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How approximate a sextic polynomial to a lower degree polynomial

by Romik
Tags: approximate, degree, polynomial, sextic
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Romik
#1
Feb20-13, 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.


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
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mfb
#2
Feb20-13, 04:20 PM
Mentor
P: 11,580
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.
Romik
#3
Feb20-13, 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.

mfb
#4
Feb20-13, 06:18 PM
Mentor
P: 11,580
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 ;).
dodo
#5
Feb20-13, 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.)
Romik
#6
Feb21-13, 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
mfb
#7
Feb21-13, 01:32 PM
Mentor
P: 11,580
I am able to truncate my original polynomial to 4th degree
Great, that has analytic solutions, and you don't need Newton.

now how should I apply Newton since I don't have numerical root and my roots are symbolical?
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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