# Maths question.

**URL:** <https://boards.straightdope.com/t/maths-question/342024>\
**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:** 17\
**Page:** 2

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**Author:** ![biqu](https://avatars.discourse-cdn.com/v4/letter/b/7feea3/32.png) [@biqu](https://boards.straightdope.com/u/biqu)\
**Post date:** [January 29, 2006, 12:58pm UTC](https://boards.straightdope.com/t/maths-question/342024/21 "2006-01-29T12:58:17Z")

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> [@severus](#):
>
> This proof is only valid if it is also proved that the sequence converges, which is what **Hari Seldon** did.

**Hari Seldon** ’s proof is good enough for the purposes of solving the OP’s problem, but one can obtain more relaxed bounds on the set of real numbers for which the sequence converges; see for example [this page by math professor Lou Talman](http://clem.mscd.edu/~talmanl/) (Fun Stuff, Item 5).

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**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\
**Post date:** [January 29, 2006, 4:13pm UTC](https://boards.straightdope.com/t/maths-question/342024/22 "2006-01-29T16:13:30Z")

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> [@ultrafilter](#):
>
> Try that with x = 2 and n = 4, see if it works out for you.

No, first you define it for any n, then you allow n to approach infinity, and find a limit. So it’s more like, try it with x=2, then let n=4, now n=5, now n=6, and what limit are you getting closer and closer and closer too? That’s the limit at infinity.

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**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, 5:04pm UTC](https://boards.straightdope.com/t/maths-question/342024/23 "2006-01-29T17:04:48Z")

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> [@TJdude825](#):
>
> No, first you define it for any n, then you allow n to approach infinity, and find a limit. So it’s more like, try it with x=2, then let n=4, now n=5, now n=6, and what limit are you getting closer and closer and closer too? That’s the limit at infinity.

That’s a counterexample for **silverfish** ’s claimed equality. I know what a limit is, thank you very much.

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**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 30, 2006, 12:56pm UTC](https://boards.straightdope.com/t/maths-question/342024/24 "2006-01-30T12:56:58Z")

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> [@severus](#):
>
> 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.

Even proving that the sequence cnverges is not enough. You actually have to show it converges to 2 and not, say to 4.

Suppose you were given the equation x^x^x^x^… = 4. You could go through the same argument I gave to show that if x#n \< 4 then x#(n+1) \< 4 so it is a bounded increasing sequence and therefore converges. You could see that x^4 = 4 and then conclude that x = sqrt(2). But that is not the solution; there is none. The rest of my post was devoted to showing that when x = sqrt(2), then the only possible values for x^x^x^… are 2 and 4 and then the fact that x#n \< 2 for all n implies the limit is 2.  
“It is not enough to succeed; others must fail” (I think this was attributed to some Hollywood director.)

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**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 30, 2006, 3:19pm UTC](https://boards.straightdope.com/t/maths-question/342024/25 "2006-01-30T15:19:55Z")

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Just for fun and interest, here’s a similar question with an interesting answer.

Solve for X:

X = sqrt ( 1 + sqrt (1 + sqrt (1 + sqrt (1 + …

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**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [January 30, 2006, 3:29pm UTC](https://boards.straightdope.com/t/maths-question/342024/26 "2006-01-30T15:29:17Z")

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> [@KarlGauss](#):
>
> Just for fun and interest, here’s a similar question with an interesting answer.
> 
> Solve for X:
> 
> X = sqrt ( 1 + sqrt (1 + sqrt (1 + sqrt (1 + …

x = sqrt(1+x)

x^2 = 1+x

x^2 - x - 1 = 0

x = (1+sqrt(5))/2

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**Author:** ![av8rmike](https://avatars.discourse-cdn.com/v4/letter/a/edb3f5/32.png) [@av8rmike](https://boards.straightdope.com/u/av8rmike)\
**Post date:** [January 30, 2006, 3:56pm UTC](https://boards.straightdope.com/t/maths-question/342024/27 "2006-01-30T15:56:11Z")

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> [@KarlGauss](#):
>
> Just for fun and interest, here’s a similar question with an interesting answer.
> 
> Solve for X:
> 
> X = sqrt ( 1 + sqrt (1 + sqrt (1 + sqrt (1 + …

That’s the nested radical form of the [Golden Ratio](http://mathworld.wolfram.com/GoldenRatio.html), as **Polerius** notes.  
The problem with the method **Bryan Ekers** and a few others suggested is that power towers are not associative like addition and multiplication; they have to be evaluated right-to-left.  
**Frylock** , you have to enable scientific mode, then exponentiation is in the lower-left center, as “x^y”.

