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How approximate a sextic polynomial to a lower degree polynomial

  1. Feb 20, 2013 #1
    Hi all,

    I have been stopped by a sextic (6th degree) polynomial in my research. I need to find the biggest positive root for this polynomial symbolically, and since its impassible in general, I came up with this idea, maybe there is a way to approximate this polynomial by a lower degree polynomial which is solvable.


    κ2/112 (A2 ) u62/16 (A2 ) u52/20 (1/2 B2+3 A2 ) u42/8 (A2+B2 ) u3-((ω2-B2 κ2)/6) u22 κ2 ω2=0

    this polynomial is come from a nonlinear PDE related to waves.
    κ, A, B, v, ω , u are not constant.

    I appreciate any helpful comment or solution.
    Thanks,
    Romik
     
  2. jcsd
  3. Feb 20, 2013 #2

    mfb

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    Staff: Mentor

    Divided by u6, this is a polynomial of 6th order in (1/u), where you look for the smallest positive root. Depending on the parameters, a taylor expansion or something similar might give some reasonable analytic approximation.
     
  4. Feb 20, 2013 #3
    Thanks for the reply,
    biggest or smallest positive root, that's not the main issue here, I need to find an approximate root based on variables, reduce from 6th degree to lets say 4th degree which I can solve it exactly.
     
  5. Feb 20, 2013 #4

    mfb

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    Well, the biggest root in your original equation would be the smallest root in my modified equation.
    On second thought, my idea with a taylor approximation around the origin would simply neglect the absolute term. The remaining polynomial can be expressed as u^2 P(u) where P has order 4, so there are analytic solutions. It might be interesting to improve this approximation with one or two steps of Newton afterwards ;).
     
  6. Feb 20, 2013 #5
    I don't know if this would help, but... you could try the substitutions[tex]\begin{align*}
    x &= A^2 u^2 \\
    y &= B^2 u^2 \\
    z &= \omega^2 v^2 \\
    s &= \omega^2 u^2
    \end{align*}[/tex]to obtain the possibly simpler equation[tex]
    \frac {\kappa^2 x} {112} u^4 + \frac {\kappa^2 x} {16} u^3 + \left( \frac {\kappa^2 (6x+y)} {40} \right)u^2
    + \frac{\kappa^2 (x+y)} 8 u - \frac{s-\kappa^2 y} 6 + \kappa^2 z = 0
    [/tex]
    If you somehow manage to obtain values for [itex]u,x,y,z,s[/itex], then [itex]\omega = \pm\sqrt{\displaystyle\frac s {u^2}}[/itex], and the values for [itex]A,B,v[/itex] can be solved for similarly.

    (I was trying to put also [itex]\kappa[/itex] into the substitutions for [itex]x,y,z[/itex], but then I can't find the original variables back. Unless you have an extra constraint on them.)
     
  7. Feb 21, 2013 #6
    thanks mfb for you helpful comments.
    can you explain more about Newton method, how could I apply it on my equation?
    I use Mathematica! with Series function, I am able to truncate my original polynomial to 4th degree, now how should I apply Newton since I don't have numerical root and my roots are symbolical?

    thank you Dodo for your reply, did you know you put your 666th post on this thread? So good luck to me :D
     
  8. Feb 21, 2013 #7

    mfb

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    Great, that has analytic solutions, and you don't need Newton.

    Let x be the approximate position of the root, f(x) be the function value there and f'(x) its derivative. Both f(x) and f'(x) are easy to express symbolically. A (hopefully) better approximation for the root is then given by x-f(x)/f'(x).
     
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