How approximate a sextic polynomial to a lower degree polynomial

In summary: Applying this iteration should converge to the root you want.In summary, the conversation discusses finding the biggest positive root for a 6th degree polynomial symbolically, and the possibility of approximating it with a lower degree polynomial. The polynomial is derived from a nonlinear PDE related to waves, and there are suggestions to use substitutions or a Taylor expansion to simplify the equation. The use of Newton's method is also mentioned as a way to find an approximate root.
  • #1
Romik
14
0
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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  • #2
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.
 
  • #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 let's say 4th degree which I can solve it exactly.
 
  • #4
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
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
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
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).
 

What is the process of approximating a sextic polynomial to a lower degree polynomial?

The process of approximating a sextic polynomial to a lower degree polynomial involves using a mathematical technique known as polynomial regression. This involves finding the best fit curve that represents the data points of the sextic polynomial, and then using that curve to create a lower degree polynomial that closely approximates the original function.

What are the benefits of approximating a sextic polynomial to a lower degree polynomial?

One of the main benefits of approximating a sextic polynomial to a lower degree polynomial is that it simplifies the original function, making it easier to understand and work with. It also allows for faster computation and can help to reduce errors in calculations.

What are the limitations of approximating a sextic polynomial to a lower degree polynomial?

While approximating a sextic polynomial to a lower degree polynomial can be useful in simplifying and representing complex functions, it is important to note that it is not always accurate. The resulting lower degree polynomial may not perfectly fit the original function, and there may be some error in the approximation.

What factors should be considered when choosing the degree of the approximating polynomial?

When choosing the degree of the approximating polynomial, it is important to consider the complexity of the original sextic polynomial, the amount of data points available, and the desired level of accuracy. Generally, a higher degree polynomial will result in a closer approximation, but it may also be more complex and require more data points.

How can I check the accuracy of the approximating polynomial?

One way to check the accuracy of the approximating polynomial is by plotting the original sextic polynomial and the resulting lower degree polynomial on a graph and visually comparing the two. Another method is to calculate the root mean square error between the two functions, which measures the overall difference between the two curves.

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