Solve a nonlinear system

I'm solving the following as:

f   = @(R01) 1/(1+R01) - .95;
R01 = fzero(f,0);
f   = @(R02) 0.08/(1+R01) + 1.08/(1+R02)^2 - .99;
R02 = fzero(f,0);

How can I solve the system in one shot, can't make it work with fsolve.

Thanks

Oleg

Answer 1

[Andrew Newell]

Is this what you're trying to do?

f   = @(x) [1/(1+x(1)) - .95; 0.08/(1+x(1)) + 1.08/(1+x(2))^2 - .99];
R = fsolve(f,[0 0]);

If so, you can't do it with fzero because it only accepts a function with a scalar input and scalar output.

[Oleg Komarov]

Ok, now I get it...Thanks

[Zulhash Uddin]

After running the program, we r getting some text with the result. How can we minimize this text?

Equation solved.

fsolve completed because the vector of function values is near zero

as measured by the default value of the function tolerance, and

the problem appears regular as measured by the gradient.

<stopping criteria details>

R =

0.0526315789063412 0.0870230886539235

[Andrew Newell]

@Zulhash, could you please submit a separate question on this?

Answer 2

[Matt Fig]

Do you mean get R01 and R02 in one shot, or do you mean find where the two functions meet (what I usually think of when someone says they want to solve a system of equations)?

If you mean, how to get to R02 in one shot,

f3 = @(R02) 0.08./(1+(1/.95-1)) + 1.08/(1+R02).^2 - .99;
R02 = fzero(f3,0)

Or,

f   = @(R01) 1/(1+R01) - .95;
f   = @(R02) 0.08/(1+fzero(f,0)) + 1.08/(1+R02)^2 - .99;
R02 = fzero(f,0)

[Oleg Komarov]

I actually wanted to use fsolve as in EXAMPLE 1 of the fcn help. Is it possible or am I attempting something senseless from a mathematical point of view?

Answer 3

[Walter Roberson]

With the symbolic toolkit, it looks like

solve(0.8e-1/(1+solve(1/(1+RO1)-.95))+1.08/(1+R02)^2-.99)

and gives the values -2.087023117, 0.08702311660

[Oleg Komarov]

Same comment as to Matt's answer.