Let $f:\Bbb{R}\to\bigcup_{n\in \Bbb Z}[2n,2n+1)$ and $f(x) = x + \left[ x \right]$. What is ${f^{ - 1}}(x)$?
Let $f(x) = x + \left[ x \right]$. What is ${f^{ - 1}}(x)$?
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calculus
floor-function
inverse-function
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0What are your thoughts? Graphed this? – 2017-01-02
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0Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – 2017-01-02
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0@Clayton - We know $\left[ {f(x)} \right] = 2\left[ x \right]$ but I can not continue. – 2017-01-02
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2What domain and co-domain are we talking about here? Because, if the function is $f:\mathbb R \to \mathbb R$, then there are holes in the range of $f$. For example, there is no real $x$ for which $f(x) = 1.6$. And so as function is not onto, inverse can't be defined. – 2017-01-02
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0@DanielW.Farlow Improved question by specifying domain and co-domain so that $f^{-1}$ exists. Request to reopen. – 2017-01-03
2 Answers
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The inverse will exist iff $f(x)$ is bijective.
$f(x)=x+[x]$ is bijective if it is defined as $$f:\Bbb{R}\to\bigcup_{n\in \Bbb Z}[2n,2n+1)$$
To find $f^{-1}$:
$y=x+[x]=2[x]+\{x\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $Note: $x=[x]+\{x\}$
$\Rightarrow y-\{x\}=2[x]$
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,
$$f^{-1}(x)=x-\frac{[x]}{2}$$
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The inverse is $g(y) = [y]/2 + \{y\} = y - [y]/2.$
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0What is $\left\{ y \right\}$? – 2017-01-02
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0@tip-lnk fractional part – 2017-01-02
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0If $y = 1.6$, then $g(y) = 1.6-1/2 = 1.1$ but $f(1.1) = 1.1+1 = 2.1$. As Shraddheya Shendre pointed out, there is no inverse $g(y)$ valid for all $y$. – 2017-01-02