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let $x^{/1/}=x+x$ (addition)

$x^{/2/}=x.x$ (multiplication)

$x^{/3/}=x^x$ (exponentiation)

$x^{/4/}={}^xx$ (tetration)

and so on.....

I have the following questions:

$(1)$ Can we define an operation between two known operations (Ex.addition and multiplication), where $n$ in $x^{/n/}$ is fractional (Ex. $x^{/1.5/}$)?

$(2)$ Is there a closed form for $x^{/x/}$?

$(3)$ Is infinitation ($x^{/\infty/}$) equal to infinite tetration $\left( x^{x^{x...}}\right)$?

I think the answer to the last question is true as infinite tetration is equal to infinite pentation (as both of them equal to $x^{x^{x^{.^{.^.}}}}$ when we convert them to their exponentiation form.) Following the same logic, any $n$-ation is equal to $x^{x^{x^{.^{.^.}}}}$, and therefore infiniation is also equal to infinite tetration.

Edit: Does this notation seem useful for defining operations?

Simply Beautiful Art
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Mathphile
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    In case you weren't aware https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation seems similar. Not sure about any of the questions though honestly. – xmq May 17 '19 at 23:41
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    Regarding 3, why would it be equal to infinite tetration and not to infinite addition, multiplication or exponentiation? – Jackozee Hakkiuz May 18 '19 at 04:41
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    @xmq i am aware of Knuth up arrow notation, but it has its limitations. Ex. You cannot represent iterated addition with Knuth up arrow notation. – Mathphile May 18 '19 at 06:49
  • @JackozeeHakkiuz as i have written in my question, any infinite $n$-ation can be represented as infinite tetration. – Mathphile May 18 '19 at 06:53
  • These definitions don't make sense if $x$ is non-integer, so you might adapt the more general notation of hyperoperations (https://en.wikipedia.org/wiki/Hyperoperation). In this notation, $a^{/n/}=a[n+1]2$. Your claim about infinite tetration and infinite pentation is vague and informal. You should note that infinite tetration is the limit of $x[4]n$ as $n\rightarrow\infty$, whereas $x^{/\infty/}$ is the limit of $a[n+1]2$ as $n\rightarrow\infty$, so these are completely different and inspection of the latter seems to suggest it rapidly diverges either way. – Thorgott May 18 '19 at 16:17
  • @Thorgott that is exactly the motivation behind my question. Is there some way to define these fractional operations. If not then why can't we try to develop a new way to define this? – Mathphile May 18 '19 at 21:48
  • @Thorgott Also infinite tetration converges for $e^{-1} \lt x \le e^{1/e}$ – Mathphile May 18 '19 at 21:55
  • I'm aware infinite tetration converges for these values. The entire point of my comment was that infinitation and infinite tetration are completely different beasts and that infinitation is rapidly divergent (except for the $x=1$ case), so the answer to $(3)$ is negative. Your comment on why you suspect them to be equal suggests you might be misunderstanding these recursions. As for the other question, you can definitely try; the difficulty is making such an extension meaningful. This seems to be very difficult and I certainly have no approach to do so, nor have I heard of one. – Thorgott May 18 '19 at 22:29
  • $(1)$ is a dupe of Continuum between addition, multiplication and exponentiation? $(2)$ is ill-defined. What does "closed form" mean? Most standard meanings would classify your $x^{/4/}$ as having no closed form. And for $(3)$, I see no reason for it to be true at all, aside from the case of $x=1$. If one excludes $x=1$, then either it is never true, or you have to actually define what this is supposed to mean when $x$ is not an integer. – Simply Beautiful Art Jul 02 '19 at 21:18
  • @Thorgott Infinitation is not divergent for all $x \neq 1$. For example, it seems that $2 \mathbin{[\infty]} 2 = 4$ and $0 \mathbin{[\infty]} 0 = 1$. See my answer below. – user76284 Oct 27 '19 at 17:05

1 Answers1

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The standard notation for your $x^{/n/}$ is $H_n(x, x)$.

(3) Is infinitation ($x^{/\infty/}$) equal to infinite tetration $\left( x^{x^{x...}}\right)$?

You're asking whether $H_\infty(x, x) = H_4(x, \infty)$. The answer is no. Notice that

\begin{align*} H_n(a, -1) &= 0 & n \geq 4 \\ H_n(a, 0) &= 1 & n \geq 3 \\ H_n(a, 1) &= a & n \geq 2 \\ H_n(2, 2) &= 4 & n \geq 1 \end{align*}

Therefore \begin{align*} H_\infty(-1, -1) &= 0 \\ H_\infty(0, 0) &= 1 \\ H_\infty(1, 1) &= 1 \\ H_\infty(2, 2) &= 4 \end{align*}

But \begin{align*} H_4(-1, \infty) &= -1 \\ H_4(0, \infty) &= \frac{1}{2} & \lim_{\varepsilon \rightarrow 0^+} \lim_{b \rightarrow \infty} H_4(\varepsilon, b) \\ H_4(1, \infty) &= 1 \\ H_4(2, \infty) & \quad \text{diverges} \end{align*}

so they disagree at 3 of these 4 points.

$(1)$ Can we define an operation between two known operations (Ex.addition and multiplication), where $n$ in $x^{/n/}$ is fractional (Ex. $x^{/1.5/}$)?

The names typically given to $H_{0.5}$ and $H_{1.5}$ are "halfation" and "sesquation", though there's little information about them.

user76284
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  • So does $H_{\infty}(x,x)$ only converge only $x=-1, 0, 1, 2$? If not then could you tell me it's radius of convergence? – Mathphile Nov 27 '19 at 02:27