I am trying to solve fsolve (multi-variable) but getting an error.

function fval = func4uo(u) 

d1=1;
n=1;
m=1;
a=1;
T=1;
PsByN_0=1; 

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

xsol = fsolve (@(u) func4uo(u), 0)

ERROR: Not enough input arguments.

[Matt J]

That's not what is shown when we run the code here,

xsol = fsolve (@(u) func4uo(u), 0)

Error using fsolve (line 300)
Objective function is returning undefined values at initial point. FSOLVE cannot continue.

function fval = func4uo(u) 

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

[Dhawal Beohar]

How it can be solved?

[Walter Roberson]

vectorize your code. u is a vector and you calculate something using it, and you / that against another vector calculated from u. The / operator is not element-by-element division, which is the ./ operator

[Dhawal Beohar]

I have corrected my code with values of variables and as you have suggested, but still I am having same error. I am not sure I have vectorized correctly.

function fval = func4uo(u)

d1=10;

n=10^-11.4;

m=2.7;

a=0.5;

T=1;

PsByN_0dB=20;

PsByN_0=10.^(PsByN_0dB/10);

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

x0 = [0];

xsol = fsolve (@(u) func4uo(u), 0)

[Torsten]

In the calculation of fval(1,1), you divide by u which is 0 for your initial guess.

Why do you prescribe x0=[0;0;0] (3 unknowns) and in the call to fsolve use x0=0 (1 unknown) ?

[Dhawal Beohar]

Hi Torsten ! I was not sure I am vectroizing the equation correctly, I have corrected now.

[Torsten]

The problem is that you divide by u (fval(1,1) = -1./u ... ) and u = 0 at the beginning.

Try

function main
  u0=1;
  usol = fsolve(@func4uo,u0)
end
function fval = func4uo(u)  
  d1=10;
  n=10^-11.4;
  m=2.7;
  a=0.5;
  T=1;
  PsByN_0dB=20;                                                              
  PsByN_0=10.^(PsByN_0dB/10); 
  fval (1,1) = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T;
end

But the solution of your problem is obvious:

u = (1-d1^m)/(d1^m*a*n*PsByN_0*T)

[Dhawal Beohar]

Thanks for all the help.

My problem is I am having two equations:

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

and

(1/u)*log(expint(-Ps*u))*exp(-Ps*u)

I need to find out for what value of u, both of the equations can be equal?

I really apreciate your help.

[Torsten]

Then determine the zero of

((-1/u)*log((d1^m)/(a*n*PsByN_0*T*u)+d1^m)*a*T)/(1-a)*T  - (1/u)*log(expint(-Ps*u))*exp(-Ps*u)

[Dhawal Beohar]

still anything is wrong?

function fval = func4uo(u) 

d1=10;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=20;                                                     

PsByN_0=10.^(PsByN_0dB/10); 

fval(1,1) = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T  - (1./u)*log(expint(-Ps*u))*exp(-Ps*u);

zero = fzero(fval,0)

[Torsten]

function main
  u0 = 1;
  u = fzero(@func4uo,u0)
end
function fval = func4uo(u)
  d1=10;
  n=10^-11.4;
  m=2.7;
  a=0.5;
  T=1;
  PsByN_0dB=20;                                                     
  PsByN_0=10.^(PsByN_0dB/10); 
  fval = log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T + log(expint(-Ps*u))*exp(-Ps*u);
end

[Dhawal Beohar]

Thanks Torsten for your help...

I have saved the file name as func4uo.m, getting an error:

>> func4uo

Error: File: func4uo.m Line: 41 Column: 1

Function 'func4uo' has already been declared within this scope.

[Torsten]

Save the file as main.m and run it after assigning a value to Ps.

[Dhawal Beohar]

Thanks ! but some other errors,

function main

u0 = 1;

u = fzero(@func4uo,u0)

end

function fval = func4uo(u)

d1=10;

n=10^-11.4;

m=2.7;

a=0.5;

T=1;

PsByN_0dB=20;

PsByN_0=10.^(PsByN_0dB/10);

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

end

Error using fzero (line 328)

Function value at starting guess must be finite and real.

Error in main (line 39)

u = fzero(@func4uo,u0)

Answer 1

[Matt J]

By choosing a=1, you are dividing by 1-a=0 for any input value, u.

f(0), f(1), f(2)

ans = -Inf

ans = -Inf

ans = -Inf

function fval = f(u) 

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

Answer 2

[Walter Roberson]

There is no zero for that function.

