Initial guess Error using fsolve

This is my code

clear all
tic
% fun= @(x)qfunc(2-0.5*x)*exp(-x^2);
% % 
P0=0.5;
P1=1-P0;
P=P0/P1;
 
rho=0.3; %covariance
 
s1=0.5; %mean of sensor 1
s2=1;%mean of sensor 2
Sm=max(s1,s2);
 
t1=-20:0.01:20 ; %threhsold for sensor 1
 
stop=0;
stop_1=0;
 
for i=1:length(t1)
 
fun= @(x2) exp(-s2.^2./2).*exp(x2.*s2).*qfunc((t1(i)-s1-rho.*x2+rho.*s2)./sqrt(1-rho.^2))-P.*qfunc((t1(i)-rho.*x2)./sqrt(1-rho.^2))
 
myfun= @(y) exp(-s2^2./2).*exp(y*s2)*(1-qfunc((t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)))./(1-qfunc((t1(i)-rho.*y)./sqrt(1-rho^2)))-P;
y0=rand;
tw0=fsolve(myfun,y0);
end

I get this error msg ''Objective function is returning undefined values at initial point. fsolve cannot

continue.''

I think the function handle becomes of 0/0 form for initial guess because of qfunc of what I have underlined. How do I put a condtion to use the approximate of qfunc whenever the fsolve counters 0/0 form?

[Walter Roberson]

Array indices must be positive integers or logical values.

Error in @(y)exp(-s2^2./2).*exp(y*s2)*(1-qfunc((t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)))./(1-qfunc((t1(i)-rho.*y)./sqrt(1-rho^2)))-P

Your t1 is a vector, but your i is not defined so it defaults to being sqrt(-1) which is not a valid index.

[Shailee Yagnik]

Hello,

I have editted the above code in question kindly go through it. It still gives the error: ''Objective function is returning undefined values at initial point. fsolve cannot continue''

Answer 1

[Walter Roberson]

NZ = @(a,b) a ./ (b + (a==0 & b == 0));
myfun=@(y) exp(-s2^2./2).*exp(y*s2).*NZ(1-qfunc((t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)),1-qfunc((t1(i)-rho.*y)./sqrt(1-rho^2)))-P;

This detects the case where numerator and denominator are both 0, and substitutes 1 for the denominator in that case, giving a 0/1 result. Perhaps you might prefer

NZ = @(a,b) (a + (a==0 & b == 0)) ./ (b + (a==0 & b == 0));

[Shailee Yagnik]

Hi, I dont understand how it detects the 0/0 case. Can u please make me understand that ?

also i would like to substitue the myfun with

@(tw0) exp(-s2.^2./2).*exp(tw0.*s2).*(1-(1./(sqrt(2.*pi).*(t1-s1-rho.*tw0+rho.*s2)./sqrt(1-rho.^2))).*exp(-((t1-s1-rho.*tw0+rho.*s2)./sqrt(1-rho.^2)).^2./2))./(1-(1./(sqrt(2.*pi).*(t1-rho.*tw0)./sqrt(1-rho.^2))).*exp(-((t1-rho.*tw0)./sqrt(1-rho.^2)).^2./2))-P

so that it can avoid the 0/0 form.

Can you please help me out with that?

[Walter Roberson]

The code

myfun=@(y) exp(-s2^2./2).*exp(y*s2).*NZ(1-qfunc((t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)),1-qfunc((t1(i)-rho.*y)./sqrt(1-rho^2)))-P;

transforms the (1-qfunc((t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)) ./ (1-qfunc((t1(i)-rho.*y)./sqrt(1-rho^2))) subexpression into NZ(1-qfunc((t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)), 1-qfunc((t1(i)-rho.*y)./sqrt(1-rho^2))) which is a call to NS with two parameters.

NZ has been defined as

NZ = @(a,b) (a + (a==0 & b == 0)) ./ (b + (a==0 & b == 0));

This checks whether both of the arguments are zero, and if so adds 1 to both the numerator and denominator, transforming it from a 0 / 0 expression to a (0+1)/(0+1) = 1/1 expression, thus substituting 1 for the 0/0

[Shailee Yagnik]

Thanks for the explanation. Is it possible to substitute an expression in the function when both the arguments of original function forms 0/0 form?

[Walter Roberson]

NZ = @(a,b) (a + (a==0 & b == 0) .* TheExpression) ./ (b + (a==0 & b == 0));

In the case of 0, 0, this would give TheExpression on the numerator and 1 in the denominator.

Note: this will not work if TheExpression evaluates to nan or inf when a is non-zero or b is non-zero.

[Shailee Yagnik]

Thank You. It solved the problem.

[Shailee Yagnik]

NEED Help!

is it correct to write and execute the function in context to the above problem:

NZ =  @(a,b,y) ( (a+ (a==0 & b==0).*(1-1./(sqrt(2.*pi).*(t1-s1-rho.*y+rho.*s2)./sqrt(1-rho.^2))).*exp(-((t1-s1-rho.*y+rho.*s2)./sqrt(1-rho.^2)).^2./2))./ (b + (a==0 & b==0 )*(1-1./(sqrt(2.*pi).*(t1-rho.*y)./sqrt(1-rho.^2))).*exp(-((t1-rho.*y)./sqrt(1-rho.^2)).^2./2)));
myfun= @(y) (exp(-s2^2./2).*exp(y*s2)*NZ(qfunc(-(t1(i)-s1-rho*y+rho*s2)./sqrt(1-rho^2)),qfunc(-(t1(i)-rho.*y)./sqrt(1-rho^2)),y))-(P);

[Walter Roberson]

Possibly. However for clarity I would probably move some of that expressions into helper functions; in its current form, it is difficult to read.