Is it possible to use Matlab to get expressions for the coefficients in polynomial regression? (I.e., not just constant coefficients)

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I'm trying to modify one section of an old code so that it uses my new data for one of its parameters. The code expresses an equation as a fourth-order power series in X and y. I know how to do polynomial regression on a curve like the one below to get an expression like:
Phi=k0+k1y+k2y2+k3y3+k4y4
However, my problem is that the code uses a more sophisticated expression for k. I.e.,
I need to express the k this way (in terms of a coefficients) because they're needed elsewhere in the code. Right now, the a coefficients are provided in a table in the code, but I have no idea how to calculate them.
I'm comfortable using Matlab and would be grateful if someone could tell me whether there is a way to do polynomial regression for a fourth-order power series (based on curves like the one in the top figure above) and get the coefficients in terms of a coefficients as shown in the figure above. I have looked around but it does not seem like it is possible. For example "multiple linear regression" and "multivariate regression" don't seem to do what I need. Maybe they do though and I'm not experienced enough to see it? Thank you for any advice.
  2 件のコメント
Torsten
Torsten 2022 年 5 月 25 日
So X and Y are independent variables and phi = f(X,Y) ?
I ask this because in your ansatz for phi
phi = k0 + k1*Y + k2*Y^2 + k3*Y^3 + k4*Y^4
X as variable does not appear.
Srh Fwl
Srh Fwl 2022 年 5 月 26 日
編集済み: Srh Fwl 2022 年 5 月 26 日
Torsten, thank you for your help and apologies for the wait (fell asleep!). Yes, they are independent variables. Yes, I also don't understand why X does not appear in my expression, but I think it's OK. Each curve (like in the top figure) is for a specific X. So the phi–Y relationship, I can figure out how to get the correct "a" coefficients," should be correct for X as well.

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採用された回答

Torsten
Torsten 2022 年 5 月 25 日
M = [X.^(-1).*Y,X.^(-1).*Y.^2,Y.^2,X.*Y.^2,X.^(-1).*Y.^3,X.*Y.^3,X.^2.*Y.^3,X.^(-2).*Y.^4,X.^(-1).*Y.^4,Y.^4];
b = PHI - 1.0;
a = M\b;
where X,Y and PHI are column vectors of the same size.
Here, a(i) = ai in your problem description.
  7 件のコメント
Torsten
Torsten 2022 年 5 月 26 日
It makes sense that the a coefficients are going to be in the form of a vector, but I'm hung up on what Bruno has done to get his "M = " expression.
M*a = b
written out gives
x^(-1)*y*a(1)+x^(-1)*y^2*a(2)+y^2*a(3)+x*y^2*a(4) ... = PHI - 1
thus your model function equation.
You want to solve this equation for a, thus
a = M\b
Srh Fwl
Srh Fwl 2022 年 5 月 26 日
編集済み: Srh Fwl 2022 年 5 月 26 日
Thank you both so much. I finally get it! Really appreciate your patience. It's great to finally understand what's going on with this.

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その他の回答 (1 件)

Walter Roberson
Walter Roberson 2022 年 5 月 25 日
For each equation divide both sides by the highest negative coefficient,
k4(Xhat)/Xhat^(-2) = a8*Xhat^(-2)/Xhat^(-2) + etc
k4(Xhat)*Xhat^2 = a8 + a9*Xhat + a10*Xhat^2
subtract left from right and you have a standard polynomial that you can do fitting on
  1 件のコメント
Srh Fwl
Srh Fwl 2022 年 5 月 26 日
編集済み: Srh Fwl 2022 年 5 月 26 日
Sorry, Walter, I don't follow (very rusty at algebra!). If you have time, would you please explain how to get the "a" values from this? Are you saying that I should use each individual k equation to fit the data in the figure? I'm embarrassed to be so clueless on this!

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