The graph of the function $f(x)= \left\{ \frac{1}{2 x} \right\}- \frac{1}{2}\left\{ \frac{1}{x} \right\} $ for $0<x<1$

Question

Let for reals $$\{x\}=\text{Frac}(x)$$ the fractional part function, take for example the more common definition, the first (there is a different definition as you see in this MathWorld's Page, from Graham et alumni). Thus I know that $$\text{Frac}(x)=x$$ for $0<x<1$.

I am interesting in basics about the graphic representation of some of the following functions for a fixed real $0<\lambda<1$, $$f_\lambda(x)= \left\{ \frac{\lambda}{ x} \right\}- \lambda\left\{ \frac{1}{x} \right\} $$ for $0<x<1$.

I am stuck because it is a difficult function, I don't know if there is good references in this site or in the literature to study some aspect of this kind of functions. Please add in your answer or in comments useful posts in this site.

Question. Can you explain us the graphic (or at least the more relevant facts) of the function $$f(x)= \left\{ \frac{1}{2 x} \right\}- \frac{1}{2}\left\{ \frac{1}{x} \right\} $$ for $0<x<1$. You can add your computations and draws that you want explain to get a good reference for this family of functions from this example $\lambda=\frac{1}{2}$. Thanks in advance.

My code for Wolfram Alpha online calculator were first

$\text{Frac(1/(2x)) for 0<x<1 }$

(respectively with $\text{(1/2 )Frac(1/x)}$ ) and after

$\text{Frac(1/(2x)) -(1/2 )Frac(1/x) for 0<x<1 }$.

Like this: It is possible an explanation of how get such difference between arcs of hyperbolas?, or, are there infinitely many straight segments in the graph of our $f(x)$?, or can you obtain a closed-form for such difference of hyperbolas as a constant (those straight segments in the graph of our $f(x)$)?...this was my Question.

Answer 1

Let us write down $\frac 1 x = n + f$, where $n\in\mathbb N_0$, $0\leq f < 1$, and let us consider two cases:

Case 1. $n = 2k$, $k\in\mathbb N_0$

We have $\frac 1 x = 2k + f$ and $\frac 1 {2x} = k + \frac f 2$, thus $f(x) =\left\{\frac 1 {2x} \right\} - \frac 1 2\left\{ \frac 1 {x}\right\} = \frac f 2 - \frac f 2 = 0$.

Case 2. $n = 2k+1$, $k\in\mathbb N_0$

We have $\frac 1 x = 2k + 1 + f$ and $\frac 1 {2x} = k + \frac {f+1} 2$, thus $f(x) =\left\{\frac 1 {2x} \right\} - \frac 1 2\left\{ \frac 1 {x}\right\} = \frac {f+1} 2 - \frac f 2 = \frac 1 2$.

Now,

$$f(x) = 0\iff 2k\leq \frac 1 x < 2k+1 \iff x\in\bigcup_{k\in\mathbb N}\left\langle \frac 1 {2k+1},\frac 1 {2k}\right]$$

$$f(x) = \frac 1 2\iff 2k-1\leq \frac 1 x < 2k \iff x\in\bigcup_{k\in\mathbb N}\left\langle \frac 1 {2k},\frac 1 {2k-1}\right]$$

so we conclude that $f$ is a half of the indicator function of the set $\bigcup_{k\in\mathbb N}\left\langle \frac 1 {2k},\frac 1 {2k-1}\right]$.

Since the length of interval $\left\langle \frac 1 {n+1},\frac 1 n\right]$ is $\frac 1 n - \frac 1{n+1} = \frac 1 {n(n+1)}$, $f$ changes from $\frac 1 2$ to $0$ and back to $\frac 1 2$ approximately at the rate of $\frac 1{n^2}$ approaching $0$ as $n$ goes to infinity.

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Let us consider case of $\lambda = \frac 1 m$, for fixed $m\in\mathbb N$ and let $l\equiv \lfloor \frac 1 x\rfloor \pmod m$.

We have $\frac 1 x = mk + l + f$, $0\leq f < 1$ and consequently $$f_{\frac 1 m}(x)= \left\{ \frac 1 {mx}\right\} - \frac 1m\left\{\frac 1 x\right\} = \frac{f+l}m - \frac fm = \frac l m$$

Answer 2

In order to visualize what's happening put $x:={1\over y}$ and consider the functions $$g_\lambda(y):=\{\lambda y\}-\lambda\{y\}=\lambda\lfloor y\rfloor-\lfloor\lambda y\rfloor\qquad(y>1)$$ instead. They satisfy $$-\lambda<g_\lambda(y)<1\ ,\tag{1}$$ have jump discontinuities at the integer points $k\geq2$ and at the points ${k\over\lambda}$ $(k\geq1)$, and are constant between two consecutive jumps. If $\lambda={p\over q}$ is rational then $g_\lambda$ is periodic with period $q$, but if $\lambda$ is irrational there is no periodic behavior. The figure below shows $g_\lambda$ for $\lambda:={\sqrt{5}-1\over2}$.

It follows that the functions $x\mapsto f_\lambda(x)$ have isolated jump discontinuities accumulating at $0$, and that $f_\lambda$ is constant within the bounds $(1)$ between successive jumps.