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What is the name for the class of functions of the form:

\begin{equation} f(x)=ax^{-b}+c\;. \end{equation}

With $c=0$ and $a$ satisfying some normalization property these functions are called power laws, because they describe a probabilistic law. Even with $c=0$ I don't know how these type of functions are called. It seems to me that 'hyperbolic' is not the right term here.

Marlo
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    not sure what power laws have to do with probabilistic laws – gt6989b Oct 02 '17 at 16:39
  • I would call it a shifted power law, even though this is ambiguous because the shift could be on $x$ as well. –  Oct 02 '17 at 16:42
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    Mathematics is about analysis and reasoning, not about calling names, in my humble opinion. If people are fond of extensive classifications with three part names in Latin (and you need those long names, due to the sheer number of objects to be named), I can recommend other subjects, like entomology. –  Oct 02 '17 at 16:48
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    @ProfessorVector: Don't you mean etymology? – David G. Stork Oct 02 '17 at 17:28
  • thanks ProfessorVector for that enlightening lecture. @gt6989b: In probability theory the law of a random variable $X$ is the function: $a\mapsto \mathbb P(X\leq a)$ ... or maybe its derivative, I don't remember. If $X$ has a density $f(a):=\frac{\partial P(X\leq a)}{\partial a}=a^{-\beta}/N$, then this law is referred to as the \textit{power law}. – Marlo Oct 02 '17 at 18:30
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    @YvesDaoust Shifted power law sounds most appropriate to me as well – gt6989b Oct 02 '17 at 19:36
  • @ProfessorVector: I can't agree with you. Without names, which are shorthands for lengthy definitions, any mathematical statement would be nightmarish. "The positive number which, when multiplied by itself, yields the value $2$ does not belong to the set of the numbers which are the result of the division of an integer number by another." And "There is only $1$ element in the set of numbers such that there are exactly $2$ integers that yield no remainder when dividing by them, which belongs to the set of integers that are the product of an integer and $2$." –  Oct 02 '17 at 20:01
  • @Yves Daoust I objected against replacing a clear, symbolic definition like $\begin{equation} f(x)=ax^{-b}+c \end{equation}$ by some vague and ambiguous "shifted power law", and you are insinuating I object against using a symbol $1$ instead of a wordy definition?! You must be kidding. –  Oct 03 '17 at 07:23
  • @ProfessorVector: who said that the expression would be used without a definition ? Please don't be bad faith. –  Oct 03 '17 at 07:34

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