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
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
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?

```

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