fzero or solve a file full of relations

Hi all-

I was playing around with fzero here: http://www.mathworks.com/help/matlab/ref/fzero.html#description

Trying to get it to vary x below until y below is 0.

z=[-300.00 1082.00 7494.20];
x=1.750871192;
zprime=[z(1)*x/10000 z(2)*x/10000 z(3)*x/10000];
a=[1 zprime(3) zprime(2) zprime(1)]';
B=[1.637510856e-06 -2.252209129e-05 0.0001949906158 9.284333582e-06
-2.252209129e-05 0.217829072 0.03330113194 0.0449503853 
0.0001949906158 0.03330113194 0.5146272212 0.03759452473 
9.284333582e-06 0.0449503853 0.03759452473 0.7076169167];
Ba = B*a;
atBa = a'*Ba;
y=atBa-1;

I think it can be done with fzero which requires no license for solve. But I do have the license for solve if that's the way to go. I hate to fight over limited licenses :)

Hope this one isn't too tough. Need to call it many times, quickly.

Thanks a ton.

Dave

Answer 1

[Torsten]

Try

function main
x0=1;
sol=fzero(@f,x0);

function y=f(x)
z=[-300.00 1082.00 7494.20];
B=[1.637510856e-06  -2.252209129e-05 0.0001949906158 9.284333582e-06
   -2.252209129e-05 0.217829072      0.03330113194   0.0449503853 
   0.0001949906158  0.03330113194    0.5146272212    0.03759452473 
   9.284333582e-06  0.0449503853     0.03759452473   0.7076169167];
for k=1:length(x)
  xk=x(k);
  a=[1 z(3)*xk/10000 z(2)*xk/10000 z(1)*xk/10000]';
  y(k)=a'*B*a-1;
end

Best wishes

Torsten.

[David Pesetsky]

That works. Matches excel's goal seek pretty close.

I don't get the use of the For loop, or how X is getting assigned.

[Torsten]

In function main, you give a starting guess for x (x0=1).

In function f, fzero wants your function to be evaluated at several values of x ( x is a vector here). The for-loop evaluates your function element-wise (in elements xk of the vector x), saves the results in the array y and returns y to fzero.

Best wishes

Torsten.

Answer 2

[Matt J]

You could use either one, but it will be overkill regardless of which one you choose. The problem that you have shown is a 1D quadratic equation in x. You could solve it by hand or with roots().

[David Pesetsky]

Sorry, I shortened the real problem here to include the linear only terms. It can go up to quadratic, and is transcendental. I guess maybe roots will still solve it? But how to get the coefficients from my matrices to input into roots?

I wouldn't mind some brute force if needed.

[Matt J]

The coefficients would be obtained by expanding out the terms of atBa. But if the true function is not a polynomial, roots will not solve it.