# Tag Info

43

Let $x_1 = x$ and by induction $x_{n+1} = x^{x_n}$: so $x_1 = x$ is rational by hypothesis, $x_2 = x^x$ is algebraic irrational, $x_3 = x^{x^x}$ is transcendental by the Gelfond-Schneider theorem, and the question is to prove that $x_4, x_5,\ldots$ are transcendental (or at least, irrational). I will assume Schanuel's conjecture and use it to prove by ...

41

Might as well... The power tower $x^{x^\ldots}$ is equivalent to the function $\exp(-W(-\log\,x))$, where $W(z)$ is the Lambert function, in the range $e^{-e}\leq x\leq e^{1/e}$ (as Norbert mentions in the comments; see also equation 13 in the MathWorld entry linked to). $\exp(-W(-\log\,x))$ can be inverted, like so: \begin{align*} ... 35 If a solution exists, you have x^2 = x^{(x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}})}=2 $$which means what you've got. (This part has been mentioned to be wrong for logarithmic properties misuse reasons) Not both of these are not solutions, since$$ 1 = \log_2(2) = \log_2(x^{x^{\dots}}) = x^{x^{\dots}} \log_2 (x) = 2 \log_2(x) $$and in the case x = -\sqrt 2, ... 33 My personal opinion is that the exponential is not naturally regarded as the next step in the progression from addition to multiplication, so there's no reason to expect it to share properties with the other two. Notice what happens if you demand that all of your quantities have units. Addition is an operation you do to two quantities with the same units: ... 30 Let's add the hypothesis that x>0 to the problem, so that it's clear your derivations are correct. Pay attention to what you've proven: If x^{x^{\cdot^\cdot}} = 2, then x = \sqrt{2} If x^{x^{\cdot^\cdot}} = 4, then x = \sqrt{2} This is very different from If x = \sqrt{2}, then x^{x^{\cdot^\cdot}} = 2 If x = \sqrt{2}, then ... 28 What you're after is called tetration (the example you computed is given here), and it has an active community of people who are interested in it (though my sense is that it is not quite in the mainstream of mathematics research at the moment, for whatever reason). The Wikipedia page indicates that the problem of extending tetration to arbitrary real powers ... 28 This is not an answer to your question but a long comment on its motivation. Multiplication is at least two conceptually distinct things, only one of which can reasonably be described as repeated addition: The natural map \mathbb{Z} \times A \to A given by (n, a) \mapsto na where A is an abelian group; this really is repeated addition, and is in ... 27 You can readily check this using an independent method. Let x_n + i y_n\in\mathbb{C} be the value of a tower of n copies of e with a single i at the top, so that x_{n+1}+iy_{n+1}=\exp(x_n+iy_n). This can be rewritten as e^{x_n}\left(\cos y_n+i \sin y_n\right), giving the recursion$$ x_{n+1}=e^{x_n}\cos y_n,\qquad y_{n+1}=e^{x_n}\sin y_n. $$... 23 Here is a thought, which is not a full answer, but too long for a comment. Addition a + b means something like: Add 1 to a, b times = 1\cdot b + a. Commutativity here means that 1\cdot b + a = 1\cdot a + b. We can see that it's only through the fortunate use of 1 that this is commutative; cb + a \not= ca+b in the general case. If we defined ... 23 Here's a question: if x = \sqrt[3]{3}, then what is the value of y = x^{x^{x^{\dots}}}? As you say, if y is a solution to this, then y \ln x = \ln y, so that$$ \frac{\ln y}{y} = \ln(x) = \frac{\ln(3)}{3} $$Now, how can we "solve this" for y? As it ends up, there are two solutions for y. The answer you are getting is the second root, y ... 19 Let us denote x=i^{i^{i^{i^\cdots}}}. Then we have$$i^x=x.$$It looks like the solution is x= \frac{2i}{\pi} W(-i\pi/2) with W Lambert's W function. Now, W is multivalued. You have to figure out which of the different branches x converges to (and if it converges at all). Numerically, you find (using the principal branch of the logarithm to ... 18 Here is a numerical result supporting Fabian's argument. Here, the complex logarithm$$ z^{w} := \exp (w \operatorname{Log} z) $$is defined via the principal value \mathrm{Log} of the logarithm, defined on \Bbb{C} \setminus (-\infty, 0]. 18 It is easier to prove that the inverse function is strictly increasing. Since the inverse function is just:$$ g(x) = \left(\frac{1}{x}\right)^{-\frac{1}{x}}$$with a change of variable everything boils down to proving that h(x)=x^x is increasing over \left[\frac{1}{e},1\right]. That is trivial since:$$ h'(x) = h(x)\cdot\frac{d}{dx}\log h(x) = (1+\log ...

17

It is a known open problem. It is also open for all equations of the form ${^n q}=2$ or ${^n q}=3$ for integer $n>3$. For $n=3$ the roots are known to be irrational, but not known to be algebraic or transcendental.

17

Ah yes, a fave topic of mine. Basically, there is no universally-agreed on way to do this. The problem is, that, in general, there isn't a unique way to interpolate the values of tetration at integer "height" (which is what the "number of exponents in the 'tower'" may be called). So in theory, you could define it to be anything. In the case of ...

