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Topic: product of tangents problem
Replies: 12   Last Post: Aug 10, 2011 2:43 AM

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Jim Ferry

Posts: 484
Registered: 12/10/04
Re: product of tangents problem
Posted: Sep 24, 2009 11:35 AM
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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 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


More generally, for k and n relatively prime, we have

prod(j=1,n-1,tan(j*(k/n)*Pi/2)) = ...

(-1)^((k-1)/2) for even n,
1 for odd n and odd k, and
n (-1)^((n-1)/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,n-1,x-tan(j*Pi/n)) = Re[(x+I)^n - I^n]/x for real x.

This identity holds because the n-1 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 n-1.



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