cubic polynomial with variables

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mark wentink
mark wentink 2012-7-4
Hello everyone,
This is probably a fairly easy question to answer.
I am trying to solve a 3rd degree polynomial with 4 variables, a b c d, each functions of k1, k2, Tr, Tf (or some of those).
Is it possible to obtain the roots of the polynomial in terms of k1, k2, Tf, Tr?
Thanks for your help,
Mark

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回答（3 个）

Mark Whirdy
Mark Whirdy 2012-7-4
Hi Mark
Can you write the polynomial out in full here?
Mark
Walter Roberson
Walter Roberson 2012-7-4
syms x k1 k2 Tr Tf
a = SomeFunction(k1,k2,Tr,Tf);
b = SomeOtherFunction(k1,k2,Tr,Tf);
c = AThirdFunction(k1,k2,Tr,Tf);
d = FourthFunction(k1,k2,Tr,Tf);
CubeRoots = solve( a*x^3 + b*x^2 + c*x + d, x);
You might also want to simplify() each of the roots.
I will caution you that the results you get out are likely to be messy to read: the generalized solution for a cubic is not fun to make mental sense of. (Quadratic can be a lot worse though!)
mark wentink
mark wentink 2012-7-4
I tried the solve function, followed by simplify, which gives me a root of about 10 lines, is there another way of getting it simplified?
The actual polynomial is:
(k1/2) * x^3 + (k1*(k2+Tf-Tr)-2*(k2 ^2))/2 * x^2 + k2*Tr*((4*k2-k1)/2) * x - k2^2*Tr^2 = 0
Also, I imagine it would help if I had an actual value for k1. I've got an average with a standard deviation. Is there any way of incorporating that error? Or should I just use the average?
  1 个评论
Walter Roberson
Walter Roberson 2012-7-5
Only about 10 lines? Then MuPAD has already done quite a good job of simplifying: the result that I find is over 4000 characters, about 30 lines each over 128 characters long.
Symbolic forms of cube roots are long, period. There isn't much in the way of simplification available. There is common sub-expression optimization, which gives you efficiencies but often isn't much more readable.

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