linear appriximation is not enough in your case, the Monter-Carlo-type computation can be used, in which you generate pseudo-randomly the cloud in your parameter space and estimate its spread within the spread of the computed data around your experimental curve.

Vladimir Sovkov 2024-3-19