The M1M1 treasure hunt, week 6

"Thankyou for the tea," said Aless, "But I really must be going now." Yet the front door remained closed.

"I'm afraid there's been a change of plan. You can't progress until after the test on Monday," explained the hatter. "Then you may leave by the x-staircase round the back. I think it goes somewhere provided the steps are the right size."

"An x-staircase?" pondered Aless. "I wonder what that is. And why does it matter how big the steps are? You'd just need more of them if they're smaller." Aless soon found the staircase in question, which climbed up as far as the eye could see. On the first step was written "x." All the other steps had "^x" written on them, so that the first two steps together read x^x and the first three x^xx and so on.

"Even three steps is confusing enough," thought Aless. "Would an infinite number of steps make any sense? I suppose if it did, then raising x to that power would make no difference. So if I call the answer y, then we would have y=x^y.
I wonder what is the maximum (positive) value of x for which that equation can be solved for y?"

x_max = ."?

"Oh I must make sure x is less than that," said Aless. "Or I may have to climb a dreadfully long way."
About half way up the staircase, Aless met the White Rabbit again, who had stopped to wind his gold watch.
"Why are you following me, child?" he asked tetchily.
"I'm not following you, sir. Especially now you've stopped. I'm just going upstairs. I'm fairly sure I'll get somewhere provided I take small enough steps."

"Are you now?" frowned the Rabbit. "Just because an equation has a solution, doesn't mean you'll find it, you know."

"Of course I will if I know where to look," snapped Aless, who was tired of being lectured to by all these strange Wonderland creatures.

"Listen, child. Suppose you've reached y_n after n steps. On your next step you reach y_n+1 = f(y_n). Then you start with this new value and take another step. You think just because c = f(c) that you'll eventually reach c? Maybe you will, and maybe you won't."

"I probably will if |y_n+1 − c| < |y_n − c|" argued Aless. "And don't call me "Child". I'm sure a child couldn't expand f(y_n) as a Taylor series about y_n = c, so that

y_n+1 − c = f(y_n) − c = f(c) + f '(c)(y_n − c) + ...− c = f '(c)(y_n − c) + ...

so I suppose I need |f '(c)| < 1. Hmmm. Maybe there's more to this staircase than I thought."

Since f(y) = x^y, and c = f(c), this means |f '(c)| < 1 provided

x_max > x > x_min ≡ ."?

"Quite right, child. If you have x = 0, for example, then x^x = 1 and x^xx = 0 and you keep hopping backwards and forwards." And to illustrate his point, the White Rabbit hopped backwards and forwards between 0 and 1, until he accidentally fell over the edge of the stairs.

Aless wasn't sorry to see him go. "He can join the tea party - I've never known the difference between rabbits and hares anyway."
"Now I wonder what value of x I should choose if I want to reach the top of this staircase as quickly as possible. In other words, what value of x minimises |f '(c)|?"

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