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Topic: x_(n+1) = sin(x_n) (Read 15481 times) 

Aryabhatta
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x_(n+1) = sin(x_n)
« on: Jul 27^{th}, 2009, 4:22pm » 
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Consider the sequence x_{n+1} = sin(x_{n}) x_{1} = 1. Show that for any t, such that 0 < t < 1/2 the sequence n^{t} x_{n} converges to 0 as n > infinity.


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Aryabhatta
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Re: x_(n+1) = sin(x_n)
« Reply #1 on: Jul 27^{th}, 2009, 6:25pm » 
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Also show that: n^{0.5} x_{n} converges to sqrt(3). (i think this is so, i hope my working is right)

« Last Edit: Jul 27^{th}, 2009, 6:55pm by Aryabhatta » 
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Eigenray
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Re: x_(n+1) = sin(x_n)
« Reply #2 on: Jul 27^{th}, 2009, 8:17pm » 
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Many moons ago, I showed the following on probability problem set. You might find it interesting. Background: In a branching process, we have a fixed probability distribution p, and in each generation, every individual independently has k offspring with probability p(k), k = 0,1,2,.... So if X_{n} is the number of individuals alive in generation n, then X_{n+1} is the sum of X_{n}many independent, identically distributed random variables. Let's assume that X_{0} = 1, p(0) > 0, and = k p(k) = E(X_{1}) 1. (a) If = 1 and ^{2} < , then there exist constants 0 < c_{1} < c_{2} < such that c_{1}/n < P( X_{n} 0 ) < c_{2}/n (b) If > 1, then there exist c,b > 0 such that P( extinction  X_{n} 0 ) ce^{bn}. If we let 1  ( sin {1x} )^{2} = p_{k} x^{k}, then part (a) applies to show c_{1}/n < x_{n} < c_{2}/n for some c_{1},c_{2}.


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Eigenray
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Re: x_(n+1) = sin(x_n)
« Reply #3 on: Jul 27^{th}, 2009, 8:54pm » 
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I would guess that under the conditions of (a) above, we have generally that n * P(X_{n} 0) 2/''(1), where (x) = p(k) x^{k}. If so this would imply n^{1/2} x_{n} 3.


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Aryabhatta
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Re: x_(n+1) = sin(x_n)
« Reply #4 on: Jul 28^{th}, 2009, 10:28am » 
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That is an interesting approach Eigenray. Do you have a proof of a) & b) written down somewhere? The approach I had in mind: I think we can show that sqrt(3)/(sqrt(n) + 1) <= x_{n} <= sqrt(3/n) using x  x^{3}/6 < sin(x) < x  x^{3}/6 + x^{5}/120

« Last Edit: Jul 28^{th}, 2009, 10:30am by Aryabhatta » 
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Aryabhatta
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Re: x_(n+1) = sin(x_n)
« Reply #5 on: Jul 28^{th}, 2009, 2:09pm » 
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I think I discovered a mistake in the 'proof' I had. Back to the board.


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Eigenray
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Background : If a_{n} = P( X_{n} = 0 ), then a_{n+1} = (a_{n}) a, where the probability of extinction, a, is the smallest positive root of (a) = a. We have a < 1 iff = '(1) > 1. It's a little awkward because it's phrased in terms of iterating instead of what would be more natural, f(t) = 1  (1t). Suppose f : [0,1] [0,1] is sufficiently smooth, with f(0) = 0, f'(0) = 1, f''(x) < 0. If x_{n+1} = f(x_{n}), x_{0}=1, then we should have n x_{n} 2/f''(0). Heuristically, we can think of f(x) ~ x(1rx), and x_{n} ~ 1/(r n).


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Aryabhatta
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Re: x_(n+1) = sin(x_n)
« Reply #7 on: Jul 28^{th}, 2009, 8:48pm » 
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Actually, I reworked the whole thing and now it looks correct to me! i.e. for all n > 0, sqrt(3)/(sqrt(n) +1) < x_{n} < sqrt(3/n) Proved using (very tedious) induction on n, and the identities x  x^{3}/6 < sin x < x  x^{3} + x^{5}/120. The LHS is pretty straightforward, but the RHS yields a cubic f(x) = 960x^{3} 520x^{2} + 111x  9 which needs to be > 0 for x >= 1. This has only one real root < 1, and so f(x) > 0 for x >= 1. Perhaps we can simplify it. I did find the result(inspite of the boring proof) to be interesting enough to post it here.


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