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3rd degree polynom

  1. Jul 27, 2003 #1

    ddr

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    [SOLVED] 3rd degree polynom

    if given Axx+Bx+C=0 then x=f(A,B,C)=(-B(+/-)sqrt(BB-4AC))/2A.
    Now given:Axxx+Bxx+Cx+D=0
    Solve:x=f(A,B,C,D)
    MATHCAD PRO 2K solves it but I don't know how.
     
  2. jcsd
  3. Jul 27, 2003 #2

    Hurkyl

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    The exact solution is not very pretty... do you know complex numbers?
     
  4. Jul 27, 2003 #3


    start with ax3+bx2+cx+d =0

    and then make the subsitution x=y-b/3a

    the quadratic term will cancel out, and divide through by a leaving you with

    y3+py+q=0

    where p=(3ac-b2)/3a2 and q=(2b3-9abc+27a2d)/27a3

    next, make the substitution y=z-p/3z, and multiply through by z3:

    z6+qz3-p3/27=0

    which is a quadratic in z3. solve it using the quadratic formula, which you listed above:

    z3=(-q±√(q2+4p3/27))/2


    take the three cube roots of each of those two solutions, giving you 6 solutions, only three of which should be independent. stick each solution into first y=z-p/3z and then x=y-b/3a to get three solutions for x.





    Edit: fixed some arithmetic errors. thanks ddr.
     
    Last edited: Jul 31, 2003
  5. Jul 28, 2003 #4

    ddr

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    Very nice solution.lethe you are the man.I'll check it out.I was circling around the same path all a long but I just couldn't do it.

    I wonder if this method can be generalized on whatever-degree polynom?
     
  6. Jul 28, 2003 #5
    lethe, how did you determine what to substitute? x=y-b/3a wasn't just some random substitution that just happens to make the quadratic term cancel?
     
  7. Jul 28, 2003 #6

    marcus

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    Maybe Lethe wont mind if I steal his thunder and butt into the conversation and say how I think he could have figured that part out. Suppose Lethe had never heard of doing this before and was inventing the method. Or suppose you Stephen were inventing it, and you wanted to remove the quadratic term by a substitution

    x = y - K

    what K would you use? It is not so hard to find out what K to use!

    In the new polynomial the only stuff quadratic in y will be
    the term you know about already---namely by2----and the one that comes from the cubic term
    a(y - Z)3

    and from the cubic term you are going to get

    -a3y2Z

    so there will be cancelation if b = 3aZ (which decides what Z to use)

    ---------------------
    Footnote you know Pascal's triangle and the old 121 and 1331 and 14641 and so on, so you expand

    (y - Z)3 = y3 - 3y2Z + 3yZ2 - Z3

    and that tells you the quadratic piece of a(y - Z)3 is going to be - a3y2Z...

    If Lethe or you were inventing this (but maybe the persian algebraist/astronomer/poet Omar Khayyam did it in year 1100 or somebody out of 1001 Nights, who knows, Ali Baba or Sheherezade?) getting rid of the quadratic term would be routine. the clever part would be to see that a cubic which is missing a quadratic term can be solved with the quadratic formula. my respect is reserved for the person who saw how to continue and do that.
     
  8. Jul 28, 2003 #7

    marcus

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    1515 Scipione del Ferro

    Wow! Britannica says that indeed Omar Khayyam did write extensively about solving cubic polynomials around 1100 (he also wrote the Rubaiyat) but that he did not use radicals (the square root formula and that).

    According to Britannica, the first solution of a cubic by radicals was in Bologna by Scipione del Ferro in 1515.

    He passed the method on to his pupils and for a while it was a close kept secret (they did not immediately run to CNN with their discoveries during the Renaissance) and then someone named Tartaglia told someone named Cardano and Cardano published it, in 1545.

    So history tells us that getting rid of the quadratic term in a cubic polynomical is the easy part. The hard part, which is solving
    the cubic once that has been removed, took this planet, with its peculiar species of evolved two-legged fish, until 1515.
     
    Last edited: Jul 28, 2003
  9. Jul 28, 2003 #8
    Unforunately, the only other degree polynomaial we can do this for is of degree 4. This is a consequence of the so-called Fundamental Theorem of Galois Theory. The fact that a polynomaial of degree greater than 4 is not soluable by radicals (i.e. there is no general formula) was proven by Ruffini in 1799.
     
  10. Jul 28, 2003 #9
    like marcus says, that is the easy part. b/3a you can guess by analogy with the quadratic equation: to complete the square of the quadratic equation (and thus solve), you substitute x=y-b/2a. when you move to the cubic equation, simply replace the 2 with a 3. and by the way, this will work again when we go to the quartic equation.

    by the way, i didn t figure this formula out. i learned it in high school. but the whole business is more rewarding when you take a course in galois theory, where the most general results are proved.
     
  11. Jul 30, 2003 #10

    ddr

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    This is all right but I think that the last substitution has to be:
    y=z-p/(3az).I checked it with MATHCAD PRO 2K and that's what I got.
     
  12. Jul 30, 2003 #11
    In the last substitution, a=1, and y=z-p/(3z)=z-p/(3az).... unless you're talking about the original a.
     
  13. Jul 31, 2003 #12
    yes, you are correct, i forgot to divide out by a in the second step. it seems i also somehow lost the d term along the way. my apologies. i have updated the expressions for p and q accordingly. thanks for picking up on that. (you shouldn t make these changes in the z substitution.)
     
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