A Question about Sums of Tangents
whei...@corunduminium.com
Can one determine whether the sum of tangents of angles of this form between 0 and pi, say where k is congruent to 1 (mod 4), is positive or negative? Supposing n = 4j, does it depend on whether an odd j is congruent to 1 or to 3 (mod 4); or whether j is odd or even?
These polynomials are monic factors of a polynomial of the form Q(x) = P(x^2) where P(x) is irreducible. Since the second term of any polynomial is minus the sum of its roots, knowing this will do the job.
We have an elementary algorithm for computing the factors, and a numerical check of one tangent will determine the correspondence; but I wonder if there is a general theoretical result that will make this unnecessary.
As an exercise, it would be fun (but tedious) to compute the minimal plynomial for the tangent of one degree, which is tan(pi/180). It would have algebraic degree 24. I hear there may be another string where this has been done.
Robert Israel
tan(x)^2 + 1 = 1/cos(x)^2. If x = k*pi/n with (k,n)=1 and n is even,
T_[n/2](cos(x)) = cos(k pi/2) = 0, where T_[n/2] is a Chebyshev polynomial.
Thus the minimal polynomial for cos(x) is a factor of T_[n/2]. If
n is divisible by 4, T_[n/2] is an even function, and I suspect the
irreducible factors of T_[n/2] are even as well. If P(c^2) is the
minimal polynomial for c = cos(x), where P has degree d, then
some factor of Q(t) = (1+t^2)^d P(1/(1 +t^2)) would be the minimal polynomial
for t = tan(x). For example, the minimal polynomial for c = cos(pi/180) is
P(c^2) =
281474976710656*c^48-3377699720527872*c^46+18999560927969280*c^44-66568831992070144*c^42+162828875980603392*c^40-295364007592722432*c^38+411985976135516160*c^36-452180272956309504*c^34+396366279591591936*c^32-280058255978266624*c^30+160303703377575936*c^28-74448984852135936*c^26+28011510450094080*c^24-8500299631165440*c^22+2064791072931840*c^20-397107008634880*c^18+59570604933120*c^16-6832518856704*c^14+583456329728*c^12-35782471680*c^10+1497954816*c^8-39625728*c^6+579456*c^4-3456*c^2+1
Q(t) = (1+t^2)^24 P(1/(1+t^2)) =
(t^24-48*t^23-564*t^22+1456*t^21+21186*t^20-12432*t^19-269412*t^18+17424*t^17+1470447*t^16+45344*t^15-3923304*t^14-51744*t^13+5407388*t^12-51744*t^11-3923304*t^10+45344*t^9+1470447*t^8+17424*t^7-269412*t^6-12432*t^5+21186*t^4+1456*t^3-564*t^2-48*t+1)*(t^24+48*t^23-564*t^22-1456*t^21+21186*t^20+12432*t^19-269412*t^18-17424*t^17+1470447*t^16-45344*t^15-3923304*t^14+51744*t^13+5407388*t^12+51744*t^11-3923304*t^10-45344*t^9+1470447*t^8-17424*t^7-269412*t^6+12432*t^5+21186*t^4-1456*t^3-564*t^2+48*t+1)
and the first of these is the minimal polynomial of tan(pi/180). The
roots of this minimal polynomial are tan(k pi/180) for
k = -83, -79, -71, -67, -59, -47, -43, -31, -23, -19, -11, -7, 1, 13, 17, 29,
37, 41, 49, 53, 61, 73, 77, 89
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
whei...@corunduminium.com
Also, how were the factors found - by a brute force computer program (we knew the factors existed, but what were they)? What we have is an algorithm where "guessing" is not required, in the sense that the two factors can be obtained inductively by elementary (precalculus) polynomial division (in a manner analogous to that used to generate the cyclotomic polynomials from (x^n)-1). What remained was how we know which factor has which class (1 or 3 (mod 4)) as its root set.
It seems you have established the sum of the tangents of "class 1" is 48 which happens to be twice the degree of the polynomial (and in particular positive). I wonder if there is a way of establishing this from elementary identities or other poperties of tangents. Perhaps, they form twelve pairs in a useful manner. Wow!
Robert Israel
> Robert, you certainly saved me a lot of work computing this polynomial!
> How did you knew it is the first factor and not the second that is the
> minimal polynomial for the case 1 degree? Was it an arithmetical
> verification?
Evaluate and simplify, using Maple.
> Also, how were the factors found - by a brute force computer program (we
> knew the factors existed, but what were they)? What we have is an
> algorithm where "guessing" is not required, in the sense that the two
> factors can be obtained inductively by elementary (precalculus) polynomial
> division (in a manner analogous to that used to generate the cyclotomic
> polynomials from (x^n)-1). What remained was how we know which factor has
> which class (1 or 3 (mod 4)) as its root set.
They were found using Maple's factor command.