How to solve singularity problem while using fsolve

Hi,I am trying to solve for x in a equation using fsolve. Below is the code and equation:

a = 0.1886;
b = 0.6886;
c = 1.1886;
m = 0.31372;
x0 = 0.1;
F = @(x)[(((x-a)/(x-c))^(m))+((x-a)/(x-b))];
options = optimset('Display','iter','MaxFunEvals',1e20,'TolFun',2e-50,'TolX',2e-50);
[xd, fval, exitflag, output]= fsolve(F, x0, options)

However, I am getting 'No solution found' with a message: fsolve stopped because the problem appears to be locally singular. I have tried changing the options parameters and initial guess still getting same output. Any help is appreciated.

Thanks, KB

Answer 1

[John BG]

Try splitting the fsolve approach with real() and imag() because with real() i get something:

..
F = @(x)[real(((x-a)/(x-c))^(m))+((x-a)/(x-b))];
..
[xd, fval, exitflag, output]= fsolve(F, x0, options)

                                           Norm of      First-order   Trust-region
   Iteration  Func-count     f(x)          step         optimality    radius
       0          2        0.366931                          1.77               1
       1          4     0.000271336       0.207172         0.0421               1
       2          6     2.03148e-07     0.00645204        0.00109               1
       3          8     1.38596e-13    0.000186725       8.97e-07               1
       4         10     7.81816e-26    1.54486e-07       6.74e-13               1
       5         12     3.08149e-33    1.16029e-13       1.34e-16               1

Equation solved, inaccuracy possible.

The vector of function values is near zero, as measured by the selected value
of the function tolerance. However, the last step was ineffective.

<stopping criteria details>

xd =
          0.30
fval =
         -0.00
exitflag =
          3.00
output = 
       iterations: 5.00
        funcCount: 12.00
        algorithm: 'trust-region-dogleg'
    firstorderopt: 0.00
          message: 'Equation solved, inaccuracy possible.…'

repeat for the imag() part.

Or alternatively use abs() and arg()

all together, you should catch at least 2 poles on b and c.

If you find this answer of any help solving your question,

please click on the thumbs-up vote link,

thanks in advance

John

[e-mail]

Answer 2

[Walter Roberson]

 x = 943/5000

The equation is exactly 0 at that point, which corresponds to x = a . However, there is no zero crossing because it goes imaginary until x = c at which point it slopes down from infinity towards 2 (and so never crosses 0 for greater x either)