x to the xtreme

[psi20]

Putting powers upon powers aren't taught in school. But this problem is interesting.

 

The problem is to find x. x^x^x^x^x^x... = 2. Find x. This was in the book The Art of the Infinite.

 

Now the question is whether or not there is such a number x. If there was, it'd have to be greater than 1. But it'd have to be less than 2. The idea is that because x^x^x^x^x... = 2, it is the same as x^x^x^x... = 2 and so you can have x^2 = 2, and x is the square root of 2. Now the steps seemed logical to me, but I was skeptical so I tried to check it on the calculator. But x can't be the square root of 2 because this diverges. I tried it on the calculator different ways, pressing Ans ^ Ans and then going step by step, but it still showed that it diverged.

 

So the question still remains. What is x?

 

I played on the calculator a bit and saw something interesting. sqr rt 2 to sqrrt 2 to sqr rt 2. cube root of 3 of to the cube root of 3 to the cube root of 3 to the cube root of 3. and so on. Interesting to look at, but I didn't make much sense of it. Oh well.

[CPL.Luke]

and what were all of these roots tell you? other than that you can raise the square root of two to the square root of two?

 

I don't think that there is any algebraic solution too your equation, I think you would just need a computer run numbers through till it found one that worked

[shmoe]

I tried it on the calculator different ways, pressing Ans ^ Ans and then going step by step, but it still showed that it diverged.

 

This isn't what x^x^x... represents. It's the limit of the sequence x^x, x^(x^x), x^(x^(x^x)),... what you are doing on your calculator appears to be x^x, (x^x)^(x^x), ((x^x)^(x^x))^((x^x)^(x^x)),... which is something different.

 

With x=sqrt(2) you'll get:

 

x=1.414213562....

x^x=1.632526919...

x^(x^x)=1.760839556...

 

You can keep going, it will be approaching 2 but of course doing a finite number of terms proves nothing at all. You should prove (when x=sqrt(2)) that a) this sequence is increasing b) this sequence is bounded above by 2 (induction for both). This will prove the limit does in fact exist, you can then do some manipulations (combined with continuity of the exponential) to show this limit is 2.

[psi20]

Ah cool.

 

Is the method to find x described in my initial post a valid one?

[matt grime]

What method? You went from x^x^x^x..=2 to x^2=2 without justifying why this is true (it is: you are just raising x to the same power, but you didn't say this explicitly enough, I feel).

[jordan]

matt, would you be able to show what an explicit proof for this problem looks like?

[matt grime]

the method has been given. Let f_n(x) be the result of x^x^x^..^x, with n terms, and let f(x) be the limit as n tends to infinity of f_n(x). This will not a priori exist for all x, but that is immaterial at this stage.

 

Now, assume that there is a solution to f(x)=2, if there is, then it must be root 2, since x^f(x)=x^2, and x^f(x)=f(x)=2. It remains to show that f_n(sqrt(2)) actually converges which is done using shmoe's hints.

[Primarygun]

Did someone mention this method?

It suits the name of the book "The Art of the Infinite"

let the left-hand side of the equation be y,

and since there are infinite x, do the "order" again.

so we have the yth power of x = 2.

As y is originally equal to 2.

We have x being the root of 2

[matt grime]

Yes, that has been mentioned (by the OP, by me and by shmoe), and doesn't address the fact that it assumes that the limit exists in the first place. You also need to prove that fact. After all, if I assume there is a largest integer N, then N^2 is larger than N but N is the largest, hence N^2=N so N=0 or 1, and since 1 is larger than 0 there are no integers larger than 1. See what problems you have if you assume a false premise?