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What (simple) function has the following properties:

  • It maps the positive reals (except for 1) to positive reals: $f: (0,1) \cup (1, +\infty) \to (0, +\infty)$.

  • It is symmetric about reciprocals: $f(x) = f(1/x)$.

  • It has limits $\lim_{x \to 0} f(x) = 0, \lim_{x \to \infty} f(x) = 0$, and $\lim_{x \to 1} f(x) = \infty$.

  • It may incorporate one arbitrary constant (a parameter), which we'll call $d$, ranging from $(0, + \infty)$. The function $f$ is linear in $d$.

Surprisingly, I'm not able to construct any function which has all of those properties.

Background: In the context of a particular problem, I've been able to show that the solution involves a function $f$. I've yet to figure out what $f$ is, but I can show it has the properties above. I'd like to use them to form a candidate function, which I can then perhaps use as an ansatz or refine to get the solution.


Clarification

Of course, I could piecewise construct such a function, but this is unlikely to be a good ansatz. I'm looking for a function which has these properties "naturally", for some definition of "naturally". As an inspiration, consider the gamma function: There are many (infinite) functions that equal factorial on the integers, but only one "natural" function.

I don't know how to define "natural" - if I did, that would probably be my answer - but piecewise is certainly not.


More information

For those interested, the problem I'm working on is the Circle of Apollonius (Interactive), which has complex equation $$|z - z_0| = \rho |z - z_1|.$$

Source: http://www.mathematicalgemstones.com/gemstones/pearl/circles-of-apollonius-and-magnetism/

I was trying to find a function that maps the constant $\rho$ to the radius of the circle, and deduced that it must have the above properties (with $d = |z_1 - z_0|$).

Indeed, it seems the answers from "Fluffy Alpaca" and aschepler are correct, in that the radius seems to be proportional to $$\frac x {(x-1)^2} = \frac 1 {x + \frac 1 x - 2}$$ where $x = \rho^2$. I've yet to determine the constant of proportionality, but assume it is linear in $d$ (because the radius must be linear in some measure of distance, which can only come from $d$), suggesting $$\text{radius} = \frac d {x + \frac 1 x - 2}.$$

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  • $\begingroup$ Well, just define something that works on $(0,1)$ and extend it by symmetry. $f(x)=\frac 1{1-x}-1$ for instance, but that choice isn't terribly important... all you need is the right behavior at $0$ and $1$. $\endgroup$
    – lulu
    Jul 11, 2023 at 12:16
  • $\begingroup$ @lulu: you beat me by 56 seconds! $\endgroup$
    – TonyK
    Jul 11, 2023 at 12:20
  • $\begingroup$ Maybe something like ${1\over|\log(x)|}$? $\endgroup$
    – abc
    Jul 11, 2023 at 12:30

3 Answers 3

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Let $$g(x)=x+\frac{1}{x}$$ $$f(x)=\frac{1}{g(x)-2}$$

This should work - try playing around a bit with it on desmos for example.

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    $\begingroup$ Thanks. To meet the positive criteria, I can use $h(x) = [f(x)]^2$. $\endgroup$ Jul 11, 2023 at 12:27
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    $\begingroup$ @SRobertJames There's no need to, because for $x>0$ we have $g(x) \geq 2$ and so $f(x)>0$ $\endgroup$ Jul 11, 2023 at 12:33
  • $\begingroup$ Interesting: For at least one (unknown) value of $d$, the answer seems to be $[\frac {1+x}{1-x}]^2 - 1$, which, if my division is correct, equals $4 \cdot f(x)$. Is that correct? $\endgroup$ Jul 11, 2023 at 12:59
  • $\begingroup$ This indeed seems to be the function! - See updated post. $\endgroup$ Jul 11, 2023 at 13:21
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One function satisfying these properties, and without "definition by cases", is

$$ f(x) = \frac{d}{(\ln x)^2} $$

This comes from noticing $\lim_{x \to 0^+} \ln x = -\infty$, $\ln 1 = 0$, and $\lim_{x \to +\infty} \ln x = +\infty$, then applying the power $-2$ to get the desired limits.

Or along the same lines,

$$ f(x) = \frac{dx}{(x-1)^2} $$

(That's $d$ times $x$, not a differential form.)

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  • $\begingroup$ Thank you. It's worth noting that $\frac x {(x-1)^2} = \frac 1 {x + \frac 1 x - 2}$. $\endgroup$ Jul 11, 2023 at 12:53
  • $\begingroup$ This indeed seems to be the function! - See updated post. $\endgroup$ Jul 11, 2023 at 13:20
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Take any function $g:(0,1)\to(0,\infty)$ that has $\lim_{x\to 0}g(x)=0$ and $\lim_{x\to 1}g(x)=\infty$. For instance, $g(x)=dx/(1-x)$ with $d\in(0,\infty)$. Then just define $$f(x) = \begin{cases} g(x), & \text{if $x\in(0,1)$} \\ g(1/x), & \text{if $x\in(1,\infty)$} \\ \end{cases}$$

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  • $\begingroup$ Yes, please see update that I'm trying to avoid piecewise. $\endgroup$ Jul 11, 2023 at 12:23

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