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38

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 ...


35

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*} ...


33

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


29

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: ...


28

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 ...


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


26

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 ...


22

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 ...


21

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 ...


20

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 ...


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 ...


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.


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

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


13

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


12

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

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 ...


11

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

$7^4 = 2401 \equiv 1 \pmod{100}$, 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 $$7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod {100}$$


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 ...


8

You have $x^a=a$ and hence $x=a^{1/a}$. It converges for all $a\in[e^{-e},e^{1/e}]$.


8

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.


7

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 ...


7

Taking $n=7$ and looking for the last three digits for an example, note that $7^m \pmod {1000}$ is periodic with period $20$. You can check this easily with a spreadsheet. So now, we only need the tower above the first $7$ to $\pmod {20}$. That has period $4$, so we only need the tower above the first two $7$'s $\pmod 4$. That has period $2$, and the ...


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 ...


7

It seems to me that before much more progress can be made in the calculus of ${}^xy$, more fundamental questions have to be answereed, such as, how to define ${}^xy$ for rational $x$? It's clear how the OP's definition works if $x$ is a non-negative integer; but how do we define ${}^xy$ if, say, $x = 7/2$? What then is "one-half" of an occurrance of $x$ in ...


7

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

Hint: since the number of digits of any number $m$ is $\lfloor \log_{10} m \rfloor +1$, you can compute the base $10$ log of $N$, which is a much smaller number.



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