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**Author:** ![cmurdough](https://avatars.discourse-cdn.com/v4/letter/c/f9ae1b/32.png) [@cmurdough](https://boards.straightdope.com/u/cmurdough)\
**Post date:** [January 30, 2006, 4:09pm UTC](https://boards.straightdope.com/t/maths-question/342024/28 "2006-01-30T16:09:43Z")

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[QUOTE=Frylock]  
I’m looking at Windows Calculator and don’t see a button for exponentiation. Where is it?

-FrL-[/QUOTE  
Change the View on your calculator from Standard to Scientific.

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**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\
**Post date:** [January 30, 2006, 4:24pm UTC](https://boards.straightdope.com/t/maths-question/342024/29 "2006-01-30T16:24:45Z")

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> [@av8rmike](#):
>
> That’s the nested radical form of the [Golden Ratio](http://mathworld.wolfram.com/GoldenRatio.html), as **Polerius** notes.  
> The problem with the method **Bryan Ekers** and a few others suggested is that power towers are not associative like addition and multiplication; they have to be evaluated right-to-left.  
> **Frylock** , you have to enable scientific mode, then exponentiation is in the lower-left center, as “x^y”.

Thanks, **av8rmike** and **cmrdough**. I didn’t realize there was a half-decent calculator to be had using that application!

So anyway, now I’ve tried out sqrt2 as an answer to the question, and as far as I can tell, it _doesn’t_ converge to 2. It hits two on step three, then continues right on past it.

But maybe I’m doing it wrong.

I’m hitting the following buttons:

# 2 x^y .5

M+

then

# MR x^y MR

then

# x^y MR

over and over again.

This, it seems to me, raises sqrt2 to the power of sqrt2, then raises the result to the power of sqrt2, then raises the _new_ result to the power of sqrt2, and so on.

Is that not the right way to understand the equation in question?

-FrL-

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**Author:** ![av8rmike](https://avatars.discourse-cdn.com/v4/letter/a/edb3f5/32.png) [@av8rmike](https://boards.straightdope.com/u/av8rmike)\
**Post date:** [January 30, 2006, 5:00pm UTC](https://boards.straightdope.com/t/maths-question/342024/30 "2006-01-30T17:00:38Z")

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> [@](#):
>
> Is that not the right way to understand the equation in question?

No, it’s not. It’s like I said, you have to evaluate the tower from _right_ to left, what you’re doing is equivalent to:  
(((((x^x)^x)^x)^x)…), which is not the same as (x^(x^(x^(x^x)…)))). Write out a few terms and you’ll see what I mean. To do it on the calculator, you’ll need to compute:  
sqrt(2)^sqrt(2),  
store that result,  
take sqrt(2), raise that to the value of the stored number,  
store that result,  
take sqrt(2), raise that to the value of the stored number,  
etc…

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**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 30, 2006, 10:15pm UTC](https://boards.straightdope.com/t/maths-question/342024/31 "2006-01-30T22:15:59Z")

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Again, you have to be more careful. It is easy to see that the golden ratio, I will call it g, is the only possible solution. What is harder is showing it is. Let x\_n be the nth interate, so x\_{n+1} = sqrt(1+x\_n). If we suppose x\_n \< g, then 1+x\_n \< 1+g. Then x\_{n+1} = sqrt(1+x\_n) \< sqrt(1+g) = g and so we see that all x\_n \< g. It is not quite evident that the sequence is monotone. So I will show that x\_n \< x\_{n+1}. In fact from x\_n \< g, we get x\_n+1 \< g+1 and then multiply the inequalities (incidentally, it is clear that all terms are positive so we can take positive square roots and multiply inequalities as lib), we get x\_n^2 - x\_n \< g^2 - g and then x\_n^2 - x\_n -1 \< g^2 - g -1 = 0. But then x\_n^2 \< x\_n +1 and, taking square roots, x\_n \< sqrt(x\_n+1) = x\_{n+1}.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [January 30, 2006, 10:28pm UTC](https://boards.straightdope.com/t/maths-question/342024/32 "2006-01-30T22:28:39Z")

