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Given $f(x)=x+[x]$ where $$f:\Bbb{R}\to\bigcup_{n\in \Bbb > Z}[2n,2n+1)$$

Find $f^{-1}(x)$ ?

My Attempt:

$y=x+[x]=2[x]+\{x\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $Since $x=[x]+\{x\}$

$\Rightarrow y-\{x\}=2[x]$

The above equation implies $y-\{x\}$ is an integer, therefore it can be written as $y-\{y\}$.Hence, it implies $\{x\}=\{y\}.$

$\Rightarrow y-\{y\}=2[x]$

$\Rightarrow [x]=\frac{[y]}{2}$

Adding {x} both sides we get,

$\Rightarrow [x]+\{x\}=\frac{[y]}{2}+\{x\}$

$\Rightarrow x=\frac{[y]}{2}+\{y\}=y-\frac{[y]}{2}$

Hence,

$$f^{-1}(x)=x-\frac{[x]}{2}$$

What are some other ways to solve this problem and to find inverses of such type of function ?

In the above context, $[x]$ is Greatest Integer Function(floor) and $\{x\}$ is fractional part of $x$.

3 Answers 3

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Another way: \begin{eqnarray} y &=& x+[x] \\ x &=& y-[x] ~~;~~(1)\\ y &=& x+[x] ~~;~~(From~main~function) \\ y &=& x+[y-[x]] ~~;~~From(1)\\ y &=& x+[y]-[x] \\ y &=& x+[y]-(y-x) \\ y &=& x+[y]-y+x \\ 2y &=& 2x+[y] \\ x &=& \frac12(2y-[y]) \\ x &=& y-\frac{[y]}{2} \end{eqnarray}

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The floor function is piecewise constant, so that $f$ is piecewise linear. Hence every piece is invertible separately, and the whole function is invertible provided there is no overlap of the codomains (which is the case here).

With $x\in[n,n+1)$, $f(x)=x+[x]=x+n$, so that $f^{-1}(y)=y-n$.

As the codomain is $[2n,2n+1)$, you can retrieve $n$ using the fact that $[y]=[x]+[x]=2n$.

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Another method is to plot the function $y=x+[x]$ and take its reflection in $y=x$ which will give you the plot of $y = x - \frac{[x]}{2}$.