Symbolic Computations for three non-linear equations solution
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Dear community members,
I am trying to find three unknown values A,fu and alpha using three equations:
fl = [1.883806, 1.883135, 1.882170]*10^9;
Qe = [161.43, 10.765, 2.5734]*10^4;
syms A fu alpha
eq1=fl(1)-(1+(A/(Qe(1)^alpha)))*fu==0;
eq2=fl(2)-(1+(A/(Qe(2)^alpha)))*fu==0;
eq3=fl(3)-(1+(A/(Qe(3)^alpha)))*fu==0;
roots = vpasolve(eq1,eq2,eq3,A,fu, alpha)
A = double(roots.A)
alpha = double(roots.alpha)
fu = double(roots.fu)
The results I get:
A: 23.111701126115183488493226207409
fu: 1883037000.0
alpha: 9.7722388976929995256469179038887
Are different from the expected ones:
A = -0.1555;
fu=1.88406316*10^9;
alpha = 0.4967;
Could you please let me know how can I adjust the results?
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Torsten
2023 年 3 月 17 日
編集済み: Torsten
2023 年 3 月 17 日
fl = [1.883806, 1.883135, 1.882170]*10^9;
Qe = [161.43, 10.765, 2.5734]*10^4;
syms A fu alpha
eq1=fl(1)-(1+(A/(Qe(1)^alpha)))*fu==0;
eq2=fl(2)-(1+(A/(Qe(2)^alpha)))*fu==0;
eq3=fl(3)-(1+(A/(Qe(3)^alpha)))*fu==0;
roots = vpasolve([eq1,eq2,eq3],[A,fu, alpha],[-0.1555;1.88406316*10^9;0.4967] )
Anum = double(roots.A)
alphanum = double(roots.alpha)
funum = double(roots.fu)
double(subs([lhs(eq1),lhs(eq2),lhs(eq3)],[A,fu,alpha],[Anum,funum,alphanum]))
double(subs([lhs(eq1),lhs(eq2),lhs(eq3)],[A,fu,alpha],[-0.1555,1.88406316*10^9,0.4967]))
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