Roots of $(x-\lfloor x\rfloor)^2+(x-\lfloor x\rfloor)\left\lfloor{1\over x -\lfloor x\rfloor}\right\rfloor=1$

Question

Can you help me to find -some analytical- roots of the following function ? I know $\sqrt2$ is a root and I think there are infinitely many roots,according to plot provided by WolframAlpha.

$$ (x-\lfloor x\rfloor)^2+(x-\lfloor x\rfloor)\left\lfloor{1\over x -\lfloor x\rfloor}\right\rfloor=1 $$

I've tried Matlab and WolframAlpha for roots.Matlab gives error code

"Warning: Unable to find explicit solution. For options, see help."

But I dont trust my knowledge about Matlab.

WolframAlpha gives this

$x ≈ 4.94605856546361519886814090×10^-17$

$x ≈ 0.0119030752000093205127913177$

$x ≈ 0.0232432500308837633008456147$

etc...For more click here

But I think these are approximations,rather than exact roots. And if you find any,can you give me some properties about it.I assume roots have to be irrational.

Thanks

Answer 1

Let $u = x-\lfloor x \rfloor$. Then you want to find the solutions to $$u^2+u\left\lfloor\frac{1}{u}\right\rfloor -1 = 0$$

The range of $u$ is $[0, 1)$. In this interval $y = u\left\lfloor\frac{1}{u}\right\rfloor$ is an infinite amount of line segments with the equation $y=nu$, where $n$ ranges over all positive integers and $\frac{1}{n+1} < u \le \frac{1}{n}$. This can be seen by noting that if $\frac{1}{n+1} < u \le \frac{1}{n}$, then $n \le \frac{1}{u} < n+1$. This means we want to find the solutions to $$u^2 + nu-1 = 0$$

This is a simple quadratic in $u$, given $n$. Solving the quadratic finds $$u = \frac{-n \pm \sqrt{n^2+4}}{2}$$ Since $u \ge 0$ must be satisfied, $$u = \frac{-n + \sqrt{n^2+4}}{2}$$

Finally, another inequality must be satisfied: $\frac{1}{n+1} < u = \frac{-n + \sqrt{n^2+4}}{2} \le \frac{1}{n}$. This can be simplified to the two inequalities $$(n+1)(-n+\sqrt{n^2+4})-2 > 0$$ $$n(-n+\sqrt{n^2+4})-2 \le 0$$ Through algebraic manipulation, it is found that for the first inequality to be true, $n$ must be positive, which is always satisfied. For the second inequality to be true, $n$ can be any real number. Since both of these are satisfied given that $n$ is a positive integer, we have found all valid $u$.

The full set of solutions for $x$ is given by adding any integer to $u$, so $$x = m+\frac{-n + \sqrt{n^2+4}}{2}$$ where $m$ is any integer, and $n$ is any positive integer.

Answer 2

See this:

A simpler one:

Defination: $\lfloor x \rfloor= n;~ \mbox{if}~~ n \le x < n+1. ~~~ \mbox{where,}~~~ x \in R,~~n \in \mathbf{Z}.$

Let's write $x$ as,

$x=\lfloor x \rfloor +\{x\}=n+r.~~~~$ where, $~~ r \in [0,1). $

Now Original Eq. becomes:

$\Rightarrow r^2+ r \lfloor \frac{1}{r} \rfloor-1=0; ~~~~~ \mbox{Let} ~~\lfloor \frac{1}{r} \rfloor = m, ~~\mbox{if} ~m \le \frac{1}{r} < m+1;~~~\frac{1}{r} \in (1,\infty] \rightarrow m \in \mathbf{Z^*} $

$\Rightarrow r^2+ m r-1=0 \Rightarrow r= \frac{-m \pm \sqrt{m^2+4}}{2}$

Since $r \ge0$; so we have only: $~~r= \frac{-m + \sqrt{m^2+4}}{2}$

Hence the solution is:

$x = n+r= n+\frac{-m + \sqrt{m^2+4}}{2}; ~~~~\mbox{where,}~~ n \in \mathbf{Z}, ~~ m \in \mathbf{Z^*} $