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

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

$$x ≈ 4.94605856546361519886814090×10^-17$$

$$x ≈ 0.0119030752000093205127913177$$

$$x ≈ 0.0232432500308837633008456147$$

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

• Is this supposed to be $$(x-\lfloor x\rfloor)^2+(x-\lfloor x\rfloor)\left\lfloor{1\over x -\lfloor x\rfloor}\right\rfloor=1?$$ – saulspatz Jul 6 '19 at 22:06
• @saulspatz Yes, I mean that. Sorry for my lack of knowledge in LaTeX. – gencayotunc Jul 6 '19 at 22:13

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.

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:

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