why I get imaginary part using solve function
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Sarah Alhabbas
am 14 Jun. 2022
Kommentiert: Walter Roberson
am 14 Jun. 2022
I am trying to use the solve function but somehow I keep getting more than one answer with imaginary parts and negative numbers

the correct answer should be the second answer = 0.85
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Torsten
am 14 Jun. 2022
Bearbeitet: Torsten
am 14 Jun. 2022
If you multiply eq4 by (1+y*m4^2)^2, you get a polynomial equation of degree 4 in m4. This equation has 4 zeros (which are listed in the output of vpasolve). Two of them are purely imaginary, two of them are real. One of the solution is the one you want (the second one).
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Walter Roberson
am 14 Jun. 2022
You have an expression of the form f(x^4)/g(x^2) + b = 0
Multiply through by g(x^2) (assuming nonzero) to get
f(x^4) + b*g(x^2) = 0
collect x terms to get a polynomial in x^4.
Solve the degree 4 polynomial, getting four solutions.
Therefore "the answer" is all four solutions, not just a single solution.
If you have constraints on the outputs, such as being real valued, then filter the results.
3 Kommentare
Torsten
am 14 Jun. 2022
y = 1.4;
to3 = 300;
t_star = 400;
syms m4
eq4 = (((2*(y+1)*m4^2*(1+(y-1)/2)*m4^2))/(1+y*m4^2)^2) - to3/t_star;
m4 = vpasolve(eq4,m4);
m4 = m4(abs(imag(m4)) < eps & real(m4) > 0)
Walter Roberson
am 14 Jun. 2022
y = 1.4;
to3 = 300;
t_star = 400;
syms m4 positive
eq4 = (((2*(y+1)*m4^2*(1+(y-1)/2)*m4^2))/(1+y*m4^2)^2) - to3/t_star;
m4 = solve(eq4,m4);
m4
vpa(m4)
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David Hill
am 14 Jun. 2022
y=1.4;
to3=300;
t_star=400;
eq4=@(m4)(((2*(y+1)*m4^2*(1+(y-1)/2)*m4^2))/(1+y*m4^2)^2)-to3/t_star;
m_4=fzero(eq4,.8)
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