How can I modify $\frac{\cos(x)+1}{2}$ so that the range does not include 0?

Question

I am trying to find a bijection between $[0,\infty)$ and $(0,1]$. All I've come up with so far is the below function which has a range of $[0, 1]$.

$$ \begin{align*} \frac{\cos(x)+1}{2} \end{align*} $$

Is there a way I can restrict the range of the function so that it does not include 0?

EDIT: I realize now that my function only worked for the domain $[0,\pi]$.

Answer 1

That which makes the result zero is when $cos(x)+1=0$. Therefore, multiply both the numerator and denominator by $cos(x)+1$ (multiplying by one) and all zero results will be holes in the graph.

Answer 2

I suggest that you consider\begin{array}{ccc}[0,+\infty)&\longrightarrow&(0,1]\\x&\mapsto&\dfrac1{1+x^2}\end{array}instead.

Answer 3

$e^{-x}$ will do the job. I don't think you can modify your function to get an answer.

Answer 4

Note that

$$\frac{\cos(x)+1}{2}$$

is not injective onto $[0,+\infty)$ therefore is not a suitable function to obtain the result.

What about $$1-\frac2{\pi}\arctan x$$

Answer 5

Your intervals are both closed in one end and open in the other. That means that it's usually a good idea to try to map the closed end to the closed end and the open end to the open end. If you just draw a graph of such a function, you should be able to get ideas for what kinds of functions let you do that; a monotonic function with $f(0) = 1$ and, if you excuse the abuse of notation, "$f(\infty) = 0$" (technically it's written $\lim_{x\to \infty}f(x) = 0$).

I would personally have gone for $\frac1{x+1}$ as a first attempt, but the other suggestions you've gotten are equally good.