If you use negative u, then the imaginary component of the function approaches negative infinity as u gets close to zero, and only reaches zero again as u gets to -infinity.

If you use positive u and floating point values, then the expint() overflows to infinity when you reach about 8, and the exp() term numerically goes to 0 in floating point, and inf*0 is nan.

If you use positive u with the symbolic toolbox, you can show that the real part of the function is negative until infinity is reached.

Or perhaps I should say that the root is u = +inf as in the limit the function does become 0.

format long g
U = linspace(5,8);
Z = func4uo(U);

figure(); plot(U, real(Z), 'k'); title('real'); xlim([0 10])

figure(); plot(U, imag(Z), 'r'); title('imaginary'); xlim([0 10])

func4uo(10)

ans = 
                        NaN +                   NaNi

func4uo(sym(10))

ans = 

vpa(ans)

ans = 

syms u
Z = func4uo(u)

Z = 

limit(Z, u, inf)

ans = 

0

vpa(ans)

ans = 

0.0

function fval = func4uo(u) 
    d1=10;
    n=10^-11.4;
    m=2.7;
    a=0.5;
    T=1;
    PsByN_0dB=20;                                                     
    PsByN_0=10.^(PsByN_0dB/10); 
    
    fval = ((-1./u).*log((d1.^m)./(a.*n.*PsByN_0.*T.*u)+d1.^m).*a.*T)./(1-a).*T  - (1./u).*log(expint(-PsByN_0.*u)).*exp(-PsByN_0.*u);
end

[Dhawal Beohar]

Thanks Walter for the help and explanation. I might need to check equation again.

Answer 3

[Walter Roberson]

Z = @(PS) arrayfun(@(ps) fzero(@(u)func4uo(u,ps), [0.6775499178144678 1e3]), PS)

Z = function_handle with value:
    @(PS)arrayfun(@(ps)fzero(@(u)func4uo(u,ps),[0.6775499178144678,1e3]),PS)

P = linspace(-5, 1);
syms u
F = func4uo(u, P(1))

F = 

string(F)

ans = "- log(692455071077987426423013376/(275018307117627*u) + 2204244764264291/4398046511104)/u - (exp(5*u)*log(expint(5*u)))/u"

%vpasolve(F)

%{
U = Z(P);
plot(P, real(U), 'k', P, imag(U), 'r');
xlabel('Ps'); ylabel('u')
%}

function fval = func4uo(u,Ps) 
    d1=10;
    n=10^-11.4;
    m=2.7;
    a=0.5;
    T=1;
    PsByN_0dB=20;                                                     
    PsByN_0=10.^(PsByN_0dB/10); 
    
    fval = ((-1./u).*log((d1^m)./(a.*n.*PsByN_0.*T.*u)+d1.^m).*a.*T)./(1-a).*T  - (1./u).*log(expint(-Ps.*u)).*exp(-Ps.*u);
end

[Dhawal Beohar]

Hi Walter, appreciate your help. I need to find out for what value of u, both of the equations can be equal?

I am still having error 'Not enough input arguments'.

[Walter Roberson]

You need to define Ps.

I am having difficulty finding Ps values that balance.

[Dhawal Beohar]

I have now corrected the equtaion.

I need to find out for what value of u, both of the equations can be equal?

ERROR: Not enough input arguments.

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

fval (1,1) = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T  - (1./u)*log(expint(-PsByN_0*u))*exp(-PsByN_0*u);

zero = fzero(fval,0)

function fval = func4uo(u) 

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

fval (1,1) = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T  - (1./u)*log(expint(-PsByN_0*u))*exp(-PsByN_0*u);

zero = fzero(fval,0)  

[Dhawal Beohar]

have you done any changes? Sorry I am not able to find any change....

I am facing below error:

Not enough input arguments.

Error in func4uo (line 47)

fval (1,1) = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T - (1./u)*log(expint(-PsByN_0*u))*exp(-PsByN_0*u);

function fval = func4uo(u)

d1=10;

n=10^-11.4;

m=2.7;

a=0.5;

T=1;

PsByN_0dB=20;

PsByN_0=10.^(PsByN_0dB/10);

fval (1,1) = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T - (1./u)*log(expint(-PsByN_0*u))*exp(-PsByN_0*u);

zero = fzero(fval,0)

[Walter Roberson]

In your other Question I show that your revised code has no root (unless you count u = infinity)