Discrete equation with two unknown variables

[EDIT: 20110523 16:16 CDT - clarify - WDR]

Hi,

I am looking for a simple way to find out the solution for this equation:

y1 = a*x1/(1+b)*x1
y2 = a*x2/(1/b)*x2

a and b are unkown but x1, x2, y1, y2 are known. I need discrete solutions for this equation and not 1 and 0 as solution. How can I compute this in MatLab, I know it is rather simple by head, but I have been cracking my head over this for several days and never am sure of what I find is right.

[Ruben Verkempynck]

Guys,

I made a huge mistake the equation should be:

y1 = a*x1/(1+b)*x1

y2 = a*x2/(1+b)*x2

Answer 1

[Arnaud Miege]

This can be rewritten as:

a*x1^2 - b*y1 = y1
a*x2^2 - b*y2 = y2

or in matrix form:

A * [a b]' = [y1 y2]'

where:

A = [x1^2 -y1; x2^2 -y2];

So:

x1 = 2;
x2 = 3.3;
y1 = 9;
y2 = 2.9;
A = [x1^2 -y1; x2^2 -y2];
RHS = [y1 y1]';
solution = A\RHS;
a = solution(1);
b = solution(2);

HTH,

Arnaud

[Andrei Bobrov]

Hi, Arnaud

little typo, need

a * x1 ^ 2 - b * y1 = y1

a * x2 ^ 2 - b * y2 = 0

[Arnaud Miege]

Oops, actually, it's more like:

a*x1^2 - b*y1 = y1

a*b*x2^2 = y2

so it's a non-linear system of equations and the matrix approach I suggested won't work. However, by substitution, you can get a second order polynomial in b and use roots:

b^2*y1 + y1 - y2*x1^2/x2^2 = 0

which doesn't require the optimization toolbox

[Clemens]

Just a note - if A got an eigenvalue equal 0 there is a subspace of solutions.

So you can construct solutions by picking values in this subspace.

I suppose by discrete solutions you mean integers. You can search for them in this subspace.

[Arnaud Miege]

Correction:

b^2*y1 + b*y1 - y2*x1^2/x2^2 = 0

[Ruben Verkempynck]

Guys,

I made a huge mistake the equation should be:

y1 = a*x1/(1+b)*x1

y2 = a*x2/(1+b)*x2

Is this possible to find unique solutions? I am actually trying to find an a and b, these are parameters for a function that are somewhere in literature, but I can't find them. y's and x's are calculated through this function. I could find all these y's and x's in a list. But not the a and b.

[Arnaud Miege]

Then, my answer above "solution = A\RHS" is correct.

[Clemens]

But it is only unique if det(A) is not 0.

[Arnaud Miege]

True

[Ruben Verkempynck]

OK, so this is what I did:

x1 = 17698;

x2 = 81516;

y1 = 37601;

y2 = 102743;

A = [x1 -y1; x2 -y2];

RHS = [y1*x1 y1*x1]';

solution = A\RHS;

a = solution(1)

b = solution(2)

a =

-3.4770e+04

b =

-3.4064e+04

>> det(A)

ans =

1.2467e+09

As you can see the det(a) is not zero, but my a and b are know negative.

[Ruben Verkempynck]

Again a mistake:

RHS = [y1*x1 y1*x1]'; should be

RHS = [y1*x1 y2*x2]';

[Arnaud Miege]

Your A and RHS are wrong:

A = [x1^2 -y1; x2^2 -y2];

RHS = [y1 y2]';

solution = A\RHS;

a = solution(1)

b = solution(2)

This gives a = 0 and b = 1:

>> a*x1^2 - b*y1

ans =

37601

>> y1

y1 =

37601

>> a*x2^2 - b*y2

ans =

102743

>> y2

y2 =

102743

>> det(A)

ans =

2.1767e+014

[Arnaud Miege]

Correction: b = -1

Answer 2

[Oleg Komarov]

x1 = 2;
x2 = 3.3;
y1 = 9;
y2 = 2.9;

f   = @(ab) [ab(1)*x1/(1+ab(2))*x1 - y1
             ab(1)*x2/(1/ab(2))*x2 - y2];
R = fsolve(f,[0 0]);

[Ruben Verkempynck]

??? Undefined function or method 'fsolve' for input arguments of type 'function_handle'.

Error in ==> Beverton_Holt at 8

R = fsolve(f,[0 0]);

>>

The example doesn't work, do I need a certain toolbox for fsolve?

[Oleg Komarov]

yes, sorry. The optimization toolbox.

[Arnaud Miege]

fsolve is part of the Optimization Toolbox, see http://www.mathworks.com/help/releases/R2011a/toolbox/optim/ug/fsolve.html

[Ruben Verkempynck]

OK, thanks, I should get that toolbox.

Answer 3

[Andrei Bobrov]

f1 = @(x,x1,x2,y1,y2)[y1-x(1)*x1.^2./(1+x(2));y2-x(1)*x2.^2/(1./x(2))];
x1 = 1;x2 = 2;y1=10;y2=15;
fsolve(@(x)f1(x,x1,x2,y1,y2),[10,10]);