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How approximate a sextic polynomial to a lower degree polynomial 
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#1
Feb2013, 01:49 PM

P: 14

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. κ^{2}/112 (A^{2} ) u^{6}+κ^{2}/16 (A^{2} ) u^{5}+κ^{2}/20 (1/2 B^{2}+3 A^{2} ) u^{4}+κ^{2}/8 (A^{2}+B^{2} ) u^{3}((ω^{2}B^{2} κ^{2})/6) u^{2}+ν^{2} κ^{2} ω^{2}=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
Feb2013, 04:20 PM

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P: 11,589

Divided by u^{6}, 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.



#3
Feb2013, 05:58 PM

P: 14

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. 


#4
Feb2013, 06:18 PM

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P: 11,589

How approximate a sextic polynomial to a lower degree polynomial
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 ;). 


#5
Feb2013, 06:28 PM

P: 688

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.) 


#6
Feb2113, 01:21 PM

P: 14

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 


#7
Feb2113, 01:32 PM

Mentor
P: 11,589




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