How approximate a sextic polynomial to a lower degree polynomial

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SUMMARY

This discussion focuses on approximating a sextic polynomial derived from a nonlinear PDE related to waves, specifically the equation κ²/112 (A²) u⁶ + κ²/16 (A²) u⁵ + κ²/20 (1/2 B² + 3 A²) u⁴ + κ²/8 (A² + B²) u³ - ((ω² - B² κ²)/6) u² + ν² κ² ω² = 0. The main goal is to reduce the polynomial's degree from 6 to 4 for easier solvability. Participants suggest using Taylor expansions and Newton's method for root approximation, emphasizing that symbolic roots can be derived using Mathematica's Series function.

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Romik
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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 ) u6+κ2/16 (A2 ) u5+κ2/20 (1/2 B2+3 A2 ) u4+κ2/8 (A2+B2 ) u3-((ω2-B2 κ2)/6) u2+ν2 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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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.
 
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 let's say 4th degree which I can solve it exactly.
 
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 ;).
 
I don't know if this would help, but... you could try the substitutions\begin{align*}<br /> x &amp;= A^2 u^2 \\<br /> y &amp;= B^2 u^2 \\<br /> z &amp;= \omega^2 v^2 \\<br /> s &amp;= \omega^2 u^2<br /> \end{align*}to obtain the possibly simpler equation<br /> \frac {\kappa^2 x} {112} u^4 + \frac {\kappa^2 x} {16} u^3 + \left( \frac {\kappa^2 (6x+y)} {40} \right)u^2<br /> + \frac{\kappa^2 (x+y)} 8 u - \frac{s-\kappa^2 y} 6 + \kappa^2 z = 0<br />
If you somehow manage to obtain values for u,x,y,z,s, then \omega = \pm\sqrt{\displaystyle\frac s {u^2}}, and the values for A,B,v can be solved for similarly.

(I was trying to put also \kappa into the substitutions for x,y,z, but then I can't find the original variables back. Unless you have an extra constraint on them.)
 
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
 
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).