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=6852009

Jim Ferry proved that if k, n are positive integers

with 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 Diophantine

approximation language.

In other words, (pi/2)/(n/k) is nearly 1,

or n/k is a "very good" Diophantine

approximation 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 the

product (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 good

Diophantine approximations to pi/2 (viz. resp. pi w.r.t. Wapner's

product of sines) with no conditions on the parities of

n and k, other than the "obvious" one that n and k not both

be 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)| ),

and

B = 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 on

integer values in 1, 2, ... n-1 just like in the product

and 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 to

provably allow infinintely many co-prime n, k to

approximate diophantinely pi/2 with the parity

constraint 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) ) ] means

the interval [a, b] on the real line, where

a = min(j, pi*j*k/(2n) )

and

b = max(j, pi*j*k/(2n) ).

We don't know that log( |tan(x)|) is differentiable and

continuous on [a, b], but we'll assume it for now to

see if there's a remote chance of progress...

So, we're assuming that we can apply the mean value theorem

of 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 to

the real-valued function

x |-> 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.