using fzero to find root of nonlinear equation

Dear Matlab user,

I try to find root of highly nonlinear equation as in

T=1;
r=0.1;
d=0.3;
sigma=0.2;

p=5545.17744447956;
b=(r-d-0.5*(sigma^2))/(sigma^2);
b2=b^2;

q1=-b+sqrt(b2+(r+p)*(2/sigma^2));         
q2=-b-sqrt(b2+(r+p)*(2/sigma^2));

x=fzero(@(x) x.^q2*(r+p-r.*q1+d.*q1)./q1./(r+p)./(d+p)+ x.*d.*(1-q1)./(p.^q2)./q1./(d+p)+ r./(p.^(1+q2))./(r+p),[0.01 10000],optimset('Display','off'));

but it doesn't work. Would anyone help me with this ? Thanks alot.

Regards,

Endah

[Matt J]

You have some expressions in there that aren't finite:

>> d.*(1-q1)./(p.^q2)./q1./(d+p)

ans =
    -Inf

>> r./(p.^(1+q2))./(r+p)

ans =
     Inf

Also q1 and q2 are very large exponents. That's going to create numerical issues.

[endah]

I have checked that. Yes you're right Matt. Thanks.. I have to fix the formula.

Answer 1

[Matt Fig]

To highlight what Matt J said in a comment, you have a problem with your function definition:

F = @(x) x.^q2*(r+p-r.*q1+d.*q1)./q1./(r+p)./(d+p)+ x.*d.*(1-q1)./(p.^q2)./q1./(d+p)+ r./(p.^(1+q2))./(r+p);
x = .1:.01:1000; % The range over which you are looking for a root.

all(isnan(F(x)))  % F is not-a-number everywhere on this range!
ans =

       1

[endah]

Yes, the value is infinite everywhere , seems the problem is in my formula..Thanks !