```Date: Apr 3, 2013 10:15 PM
Author: David Bernier
Subject: more on |tan(1) tan(2) tan(3) ... tan(m)|   m = 1 ... oo  series

In:http://mathforum.org/kb/message.jspa?messageID=6852009Jim Ferry proved that if  k, n are positive integerswith gcd(k,n)=1, and n is odd while k is even, then:product_{j=1 ... n-1} |tan(pi*j*k/(2n))|  = n.The working heuristic is that:pi*k/(2n) is unusually close to 1, in Diophantineapproximation language.In other words, (pi/2)/(n/k) is nearly 1,or  n/k  is a "very good" Diophantineapproximation to pi/2.Illustration:  n = 355, k = 226 ;    then  |pi/2 - 355/226| ~= 0.006812 * 226^(-2) .Unlike with the Leonard Wapner's problem on theproduct (2sin 1)(2sin 2) ... (2sin m) from 1/1/2007,<  http://mathforum.org/kb/message.jspa?messageID=5464104 > ,we don't have the luxury of choosing very goodDiophantine approximations to pi/2   (viz. resp. pi w.r.t. Wapner'sproduct of sines) with no conditions on the parities ofn and k,  other than the "obvious" one that n and k not bothbe even ...The period of x |-> |tan(x)| is pi/2.If n and k are thought of as fixed parameters,we can write (in shorthnad)A = sum_{j=1 ... n-1} log( |tan(j)| ),andB = sum_{j=1 ... n-1} log( |tan(pi*j*k/(2n))| ) .Variation = A -B .We know that B = log(n) from Jim Ferry's proof.Let   s_j =  log( |tan(j)| ) - log( |tan(pi*j*k/(2n))| ) .(It's best to think of k and n as fixed and defined forever  at the beginning somewhere ...).  The j  takes oninteger values in 1, 2, ... n-1   just like in the productand sum expressions.Then,A - B  = sum_{j=1 ... n-1} s_j  .Reminder:  Variation = A-B is the offset from B going to A.j ~= pi*j*k/(2n)   because  |n/k - pi/2| < C/k^2 ,with C>0 a real number just big enough toprovably allow infinintely many co-prime n, k toapproximate diophantinely pi/2 with the parityconstraint forced on us, namely n odd and even.Finding C is for another day ...[ min(j, pi*j*k/(2n) ), max(j, pi*j*k/(2n) ) ]  meansthe interval [a, b] on the real line, wherea = min(j, pi*j*k/(2n) )andb = max(j, pi*j*k/(2n) ).We don't know that log( |tan(x)|) is differentiable andcontinuous on [a, b], but we'll assume it for now tosee if there's a remote chance of progress...So, we're assuming that we can apply the mean value theoremof calculus on [a, b]with[a, b] =  [ min(j, pi*j*k/(2n) ), max(j, pi*j*k/(2n) ) ]for all integer j such that 1<= j <= n-1  tothe real-valued functionx |->  log( |tan(x)| ).d/dx  log( |tan(x)| ) ={ - sec^2(x)/tan(x)   for  0< x< pi/4{  sec^2(x)/tan(x)    for  pi/4 < x < pi/2 .To be continued ...dave-- Jesus is an Anarchist.  -- J.R.
```