Whilst reading a book, a question was posed. What is the answer to x^x^x… (i.e. x raised to iself an infinite number of times) = 2. Does this equation have an answer? If so, what is it, and how would one go about finding it?
The first question to address is whether x -> x -> n has a limit as n increases without bound for any real x. My guess is that it does for any x in (1, 2), but I have nothing to back that up. The next thing to do is to find a way to calculate that limit, either through a closed form or a numerical approximation. Since the value in question increases monotonically (not proved, but obvious), interpolation could give you a value for which it’s equal to 2.
What book was it?
“The Art of the Infinite, Our Lost Language of Numbers”.
Wrong arrow notation. See the arrow notation linked to from that page, and imagine I wrote x (up arrow) n.
Curried functions, power towers etc. Too many arrows 
Just for laughs, I typed 1.0001 in windows calculator, hit the X^Y button and rested a weight on the ENTER button. The values increased slowly but steadily and I can’t see a reason any arbitrarily low 1.0000…0001 wouldn’t increase without limit.
Maybe, maybe not. I haven’t shown anything. Clearly, 1^1^1^… = 1, and 2^2^2^… doesn’t converge. What happens in between is a matter for investigation.
I think that if A(n) = x^x^x… n times: A(n) = x^(x^n)
Then, I think it’s fairly obvious that if x > 1, x^n goes to infinity, and so A(n) goes to infinity, as n -> infinity.
Try that with x = 2 and n = 4, see if it works out for you.
A^B^C=A^(BC) so we have x^x^x=x^(2x). For x>1 that does not converge.
Try A = 2, B = 2, C = 4.
The solution is sqrt(2). Let me write x#n for x to the x to the x … interated n times. Clearly x#(n+1) = x^(x#n). Let x = sqrt(2) and it is clear that if x#n < 2, then x#(n+1) = sqrt(2)^(x#n) < sqrt(2)^2 < 2. Thus starting with x = sqrt(2) the sequence is bounded above and monotone so it converges to something. As for what it converges to, call it y and it is clear that sqrt(2)^y = y which has at least two solutions y = 2 and y = 4. The latter is too large since every term is less than 2. Are there any other solutions? Well, the equation comes to (log 2)/2 = (log y)/y and the derivative of the right hand side is (1 - log y)/y^2, which is positive for y < e and negative for y > e. This means the function is increasing between 1 and e and decreasing thereafter and hence can take on the value (log 2)/2 at most twice.
Wait a sec, at what point did an infinite series turn into a limited one? x^x^x = 2 if x = root 2, but x^x… infinite certainly is not.
Let r be a strictly positive, real number.
Case #1: 0 < r < 1: lim(n){r^n} = 0
Case #2: r = 1: lim(n){r^n} = 1
Case #3: r > 1: lim(n){r^n} does not converge: the sequence increases without bound.
Other Cases:
Case #4: r=0 : lim(n){r^n} = 0
Let r be a strictly negative, real number.
Case #5: -1 < r < 0: lim(n){r^n} = 0
Case #6: r = -1: lim(n){r^n} does note converge, the sequence flips between -1 and +1.
Case #3: r < -1: lim(n){r^n} does not converge: the sequence increases in absolute magnitude without bound, and alternates in sign.
Having said that, there are sequences like this:
{2^(1+(1/n))}, which converges to 2 as n increases without bound.
Hari Seldon is correct, of course, but I think there’s a simpler solution.
We have X^X^X^X^X^X … = 2
Hence X^(X^X^X^X^X …) = 2
Therefore, by substitution X^2 = 2
(i.e. replace all but of one of the exponentiated X’s with 2)
The answer then is sqrt(2)
I’m looking at Windows Calculator and don’t see a button for exponentiation. Where is it?
-FrL-
This proof is only valid if it is also proved that the sequence converges, which is what Hari Seldon did. Without it, it’s meaningless to assume that the sequence converges to 2.
I think Bryan Ekers is looking at a different problem. Instead of looking at the problem x^x^x^x^…, what he is computing is (((x^x)^x)^x)^… = (x^3x)^…, which does not converge for any x > 1.
Note: the 3 in my term (x^3x)^… in my previous post is only there because I had three x as exponents in the term to the left of the equality sign. What I should have written would be lim n->infinity (x^nx). Which does not converge for x > 1.
My earlier algebra addresses a different sequence.
My numerical investigations suggest that if 0<x<1, then the sequence eventually creeps to 1. If x=1, then the result is trivial. The sequence doesn’t make sense for x=0. The sequence blows up for x>1.
For x < -1, the sequence swings wildly between “negative infinity” and 0. For x=-1, the sequence is trivilally constant at -1. For -1 < x < 0, the sequence isn’t always defined.
Symbolically:
If x=1, then
x1=x^x=1^1 =1, and furthermore xN=1.
If x=0, then the entire sequence is undefined.
If 0 < x < 1 then…
x^x > x, since fractional power laws increase positive fractions. In fact, Each layer (*)^x increases the value, but all the values are bounded strictly above by 1. So therefore, all of the sequences with 0<x<1 converge to 1.
If x>1, however, x^x > x > 1, and in fact each layer (*)^x increases the value: 1 < x < x^x < x^x^x < ….
Define x2=x^x, x3=x^x2, x3=x2^x, …xn=x^xn-1.
Consider the difference D=(xN)-(xN-1)=x^(xN-1)-(xN-1) = x^(xN-1) - x^(xN-2) = x^(xN-2)[x-1]. The limit of this difference is not zero, which indicates that the sequence cannot converge for x>1.
Ah, cool. I’d have never have guessed it would have had an answer as simple as root(2).