Asked by Nuno
on 2 Dec 2012

Problem:

syms w; % variable, that define the y axis

lam = 4*pi()^2 % constant, that define the x axis

tan(w/2) == w/2 - (4/(lam))*(w/2)^3 % equation

That equation with that constant has many solutions. What I pretend is to resolve the equation (maybe with solve) for lam = 4*pi()^2, in a certain interval of y, in order to catch a specific solution. For example, I know that for that constant there is a solution of w/pi() = 2 in the interval w/pi()=[1.5 2.5], (y axis).

I was trying with this:

syms w;

lam=4.*pi().^2;

x = fsolve(@(w) tan(w/2) == w/2 - (4./(lam)).*(w/2).^3, [1.5*pi() 2.5*pi()]);

X=x/pi()

But is giving me some problems that I can´t resolve.

On Maple software the structure of the program is more or less the same, but it gives me the solutions that I want in specific intervals of y axis.

Matlab has to do it too or it doesn't?

Thanks to all.

*No products are associated with this question.*

Answer by Walter Roberson
on 2 Dec 2012

Accepted answer

The function you provide to fsolve() needs to have a numeric result rather than a logical result. Convert your form A==B to (A)-(B) or (B)-(A)

Show 1 older comment

Nuno
on 8 Dec 2012

Now I'm confused again.

%Problem:

syms x;

a=fsolve(@(x) x.^2-4,[5 8])

%results:

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.

a =

2.0000 2.0000

Why the "fsolve" is returning 2 values, when I said that I want values in the interval [5 8]?

If I put an interval like [0 5] the fsolve gives me: a= 0 and 2, that is strange, right?

Walter Roberson
on 8 Dec 2012

Opportunities for recent engineering grads.

## 0 Comments