Generating Multivariate Polynomial Coeffients with known Constants

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?

Answer 1

[Steven Lord]

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.

[Eric]

x1 = [20]';
x2 = [250]';
x3 = [20.*20]';
x4 = [20.*250]';
x5 = [20.*20.*250]';
y = [62.5306]';

X = [ones(size(x1))  x1  x2  x3  x4  x5];

a = X\y

x1 = [20]';
x2 = [250]';
y = [62.5306]';

X = [ones(size(x1))  x1  x2  x1.*x1  x1.*x2  x1.*x1.*x2];

a = X\y

I must be missing something.

I tried both ways and it gives me 0 for the first 4 coefficients.

[Steven Lord]

x1 = [20]';
x2 = [250]';
x3 = [20.*20]';
x4 = [20.*250]';
x5 = [20.*20.*250]';
y = [62.5306]';

X = [ones(size(x1))  x1  x2  x3  x4  x5];

v = X\y

v = 6x1

1.0e-03 * 0 0 0 0 0 0.6253

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This means that y = s.^2*a*v(6). With one data point, how do you expect to compute more than one non-zero coefficient? It is a valid solution; there are others.

[v(6)*x5; y]

ans = 2x1

62.5306 62.5306

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svalue = 20;
avalue = 250;
v(6)*svalue.^2.*avalue - y

ans = -7.1054e-15

That looks like when I evaluate the polynomial at the specified value of s and a I get the correct value of y.

Try fitting with more sets of (s, a, y) values at once and the coefficients may not be 0 values.

[Eric]

I will do some more testing/learning. I was trying to learn on a given data point that I had all the answers for before I try my 10400x4 matrix. I have what the coefficients "should be" for that example I am working with, kind of a "reverse engineering" for me.

I have:

if

x1 = [20]

x2 = [250]

x3 = [20.*20]

x4 = [20.*250]

x5 = [20.*20.*250]

62.53059= (-240.3125) + (x1 * 7.119140625) + (x2 * 0.884765625) + (x3 * -0.2564697265625) + (x4 * -0.0068206787109375) + (x5 * 0.000759601593017578)

[Steven Lord]

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

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This is another solution.

coeffs = [lhs; 0; 0; 0; 0; 0];
check = [rhs(coeffs); lhs]

check = 2x1

62.5306 62.5306

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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

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check = [rhs(coeffs); lhs]

check = 2x1

62.5306 62.5306

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

[Eric]

I read Cleve Molers article. That helps put it in perspective. Thank you!

I think my next project is to add lower and upper constraints to the coefficients.

Thank you again.

[Steven Lord]

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

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checkFirstEquation = A(1, :)*solutionsToFirstEquation

checkFirstEquation = 1x4

3 3 3 3

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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

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satisfiesBothEquations = A*solutionsToFirstEquation % Only 2nd is right

satisfiesBothEquations = 2x4

3.0000 3.0000 3.0000 3.0000 0 2.0000 6.0000 19.7168

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satisfiesBothEquations2 = [A*x, b]

satisfiesBothEquations2 = 2x2

3 3 2 2

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[Eric]

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