
Re: product of tangents problem
Posted:
Sep 24, 2009 11:35 AM


On Sep 24, 4:11 am, David Bernier <david...@videotron.ca> wrote: > Jim Ferry wrote: > > On Sep 21, 3:03 am, David Bernier <david...@videotron.ca> wrote: > >> Leonard Wapner asked about the behaviour of > >> a(n) = (tan 1)(tan 2) ... (tan n) > >> in 2006: > > >> <http://groups.google.com/group/sci.math/msg/fa3878fa776a992b> . > > >> If n is very close to an odd multiple of pi/2, the > >> tangent of n will be large in absolute value. > > >> One of the convergents in the continued fraction expansion > >> of pi/2 is 52174/33215 . > > >> I find a(52174) ~= 476029.121 . > > >> On the other hand, if we exclude the tan(52174) term, > >> we get a(52173) ~= 2.6217 > > >> while tan(52174) ~= 181570.2957 . > > >> Since pi/2 is very close to 52174/33215 , > >> (33215/52174)*pi/2 is very close to 1. > > >> It seems that if x = (33215/52174)*pi/2 , then > >> (tan x)(tan 2x)... (tan 52173x) might be 1 : > > >> ? prod(X=1,52173,tan(X*(33215.0/52174.0)*Pi/2)) > >> %37 = > >> 1.00000000000000000000000000000000000000000000000000000000000000000000000 > >> (using PARIgp ). > > >> David Bernier > > > Yes, it's 1: in general, for any integers n,k > 0 with n even and k > > relatively prime to n, we have > > > prod(X=1,n1,tan(X*(k/n)*Pi/2)) = > > > prod(X=1,n/21,tan(X*(k/n)*Pi/2)) * tan((n/2)*(k/n)*Pi/2) * prod(X=1,n/ > > 21,tan((nX)*(k/n)*Pi/2)). > > > Because k is odd we have tan((n/2)*(k/n)*Pi/2) = tan(k*Pi/4) = (1)^ > > ((k1)/2), and > > > tan((nX)*(k/n)*Pi/2) = tan(k*Pi/2  X*(k/n)*Pi/2) = cot(X*(k/n)*Pi/ > > 2). > > > Because k and n are relatively prime, each of the tangent and > > cotangent factors is finite, and the > > product of each pair is 1 for X = 1 to n/21 so > > > prod(X=1,n1,tan(X*(k/n)*Pi/2)) = (1)^((k1)/2). > > Thanks. So 'tans' get transformed to 'cotans', with > cancellation among those ... > > Perhaps there is also a result when n and k are odd, coprime, > positive: > > ? prod(X=1, 260514, tan(X*((165849/260515)*Pi/2))) > %17 = > 1.000000000000000000000000000000000000000000000000000000000000000000000000000000000\\ > 000000000000000000000000000000000000000000000000000000000000000000000000000000000000\\ > 0000000000000000000000000000000000 > > Here, k = 165849, n = 260515, n1 = 260514 so, > following your method, > > prod(X=1, 260514 ,tan(X*(165849/260515)*Pi/2)) = > prod(X=1, 130257, tan(X*(165849/260515)*Pi/2)) * prod(X=130258, 260514, > tan(X*(165849/260515)*Pi/2)) > > prod(X=130258, 260514, tan(X*(165849/260515)*Pi/2)) > = > prod(X=1, 130257, tan((nX)*(165849/260515)*Pi/2)) > = prod(X=1, 130257, tan(165849*Pi/2  X*(165849/260515)*Pi/2)) > = > prod(X = 1, 130257, cot( X*(165849/260515)*Pi/2 ) ). > > So, > prod(X=1, 260514 ,tan(X*(165849/260515)*Pi/2)) = > = > prod(X=1, 130257, tan(X*(165849/260515)*Pi/2)) *\ > prod(X = 1, 130257, cot( X*(165849/260515)*Pi/2 ) ) > > = 1 . > >  > > If we represent a convergent of pi/2 as > n/k (for example, n = 52174, k = 33215), gcd(n, k) = 1 assumed > below ... > > it seems to me that when k is odd, we have a simple result > > for prod(X=1,n1,tan(X*(k/n)*Pi/2)) [ maybe +/ 1 ]. > > In case k is even and n odd, k*pi/2 is very close > to an integer multiple of pi. > > Maybe we get something too, I'm not sure: > > ? prod(X=1, 354, tan(X*(226/355)*(Pi/2))) > %18 = 355.000000000000000000000000000000000000000000000\ > 000000000000000000000000000000000000000000000000000000\ > 0000000000000000000000000000000000000000000000000000000\ > 0000000000000000000000000000000000000000000 > > (For n = 355, k = 226.) > > David Bernier
More generally, for k and n relatively prime, we have
prod(j=1,n1,tan(j*(k/n)*Pi/2)) = ...
(1)^((k1)/2) for even n, 1 for odd n and odd k, and n (1)^((n1)/2) for odd n and even k.
The odd n and odd k case uses the same trick of matching tans and cots (with no factor left over, hence the result 1).
For odd n and even k, it is sufficient to prove the result for the k=2 case, as larger values of k simply permute the factors. This result is a simple corollary of
prod(j=1,n1,xtan(j*Pi/n)) = Re[(x+I)^n  I^n]/x for real x.
This identity holds because the n1 distinct values tan(j*Pi/n) all make (x+I)^n purely imaginary (as does 0, which is factored out), which follows from n being odd and cis(j*Pi/n)^n being real for each j = 1 to n1.

