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Topic: Re: Infinite exponent (Read 576 times) |
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Icarus
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Re: Infinite exponent
« on: Feb 18th, 2006, 6:39am » |
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This has been posted before, but I am too lazy to look it up. Particularly since the solution is simple: Note that xA = A, so x = A1/A.
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| « Last Edit: Feb 18th, 2006, 6:39am by Icarus » |
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Icarus
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Re: Infinite exponent
« Reply #1 on: Feb 18th, 2006, 7:23am » |
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For that it is easier to find the other thread (which also dealt with a number of variations). The post I linked to is the first to directly deal with convergence. The final concensus was that the sequence converges for e-e <= x <= e1/e.
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| « Last Edit: Feb 18th, 2006, 7:38am by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
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Posts: 4863
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Re: Infinite exponent
« Reply #2 on: Feb 18th, 2006, 8:05am » |
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Looking more carefully through the other thread, I see that it is a bit disorganized, but here are the major points:
(1) If the sequence converges, then x = A1/A. But the maximum value for A1/A is e1/e. So the sequence cannot converge for x > e1/e.
(2) When x >= 1, the sequence is increasing, and Fn(x) is an increasing function of x for all n. So, if x <= e1/e, then Fn (x) <= Fn(e1/e) <= e for all n. Since it is a bounded increasing sequence, Fn(x) must converge.
(3) When 0 < x < 1, F2n(x) is a decreasing sequence in n, while F2n+1(x) is an increasing sequence in n. Since both are trapped between 0 and 1, they both converge.
The full sequence converges when the limits of these two subsequences are the same. Let M = limn F2n(x) and L = limn F2n+1(x).
Then xL = M and xM = L. This equation can be solved to give M ln(M) = L ln(L).
The equation z ln(z) = k has only one solution in z when k >= -1/e. This corresponds to the condition that x >= e-e, so the sequence must converge for those x.
(4) When k < -1/e, z ln(z) = k has multiple solutions, and it can be seen that M will be the upper solution, while L will be the lower solution. So for values of x < e-e, the total sequence does not converge.
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| « Last Edit: Feb 18th, 2006, 8:24am by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
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Re: Infinite exponent
« Reply #3 on: Feb 19th, 2006, 12:23pm » |
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The equation y=x1/x looks something like this when graphed (with a logarithmic scale on the x axis - the decrease on the right is very slow):
. . . . . . x . . . . . . . . . . . . . .
. . . . x . . . x . . . . . . . . . . . .
. . x . . . . . . . . x . . . . . . . . .
. x . . . . . . . . . . . . . x . . . . .
x . . . . . . . . . . . . . . . . . . . x
So when you enter a value > 1 for A into x = A1/A, you do indeed get a value for x back. But unless A = e, there is a second value of A that gives the same value for x. x^x^x^... can only return one of the two values. It will in fact alway return the one <= e.
In other words, f(x) = x^x^x^... is defined on the interval [e-e, e1/e] and has as range the interval [L, e], where L satisfies e-e = L1/L. Since 1000 is outside the range of f, there is no value of x that returns it.
If we calculate x=10001/1000 = 1.0069316..., then calculate f(x) = 1.0069802..., what we get is the other value of A that gives the same x:
10001/1000 = (1.0069802...)1/1.0069802...
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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