Not able to find fzero

--> difference.m

function y = difference(u,d1,n,a,m,T,PsByN_0,UmaxN_0)

d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=5;                                                     
PsByN_0=10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0=10.^(UmaxdB/10);


fun1 = (-1./u)*log(((d1^m)./(a*n*PsByN_0*T*u+d1^m)*a)./(1-a));

fun2 = (1./u)*log(((-exp(u*UmaxN_0)*(exp(-PsByN_0*u)))./(u*UmaxN_0+PsByN_0*u))*(PsByN_0*u)-(PsByN_0*u*(exp(-PsByN_0*u)))*(expint(u*UmaxN_0+PsByN_0*u))+(exp(-PsByN_0*u))+((PsByN_0*u)*(exp(-PsByN_0*u)))*(expint(PsByN_0*u))+(exp(u*UmaxN_0))./((UmaxN_0/PsByN_0)+1));

y = (fun1 - fun2);

g0=fzero(@(u) difference(u,d1,n,a,m,T,PsByN_0,UmaxN_0), 10);

end









        

Answer 1

[Torsten]

d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=5;                                                     
PsByN_0=10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0=10.^(UmaxdB/10);
fun1 = @(u) (-1./u)*log(((d1^m)./(a*n*PsByN_0*T*u+d1^m)*a)./(1-a));
fun2 = @(u) (1./u)*log(((-exp(u*UmaxN_0)*(exp(-PsByN_0*u)))./(u*UmaxN_0+PsByN_0*u))*(PsByN_0*u)-(PsByN_0*u*(exp(-PsByN_0*u)))*(expint(u*UmaxN_0+PsByN_0*u))+(exp(-PsByN_0*u))+((PsByN_0*u)*(exp(-PsByN_0*u)))*(expint(PsByN_0*u))+(exp(u*UmaxN_0))./((UmaxN_0/PsByN_0)+1));
fun =@(u) (fun1(u) - fun2(u));
g0=fzero(fun, 10);

[Dhawal Beohar]

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

because complex function value encountered during search.

(Function value at -2.8 is 9.2614+1.122i.)

Check function or try again with a different starting value.

This is what I am getting.Not sure how to proceed further.

[Torsten]

Plotting the function, I don't see a real root.

Try

d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=5;                                                     
PsByN_0=10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0=10.^(UmaxdB/10);
fun1 = @(u) (-1./u).*log(((d1^m)./(a*n*PsByN_0*T*u+d1^m)*a)./(1-a));
fun2 = @(u) (1./u).*log(((-exp(u*UmaxN_0).*(exp(-PsByN_0*u)))./(u*UmaxN_0+PsByN_0*u)).*(PsByN_0*u)-(PsByN_0*u.*(exp(-PsByN_0*u))).*(expint(u*UmaxN_0+PsByN_0*u))+(exp(-PsByN_0*u))+((PsByN_0*u).*(exp(-PsByN_0*u))).*(expint(PsByN_0*u))+(exp(u*UmaxN_0))./((UmaxN_0/PsByN_0)+1));
fun =@(u) (fun1(u) - fun2(u));
u=0.01:0.01:10;
plot(u,fun(u))

[Dhawal Beohar]

How to analyse the graph...i got this..should I try different for values of u?

[Torsten]

No. For that a solution u for your equation exists, the graph had to cross the line y=0, but it doesn't.

[Dhawal Beohar]

Yes, it's not crossing 0. So, what possible mistake I am doing. What do you suggest doing?

[Torsten]

Check the parameters and the functions.

[Dhawal Beohar]

I changed parameters and now it seems to be crossing y=0.

[Torsten]

Then try

g0=fzero(fun, [0.001 0.1]);

[Dhawal Beohar]

giving this error

Error using fzero

Function values at the interval endpoints must differ in sign.

[Torsten]

Use the plot to determine ul, ur such that fun(ul) * fun(ur) < 0. Then call fzero as

g0=fzero(fun, [ul,ur]);

[Dhawal Beohar]

Version - R2022a

d1=20;

n=10^-11.4;

m=2.7;

a=0.5;

T=1;

PsByN_0dB=15;

PsByN_0=10.^(PsByN_0dB/10);

UmaxdB = 5;

UmaxN_0=10.^(UmaxdB/10);

[Dhawal Beohar]

really sorry but I did'nt understood what you said. What is ul and ur?

