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   x_(n+1) = sin(x_n)
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   Author  Topic: x_(n+1) = sin(x_n)  (Read 15481 times)
Aryabhatta
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x_(n+1) = sin(x_n)  
« on: Jul 27th, 2009, 4:22pm »
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Consider the sequence
 
xn+1 = sin(xn)
 
x1 = 1.
 
Show that for any t, such that 0 < t < 1/2
 
the sequence
 
nt xn 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 27th, 2009, 6:25pm »
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Also show that:
 
n0.5 xn converges to sqrt(3).
 
(i think this is so, i hope my working is right)
« Last Edit: Jul 27th, 2009, 6:55pm by Aryabhatta » IP Logged
Eigenray
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Re: x_(n+1) = sin(x_n)  
« Reply #2 on: Jul 27th, 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 Xn is the number of individuals alive in generation n, then Xn+1 is the sum of Xn-many independent, identically distributed random variables.
 
Let's assume that X0 = 1, p(0) > 0, and = k p(k) = E(X1) 1.
 
(a)  If = 1 and 2 < , then there exist constants 0 < c1 < c2 < such that
c1/n < P( Xn 0 ) < c2/n
 
(b) If > 1, then there exist c,b > 0 such that P( extinction | Xn 0 ) ce-bn.
 
 
If we let
1 - ( sin {1-x} )2 = pk xk,
then part (a) applies to show
c1/n < xn < c2/n
for some c1,c2.
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Eigenray
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Re: x_(n+1) = sin(x_n)  
« Reply #3 on: Jul 27th, 2009, 8:54pm »
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I would guess that under the conditions of (a) above, we have generally that
n * P(Xn 0)   2/''(1),
where (x) = p(k) xk.
If so this would imply n1/2 xn 3.
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Aryabhatta
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Re: x_(n+1) = sin(x_n)  
« Reply #4 on: Jul 28th, 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) <= xn <= sqrt(3/n)
 
using  
 
x - x3/6 <  sin(x) < x - x3/6 + x5/120
« Last Edit: Jul 28th, 2009, 10:30am by Aryabhatta » IP Logged
Aryabhatta
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Re: x_(n+1) = sin(x_n)  
« Reply #5 on: Jul 28th, 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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Re: x_(n+1) = sin(x_n)   branchingprocess.pdf
« Reply #6 on: Jul 28th, 2009, 2:21pm »
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Background : If an = P( Xn = 0 ), then an+1 = (an) 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 - (1-t).
 
Suppose f : [0,1] [0,1] is sufficiently smooth, with f(0) = 0, f'(0) = 1, f''(x) < 0.  If xn+1 = f(xn), x0=1, then we should have n xn -2/f''(0).  Heuristically, we can think of f(x) ~ x(1-rx), and xn ~ 1/(r n).
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Aryabhatta
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Re: x_(n+1) = sin(x_n)  
« Reply #7 on: Jul 28th, 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) < xn < sqrt(3/n)
 
Proved using (very tedious) induction on n, and the identities
 
x - x3/6 < sin x < x - x3 + x5/120.
 
The LHS is pretty straightforward, but the RHS yields a cubic
 
f(x) = 960x3 -520x2 + 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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