Re: MINIMAL POLYNOMIALS FOR TANGENTS
- From: Will Heierman <williamh@xxxxxxxxxxxxxxxxx>
- Date: Mon, 12 Jan 2009 21:01:23 EST
Thanks to some valuable hints posted on this string, I have managed to prove almost everything asserted earlier regarding the minimal polynomials for tan(k*pi/n). We are assuming that the fraction is in lowest terms in this representation; i.e., that (k,n)=1. The problem is trivial if n=1 or n=2, so from now on let us assume n>2.
The assertion was that the algebraic degrees of these values are phi(n) unless n is a multiple of 4 (in which case it is [phi(n)]/2).
Polynomials T(sub n) of degree phi(n) have been constructed which have these tangents as their root sets. By restricting the angles to the interval (0, pi), we see they are uniquely represented, and clearly there are phi(n) distinct such values.
When n is a multiple of 4, the tangents can be partitioned to form two sets (according to whether k is congruent to 1 or to 3 (mod 4)), each containing half of them, and these have been shown to be the root sets of two irreducible polynomials over the integers whose product is T(sub n). We also have an inductive way to construct these factors; meaning we actually have access to the minimal polynomials in this case.
The only remaining point to be rigorously established (yes, I am convinced it's true) is that T(sub n) is irreducible if n is not a multiple of 4. It will suffice to show T(sub n) is irreducible if n is odd, because in this case irreducibility of
T(sub 2n) is an elementary consequence.
I do have some nice closed forms and recursive relationships for these T(sub n) (and the polynomials themselves for n<128), but so far none of these have produced inspirations that lead to irreducibility. I am of the opinion that it is a peculiar property of the tangent function, rather than some Galois-theoretic property of polynomials in general, that will be the key.
I am also convinced that this nagging detail may be the reason the problem may in fact be unsolved (meaning literature searches have not revealed any published solution). If you have any ideas, please let me know. You may post here or send message to my college email which is williamh(at)wcjc(dot)edu. Assuming we get it all done, I will publish the full paper on my website (with URL given in an announcement on this string).
Oh, yeah:
T(sub n)(x) = [1+x^2]^{[phi(n)]/2}* C(sub n)([1-x^2]/[1+x^2]),
where C(sub n) is the minimal polynomial for cos(2k*pi/n).
.
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