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I heard this problem from my friend today, but unfortunately we could not figure out what's happening here.

$$x^{x^{\large x^{x^{\unicode{x22F0}}}}}=2$$ $$x^2=2$$ $$x=\sqrt2$$

but

$$x^{x^{\large x^{x^{\unicode{x22F0}}}}}=4$$ $$x^4=4$$ $$x=\sqrt2$$

I have a feeling that this might be a duplicate, but I could not find this problem (basically I couldn't figure out what to search). So what's happening here, and what's the real value of $$\sqrt2^{\sqrt2^{\large \sqrt2^{\sqrt2^{\unicode{x22F0}}}}}$$

?

avz2611
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3 Answers3

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Finding the value of an infinitely iterated function can be tricky. Usually, if $z$ is a fixed point of a function $f$ (that is, if $f(z)=z$), then it often happens that $$\lim_{n\to\infty} f^n(x)=z, \forall x$$ ...however, things tend to get a little bit complicated when $f$ has more than one fixed point, which is the case with your function $$f(x)=\sqrt 2^x$$ that has fixed points $$f(2)=2$$ $$f(4)=4$$ In this case, the value of $f^\infty(x)$ depends on $x$, which doesn't really make sense given the way you wrote it. Basically this means that its value depends on the number "on top" of your power tower. For example, if you start with $4$, and observe the sequence $$a_0=4$$ $$a_{n+1}=\sqrt 2^{a^n}$$ then each $a_n=4$, and so $$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=4$$ However, if you take $a_0=1$, $$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=2$$ However, things are a little bit less complicated in this case, since $x=2$ turns out to be something called an "attracting fixed point". Basically, we end up with $$f^\infty\left(x\right) = \left\{ \begin{array}{lr} 2 & : x \lt 4\\ 4 & : x = 4\\ \infty & : x \gt 4\\ \end{array} \right.\\$$

Franklin Pezzuti Dyer
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    +1 This is the best answer I've seen to this kind of question in a while. – Carl Schildkraut Oct 07 '17 at 18:15
  • @CarlSchildkraut Glad to hear it! :D – Franklin Pezzuti Dyer Oct 07 '17 at 18:15
  • @Nilknarf an awesome answer man really insightful, Thanks! will accept after threshold time lapses – avz2611 Oct 07 '17 at 18:18
  • @avz2611 I'm happy to help! If you have any more questions about it, feel free to ask. – Franklin Pezzuti Dyer Oct 07 '17 at 18:19
  • @Nilknarf just a general question, when do i need to know to tip-toe around these situations, cause in our daily life if someone said me to solve the equation with 4 involved, i dont think i would be thinking about fixed points. Does it come down to practice/intuition or you can tell something about it before hand – avz2611 Oct 07 '17 at 18:23
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    In general, when there is an infinite number of anything involved, there is room for fallacy. When that happens, you need to keep in mind that you aren't actually looking at an infinite number of objects, but the limit of a sequence: $$\lim_{n\to\infty}a_n$$ Once you set up this sequence and define it formally, you will be forced to ask yourself questions like "what is the first term"? This will help you avoid fallacy and confusion. – Franklin Pezzuti Dyer Oct 07 '17 at 18:26
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    @avz2611 When the problem has to do with iteration (a certain function being applied many times) you will almost always have to look at the fixed points. Read this Wikipedia article for the basics: https://en.wikipedia.org/wiki/Fixed_point_(mathematics) – Franklin Pezzuti Dyer Oct 07 '17 at 18:27
  • @Nilknarf learn't something very new today, Thanks man!! – avz2611 Oct 07 '17 at 18:29
  • @avz2611 No problem! I love problems about functional iteration, so thanks for the interesting question! :) – Franklin Pezzuti Dyer Oct 07 '17 at 18:30
  • @Nilknarf Erm, which function do we iterate here? – Friedrich Philipp Oct 07 '17 at 18:47
  • @FriedrichPhilipp $f(x)=\sqrt 2 ^x$, as stated. – Franklin Pezzuti Dyer Oct 07 '17 at 18:48
  • @Nilknarf Ok, but the power tower (as a function of $x$) is not an iteration of a function, right? – Friedrich Philipp Oct 07 '17 at 18:53
  • @FriedrichPhilipp Yes, it is an iteration of a function. On its own, the expression $$\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}$$ has no meaning, since it can take on multiple values (as demonstrated). Thus, the only way to express it is as a limit of iterations of the function $f(x)=\sqrt 2^x$ – Franklin Pezzuti Dyer Oct 07 '17 at 18:56
  • @Nilknarf Then please give me the function $f$ such that the power tower $x^{x^{x^{\dots}}}$ can be expressed by $f(f(f(\dots f(x)\dots)$. – Friedrich Philipp Oct 07 '17 at 18:58
  • @FriedrichPhilipp If $f(y)=x^y$, then one can interpret $$x^{x^{x^{x^{...}}}}=f^{\infty}(y)$$ However, the expression $$x^{x^{x^{x^{...}}}}$$ has no standalone value, because its value depends on the value of $y$. Trying to assign it a value without knowing $y$ is like trying to find the limit of a recursively-defined sequence without knowing the first term. It simply doesn't make sense. – Franklin Pezzuti Dyer Oct 07 '17 at 19:01
  • @Nilknarf So, you agree with me that there is no such function $f(x)$? – Friedrich Philipp Oct 07 '17 at 19:02
  • Right... however, the value of the power tower $x^{x^{x^{...}}}$ can be expressed as a function in two variables: $$f(x,y)=\lim_{n\to\infty} a_n$$ where $$a_0=y$$ $$a_{n+1}=x^{a_n}$$ – Franklin Pezzuti Dyer Oct 07 '17 at 19:04
  • @Nilknarf Of course, I know that (well, $y$ should be $x$). So, yes, we have to add a second variable. When the power tower is $F(x)$, then we have$$F(x) = \lim_{n\to\infty}f_x^n(x),$$where $f_x(t) = x^t$. – Friedrich Philipp Oct 07 '17 at 19:07
  • @FriedrichPhilipp Right, that is what I am asserting. – Franklin Pezzuti Dyer Oct 07 '17 at 19:08
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While some people in this thread may disagree, I believe the most obvious definition of a power tower $x^{x^{x^{\dots}}}$ is as the limit of the series $x, x^x, x^{x^x}, x^{x^{x^x}},\dots$

