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?
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3Why do you not consider the form $x^2\cdot\left\lfloor\frac{1}{x^2}\right\rfloor$ "closed"? – 5xum Feb 27 '19 at 10:15
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My teacher claimed that this function can be expressed through a more "beautiful" relation. I agree that the term "closed" is not the most appropriate. – MathEnthusiast Feb 27 '19 at 11:27
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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.
5xum
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1@MathEnthusiast I considered the question "when is the function $\left\lfloor\frac1{x^2}\right\rfloor$ constant?" – 5xum Feb 27 '19 at 11:29
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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.