How can I solve this?

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CHENG WEI LI
CHENG WEI LI el 23 de Oct. de 2022
Respondida: Torsten el 24 de Oct. de 2022
clc
clear
syms PL c k
eqn = [PL * exp(-c*exp(-k*0)) == 179323 ; PL * exp(-c*exp(-k*10)) == 203302 ; PL * exp(-c*exp(-k*20)) == 226542]
S = solve(eqn , [PL;c;k] )
S =
struct with fields:
PL: [0×1 sym]
c: [0×1 sym]
k: [0×1 sym]
  2 comentarios
Muhammad Usman
Muhammad Usman el 23 de Oct. de 2022
the set of equations are non linear in nature that's why you can't use solve to compute the solution
Walter Roberson
Walter Roberson el 23 de Oct. de 2022
The equations have no solution over reals.

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Respuestas (2)

Muhammad Usman
Muhammad Usman el 23 de Oct. de 2022
% Solve the system of equations starting at the point [0,0,0].
% PL = x(1); c = x(2); k = x(3);
% Initial guess is [0,0,0], you can change it accordingily
fun = @root2d;
x0 = [0,0,0];
x = fsolve(fun,x0)
Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt algorithm instead.
No solution found. fsolve stopped because the relative size of the current step is less than the value of the step size tolerance, but the vector of function values is not near zero as measured by the value of the function tolerance.
x = 1×3
1.0e+05 * 2.0192 -0.0000 0
function F = root2d(x)
F(1) = x(1) * exp(-x(2)*exp(-x(3)*0)) - 179323;
F(2) = x(1) * exp(-x(2)*exp(-x(3)*10)) - 203302;
F(2) = x(1) * exp(-x(2)*exp(-x(3)*20)) - 226542;
end
  2 comentarios
CHENG WEI LI
CHENG WEI LI el 23 de Oct. de 2022
The function has correct answer.
why this answer is wrong
Alex Sha
Alex Sha el 24 de Oct. de 2022
There are actually numerical solutions like below:
x1: 446505.431672107
x2: 0.912262916225993
x3: 0.0148006249649759

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Torsten
Torsten el 24 de Oct. de 2022
syms PL c k
eqn1 = PL * exp(-c*exp(-k*0)) == 179323;
eqn2 = PL * exp(-c*exp(-k*10)) == 203302;
eqn3 = PL * exp(-c*exp(-k*20)) == 226542;
sPL = solve([eqn1 eqn2],[c,k]);
Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.
PLsol = solve(subs(eqn3,[c,k],[sPL.c sPL.k]),PL);
sck = solve([subs(eqn1,PL,PLsol),subs(eqn2,PL,PLsol)],[c,k]);
csol = sck.c;
ksol = sck.k;
vpa(PLsol)
ans = 
446505.43167210849515476976968181
vpa(csol)
ans = 
0.91226291622599614437018856917074
vpa(ksol)
ans = 
0.014800624964975891465693800088887

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