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What **Hari Seldon** is distinguishing here is the difference between existence and uniqueness proofs. For both of the problems mentioned, it’s easy to show that if a solution exists, it must be some particular value. Showing that a solution does exist is a little harder.

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**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\
**Post date:** [January 31, 2006, 4:41pm UTC](https://boards.straightdope.com/t/maths-question/342024/33 "2006-01-31T16:41:24Z")

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Sqrt2^(Sqrt2^(Sqrt2…))) seems to come to both 2 and 4.

Its clear it comes to 2 from comments on this thread. But by a similar argument, it seems like we can show that it comes to 4 since if we set x^(x^(x…)))=4, we get x^4=4 which means x=Sqrt2.

So what I presently don’t understand/don’t know is:

1. Why does 2 get to be the limit and not 4?
2. Is it true that Sqrt2^(Sqrt2^(Sqrt2…))) in some way _really_ “equals” (or anyway “comes to” in some sense) both 2 and 4? Or is the 4 result somehow deceptive?

I have a feeling **Hari Seldon’s** posts answer this issues I’ve asked about, but I can’t follow them. He skips too many steps for my level of knowledge. And I dont’ know what “monotone” means in this context.

Is there a free graphing utility? Thinking about this has made me curious as to what the graph for y = the “xth root” of x would look like.

-FrL-

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [January 31, 2006, 5:27pm UTC](https://boards.straightdope.com/t/maths-question/342024/34 "2006-01-31T17:27:11Z")

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The argument that x^(x^(x^(…))) = 4 implies x = sqrt(2) shows that if the original equation has a solution, the only solutions (in **R** , anyway) are +sqrt(2). What still needs to be shown is that there is a solution to that equation, and that’s what **Hari Seldon** has disproved.

We say that a function f is monotone increasing iff x \> y implies that f(x) \> f(y).

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**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\
**Post date:** [January 31, 2006, 7:00pm UTC](https://boards.straightdope.com/t/maths-question/342024/35 "2006-01-31T19:00:03Z")

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> [@ultrafilter](#):
>
> What still needs to be shown is that there is a solution to that equation, and that’s what **Hari Seldon** has disproved.

Right. As I tried to indicate, I figured this was what **Hari** was up to. I was just hoping there could be a more complete explanation as to how his proof works. But I recognize this may not be possible (reasonably feasible) in a venue such as this.

-FrL-

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**Author:** ![pmwgreen](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@pmwgreen](https://boards.straightdope.com/u/pmwgreen)\
**Post date:** [January 31, 2006, 11:20pm UTC](https://boards.straightdope.com/t/maths-question/342024/36 "2006-01-31T23:20:08Z")

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Playing around with this, it appears that x^(x^(x^(…))) = y only has a solution if y is less than (or equal?) to e (i.e. 2.71828…).

Can anyone offer a proof? I’m headached out.

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**Author:** ![Rachm\_Qoch](https://avatars.discourse-cdn.com/v4/letter/r/a8b319/32.png) [@Rachm\_Qoch](https://boards.straightdope.com/u/Rachm_Qoch)\
**Post date:** [February 1, 2006, 12:09am UTC](https://boards.straightdope.com/t/maths-question/342024/37 "2006-02-01T00:09:25Z")

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> [@pmwgreen](#):
>
> Playing around with this, it appears that x^(x^(x^(…))) = y only has a solution if y is less than (or equal?) to e (i.e. 2.71828…).
> 
> Can anyone offer a proof? I’m headached out.

x^(x^(x^(…))) converges for (1/e)^e \<= x \<= e^(1/e). If x=e^(1/e) the sequence converges to e and satifies the equation for y=e.

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