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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:






I'm asking about the first sort of thing.
Mee Ming Wong's profile photoManoel Rodríguez Calvo's profile photoJohn Baez's profile photoYao Liu's profile photo
I'm going to wild-guess '2' before I even think about it ;-)
2 and 4 are both reasonable candidates.
And something called Lambert's W function is related to this.
Here's what I did.  I figured that sqrt(2)^x == x.  So I plotted the two sides of the equation and saw that they crossed at x==2 and x==4. I think that's an okay way to solve it. I checked both answers.
I'm scared of math again suddenly. Obviously it isn't like riding a bike. Last math class as DiffEQ/Calc4
If x is the solution, taking logarithms it is easy to show that ln(x)/x=ln(2)/2
+Douglas Summers-Stay - so did you figure out which answer this number is closer to, either by calculation or pure thought?  If you're trying not to spoil the puzzle, that's fine.
Just less than 2, and I believe the limit as the tower grows larger is actually 2.  (It is easy to show that sqrt(2) ^^ n is less than 2 for all n and is monotonically increasing.)
+Jon Lightner - yes, this tower of powers is sort of scary looking.  I hope it didn't cause a major relapse.
+John Baez  When I said I checked both answers, that's what I was referring to.  But it made me think, what if I wasn't sure for each of these sqrt(2) whether it was referring to the positive or negative square root? What possible answers could there be?
Well, really 2 has two roots, right? I mean, we usually take sqrt(2) to be the positive answer, but both are true.
I'm not saying this was part of the puzzle, just that it made me think of a bigger puzzle.
+Douglas Summers-Stay - Good point: if you take sqrt(2) to mean ±sqrt(2) then this puzzle has 512 answers - perhaps not all distinct; I haven't checked.  But of course I meant the positive square root. 
Plotting this graph for x^(x(^x...  is quite interesting. If I remember correct it is defined in the interval ]1/e, e[. Before 1/e it does not converge but alternate between two values. (just like the figenbaum-tree). I played with this function before the internet existed and my results are gone and I could not find the graph on the internet.
Someone should make a wikipedia page for this function :)
I like that word, infinite tetration. Sounds like a Tetris marathon to me. :D
Yes, tetration sounds and is cool.  It works like this:

2 ↑↑ 4 = 2^(2^(2^2))

So, my puzzle is closely related to the question

sqrt(2) ↑↑ ∞ = ????
+Thomas Mach Consider the sequence x[0]=1, x[n+1]=√(2)^x[n). This is just iteration of the function f(x)=√(2)^x, which is convex and has two fixed points x=2 and x=4. If you draw a graph of the iteration procedure you can see why the fixed point at x=2 is attractive, while the one at x=4 is repulsive. (It has to do with f'(2)<1 while f'(4)>1.)
Fascinating. Had done some math question like this long ago! I guess one can also generalize this to cubic or higher roots.

It seems the series diverges if we increase 2 to 2.09 (approximately) or above. Is there anything general or interesting about when it begins to diverge?
+André Schemaitat - I meant what I wrote: a tower of nine sqrt(2)'s.  However, I said to estimate it!   So, it's possible that an infinite tower would give a reasonable approximation that's much easier to compute.  And, some people here claim that's true.  :-)
+Alexander Kruel - I'm not sure how you're getting that and I'm too lazy to figure out if it's right.  If you want, you can compare that expression to the answer +Dan Smith worked out using a (fancy) calculator.
I thought about something like this: you could start from the top and add √2's below. √2^1 < √2^√2 < √2^2 since 1<√2<2. So √2^√2 is somewhere between √2 and 2. Let's say it's about 1.5. Now add √2 below that and repeat. The process is increasing but can never exceed 2, so that would point to 2 as well I guess...
This is a really fun interesting problem. And your tetration notation is useful.  The sequence sqrt(2) ↑↑ n is clearly increasing in n because sqrt(2) > 1 .  Also, however in each term if you replace the last(topmost) sqrt(2) by 2 you get 2.  So sqrt(2) ↑↑ n  < 2 for all n, so there must be a limit which you can calculate by using the trick, sqrt(2) ↑↑ ∞ = x and sqrt(2)^x = x which gives x=2,4 but the limit must be 2 because of the inequality.

To get an estimate of the finite tetration let's see what happens to sqrt(2)^(2 - ϵ) = exp[ln(2).(2-ϵ)/2] ~ 2 - ϵln(2) +O(ϵ^2)

So to perform sqrt(2) ↑↑ 9 we start with ϵ = 2-sqrt(2) and exponentiate it 8 times to end with sqrt(2) ↑↑ 9 ~ 2 - (2-sqrt(2)).(ln2)^8 ~ 1.969... which is really close to the actual 1.976...
Very very nice, +Shanthanu Bhardwaj.  The really cool new part - to me - is your analysis of how much the answer differs from 2. 
Latecomer, but I'll give it a try before seeing other answers. If this is an infinite sequence, we have sqrt(2)^x = x => 2^(x/2) = x => exp(x*ln(2)/2) = x => (-x*ln(2)/2)*exp(-x*ln(2)/2) = -ln(2)/2. Here I'll use the Lambert W function, that gives the inverse of x*e^x  => -x*ln(2)/2 = W(-ln(2)/2) = -ln(2) => x = 2. Wow, so much work for such a simple solution...
+Thomas Mach Where the graph crosses the diagonal $y=x$ from above to below, the derivative must be ≤1. Where it crosses the other way, the derivative must be ≥1. Where it crosses transversally, the inequality becomes strict. There is no need to plot the derivative to see this?
Clearly wrong ( where is error? is it possible to correct this? ) but as I think funny estimation with elementary calcullus only ( Sum(y,k) = \Sum^{y}_{k) , C(y,k) = Newton symbol "n above k" =number of combinations )
-1. ln(1+x) ~x ( x<1)
0. sqrt(2)^y =y
1. sqrt(2)^y = (1+ s)^y
2. (1+s)^y = Sum(y,k) C(y,k) s^(y-k) ~ s^y + ys^(y-1)
3. (1) = s^y + ys^(y-1) = y = (0)
4. (3) = s^(y-1)(s + y) = y
5. (y-1)ln(s) + ln(s + y) = ln(y)
6. (y-1)ln(s) + ln(y(1+s/y)) = ln(y)
7. (y-1)ln(s) + ln(y) +  ln(1+s/y) = ln(y)
8. (y-1)ln(s) +ln(1+s/y) = (y-1)ln(s) + s/y =0
9. y^2 -y + s/ln(s) =0
10.s = 0.42..  \delta = 1 - 4(s/ln(s)) >0
11. x1 ~ -1/2 (is <0 so clearly not correct answer)
12. x2 ~ 3.7/2 = 2 (that is it!!!)

dropped term in (2) gives error ~ 1/2 y*(y-1)s^(y-2) ~ 1 ;-)
+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.
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)
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. 
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).
Yao Liu
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.
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