3
$\begingroup$

While engaging in personal study with regards to metric spaces I wished to construct a metric space of sequences. As a part of this I need a function that normalized positive reals into the range between 0 and 1. The idea being that I can define my metric as.

$ d(x,y) = \sum_{n=1}^\infty f( |x_n - y_n| )10^{-n} $

I don't see any reason why such a function shouldn't exist but I cannot think of one.

$\endgroup$
3
  • $\begingroup$ You could use something resembling the function $1/x$. $\endgroup$
    – David
    Commented Jan 29, 2016 at 4:39
  • 1
    $\begingroup$ What do you mean by "isomorphism"? $\endgroup$
    – user228113
    Commented Jan 29, 2016 at 4:39
  • 1
    $\begingroup$ @G.Sassatelli Likely an order isomorphism. $\endgroup$
    – David
    Commented Jan 29, 2016 at 4:39

3 Answers 3

Reset to default
3
$\begingroup$

$f(x) = 1 - \frac{1}{1 + x} $.

$\endgroup$
1
$\begingroup$

There are multiple functions that can do this with slight modifications, including arctan and 1/x.

Shifting $R^+$ to $R^{>1}$ by $x\to x+1$ will work for $1/x$, giving us $\frac{1}{x+1}$

Arctan just needs to be scaled to fit in the range: $\frac{2}{\pi}\arctan(x)$

$\endgroup$
3
  • $\begingroup$ I thought of 1/X but 1/X is outside my codomian when X< 1 $\endgroup$
    – Q the Platypus
    Commented Jan 29, 2016 at 4:41
  • $\begingroup$ That's what the composition I mentioned is for. $x\to x+1$ moves R^+ to R>1, so then you can apply it. I'll edit to make this clearer $\endgroup$
    – Stella Biderman
    Commented Jan 29, 2016 at 4:43
  • $\begingroup$ @QthePlatypus is that clearer $\endgroup$
    – Stella Biderman
    Commented Jan 29, 2016 at 4:48
0
$\begingroup$

$$\quad\frac2\pi\arctan\quad$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .