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Topic:
more on tan(1) tan(2) tan(3) ... tan(m) m = 1 ... oo series
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more on tan(1) tan(2) tan(3) ... tan(m) m = 1 ... oo series
Posted:
Apr 3, 2013 10:15 PM


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 ... n1} 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 ... n1} log( tan(j) ), and
B = sum_{j=1 ... n1} 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, ... n1 just like in the product and sum expressions.
Then, A  B = sum_{j=1 ... n1} s_j .
Reminder: Variation = AB 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 coprime 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 <= n1 to the realvalued 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.



