## How to solve an equation in a certan interval.

on 2 Dec 2012

### Walter Roberson (view profile)

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.

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### Walter Roberson (view profile)

on 2 Dec 2012

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)

Nuno

### Nuno (view profile)

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.

criteria details

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?

Nuno

### Nuno (view profile)

on 8 Dec 2012

In "fsolve" can you use an interval? because if you put a number in the place of the interval, the fsolve gives you one number, but if you put [number1 number2], it gives you two numbers. It is ridiculuos when you should get just one, the solution.

Walter Roberson

### Walter Roberson (view profile)

on 8 Dec 2012

No, fsolve() does not take an interval. When you provide multiple values for x0 it becomes an equation with multiple variables. fzero() is the one that takes an interval.

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