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**Puzzle:**estimate this number!

If you're good at math, you can do this by just thinking. If you're not, you can use a calculator and see what you get. But in that case, beware: a 'tower of powers' like this:

x^x^x^x^x

means

x^(x^(x^x))

not

((x^x)^x)^x

I'm asking about the first sort of thing.

View 47 previous comments

- +Mee Ming Wong - exponentiation is not associative so you get different answers depending on how you group the terms. If you click on "Expand this post" you'll see my explanation of this issue.Nov 8, 2012
- Thanks +John Baez.Nov 8, 2012
- I have a question. In the comments is argued that as sqrt(2)^x=x has solutions x=2,4 , and the the (n+1)th term is bigger or equal to the nth term, then the limit of the progression is 2. Is that enough? I mean isn´t it possible that the progression converges to, let´s say 1.98? (of course supposing that no term is equal to 1.98, 1.98 being just the limit)Nov 8, 2012
- With some extra work you can show that if the progression converged to 1.98 then we would necessarily have sqrt(2)^{1.98} = 1.98.Nov 8, 2012
- The "extra work" is proving this: if we have a continuous function f from the real numbers to the real numbers, and we have any sequence with

x_{i+1} = f(x_i)

that converges to some point x, then

x = f(x).Nov 9, 2012 - Convergence is best proved by graphing y=sqrt2^x and the y=x line. Hope I'm not spoiling it. If you want (-sqrt2), that's undefined for freshmen, and need a branch cut for upperclassmen.Nov 9, 2012