17

We can define $x=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}$ as follows: Let $x_1 = \sqrt 2$ and $x_{n+1} = (\sqrt 2)^{x_{n}}$ We can show $x_n \lt 2\ \forall n$ by induction, since if $y \lt 2$, then $(\sqrt 2)^y \lt 2$. And $x_n$ is clearly monotonically increasing, so $x_n \to x$. But $$x_{n+1} = (\sqrt 2)^{x_{n}}$$ so taking limits, we ...

16

It might be not a direct answer to your question, but it is possible that there is no 2nd term of the continued fraction in question. I believe it is a long-standing open problem if $\,{^5 e}\in\mathbb{N}$, and, in general, for every integer $n \ge 5$, if $\,{^n e}\in\mathbb{N}$ (and also, for every integer $n \ge 4$, if $\,{^n \pi}\in\mathbb{N}$). It is ...

15

$\textbf{hint:}$ $$f(x) = x^{f(x)}$$ Then proceed

15

I believe, it is a known open problem. Ditto for ${^3 x}=3$, ${^3 x}=4$, ${^3 x}=5$ (left superscript denotes tetration).

15

$$x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}=y$$ $$\ln{x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}}=\ln y$$ $$x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}\ln{x}=\ln y$$ $$y\ln{x}=\ln y$$ $$\ln{x}=(\ln{y})/y$$ $$\ln{x}=\ln({y^{1/y}})$$ $$x= y^{1/y}$$ ...

13

A quick hand calculation gives \begin{align} 7^1 &\equiv 7 \pmod{100} \\ 7^2 &\equiv 49 \pmod{100} \\ 7^3 &\equiv 43 \pmod{100} \\ 7^4 &\equiv 1 \pmod{100} \end{align} So it reduces to the problem of calculating the value of $7^{7^{7^{7^{7^7}}}} \pmod 4$. And $7^2 \equiv 1 \pmod 4$, so it reduces to the problem of calculating ...

12

$7^4 = 2401 \equiv 1 \pmod{100}$, so you only need to calculate $7^{7^{7^{7^{7^7}}}} \pmod{4}$. We know that $7 \equiv -1 \pmod 4$ and $7^{7^{7^{7^7}}}$ is odd, so $7^{7^{7^{7^{7^7}}}} \equiv -1 \equiv 3 \pmod 4$, and then $$7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod {100}$$

12

Letting $h(x)$ be your infinite power tower, one can solve the functional equation $h(x)=x^{h(x)}$ in terms of the Lambert function $W(x)$, the inverse function of $x\exp\,x$. More specifically, we have $$h(x)=\exp(-W(-\log\,x))$$ One can then apply the chain rule as usual. The formula $$W^\prime(x)=\frac{\exp(-W(x))}{1+W(x)}$$ is easily derived through ...

11

Clearly $x=-1$ is a solution. Here I'll prove that it's the only real solution, complex solutions are a different matter. Given $z,\alpha \in \mathbb{C}$ , we have $$z^{\alpha} = \exp(\alpha [\log |z| + (\arg z)i])$$ So let $x = re^{i\theta} \in \mathbb{C}$. Then \begin{align} x^{x^x} &= \exp(x^x[\log r + \theta i]) \\ &= \exp\big( \exp(x\log r ... 11 With Newton-iteration in Pari/GP I get [update:] 12 solutions: x ------------------------------------------ x0 = -1 (pure red colour) x1 = -0.158908751582 + 0.0968231909176*I (pure blue colour) x2 = 2.03426954187 + 0.678025662373*I (pure green colour) x3 = ... 11 Here's the proof of a theorem due to Thron (1956), extracted from a article of Laurent Bonavero (available at his webpage). Theorem. There is no entire function f (that is f:\mathbb C \to \mathbb C holomorphic) such that \exp = f \circ f. Proof. If such a function f exists, then f(\mathbb C)= \mathbb C^*. Indeed, f(\mathbb C) \supset ... 11 Let a_n be a tower of n x, e.g. a_1=x, a_2=x^x, for some fixed x. Suppose the sequence (a_n) converges to a limit L. Then, the sequence (a_{n+1}) also converges to L. But also, a_{n+1}=f(a_{n}) for any n, where f(y)=x^y. Since f is a continuous function in y for any x>0, (a_{n+1}) converges to f(L). By the uniqueness of ... 10 Look at this answer: http://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/44727#44727 In short, the analytic solution isg^{[1/2]}(x)=\phi(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}g^{[k]}(x)g^{[1/2]}(x)=\lim_{n\to\infty}\binom {1/2}n\sum_{k=0}^n\frac{1/2-n}{1/2-k}\binom nk(-1)^{n-k}g^{[k]}(x)$$... 9 Before proving your observations, we need to know a few things; I'll present them as lemmas. Let's denote$$a_b=\underset{b\text{ times}}{\underbrace{a^{a^{a^{\cdots^a}}}}} for $a,b\in\mathbb Z^+$. Lemma 1. $2^{k+1}\mid 7_{k+1}-7_k$ for $k\geq0$. Proof. By induction. For $k=0$ this is trivial. Let $k\geq0$ and assume $7_k\equiv7_{k-1}\pmod{2^k}$. By ...

9

You have merely shown that the equation $\sqrt 2^y = y$ has more than one solution. Then you assumed that $x^{x^{x^\ldots}}$ somehow made sense and tried to talk about it as if it meant "the solution $y$ of $x^y = y$". Which of course is nonsense when the equation has several solutions.

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