Azzera filtri
Azzera filtri

How can I solve this eqn

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CHENG WEI LI
CHENG WEI LI il 23 Ott 2022
Risposto: Walter Roberson il 23 Ott 2022
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;
  1 Commento
MisterQuax
MisterQuax il 23 Ott 2022
Hello, do you want to determine the coeffients x1,x2,x3 or do you have them and want to determine the values f1,f2,f3?

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Risposte (1)

Walter Roberson
Walter Roberson il 23 Ott 2022
Assuming that you want the third to be F(3) then there is no solution over reals.
With the symbolic toolbox you can solve() the first two equations for x(2) x(3). substitute the solutions into the third equation to get a single equation in x(1)
This equation will have a non-zero imaginary part over negative x1, so there is no solution over negative x1
The limit of the equation at 0 is 0, but if you substitute 0 into the original equations you will get a vector of negative constants, not a solution. The limit is not really usable because getting to this equation involves division by 0 for x1=0
If you plot equation over x1 you will see that it is negative until about 1.8e5 at which point it goes to negative infinity, not having crossed 0. It picks up again from positive infinity at roughly 2.03e5 and has an asymptope of 0 at infinity, without crossing 0. In-between the negative infinity and positive infinity the value of the function is complex.
So, no negative real solutions, no solution at zero (division by 0), no zero crossing before two infinity, no zero crossing after the infinities, and complex in-between the infinities. No solution over reals.
I did not investigate solutions over complex numbers.

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