[Torsten]

d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=15;                                                     
PsByN_0=10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0=10.^(UmaxdB/10);
fun1 = @(u) (-1./u).*log(((d1^m)./(a*n*PsByN_0*T*u+d1^m)*a)./(1-a));
fun2 = @(u) (1./u).*log(((-exp(u*UmaxN_0).*(exp(-PsByN_0*u)))./(u*UmaxN_0+PsByN_0*u)).*(PsByN_0*u)-(PsByN_0*u.*(exp(-PsByN_0*u))).*(expint(u*UmaxN_0+PsByN_0*u))+(exp(-PsByN_0*u))+((PsByN_0*u).*(exp(-PsByN_0*u))).*(expint(PsByN_0*u))+(exp(u*UmaxN_0))./((UmaxN_0/PsByN_0)+1));
fun =@(u) (fun1(u) - fun2(u));
g0 = fzero(fun,[0.01,0.02])

% gives g0 = 0.016736

[Dhawal Beohar]

By keeping this value as u in equation fun1 and fun2, I am getting these values

fun1 =

0

fun2 =

0.5825

not equal, but close.

[Torsten]

If I add the two lines

fun1(g0)
fun2(g0)

to the code above, I get as output

ans = 1.9901e-14
ans = 1.3268e-14

thus both fun1 and fun2 seem to be approximately 0.

[Dhawal Beohar]

I am getting these values, not sure how it's different for fun2

ans = 1.9901e-14

ans = 2.6535e-14

[Torsten]

You want to find u that makes fun(u) = 0.

fun(u) is equal to fun1(u) - fun2(u).

Now you found an u for which fun(u) = fun1(u) - fun2(u) = 1.9901e-14 - 2.6535e-14 = -6.634e-15.

I think you will admit that this is approximately zero.

[Dhawal Beohar]

I agree Torsten, Thank you for the help. You were so much help to me and learnt a lot from you.

[Dhawal Beohar]

@Torsten Hi! quick question, we found for what value of 'u' fun1 = fun2. But how we can find optimal range of 'u' if we want to plot fun1 or fun2 against 'u'.

Thanks in advance.

[Torsten]

I suggest you try an interval around the zero you found, e.g.

u = fzero(fun,[0.01,0.02])
U = linspace(0.01,0.02,20);
fU = fun(U);
plot(U,fU)

[Dhawal Beohar]

ok, thanks!

Answer 2

[Sam Chak]

Hi @Dhawal Beohar

Guess the problem that you want to solve is a complex-valued function.

function y = difference(u)

% parameters

d1 = 20;

n = 10^-11.4;

m = 2.7;

a = 0.5;

T = 1;

PsByN_0dB = 5;

PsByN_0 = 10.^(PsByN_0dB/10);

UmaxdB = 5;

UmaxN_0 = 10.^(UmaxdB/10);

% functions

fun1 = (-1./u)*log(((d1^m)./(a*n*PsByN_0*T*u + d1^m)*a)./(1 - a));

fun2 = (1./u)*log(((- exp(u*UmaxN_0)*(exp(-PsByN_0*u)))./(u*UmaxN_0 + PsByN_0*u))*(PsByN_0*u) - (PsByN_0*u*(exp(-PsByN_0*u)))*(expint(u*UmaxN_0 + PsByN_0*u)) + (exp(-PsByN_0*u)) + ((PsByN_0*u)*(exp(-PsByN_0*u)))*(expint(PsByN_0*u)) + (exp(u*UmaxN_0))./((UmaxN_0/PsByN_0) + 1));

y = (fun1 - fun2);

end

Let's try with fzero first.

[u, fval, exitflag, output] = fzero(@(u) difference(u), 10)

The exitflag = -4 indicates that complex function value was encountered while searching for an interval containing a sign change.

Next, fsolve is used.

[u, fval, exitflag, output] = fsolve(@(u) difference(u), 10)

The exitflag = -2 means that the Equation is not solved. The exit message shows that fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the default value of the function tolerance.

[Dhawal Beohar]

then how it can be further solved? what can be the solution.

[Sam Chak]

Apparently, the function does not have any real root, for which the function equals zero.

Thus, it is not surprising that fzero and fsolve cannot solve the given equation.

What exactly are you trying to solve? Or, put it another way, what is the expected solution?

Perhaps there are missing information in the given difference function. Please double check.

[Dhawal Beohar]

i wanted to know that for what value of u both the equations i.e., fun1 and fun2 becomes equal....

thank you much for the inputs but my equations are correct but dont know why the solution is not coming.

[Torsten]

Yes, as said, I plotted the difference and there is no point where the difference is 0.

[Dhawal Beohar]

@Torsten you haven always big help. Can you please look into this problem and help me. Thank you!

https://uk.mathworks.com/matlabcentral/answers/1775175-how-to-store-the-value-of-intersection-points-from-fzero-into-a-vector