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Topic: Convergence of product of tangents (Read 708 times) |
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Aryabhatta
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Convergence of product of tangents
« on: Jul 21st, 2009, 12:05pm » |
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For a real x, define
p0 = |tan(x)|
pn = pn-1 |tan(2nx)|1/2^n
Assuming that x is such that the sequence is well defined, is it possible that the statement
Limit pn as n -> infinity = (2sin(x))2 is false?
(The statement is considered false if pn does not converge or it converges to a limit other than (2sin(x))2)
(Sorry for putting a "mathy" problem in medium)
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| « Last Edit: Jul 21st, 2009, 12:05pm by Aryabhatta » |
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Aryabhatta
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Re: Convergence of product of tangents
« Reply #1 on: Jul 22nd, 2009, 6:16pm » |
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Boring? Perhaps we can move it to Putnam then...
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Eigenray
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Re: Convergence of product of tangents
« Reply #2 on: Jul 22nd, 2009, 6:34pm » |
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I think we can just take x =   /21+f(k) for f sufficiently fast-growing: letting n=f(k),
2nx  /2 +  mod ,
with < /2f(k+1)-f(k). So we can pick f(k+1) > f(k) such that pn > n. Of course then |tan(2n+1 x)| will be very small. So we can get {pn} to diverge, but can it converge to anything other than (2sin(x))2?
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| « Last Edit: Jul 22nd, 2009, 6:39pm by Eigenray » |
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Aryabhatta
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Re: Convergence of product of tangents
« Reply #3 on: Jul 22nd, 2009, 10:22pm » |
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Good question. I am guessing, yes, there would be such an x.
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Aryabhatta
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Re: Convergence of product of tangents
« Reply #4 on: Jul 23rd, 2009, 6:11pm » |
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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.
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| « Last Edit: Jul 23rd, 2009, 8:05pm by Aryabhatta » |
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Aryabhatta
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Re: Convergence of product of tangents
« Reply #5 on: Jul 23rd, 2009, 8:05pm » |
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Here is a proof:
It is enough to consider the convergence of Sn = |sin(2nx)|1/2^n.
(Basically pn = (2sin(x))^2/Sn * cn where cn -> 1)
We show that if Sn 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, ..., im, .... digits. (i.e the first two binary digits of the fractional part of 2i_ky is 10).
For this x, we can see that
1/sqrt(2) < |sin(2i_kx)| < 1
Thus the subsequence {Si_n} converges to 1.
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jpk2009
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Re: Convergence of product of tangents
« Reply #6 on: Jul 24th, 2009, 9:37am » |
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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.
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Aryabhatta
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Re: Convergence of product of tangents
« Reply #7 on: Jul 24th, 2009, 10:14am » |
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Write tan as sin/cos and use the fact that
sin(2nx) = 2nsin(x)cos(x)cos(2x)cos(4x)....cos(2n-1x)
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| « Last Edit: Jul 24th, 2009, 10:14am by Aryabhatta » |
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