or, in other words, $\lim_{n\to \infty} a_n$ where $a_0 = x$ and $a_n = x^{a_{n-1}}$.

Now to answer your two questions:

  1. Why does $x=\sqrt{2}$ seemingly solve both of the following equations: $x^{x^{x^{\dots}}}=2$ and $x^{x^{x^{\dots}}}=4$ ?

The answer is that you need to be very careful when working with infinite sequences. Lots of properties that are normally true about every day numbers are no longer true about infinite things. For example, this reminds me of the following bogus proof that $1+2+4+8+\cdots = -1$

$S = 1+2+4+8+\cdots$

$S - 1 = 2 + 4 + 8 + \cdots$

$\frac{S-1}{2} = 1 + 2 + 4 + 8 + \cdots = S$

$S-1 = 2S \Rightarrow S = - 1$

The issue with the above proof is that it assumes a limit exists. If a limit were to exist, then it would be $-1$, but since no limit exists, that answer is moot.

Now back to your question. When you take $x^{x^{x^{\dots}}}=b$, plug in $b$ to get $x^b=b$, and realize you can solve for $x = \sqrt{2}$, what you are really showing is that if the limit $\lim_{n \to \infty} a_n = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}$ exists, then that limit is equal to one of those values $b$.

  1. What does $\lim_{n \to \infty} a_n = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}$ truly equal?

Lemma 1: $a_n$ is an increasing series.

Proof by induction:

Since $\sqrt{2} > 1$, we have that $a_1 = \sqrt{2}^\sqrt{2} > \sqrt{2}^1 = a_0$. For the inductive step, assuming that $a_{n-1} > a_{n-2}$, it then follows that $a_n = \sqrt{2}^{a_{n-1}} > \sqrt{2}^{a_{n-2}} = a_{n-1}$.

Lemma 2: $a_n < 2$ for all $n$.

Proof by induction:

Obviously, $a_0 =\sqrt{2} < 2$. Since $a_{n-1} < 2$, it follows that $a_n = \sqrt{2}^{a_{n-1}} < \sqrt{2}^2 = 2$.

From these two lemmas, we have that $a_n$ is bounded above by $2$ and is increasing, which means $\lim_{n \to \infty} a_n$ exists. We know from above that if such a limit exists, it must be one of the fixed points 2 or 4. And since our series $a_n$ never gets near $4$, the limit must be $2$.

Community
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user35734
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The problem here is that $4$ is not in the range (image) of the power tower function (which is only defined for $x\in [e^{−e},e^{1/e}]$). So, the assumption that the power tower equals 4 is false and thus gives a false "solution".

Friedrich Philipp
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