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31

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 ...


31

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*} ...


31

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 $x = -\sqrt 2$, ...


28

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: ...


26

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 $$ x_{n+1}=e^{x_n}\cos y_n,\qquad y_{n+1}=e^{x_n}\sin y_n. $$ ...


22

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 it 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 ...


19

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 ...


18

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 ...


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 ...


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 ...


10

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 ...


10

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 ...


7

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 ...


7

When I first read your question, I expected that it must mean that addition would possess some obscure property that multiplication lacks, after all, both the additive structure and multiplicative structure are abelian groups, so you'd expect something like this to just generalize. But after some thinking, I realized that this wasn't the case, and instead ...


6

Note that $$7^{(7^k)}\equiv 6\bmod 13\iff 7^k\equiv 7\bmod 12\iff k\equiv 1\bmod 2$$ and that for any $n\geq 2$, $$7\uparrow\uparrow n=\underbrace{7^{7^{.^{.^{.^{7}}}}}}_{n\text{ 7's}}=7^{(7^{k})}$$ for some $k\equiv 1\bmod 2$ - specifically, $k=7\uparrow\uparrow (n-2)$. Here $\uparrow\uparrow$ is Knuth's up-arrow notation. Thus, for any $n\geq 2$, ...


6

Reading the other answers, I realize this is a longer way than necessary, but it gives a more general approach for when things are not as convenient as $7^4\equiv 1\bmod 100$. Note that, for any integer $a$ that is relatively prime to $100$, we have $$a^{40}\equiv 1\bmod 100$$ because $\varphi(100)=40$, and consequently $$a^m\equiv a^n\bmod 100$$ whenever ...


6

Look at this answer: http://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/44727#44727 In short, the analytic solution is $$g^{[1/2]}(x)=\phi(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}g^{[k]}(x)$$ $$g^{[1/2]}(x)=\lim_{n\to\infty}\binom {1/2}n\sum_{k=0}^n\frac{1/2-n}{1/2-k}\binom nk(-1)^{n-k}g^{[k]}(x)$$ ...


6

Daniel, One solution would be the half iterate generated from real valued tetration, but one can also start with the $z_0\approx0.318+1.337i$ fixed point, and develop the half iterate directly from there. Of course, such a solution is not real valued at the real axis. Then, $z_0$ is defined such that $\exp(z_0)=z_0$. Then there is a Schroeder function ...


6

First, I would recommend reading "Exponentials Reiterated" by R.A.Knoebel, http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=3087&pf=1. It is by far the most comprehensive resource on infinite tetration. Second, your question might have been answered sooner if you had posted on the Tetration Forum, ...


5

We will need the following slight generalization of the totient theorem: $a^{k + \varphi(n)} \equiv a^k \bmod n$ for all $a$ provided that $k$ is at least as large as the largest exponent in the prime factorization of $n$. It follows that $$a^b \equiv a^{b \bmod 4} \bmod 5$$ provided that $b \ge 1$, and $$b^c \equiv b^{c \bmod 2} \bmod 4$$ provided that ...


5

This may be helpful. Let $$ f(x) = \frac{-1 + \sqrt{1 + 4 x}}{2}, \; \; x > 0 $$ We use a technique of Ecalle to solve for the Fatou coordinate $\alpha$ that solves $$ \alpha(f(x)) = \alpha(x) + 1. $$ For any $x > 0,$ let $x_0 = x, \; x_1 = f(x), \; x_2 = f(f(x)), \; x_{n+1} = f(x_n).$ Then we get the exact $$ \alpha(x) = \lim_{n \rightarrow ...


5

There's a pretty good heuristic that comes from set theory. If you have two sets, $X,Y$, then: $X\times Y=$ all ordered pairs $(x,y):x\in X, y\in Y$ $Y^X=$ all functions $f:X\rightarrow Y$ Now if $X$ and $Y$ happen to be finite sets, with $|X|=m$ and $|Y|=n$, then we have $|X\times Y|=mn$ and $|Y^X|=n^m$ as usual. So it should be clear that $|X\times ...


5

Tetration is a natural extension only of the integer-valued notions of addition, multiplication, etc. For example, the notion that multiplication is repeated addition fails when you start multiplying by non-integers, and even if you hack together an explanation for the rationals, it requires even more care for the irrationals. The same is also true for ...


5

As Will Jagy already pointed out, there is no accepted solution for this. There is a formal procedere which can sometimes lead to a meaningful/approximate answer; but this is then dependent on the convergence of some series, which occur in that procedere, and also, in which way we want to make sense of noninteger powers of negative or complex numbers. I mean ...


4

Well, why not. The goal of such topics is to take a function $f(z)$ defined in some region of the complex plane, often including the real line or part of it, then defining $f_1(z) = f(z)$ and $f_0(z) = z.$ The desire is to be able to define $f_s(z)$ for real or even complex $s,$ such that we get a group $$ f_{s+t}(z) = f_t(f_s(z)) = f_s(f_t(z)). $$ Now, ...


4

When dealing with power towers with bases not relatively prime to the modulus, it's useful to employ the Chinese Remainder Theorem. And then repeatedly apply the Euler's Theorem. $2 \uparrow \uparrow 10 \pmod {2^9} = 0$, so we only need to calculate $2 \uparrow \uparrow 10 \pmod{5^9}$. By Euler's Theorem, we need to first study $2 \uparrow \uparrow 9 ...


4

This might be slightly circular since the definition of exponentiation and logarithm comes after addition and multiplication, but consider: If we perform a certain unary operator on the two arguments, perform addition on them, and then take the inverse, we get $e^{\log a + \log b} = ab$. If we do the same with multiplication, $e^{\log a \log b} = a^{\log b} ...


4

I think I can dress Qiaochu's answer up in geometric terms. Commutativity of addition and multiplication (in $\bf N$) can be seen as an observation of symmetry of counting blocks that are arranged in certain ways. I recall in elementary school seeing big numbers depicted concretely in our textbooks as piles of blocks. (Their system involved $a_3$-many ...


4

Is my proof correct for the cases n=0 and n=1 ? Yes, it is correct. Can this process be continued to enable an induction proof ? Sure. Let's write $T(a,n) = a \uparrow\uparrow n$. Then $$\begin{align} T(a,n+2) - T(a,n+1) &= a^{T(a,n+1)} - a^{T(a,n)}\\ &= a^{T(a,n)}\left(a^{T(a,n+1)-T(a,n)}-1\right). \end{align}$$ The ...


4

It is a non-converging asymptotic series. See Will Jagy's answer on mathoverflow: http://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765 There is a singularity at the origin. I would add a little bit to Will Jagy's answer. There are actually four different disconnected half iterate functions, in the four different ...



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