I was working on beautiful inverse function problem ,I solved it like below but I am looking for new idea(s) to solve it .can any one help me ? $f(x)=x+\lfloor x\rfloor \to f^{-1}(x)=?$ I listed my tries below as answers .It will be appreciated any help .
Inverse of a floor function
-1
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calculus
functions
alternative-proof
floor-function
inverse-function
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0Have you got a typo in your post or does $f(x) = x + f(x)$? – 2017-02-24
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0What is the domain of $f\,$? – 2017-02-24
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0It was corrected .thank you both – 2017-02-24
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0@dxiv :Domain is $\forall x \in R$ – 2017-02-24
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1$f$ is not surjective on $\mathbb{R}$ so it has no inverse. Try to find an $x$ such that $f(x)=1.5$ for example. – 2017-02-24
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0@dxiv :NOte that $f(x)$ is $1-1$ function , you can trace it by a figure . – 2017-02-24
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2$f$ is injective, but not surjective. Whether it's invertible or not depends on how you define its codomain. – 2017-02-24
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0$f (x)=x+[x]=2 [x]+\{x\} $. So $f^{-1}(x)=[x]/2 +\{x\} $ if $[x] $ is even, the inverse does not exist otherwise.. Example. $f (13.7)=26.7$ and $f^{-1}(26.7)=26/3 +.7=13.7$. – 2017-02-24
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0Note if $[x] $ is odd then $f^{-1}(x)$ doesn't exist and isn't defined. – 2017-02-24
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0$f(x) $ most certainly is ***NOT*** 1-1. The is no x so that f (x)=1.5. It's injective but not surjective. – 2017-02-24
2 Answers
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$$y=f(x)=x+\lfloor x\rfloor \\\to \lfloor y\rfloor=\lfloor x+\lfloor x\rfloor \rfloor\\ \lfloor y\rfloor=2\lfloor x\rfloor \to \\ \lfloor x\rfloor=\frac12 \lfloor y\rfloor \to\\y=x+\frac12 \lfloor y\rfloor \\x=y-\frac12 \lfloor y\rfloor \to\\(x\leftrightarrow y)\\f^{-1}(x)=x-\frac12 \lfloor x\rfloor$$
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3$f\left(f^{-1}(1.5)\right)=f(1.5-0.5)=f(1)=2 \ne 1.5\,$. You need to restrict the domain of $f^{-1}$ more carefully. – 2017-02-24
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0Please do not attempt to clarify a horribly written post by posting two wildly different answers. Either you answered it or you didn't. – 2017-02-24
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$$x=n_x+p_x ,y=n_y+p_y\\0\leq p<1\\ y=x+\lfloor x \rfloor \to n_y+p_y=n_x+p_x+n_x \\\to \begin{cases}n_x=\frac12 n_y\\p_x=p_y\end{cases}\\x=n_x+p_x=\frac12 n_y+p_y=\frac12 (\lfloor y\rfloor)+(y-\lfloor y\rfloor)\\x=y-\frac12 \lfloor y\rfloor \\ f^{-1}(x)=x-\frac12 \lfloor x\rfloor$$ and third try was
since $f(x)$ passes $(0,0)$ guess that $$f(x)=x+\lfloor x\rfloor \to f^{-1}(x)=ax+b\lfloor x\rfloor \\f(1.5)=2.5\\f(1)=2 \to (2,1),(2.5,1.5) \in f^{-1}\\ \begin{cases}2a+2b=1\\2.5a+2b=1.5\end{cases}\\ \begin{cases}2a+2b=1\\2.5a+2b=1.5 & 0.5 a+2b-2b=1.5-1 \end{cases}\\a=1\\b=-0.5 \\\to \\f^{-1}(x)=ax+b\lfloor x\rfloor=1x+(-0.5)\lfloor x\rfloor$$
any more idea ?
thanks in advanced .
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0Please do not attempt to clarify a horribly written post by posting two wildly different answers. Either you answered it or you didn't. – 2017-02-24