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Let $\mathbb R^+$ be the set of positive real numbers and let the natural numbers be $\mathbb{N} = \{ 1,2,3,\ldots \}$. Let $\star_x$ be an operator on $\mathbb{R}^+$ such that

\begin{align} a \star_0 b &= a + b \\ a \star_1 b &= a \cdot b \\ a \star_2 b &= a^b \end{align}

And so on. In particular, if $b\in \mathbb{N}$ and $x \in \mathbb{N}$, then $\star_x$ satisfies the recursive property

$$ y \star_x b = \underbrace{y \star_{x-1} y \star_{x-1} \cdots \star_{x-1} y}_{b \text{ times}} $$

Is there a natural way of extending $\star_x$ for any $x \in \mathbb{R}^+$ so that the map $(a,b) \mapsto a \star_x b$ is a continuous function of $x$? (EDIT: And, hopefully, a continuous function of $a$ and $b$?)

eepperly16
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  • What would it be for $x=1/2$ [midway between sum and product]? – coffeemath Jan 18 '17 at 17:07
  • Um.... $a \star_x b = a\star_{\lfloor x \rfloor}b + (a\star_{\lceil x \rceil}b - a\star_{\lfloor x \rfloor}b)*(x - \lfloor x \rfloor)$? i.e. just connect $a \star_x b$ to $a\star_{x+1} b$ with a straight line. You don't have any other condition to be satisfied. – fleablood Jan 18 '17 at 17:29
  • @fleablood That certainly works, and is arguably "natural". I was hoping for something more elegant (in the sense that it might satisfy some nice recursive properties), but that definitely works. Thanks! – eepperly16 Jan 18 '17 at 17:32
  • But it's not differentiable or smooth which would be tighter conditions. And that would make for a very difficult problem which.... I'll let someone else contemplate. – fleablood Jan 18 '17 at 17:35

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