# Find the value of x in this equation:

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let (X)^X^X^X^X^X^X^X... = A ----- (1)

if (X)^(X^X^X^X^X^X^X^X...) = A

Therefore using Eq(1) (X)^(Eq(1))=A

Then (X)^A=A

Therefore X=A^(1/A) <= is your generalized form

But If you tried this with say A=4, then the iteration breaks down, where Eq(1) does not make sense in that if A=4 then X=4^0.25, then Eq(1), A=/=X^X^X^X^X^X^X...

Hence I think theres something wrong with this 'proof' and that it seems to work only because its a 2?

Please correct me if im wrong

Define a function $f(x)$ to be the infinite power tower of x evaluated right to left,

$f(x) = x^{\left(x^{\left(\cdots^x\right)}\right)}$

This function is the limit as $n\to\infty$ of the sequence $\{f_n(x)\}$ defined by

$f_1(x) = x$

$f_{n+1}(x) = x^{f_n(x)}$

The generalization of the problem $f(x)=2$ is $f(x)=a$. It is easy to show that the solution to $f(x)=a$ is $x=a^{1/a}$. There is one caveat, and it is an important one. The solution to $f(x)=a$ is $x=a^{1/a}$ if a solution exists.

The generalized problem does work for a=2.5, for example. It does not for a=4. For a=4, the solution should be $x=4^{1/4}=\surd 2$. This is exactly the solution for $a=2$. So why does the sequence $\{f_n(\surd 2)\}$ converge to 2 rather than 4 (or some other number)?

One can show that the sequence converges as expected for all $a\in(1/e,e)$. One can also show that for any number $x>1$, there are two numbers $a$ such that $a^{1/a}=x$. One solution will be smaller and one larger than $e$. The sequence $\{f_n(x)\}$ converges to the smaller of the two. The convergence gets slower and slower as $a\to e$.

Things are even more interesting for $x<1$. Here, the solution to $a^{1/a}=x$ is unique. However, the series $\{f_n(x)\}$ still fails to converge for $x<1/e$. Instead, it settles to an oscillation among a set of values. For example, the sequence $\{f_n(0.1)\}$ alternates between 1-2e-9 and 1e-10. The number of alternating elements grows as $x\to 1/e$ from below. This alternation can also be seen in the sequence as $x\to 1/e$ from above, making the convergence very slow the closer x gets to 1/e.

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• 2 weeks later...
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Define a function $f(x)$ to be the infinite power tower of x evaluated right to left,

$f(x) = x^{\left(x^{\left(\cdots^x\right)}\right)}$

This function is the limit as $n\to\infty$ of the sequence $\{f_n(x)\}$ defined by

$f_1(x) = x$

$f_{n+1}(x) = x^{f_n(x)}$

The generalization of the problem $f(x)=2$ is $f(x)=a$. It is easy to show that the solution to $f(x)=a$ is $x=a^{1/a}$. There is one caveat, and it is an important one. The solution to $f(x)=a$ is $x=a^{1/a}$ if a solution exists.

The generalized problem does work for a=2.5, for example. It does not for a=4. For a=4, the solution should be $x=4^{1/4}=\surd 2$. This is exactly the solution for $a=2$. So why does the sequence $\{f_n(\surd 2)\}$ converge to 2 rather than 4 (or some other number)?

One can show that the sequence converges as expected for all $a\in(1/e,e)$. One can also show that for any number $x>1$, there are two numbers $a$ such that $a^{1/a}=x$. One solution will be smaller and one larger than $e$. The sequence $\{f_n(x)\}$ converges to the smaller of the two. The convergence gets slower and slower as $a\to e$.

Things are even more interesting for $x<1$. Here, the solution to $a^{1/a}=x$ is unique. However, the series $\{f_n(x)\}$ still fails to converge for $x<1/e$. Instead, it settles to an oscillation among a set of values. For example, the sequence $\{f_n(0.1)\}$ alternates between 1-2e-9 and 1e-10. The number of alternating elements grows as $x\to 1/e$ from below. This alternation can also be seen in the sequence as $x\to 1/e$ from above, making the convergence very slow the closer x gets to 1/e.

Which shows that the so called proof proposed by the OP is wrong, yet people have agreed on it.

That proof only worked because it was the magic number '2'. and we know many cases where coincidental relationships and operators can be obtained that is mathematically incorrect, when dealing with the simple number 2. so his proof was not proof but merely something that seemed to be working out in his favor

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That proof only worked because it was the magic number '2'. and we know many cases where coincidental relationships and operators can be obtained that is mathematically incorrect, when dealing with the simple number 2. so his proof was not proof but merely something that seemed to be working out in his [favour]
He wasn't claiming the general case though.
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• 4 weeks later...
He wasn't claiming the general case though.

sorry but thats just weak. I cant accept that.

If it doesnt work, it doesnt work. Theres nothing 'case specific' about this. If what he did was correct then it should be correct for all cases. Obviously it wasnt so his method was definately wrong. again i proposed its probably cause it was the magic number to.

simple example what is x^2*x^2,

is the answer x^2+2 = x^4

or x^2*2 = x^4

theres only 1 right way to do it. its x^2+2

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sorry but thats just weak. I cant accept that.

If it doesnt work, it doesnt work. Theres nothing 'case specific' about this. If what he did was correct then it should be correct for all cases.

There are many things in math that only work over a limited domain. For example, the expansion

$\frac 1 {1-x} = 1 + x + x^2 + x^3 + x^4 + \cdots$

is valid only if $|x| < 1$. The interval $(-1,1)$ is the interval of convergence for this particular series.

In the problem at hand, the only thing magical about 2 is that it is in the interval $(1/e,e)$. This is the interval of convergence for the generalization of the problem as posted by the OP. The solution as outlined by the OP is not incorrect, it just lacks rigor.

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nah, not good enough.. the proof works in the range of ~ 0.37 to 2.7 only

hardly a proof of concept.. its a case specific proof

does 10^2*10^2=10^4=10^(2+2)=10^(2*2) just becasue it works?

no

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Stob being an obnoxious noob. The only thing wrong with the "proof"

$x^{x^{\cdots^x}} = a \:\: \Rightarrow \:\: x^a = a \:\: \Rightarrow \:\: x = a^{1/a}$

is that it assumes that a solution x exists. Moreover, if a solution does exist it can only be of this form. All it would have taken to complete the proof for $a=2$ is to show that the sequence $\{\surd 2, \surd 2^{\surd 2}, \surd 2^{\surd 2^{\surd 2}}, \cdots\}$ converges to 2.

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stop attempting to be a know it all. all youre saying re so general and trying to make out as some sort of pro

you are talking as if you could easily have solved the original X^X^X^X^X^X^X^X=2 problem prior to knowing. If one coud not mathematically solve it without the use of trial and error, and that the proof they gave is not a general one, then this means this problem here is simply a case specific solution

sa ive said, the 'a' can only be between ~0.37-2.7 to work and it doesnt converge anymore to the answer a

i also saw this post from the riddles forum, so its a pretty cruddy riddle, its like me telling you that

A * 0 = 0 where A is any number

then if i told you

0/0=?

Guess what is A

yea its kinda like that

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Here's my only insight, as primitive as it may be:

If x = 1, the result is 1 = 2.

If x > 1, the result is infinity = 2.

The answer would have to lie between x = 1 and x > 1. That is impossible, and thus there is no answer.

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