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eaihua
Posted: Fri Jul 11, 2008 8:41 pm    Post subject: Generalized Least Square Method Reply with quote

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When studying the relating relations between the two variable numbers (x, y), we can get a series of binate data(x1, y1, x2, y2.....xm, ym), describing the data into the x - y orthogonal coordinate system(Chart), the points are found near a curve. suppose the one-variant non-linear variant of the curve such as (Formula 1)



y = a0 + a1 xi ^k (1)

There into, a0, a1 and k are arbitrary real numbers

To set the curve variant, the numbers of a0, a1 and k must be set. Use the same way with ā€œThe Least Square Method data regressionā€ based on the square sum of the deviation of the true measure value yi and computing value.

Order:
φ = ∑(yi - y)^2 (2)

Take (Formula 1) to (Formula 2) to get :

φ = ∑(yi - a0 - a1 xik )^2 (3)

When the square of ∑(yi - a0 - a1 xik ) is the smallest, we can use functionφ to get the partial differential coefficients of a0, a1 and k, make the three partial differential coefficients zero.

(4)

(5)

(6)

Get three variant groups about a0,a1 and k which are the unknown numbers, solve the groups can get mathematical model.

Also, we can judge the right of the mathematical model with the help of correlation coefficient R, statistical variable F, residual standard deviation S to judge, it is better that R tends to 1,the absolute value of F is bigger and S tends to 0. the validation is good, but error of model computing sometimes are big, to improve the mathematical model farther, the biggest error, equal error and the equal relating error of computing model are computed to validate the model.

The Method to do multivariate nonlinear model chart



Web address: http://www.eaihua.com
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Faldo_Elrith
Posted: Sat Jul 12, 2008 7:47 pm    Post subject: Reply with quote

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Wouldn't it be easiest just to take logarithms on both sides?

y=a_0+a_1x^k

y-a_0=a_1x^k

\ln(y-a_0)=\ln a_1+k\ln x

Now you have a linear relation between \ln(y-a_0) and \ln x!
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eaihua
Posted: Sun Jul 13, 2008 3:58 am    Post subject: Sir, Faldo_Elrith Reply with quote

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Ln(y-a0) = Ln a1 + k Ln x

Such changes can not be calculated K, can not be regression on the nonlinear data.
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Faldo_Elrith
Posted: Sun Jul 13, 2008 4:10 am    Post subject: Reply with quote

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Of course they can be calculated. Let Y=\ln(y-a_0), X=\ln x, C=\ln a_1. Then you have

Y=kX+C

which is the equation of a straight line.
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eaihua
Posted: Sun Jul 13, 2008 4:33 am    Post subject: Reply with quote

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Mr, Faldo_Elrith

a0 how to calculate , Does not calculate a0 , Will not be able to achieve the regression of nonlinear data.

Thank,
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