{$\frac{1}{x}$} , $0<x\leq1$ in terms of {$x$}, $x\geq 1$, {$x$} is fractional part of $x$

Question

$f(x)$ $=$ {$\frac{1}{x}$}, $0<x\leq1$ where {$x$} denotes the fractional part of $x$.

$g(x) =$ {$x$}, $x\geq 1$

I want an expression for $f(x)$ in terms of x and $g(x)$.

My try-

If $x\in \mathbb{Z} , x\geq 1$ , g(x)={x}=0

f(x)={$\frac{1}{x}$}=$\frac{1}{x}$ $$f(x)=\frac{1}{x}+g(x)$$ What happens when x is a fraction? Jacobi theta function $\theta(x)= \sum_{n=-\infty}^\infty e^{-n^2\pi x}$ This function satisfies the functional equation $\theta(\frac{1}{x})=\sqrt{x}\theta(x)$. I want a similar functional equation of f(x) in terms of g(x)

Answer 1

There is no expression for $f$ solely in terms of $g(x)$.

Proof:

Take $x_1 = 2, x_2=3$. Then, $g(x_1)=g(x_2)$, but $f(x_1)\neq f(x_2)$.

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Intuitive explanation: an expression "in terms of $g(x)$" is an expression that only depends on the value of $g(x)$ and does not directly depend on the value of $x$. Therefore, the expression, whatever it is, will have the same value for $x_1$ and $x_2$. But since $f(x_1)\neq f(x_2)$, the expression cannot possibly capture both values, $f(x_1)$ and $f(x_2)$.

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Rigorous explanaiton: An expression of $f$ in terms of $g(x)$ is, mathematically speaking, a function $h$ such that $f(x)=h(g(x))$.

Assuming that such a function $h$ exists, however, brings us to a contradiction, since any such function means that

$$f(x_1)=h(g(x_1))=h(g(x_2))=f(x_2)$$ which is a contradiction, because we know that $f(x_1)\neq f(x_2)$.

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An expression "in terms of $g(x)$ and $x$", on the other hand, trivally always exists, since $f(x)$ itself is already an expression in terms of $x$.

Answer 2

For $x>1$,

$$\left\{\frac1x\right\}=\frac1x$$ and you are asking for $\left\{\frac1x\right\}$ as an expression involving $\{x\}$ and $x$ as a rational expression. I just wrote it above. Knowing $\{x\}$ is of no help.

For $0\le x<1$,

$$\{x\}=x$$ and you are in fact asking for $\left\{\frac1x\right\}$ as an expression involving $x$ alone as a rational expression. This is not possible due to the discontinuities in this function.

Answer 3

If $f(x)=\left\{\frac{1}{x}\right\}$ and $g(x)=\{x\}$, then $f(x)=g(\frac{1}{x})$, i.e. $f=g\circ h$ where $h(x)=\frac{1}{x}$. Perhaps this is what the author of the problem had in mind, and you have been overthinking it?