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YADD · 10-19-2007, 06:49 AM

1) A cute trick is compute

$\int_0^\infty e^{-a\frac{W^2}{2}}dW$

then take drivative w.r.t. a thrice.

3) It is a bacterium, Khoa. An insect cannot duplicate that fast. The tougher part of the puzzle is justify the solution (i.e., discarding the probability = 100% answer). A clue is realted to puzle 5) below.

5) Let me repharse the question. Define $y = x^{x^{x^{x^{...}}}}$ (infinite times). For $x = \sqrt{2}$ , find y.

The paradox comes about like this: Rewrite $y = x^y$ ; then with $x=\sqrt{2}$ , both $y=2$ and $y=4$ satisfy this equation.

The paradox is resolved as follows: Define $y_n = x^{x^{x^{x^{...}}}}$ (n times), equivalently $y_{n+1} = x^{y_n}$ . Plot $y_{n+1}$ against $y_n$ on a two-dimensional axes, the curve (for $x=\sqrt 2$ ) is a monotonically increasing function crossing the 45-degree line at two points A (y=2) and B (y=4). The slope of the curve at A is smaller than unity [one], so A represents an attractive (stable) fixed point. The slope at B point is greater than unity => B is a repulsive (unstable) fixed point. So $y_\infty \rightarrow 2,$ while point B (y=4) is not attainable.

[y=4 is a spurious root since some "non-linearity" was introduced in rewriting the definition of y.]

An aside observation from the resolution: The maximum value of x such that $y = x^{x^{x^{x^{...}}}}$ (infinite times) is well-defined is $x^\star = e^{\frac{1}{e}}$ , corresponding to $y = e$ . Beyond $x^\star$ , y explodes.