Azzera filtri
Azzera filtri

How can I solve this?

2 visualizzazioni (ultimi 30 giorni)
CHENG WEI LI
CHENG WEI LI il 23 Ott 2022
Risposto: Torsten il 24 Ott 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 Commenti
Muhammad Usman
Muhammad Usman il 23 Ott 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 il 23 Ott 2022
The equations have no solution over reals.

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Risposte (2)

Muhammad Usman
Muhammad Usman il 23 Ott 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 Commenti
CHENG WEI LI
CHENG WEI LI il 23 Ott 2022
The function has correct answer.
why this answer is wrong
Alex Sha
Alex Sha il 24 Ott 2022
There are actually numerical solutions like below:
x1: 446505.431672107
x2: 0.912262916225993
x3: 0.0148006249649759

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Torsten
Torsten il 24 Ott 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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