Find closed form of $f(x)=x^2 \cdot \lfloor {\frac{1}{x^2}}\rfloor$ Yesterday I asked a question about the continuity of $f(x)=x^2 \cdot \lfloor {\frac{1}{x^2}}\rfloor$ and we found out that $f(x)=0$, $\forall x>1$. Now I want to find its closed form over the reals. Now, since $f$ is even, suffice it to find its closed form on $(0,1)$ and we are done. How to do this? 
 A: The term "closed form" is a little soft, but I'll go with the Wikipedia entry which states

In mathematics, a closed-form expression is a mathematical expression
  that can be evaluated in a finite number of operations.

Using this definition, $x^2\cdot\left\lfloor\frac{1}{x^2}\right\rfloor$ is a closed form.

However, for analyzing $f$, perhaps a better way to represent the function is to look at its values on the interval $\left(\frac{1}{\sqrt{n+1}},\frac{1}{\sqrt{n}}\right]$. If $x$ is on this interval, then $f(x)$ will equal $n\cdot x^2$ which tells you both that the function is continous on each such interval, but is not continuous on the edges.
A: Let $$\frac1{x^2}=n+t$$ where $n$ is a natural and $t$ a fractional part (in $[0,1)$).
Then
$$\begin{cases}x_n(t)=\dfrac1{\sqrt{n+t}},
\\f_n(t)=\dfrac n{n+t}\end{cases}$$ are parametric equations of the curve pieces, in terms of the parameter $t$, indexed by $n$.
The endpoints of the curves are $\left(\dfrac1{\sqrt{n}},1\right)$ and $\left(\dfrac1{\sqrt{n+1}},\dfrac n{n+1}\right)$. They are aligned on the horizontal $y=1$ and the parabola $y=1-x^2$.
All these expressions are closed-forms stricto sensu.
