Re: How to find Minimal Poly of a possible algebraic number?
- To: mathgroup at smc.vnet.net
- Subject: [mg72222] Re: How to find Minimal Poly of a possible algebraic number?
- From: bobbaillie at frii.com
- Date: Fri, 15 Dec 2006 07:05:52 -0500 (EST)
- References: <eljbiu$7ht$1@smc.vnet.net>
The original x is given as
x = 19.05962891397391285670091722808301086216...
This x has 40 digits. If you obtain a possible polynomial where the
total number of digits in all its coefficients is about 40, then there
is nothing special about this polynomial, and it may not be the "true"
polynomial you are looking for.
For example, given any N-digit floating point number x, you can always
find a linear equation
A*t + B for which x is a solution, where A and B have a total of about
N digits. But this tells you nothing about x.
Watch what happens when you start with an x that really is the solution
to a quadratic:
Take x2 = N[Sqrt[2], 40].
Then Table[Recognize[x2, n, t], {n, 1, 4}] gives this list:
40114893348711941777 - 28365513113449345692*t,
-2 + t^2,
-2 + t^2,
-2 + t^2
This tells you that x2 is probably a solution to the quadratic t^2 - 2
= 0. Since this kind of thing does not hapoen with your x, then it is
probably not the root of a quadratic or a quintic
Also, if you have a possible polynomial, what happens if you calculate
x to more decimal places? Do you get the same solution polynomials?