Generating Multivariate Polynomial Coeffients with known Constants

Question

Eric 2024-7-2

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2133666-generating-multivariate-polynomial-coeffients-with-known-constants

评论： Eric 2024-7-4

采纳的回答： Steven Lord

I am extremely new to MatLab and am having a hard time trying to generate Coefficients from this equation:

t= (A * a) + (B * s) + (C * s * a)+ (D * s^2) + (E * s^2 * a) + (F)

I have the data for t,s,a.

Trying to find (A,B,C,D,E,F)

I can remove the Coefficient (F) if need be, being it is just an offset, and I have (t).

I could post what I have tried, but I usually dont save it because I am not getting close.

I do have Curve Fitting, Statistics, and Optimization Toolboxes thinking it would help.

I have also tried polyfit and polyfitn.

Also, if this has been answered a hundred times, what would best describe that equation for searching?

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Answer 1

Steven Lord 2024-7-2

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https://ww2.mathworks.cn/matlabcentral/answers/2133666-generating-multivariate-polynomial-coeffients-with-known-constants#answer_1479526

See the Multiple Regression section on this documentation page for an example you can adapt to your equation. Make sure you use array operators when constructing the columns of the design matrix.

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Steven Lord 2024-7-3

在 MATLAB Online 中打开

You seem to be assuming that the solution is unique. It need not be, as explained with this simpler example Cleve Moler wrote about years ago. Cleve's problem has 1 equation in 2 unknowns; yours (with that one data point) has 1 equation in 6 unknowns.

s = 20;
a = 250;
x1 = s;
x2 = a;
x3 = s*s;
x4 = s*a;
x5 = s*s*a;

This is one solution.

coeffs = [-240.3125;
           7.119140625;
           0.884765625;
          -0.2564697265625;
          -0.0068206787109375;
           0.000759601593017578];
rhs = @(coeffs) coeffs(1) + (x1 * coeffs(2)) + (x2 * coeffs(3)) + (x3 * coeffs(4)) + (x4 * coeffs(5)) + (x5 * coeffs(6));
lhs = 62.53059;
check = [rhs(coeffs); lhs]    
check = 2x1
   62.5306
   62.5306
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

This is another solution.

coeffs = [lhs; 0; 0; 0; 0; 0];
check = [rhs(coeffs); lhs]
check = 2x1
   62.5306
   62.5306
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

Here is a third.

coeffs = [0; 0; 0; 0; 0; lhs./(s.^2*a)]
coeffs = 6x1
1.0e-03 *

         0
         0
         0
         0
         0
    0.6253
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
check = [rhs(coeffs); lhs]
check = 2x1
   62.5306
   62.5306
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

So which one is "the" right one for this single point? Any of the above.

Steven Lord 2024-7-3

在 MATLAB Online 中打开

If you use more information (and so more equations) that will also affect the solution. To use Cleve's example, if I told you that the two numbers have an average of 3 there are multiple solutions (3 and 3, 4 and 2, 6 and 0, etc.)

A = [1/2 1/2; 1 -1];
b = [3; 2];
solutionsToFirstEquation = [3 4 6 16-pi; 3 2 0 -10+pi]
solutionsToFirstEquation = 2x4
    3.0000    4.0000    6.0000   12.8584
    3.0000    2.0000         0   -6.8584
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
checkFirstEquation = A(1, :)*solutionsToFirstEquation
checkFirstEquation = 1x4
     3     3     3     3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

If I then add the additional information that the difference of the two numbers is 2 the solution is unique. The solution is the second column of solutionsToFirstEquation.

x = A\b
x = 2x1
     4
     2
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
satisfiesBothEquations = A*solutionsToFirstEquation % Only 2nd is right
satisfiesBothEquations = 2x4
    3.0000    3.0000    3.0000    3.0000
         0    2.0000    6.0000   19.7168
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
satisfiesBothEquations2 = [A*x, b]
satisfiesBothEquations2 = 2x2
     3     3
     2     2
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

Eric 2024-7-4

I can't tell if constraints on the coefficients would help me or not.

I understand (now) that if I am looking for a specific outcome, then I need to apply specific cirtieria for the output to meet.

I can't seem to find how to apply constraints to the coefficient data as it is calculated. Also, I would assume this would be linear constraints:

A= >= -16 & <= 16

B= >= -64 & <= 64

C= >= -.25 & <= .25

D= >= -2.000 & <=1 .999

E= >= -.0078 & <= .0078

F= >= -2048 & <= 2048

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Generating Multivariate Polynomial Coeffients with known Constants

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