Is it possible to define a bijection $f:\mathbb{R}\setminus\mathbb{R}^{-}\rightarrow[0,1)$ such that $f$ is continuously differentiable on its entire domain?
Does my function exist?
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$\begingroup$
analysis
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0It means the set of all nonnegative real numbers. (see: http://en.wikipedia.org/wiki/Set_subtraction#Relative_complement) – 2011-03-11
3 Answers
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Let us consider $f\,:\,\left[0;+\infty\right[\rightarrow\left[0;1\right[$ defined as: $f = \frac{2}{\pi} \arctan$
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If your domain is the reals greater than or equal to zero, $1-\exp(-x)$ seems to fill the bill.
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Yes. For example, $f(x)=1-\frac{1}{1+x}$.