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        riddles >> medium >> Convergence of product of tangents
        
(Message started by: Aryabhatta on Jul 21^st, 2009, 12:05pm)

Title: Convergence of product of tangents
Post by Aryabhatta on Jul 21^st, 2009, 12:05pm

For a real x, define

p₀ = |tan(x)|
p_n = p_n-1 |tan(2ⁿx)|^{1/2^n}

Assuming that x is such that the sequence is well defined, is it possible that the statement

Limit p_n as n -> infinity = (2sin(x))² is false?

(The statement is considered false if p_n does not converge or it converges to a limit other than (2sin(x))²)

(Sorry for putting a "mathy" problem in medium)

Title: Re: Convergence of product of tangents
Post by Aryabhatta on Jul 22^nd, 2009, 6:16pm

Boring? Perhaps we can move it to Putnam then...

Title: Re: Convergence of product of tangents
Post by Eigenray on Jul 22^nd, 2009, 6:34pm

I think we can just take x = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2^1+f(k) for f sufficiently fast-growing: letting n=f(k),
2ⁿx http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif mod http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif,
with http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2^f(k+1)-f(k). So we can pick f(k+1) > f(k) such that p_n > n. Of course then |tan(2ⁿ⁺¹ x)| will be very small. So we can get {p_n} to diverge, but can it converge to anything other than (2sin(x))²?

Title: Re: Convergence of product of tangents
Post by Aryabhatta on Jul 22^nd, 2009, 10:22pm

Good question. I am guessing, yes, there would be such an x.

Title: Re: Convergence of product of tangents
Post by Aryabhatta on Jul 23^rd, 2009, 6:11pm

Actually, come to think of it, I now think that if it converges, it must converge to (2sin(x))^2.

I think I have a 'sort' of proof, I will post that later.

Title: Re: Convergence of product of tangents
Post by Aryabhatta on Jul 23^rd, 2009, 8:05pm

Here is a proof:

It is enough to consider the convergence of S_n = |sin(2ⁿx)|^{1/2^n}.

(Basically p_n = (2sin(x))^2/S_n * c_n where c_n -> 1)

We show that if S_n converges to a non-zero value, then it has to converge to 1.

Consider the binary representation of y = x/pi after the decimal(binary?) point.

We can assume that the sequence '10' appears infinitely often.

Say it appears after i₁, i₂, ..., i_m, .... digits. (i.e the first two binary digits of the fractional part of 2^i_ky is 10).

For this x, we can see that

1/sqrt(2) < |sin(2^i_kx)| < 1

Thus the subsequence {S_{i_n}} converges to 1.

Title: Re: Convergence of product of tangents
Post by jpk2009 on Jul 24^th, 2009, 9:37am

A slick trick using double angle formulae would be cool if it would cause some cancellations. 2ⁿx fits the pattern. I am not sure how to compose it.

Title: Re: Convergence of product of tangents
Post by Aryabhatta on Jul 24^th, 2009, 10:14am

Write tan as sin/cos and use the fact that
sin(2ⁿx) = 2ⁿsin(x)cos(x)cos(2x)cos(4x)....cos(2^n-1x)