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        riddles >> putnam exam (pure math) >> Re: Infinite exponent
        
(Message started by: Icarus on Feb 18^th, 2006, 6:39am)

Title: Re: Infinite exponent
Post by Icarus on Feb 18^th, 2006, 6:39am

This has been posted before, but I am too lazy to look it up. Particularly since the solution is simple: Note that x^A = A, so x = A^1/A.

Title: Re: Infinite exponent
Post by Icarus on Feb 18^th, 2006, 7:23am

For that it is easier to find the other thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1039112346;start=0#5) (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 <= e^1/e.

Title: Re: Infinite exponent
Post by Icarus on Feb 18^th, 2006, 8:05am

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 = A^1/A. But the maximum value for A^1/A is e^1/e. So the sequence cannot converge for x > e^1/e.

(2) When x >= 1, the sequence is increasing, and F_n(x) is an increasing function of x for all n. So, if x <= e^1/e, then F_n (x) <= F_n(e^1/e) <= e for all n. Since it is a bounded increasing sequence, F_n(x) must converge.

(3) When 0 < x < 1, F_2n(x) is a decreasing sequence in n, while F_2n+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 = lim_n F_2n(x) and L = lim_n F_2n+1(x).

Then x^L = M and x^M = 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.

Title: Re: Infinite exponent
Post by Icarus on Feb 19^th, 2006, 12:23pm

The equation y=x^1/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 = A^1/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, e^1/e] and has as range the interval [L, e], where L satisfies e^-e = L^1/L. Since 1000 is outside the range of f, there is no value of x that returns it.

If we calculate x=1000^1/1000 = 1.0069316..., then calculate f(x) = 1.0069802..., what we get is the other value of A that gives the same x:
1000^1/1000 = (1.0069802...)^{1/1.0069802...}