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

2 views (last 30 days)
Dhawal Beohar on 16 Feb 2022
Commented: Walter Roberson on 17 Feb 2022
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.
Dhawal Beohar on 17 Feb 2022
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)

Matt J on 16 Feb 2022
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

Walter Roberson on 17 Feb 2022
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 on 17 Feb 2022
Thanks Walter for the help and explanation. I might need to check equation again.

Walter Roberson on 17 Feb 2022
Edited: Walter Roberson on 17 Feb 2022
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
Walter Roberson on 17 Feb 2022
In your other Question I show that your revised code has no root (unless you count u = infinity)

R2020a

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