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1

As you have probably surmised, $x \uparrow^n 2 = x \uparrow^m 2$ and $2 \uparrow^p x = 2 \uparrow^q x$ have solutions at $x=1$ and $x=2$. If $m=n$ or $p=q$ then of course any $x$ will be a solution; if $m \neq n$ or $p \neq q$ then there are no other solutions. First, I will prove the following useful theorems: Theorem. For $a \ge 2$ and $b \ge 1$ a. $a ...


0

Yes. Here is a simple proof by induction: Theorem. $a_{n+2} > 4 b_n$ If $n=1$, then $a_3 = 16 > 4*3 = 4b_1$. Now assume $a_{n+2} > 4 b_n$: $$a_{n+3} =2^{a_{n+2}} > 2^{4b_n} = 16^{b_n} > 4^{b_n} * 3^{b_n} > 4 * 3^{b_n} = 4b_{n+1}$$ and the result follows.


3

Knut's up-arrow notation works as follows : $a\uparrow b=a^b$ $a\uparrow \uparrow b=a\uparrow a\uparrow ...\uparrow a \uparrow a$ with $b$ $a's$ $a\uparrow \uparrow \uparrow b=a\uparrow \uparrow a \uparrow \uparrow ... \uparrow \uparrow a \uparrow \uparrow a$ with $b$ $a's$ and so on. Note that the calculation is done from RIGHT , so $3\uparrow ...


0

Let $a = 9^{9^9} $ Then $$ a^a = (9^{9^9})^{9^{9^9}} = 9^{9^9 \cdot 9^{9^9}} \overset{?}{<} 9^{9^{9^{9^{9^9}}}} $$ Take logarithm to base 9 $$ {9^9 \cdot 9^{9^9}} \overset{?}{<} {9^{9^{9^{9^9}}}} $$ Take logarithm to base 9 $$ {9 +{9^9}} \overset{?}{<} {{9^{9^{9^9}}}} $$ Take logarithm to base 9 making the solution perfectly obvious: $$ \log_9({9 ...


0

In fact $^2(^{n-3}9) < {^{n-1}}9$ for all $n \ge 3$. This is easy to check for $n = 3,4$, and for $n \ge 5$ we have $$ ^2(^{n-3}9) = (^{n-3}9)^{^{n-3}9} = 9^{(^{n-4}9)(^{n-3}9)} = 9^{9^{^{n-5}9 + ^{n-4}9}} < 9^{9^{^{n-3}9}} = {^{n-1}}9. $$ In your argument, you incorrectly state that $^2(9^{9^9}) = 9^{9^{9 * 9^{9^9}}}$, when in fact $^2(9^{9^9}) = ...


1

You cannot have a closed form but you can have this nice expansion $$x^{x^{x^{x\,\cdots}}}=-\frac{W(-\ln(x))}{\ln(x)}=\sum\limits_{n=0}^{+\infty} \frac{(n+1)^n}{(n+1)!}\ln^{n}(x)=1+\ln(x)+\frac{3^2}{3!}\ln^2(x)+\frac{4^3}{4!}\ln^3(x)+...$$ Knoebel, 1981, Exponentials reiterated Eisenstein, 1844, Entwicklung von $a^{a^{a^{.^{.^{.}}}}}$ This gives: $$\int ...


2

$$\Large \int_{a}^{b} x^{x^{x^{x\,\cdots}}} \, dx = -\Large \int_{a}^{b} \frac{W\left(-\ln(x) \right)}{\ln(x)}\, dx $$ Where W is the Lambert W function. The integral is convergent on $\quad e^{-e}\leq a\leq e^{1/e} \quad \text{and} \quad e^{-e}\leq b\leq e^{1/e}$ There is no closed form with a finite number of standard functions. Example of serie ...


2

As Gottfried hints, there is yet another solution to $^{0.5}i \approx 1.07571355731 + 0.873217399108i$ I will use this question to describe a unique Abel function for $f(z)=i^z$. I wrote a pari-gp complex base tetration program available for download at math.eretrandre.org. The results posted here were generated with that program. I will use this ...


2

The number $ ^\infty i \approx 0.4383+0.3606i$ where $ i^{^\infty i} = {^\infty i}$ is a hyperbolic fixed point. Let $\epsilon$ be a very small number such that $\epsilon^2 \approx 0$. Then $\large ^\infty i + \epsilon \rightarrow i^{^\infty i+ \epsilon} = {^\infty i} \times i^\epsilon = {^\infty i} \times e^{Ln(i) \epsilon} = {^\infty i}(1 + Ln(i) ...


5

You can find a non-trivial interpolation for the fractional iteration-height when you write down the consecutive iterates in log-polar-form (with center at the fixpoint). The nearer you come to the final fixpoint the log of the distance as well as the angle come nearer and nearer to a linear relation with the index and this suggests an obvious method of ...


2

Consider the functions $$f_n(z) = \frac{\sin \pi z}{z - n}$$ for $n\in\Bbb Z$. Note that $f_n(k) = 0$ if $n \ne k$ and $f_n(n) = 1$. Define $$f(z) := \sum_{n=-\infty}^{\infty} a_{|n|}f_n(z) = \sin \pi z \sum_{n=-\infty}^{\infty} \frac{a_{|n|}}{z - n} = \sin \pi z\left(\frac{a_0} z + 2z\sum_{n=1}^{\infty} \frac{a_n}{z^2 - n^2}\right)$$ (for the appropriate ...


1

I think you can find in WP's "tetration" the criterion for the powertower over the reals, which was established by L. Euler (but independently by others) Over the complex numbers the range of convergence of the infinite tetration was determined by W. Thron(1957) "Convergence of Infinite Exponentials with Complex Elements" and based on his work ...


0

I don't know any theorems for convergence of power towers, but here is food for thought. Let's define pt(a, n) (pt as short for "power tower") of a sequence a as: pt(a, 1) = a_1 pt(a, 2) = a_1^(a_2) pt(a, 3) = a_1^(a_2^(a_3)) ... pt(a, n) = a_1^(a_2^(a_3^...^(a_n)...)) For lim [n -> oo] pt(a, n) to exist, pt(a, n) should be very near to pt(a, n+1). This ...


0

Clarification request: are you defining your series as: $s(n)=s(n-1)^{a(n)}$ If so, you'll just get $s(n)=a(1)^{\prod _{i=2}^n a(i)}$ Or are you defining it so that you replace a(n-1) with $a(n-1)^{a(n)}$ in the previous term? In other words, $s(n)=\left(s(n-1)^{\frac{1}{a(n-1)}}\right)^{(a(n-1)^{a(n)})}$ It looks like you're doing the latter ...


1

You made a small mistake: $$c=z^{z^{z^{\cdots}}}=i^{i^{i^{\dots}}}$$ Hence: $$z=i\ne i^i$$ $$i=e^{i\pi/2}$$ So, reevaluating, manipulating your formula with a Lambert W function identity: $$c=e^{-W(-\ln(e^{i\pi/2}))}=e^{-W(-i\pi/2)}=\frac{W(-i\pi/2)}{-i\pi/2}$$ This should then evaluate to the actual answer.


0

There is no universally accepted way to extend hyperoperations to the Real numbers. Refer to the very good answer to this questions Example $x$, $y$ and $z$ values for $x\uparrow^\alpha y=z$ where $\alpha\in \Bbb R-\Bbb N$ But there are many others, like How to evaluate fractional tetrations?



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