### John Baez

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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

32
13

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.﻿

Equivalent to y = sqrt(2)^y, and its easy to see that y = 2 solves it. ﻿

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

That was (and is) also my first wild guess. If there's a solution that can be calculated just by thinking, it has to be 2 (without thinking).﻿

- 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.)﻿

- yes, this tower of powers is sort of scary looking.  I hope it didn't cause a major relapse.﻿

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.﻿

Why is the other fix point not attractive?﻿

- 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.

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) ↑↑ ∞ = ????﻿

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?﻿

I draw the graph but not the derivative. Without the derivative I was not able to see it.﻿

- 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.  :-)﻿

- 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 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...﻿

Wops did that wrong.

Guessing 2.﻿

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, .  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...﻿

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?﻿

- right, the more math we know the less tiring work we need to do!  :-)﻿

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 ;-)﻿

- 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.﻿