How can I obtain two unknown with one equation through optimization?

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Parham Babakhani Dehkordi
Parham Babakhani Dehkordi 2016 年 7 月 7 日
編集済み: John D'Errico 2016 年 7 月 7 日
Hi, I have one equation with two unknowns(A and y) as follows;
ls/D=A [Jg+(y^2/Jg)] input parameters are: D=0.04; Jg and ls are vectors as;
Jg=[0.574 0.765 0.957 1.148 1.340 1.531 1.722 1.914 2.10]; ls=[0.298 0.304 0.341 0.434 0.473 0.504 0.542 0.559 0.594];
any help would be appreciated.

回答 (1 件)

John D'Errico
John D'Errico 2016 年 7 月 7 日
編集済み: John D'Errico 2016 年 7 月 7 日
You do not have ONE equation. You have a set of points, and an equation that relates them, for some unknown set of parameters, A and y. So you can think of this as many equations, with two unknowns.
The technique that is used to solve for those unknown parameters is called curve fitting. In fact, you can (logically) use the curve fitting toolbox. Or you can use tools like lsqnonlin or lsqcurvefit, from the optimization toolbox. Or nlinfit from the stats toolbox.
mdl = fittype('A*(Jg+(Y^2/Jg))','coeff',{'A','Y'},'indep','Jg');
>> res = fit(Jg',ls'/D,mdl)
Warning: Start point not provided, choosing random start point.
> In curvefit.attention.Warning/throw (line 30)
In fit>iFit (line 299)
In fit (line 108)
res =
General model:
res(Jg) = A*(Jg+(Y^2/Jg))
Coefficients (with 95% confidence bounds):
A = 7.003 (6.436, 7.571)
Y = 0.5455 (0.419, 0.672)
>> res.A
ans =
7.0034
>> res.Y
ans =
0.54553

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