# Maths question. **URL:** \ **Category:** Factual Questions\ **Created:** [January 29, 2006, 12:37am UTC](https://boards.straightdope.com/t/maths-question/342024 "2006-01-29T00:37:54Z")\ **Posts on this page:** 20\ **Page:** 1

**Author:** ![Capt.Ridley\_s\_Shooting\_Party](https://avatars.discourse-cdn.com/v4/letter/c/cc9497/32.png) [@Capt.Ridley\_s\_Shooting\_Party](https://boards.straightdope.com/u/Capt.Ridley_s_Shooting_Party)\ **Post date:** [January 29, 2006, 12:37am UTC](https://boards.straightdope.com/t/maths-question/342024/1 "2006-01-29T00:37:54Z")

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? ---

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 29, 2006, 12:46am UTC](https://boards.straightdope.com/t/maths-question/342024/2 "2006-01-29T00:46:06Z")

The first question to address is whether [x -\> x -\> n](http://mathworld.wolfram.com/ChainedArrowNotation.html) 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? ---

**Author:** ![Capt.Ridley\_s\_Shooting\_Party](https://avatars.discourse-cdn.com/v4/letter/c/cc9497/32.png) [@Capt.Ridley\_s\_Shooting\_Party](https://boards.straightdope.com/u/Capt.Ridley_s_Shooting_Party)\ **Post date:** [January 29, 2006, 12:51am UTC](https://boards.straightdope.com/t/maths-question/342024/3 "2006-01-29T00:51:00Z")

“The Art of the Infinite, Our Lost Language of Numbers”. ---

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 29, 2006, 12:52am UTC](https://boards.straightdope.com/t/maths-question/342024/4 "2006-01-29T00:52:38Z")

Wrong arrow notation. See the arrow notation linked to from that page, and imagine I wrote x (up arrow) n. ---

**Author:** ![Capt.Ridley\_s\_Shooting\_Party](https://avatars.discourse-cdn.com/v4/letter/c/cc9497/32.png) [@Capt.Ridley\_s\_Shooting\_Party](https://boards.straightdope.com/u/Capt.Ridley_s_Shooting_Party)\ **Post date:** [January 29, 2006, 12:59am UTC](https://boards.straightdope.com/t/maths-question/342024/5 "2006-01-29T00:59:32Z")

Curried functions, power towers etc. Too many arrows 😛 ---

**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [January 29, 2006, 1:02am UTC](https://boards.straightdope.com/t/maths-question/342024/6 "2006-01-29T01:02:05Z")

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. ---

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 29, 2006, 1:07am UTC](https://boards.straightdope.com/t/maths-question/342024/7 "2006-01-29T01:07:42Z")

> [@Bryan Ekers](#): > > 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. ---

**Author:** ![silverfish](https://avatars.discourse-cdn.com/v4/letter/s/e480ec/32.png) [@silverfish](https://boards.straightdope.com/u/silverfish)\ **Post date:** [January 29, 2006, 1:11am UTC](https://boards.straightdope.com/t/maths-question/342024/8 "2006-01-29T01:11:33Z")

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. ---

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 29, 2006, 1:22am UTC](https://boards.straightdope.com/t/maths-question/342024/9 "2006-01-29T01:22:33Z")

> [@silverfish](#): > > I think that if A(n) = x^x^x… n times: A(n) = x^(x^n) Try that with x = 2 and n = 4, see if it works out for you. ---

**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [January 29, 2006, 1:47am UTC](https://boards.straightdope.com/t/maths-question/342024/10 "2006-01-29T01:47:16Z")

A^B^C=A^(B_C) so we have x^x^x=x^(2_x). For x\>1 that does not converge. ---

**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 29, 2006, 2:01am UTC](https://boards.straightdope.com/t/maths-question/342024/11 "2006-01-29T02:01:07Z")

> [@treis](#): > > A^B^C=A^(B\*C)… Try A = 2, B = 2, C = 4. ---

**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [January 29, 2006, 2:14am UTC](https://boards.straightdope.com/t/maths-question/342024/12 "2006-01-29T02:14:44Z")

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. ---

**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [January 29, 2006, 2:36am UTC](https://boards.straightdope.com/t/maths-question/342024/13 "2006-01-29T02:36:28Z")

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. ---

**Author:** ![cerberus](https://avatars.discourse-cdn.com/v4/letter/c/71e660/32.png) [@cerberus](https://boards.straightdope.com/u/cerberus)\ **Post date:** [January 29, 2006, 3:03am UTC](https://boards.straightdope.com/t/maths-question/342024/14 "2006-01-29T03:03:50Z")

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. ---

**Author:** ![KarlGauss](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/karlgauss/32/3713_2.png) [@KarlGauss](https://boards.straightdope.com/u/KarlGauss)\ **Post date:** [January 29, 2006, 4:13am UTC](https://boards.straightdope.com/t/maths-question/342024/15 "2006-01-29T04:13:22Z")

**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) ---

**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [January 29, 2006, 4:51am UTC](https://boards.straightdope.com/t/maths-question/342024/16 "2006-01-29T04:51:00Z")

> [@Bryan Ekers](#): > > 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. I’m looking at Windows Calculator and don’t see a button for exponentiation. Where is it? -FrL- ---

**Author:** ![Hypnagogic\_Jerk](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hypnagogic_jerk/32/4252_2.png) [@Hypnagogic\_Jerk](https://boards.straightdope.com/u/Hypnagogic_Jerk)\ **Post date:** [January 29, 2006, 6:39am UTC](https://boards.straightdope.com/t/maths-question/342024/17 "2006-01-29T06:39:15Z")

> [@KarlGauss](#): > > **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) 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. ---

**Author:** ![Hypnagogic\_Jerk](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hypnagogic_jerk/32/4252_2.png) [@Hypnagogic\_Jerk](https://boards.straightdope.com/u/Hypnagogic_Jerk)\ **Post date:** [January 29, 2006, 6:43am UTC](https://boards.straightdope.com/t/maths-question/342024/18 "2006-01-29T06:43:21Z")

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. ---

**Author:** ![cerberus](https://avatars.discourse-cdn.com/v4/letter/c/71e660/32.png) [@cerberus](https://boards.straightdope.com/u/cerberus)\ **Post date:** [January 29, 2006, 8:19am UTC](https://boards.straightdope.com/t/maths-question/342024/19 "2006-01-29T08:19:47Z")

My earlier algebra addresses a different sequence. My numerical investigations suggest that if 0\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\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. ---

**Author:** ![Capt.Ridley\_s\_Shooting\_Party](https://avatars.discourse-cdn.com/v4/letter/c/cc9497/32.png) [@Capt.Ridley\_s\_Shooting\_Party](https://boards.straightdope.com/u/Capt.Ridley_s_Shooting_Party)\ **Post date:** [January 29, 2006, 11:39am UTC](https://boards.straightdope.com/t/maths-question/342024/20 "2006-01-29T11:39:13Z")

Ah, cool. I’d have never have guessed it would have had an answer as simple as root(2). [Next page](https://boards.straightdope.com/t/maths-question/342024.md?page=2)