I'm going to wild-guess '2' before I even think about it ;-)

### John Baez

Shared publicly -**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.

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2?

phi

Scott Anderson

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42

2 and 4 are both reasonable candidates.

And something called Lambert's W function is related to this.

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*+Mugizi Rwebangira 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).

sqrt(2)

I say 1/4!

+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.

it converges to 2

+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?

isn't sqrt(2) positive by definition?

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.

2^((1/2)*(sqrt(2)^8))

+Douglas Summers-Stay well, yes that is an interesting point indeed

+Douglas Summers-Stay Why is the other fix point not attractive?

John Baez

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+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.

+John Baez then my best bet is 2

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 :)

Someone should make a wikipedia page for this function :)

Dan Smith

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The equation in the image comes to a bit less than 2: http://www.wolframalpha.com/input/?i=sqrt%282%29**sqrt%282%29**sqrt%282%29**sqrt%282%29**sqrt%282%29**sqrt%282%29**sqrt%282%29**sqrt%282%29

John Baez

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+Thomas Egense - That function x^(x(^x... is called

http://en.wikipedia.org/wiki/Lambert_W_function#Example_3

**infinite tetration**, and it's related to the 'Lambert W function':http://en.wikipedia.org/wiki/Lambert_W_function#Example_3

Liz Krane

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I like that word,

**infinite tetration**. Sounds like a Tetris marathon to me. :DReminds of the classic infinite resistor problem http://www.mathpages.com/home/kmath668/kmath668.htm

Yes, tetration sounds and

2 ↑↑ 4 = 2^(2^(2^2))

So, my puzzle is closely related to the question

sqrt(2) ↑↑ ∞ = ????

*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?

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?

+Harald Hanche-Olsen I draw the graph but not the derivative. Without the derivative I was not able to see it.

It does not diverge until x >=e

+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. :-)John Baez

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+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.

Heikki Arponen

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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.

Guessing 2.

This is a really fun interesting problem. And your tetration notation is useful. The sequence

To get an estimate of the finite tetration let's see what happens to

So to perform

**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?

John Baez

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

Clearly

-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 ;-)

**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 ;-)

16, evaluate in groups of 3's

+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.

Thanks +John Baez.

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)

John Baez

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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.

John Baez

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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).

x_{i+1} = f(x_i)

that converges to some point x, then

x = f(x).

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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