Matlab giving the warning of Exiting fzero!

Exiting fzero: aborting search for an interval containing a sign change

because complex function value encountered during search.

(Function value at -2.8 is 10.0875+1.122i.)

Check function or try again with a different starting value.

I am getting the complex value for any initial value i am giving. How can i solve this issue. I have tried using different values for initial guess and changing some of my input values as well, but no success. Here is my code below. Any help would be really appreciated.

function fval = fun(u)
d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=25;                                                     
PsByN_0=10.^(PsByN_0dB/10);

fun1 = @(u) (((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T);
initialguess = 10;
zero = fzero(fun1,initialguess);

[Walter Roberson]

Nothing in your code assigns to fval .

Your code ignores the u that was passed in to fun. The u in the @(u) definition of fun1 "shadows" the u that was passed in to fun

Answer 1

[Walter Roberson]

d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=25;                                                     
PsByN_0=10.^(PsByN_0dB/10);
fun1 = @(u) (((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T);

syms U
F = fun1(U)

F = 

In order for that to be 0, either is infinite or the log() has to be 0 . But for the log() to be 0, the value inside would have to be 1

inner = children(children(F, 2),1)

inner = 

Ubest = solve(inner == 1)

Ubest = 

fun1(Ubest)

ans = 

0

vpa(Ubest)

ans = 

[Dhawal Beohar]

Hi, thank you so much for your reply. I request you please tell me what exactly i need to change in my code. I am confused.

[Walter Roberson]

d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=25;                                                     
PsByN_0=10.^(PsByN_0dB/10);
fun1 = @(u) (((-1./u).*log((d1.^m)./(a.*n.*PsByN_0.*T.*u)+d1.^m).*a.*T)./(1-a).*T);

syms U
F = fun1(U)

F = 

inner = children(children(F, 2),1)

inner = 

Ubest = solve(inner == 1)

Ubest = 

fun1(Ubest)

ans = 

0

vpa(Ubest)

ans = 

t = double(linspace(Ubest-1, Ubest+1, 20));
y = fun1(t);
plot(t, real(y), 'b', t, imag(y), 'r')

format long g
R = [-1600000000 -1589144000];
fzero(fun1, R)

ans = 
         -1589144427.37938

If you make the right edge too positive, then you get the error you were seeing.

[Dhawal Beohar]

Hi !

When I am using

inner = children(children(F, 2),1)

it is giving me error as,

Error using sym/children

Too many input arguments.

Error in fun (line 40)

inner = children(children(F, 2),1)

[Walter Roberson]

In your release, children() does not take an optional second argument. You will need to use temporary variables.

temp = children(F);
temp2 = children(NF(2));
inner = temp2(1);

something like that.

[Dhawal Beohar]

That worked! I updated my release. Thanks

But, I am using the same procedure for my second fun2, I am not sure how to get the limit for R = [];

fzero(fun2, R) ? This is what I am getting :

fun2 = @(u) ((1./u).*log(expint(-PsByN_0.*u)).*exp(-PsByN_0*u));
syms u
F2 = fun2(u)
inner = children(children(F2, 2),1)
Ubest = solve(inner == 1)
fun2(Ubest)
vpa(Ubest)

F2 =

(exp(-100*10^(1/2)*u)*log(expint(-100*10^(1/2)*u)))/u

inner =

-100*10^(1/2)*u

Ubest =

-10^(1/2)/1000

ans =

-100*10^(1/2)*log(-ei(-1))*exp(1)

ans =

-0.0031622776601683793319988935444327

Please advise. Thank you so much for your help.

[Torsten]

For F2 to be zero, expint(-100*sqrt(10)*u) has to be 1:

fun=@(u) expint(-100*sqrt(10)*u) - 1 
u0 = -1.0e-4;
U = fsolve(fun,u0)

[Dhawal Beohar]

Thanks, How it can be done using fzero? It is giving answer as NaN

[Torsten]

Then you'll have to vary the initial u0 until you get a reasonable solution.

If it works with "fsolve", why don't you use it ?