fzero line 398 error; debugger not solving...

I am having a problem with fzero in this code :

prueba = r_teor (8e-3) 

function Arg = r_teor (t0)  

FC = 4.24E-1;     
a  = 6.05;         
FD = 8.17E-4;
Io = 15.53;
V  = 25.00;
texp=0.009:0.0001:0.0099;

      Arg         = zeros (N,1); % prellocating

      for i       = 1 : N
      options=optimset('Display','iter','TolX',1e-5);
      % Display: off / iter / final / notify
      % TolX: Default 2.204 e-16
      Arg   (i)   = fzero (@(r) trapecios (r) - (texp(i)-t0), 5e-3, options);
      end

          function Int = trapecios(r)

          if r <= 1    
            Int = 0;
            return
          end 

          F         = @(x) FC .* exp( -(V.*a) ./ (30.*x.*log(x)) );
          Int       = 0;
          part      = 1e-3;
          L         = 1:part:r;
          me        = F(L);
          sum_me    = sum (me);
          ue        = F( max( L ) );
          Int       = zeros (N,1); % prellocating
          i         = 1:N;
          Int   (i) = part.* (sum_me - (ue/2));

          end % trapecios

end % r_teor

The error is:

Operands to the || and && operators must be convertible to logical scalar values.

Error in fzero (line 398)
        if ~isfinite(fb) || ~isreal(fb) || ~isfinite(b)

, but the debugger sais that both b and fb are real and finite:

K>> b, fb, whos b, whos fb

b =

     1.163523750296039

fb =

     1.0e-03 *

    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276
    -0.999999999999276

    Name      Size            Bytes  Class     Attributes

    b         1x1                 8  double              

    Name       Size            Bytes  Class     Attributes

    fb        10x1                80  double

My function trapecios has no problem when executing it on b = 1.163523750296039.

Could it be because fzero cannot find zeros if the function does not change it sign (as x^2) ? Could it be because I must enter t0 as a 1:10 array (t0 is the variable of r_teor, so how could I do it) ?

Any idea? Thanks !!!

Answer 1

[Steven Lord]

What requirement does fzero impose on the function you pass to it?

Function to solve, specified as a handle to a scalar-valued function or the name of such a function. fun accepts a scalar x and returns a scalar fun(x).

Does your function satisfy that requirement? In particular, does it return a scalar fun(x)?

If you want to solve a nonlinear system of equations, see fsolve in Optimization Toolbox.

[Sergio Quesada]

Thanks, Steven Lord!!

I used fzero because I am not trying to solve a system of equations, but an array of single equations...What is the right way to declaring it?

I have also tried to enter "t0" as an 1:10 array, but the same problem does not change...

I have also changed fzero with fsolve, but it solve the equations at the starting point:

                                           Norm of      First-order   Trust-region
   Iteration  Func-count     f(x)          step         optimality    radius
       0          2        3.61e-06                             0               1

Equation solved at initial point.

fsolve completed because the vector of function values at the initial point
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>

prueba =

     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000
     0.005000000000000

Thanks !!