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 件のコメント

J. Alex Lee
J. Alex Lee 2021 年 8 月 12 日
you can only solve in a least squares sense because you have too many equations for the number of unknowns
Cris LaPierre
Cris LaPierre 2021 年 8 月 12 日
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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 採用された回答

Star Strider
Star Strider 2021 年 8 月 12 日

0 投票

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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