product of tangents problem
in [Math]
|
From: David Bernier on 21 Sep 2009 03:03 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 PARI-gp ). David Bernier
From: Henry on 21 Sep 2009 08:33 On 21 Sep, 08:03, David Bernier <david...(a)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. But similarly, if n is close to a multiple of pi (i.e. an even multiple of pi/2) then tan(n) will be small in absolute value. > One of the convergents in the continued fraction expansion > of pi/2 is 52174/33215 . > > I find a(52174) ~= 476029.121 . It might also be worth looking at a(26251), a(78425) and a(130599), which are slightly (164) more than 0.5, 1.5 and 2.5 times 52174.
From: Jim Ferry on 21 Sep 2009 11:03 On Sep 21, 3:03 am, David Bernier <david...(a)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 PARI-gp ). > > 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,n-1,tan(X*(k/n)*Pi/2)) = prod(X=1,n/2-1,tan(X*(k/n)*Pi/2)) * tan((n/2)*(k/n)*Pi/2) * prod(X=1,n/ 2-1,tan((n-X)*(k/n)*Pi/2)). Because k is odd we have tan((n/2)*(k/n)*Pi/2) = tan(k*Pi/4) = (-1)^ ((k-1)/2), and tan((n-X)*(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/2-1 so prod(X=1,n-1,tan(X*(k/n)*Pi/2)) = (-1)^((k-1)/2).
From: jillbones on 21 Sep 2009 17:17 On Sep 21, 12:03 am, David Bernier <david...(a)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 PARI-gp ). > > David Bernier If n is measured in degrees, then TAN(90) = INFINITY The product increases from 1.745506492821759E-02 at n=1 degree to 3.070591331174723E-22 at n= 45 degrees; then increases to 1 at n=89 degrees. At n = 90, TAN(90) = oo and for all n > 89 the product is infinity. If n is measured in radians, then n is always an even multiple of pi/2. regards, Bill J
From: David Bernier on 24 Sep 2009 04:11
Jim Ferry wrote: > On Sep 21, 3:03 am, David Bernier <david...(a)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 PARI-gp ). >> >> 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,n-1,tan(X*(k/n)*Pi/2)) = > > prod(X=1,n/2-1,tan(X*(k/n)*Pi/2)) * tan((n/2)*(k/n)*Pi/2) * prod(X=1,n/ > 2-1,tan((n-X)*(k/n)*Pi/2)). > > Because k is odd we have tan((n/2)*(k/n)*Pi/2) = tan(k*Pi/4) = (-1)^ > ((k-1)/2), and > > tan((n-X)*(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/2-1 so > > prod(X=1,n-1,tan(X*(k/n)*Pi/2)) = (-1)^((k-1)/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, n-1 = 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((n-X)*(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,n-1,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 |