I want to solve the below 3 simultaneous exponential equation
3.43 X^y =1
4.6 X^y =2
5.86 X^y =3
I need your help to tabulate my lab data...ASAP

2 commentaires

J. Alex Lee
J. Alex Lee le 12 Août 2021
you can only solve in a least squares sense because you have too many equations for the number of unknowns
Cris LaPierre
Cris LaPierre le 12 Août 2021
We are happy to help answer your MATLAB questions, but generally not willing to do your homework for you. Share what you have tried and where you are stuck.

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 Réponse acceptée

The ‘X^y’ term is essentially just a slope in a linear relationship, and any values of ‘X’ and ‘y’ that equate to it will work. So there is no unique solution.
To illustrate —
% 3.43 * X^y = 1
% 4.6 * X^y = 2
% 5.86 * X^y = 3
L = [3.43; 4.6; 5.86];
R = [1; 2; 3];
Slope1 = L \ R
Slope1 = 0.4491
fcn = @(b,v) v.*b(1).^b(2);
[B,resnrm] = fminsearch(@(b) norm(R - fcn(b,L)), rand(2,1)*10)
B = 2×1
24.8566 -0.2491
resnrm = 0.6573
Slope2 = B(1).^B(2)
Slope2 = 0.4491
% figure
% plot(L, R, 'pb')
% hold on
% plot(L, fcn(B,L), '-r')
% hold off
% grid
% axis([3 6 0 4])
% text(3.75,3.5, sprintf('$L \\times %.3f^{\\ %.3f} = R$',B), 'Interpreter','latex')
It is possible to run the nonlinear approach an infinity of times and ‘Slope1’ will always equate to ‘Slope2’ (within the bounds of floating-point approximation error).
A new model is necessary if there is a relationship that defines these data that needs to have parameters estimated for it.
.

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