A function that creates a partition of values such that the sum is 1

Question

I'm trying to find a function $g:\mathbb N\cup\left\{0\right\} \rightarrow\left(0,1\right)$, such that, given a (real) value $k \in \left(0,1\right)$ and an integer $i>1$, allows me to calculate a partition of the interval $(0,1)$ that fulfills the following conditions:

$k$ is a parameter of the function.
The function only has a positive value if $0 \leq j\leq i$, and zero on every other case.
The sum of the function results for every $j \in \left\{ 0,1,\dots,i\right\}$ is 1: $$\sum_{j=0}^{i}g_k\left(j\right)=1$$
For every $j \in \left\{1,\dots,i\right\}$:$$g_k\left(j\right)=k·g_k\left(j-1\right)$$

Condition (3) implies the following: $$\underset{k\rightarrow1}{\lim}g_k\left(j\right)=g_k\left(j-1\right)\ ,\ j\in\left\{ 1,2,\dots,i\right\} $$

I know such function exists, but I don't remember the function (and I don't find any reference to it), and I've been struggling with this for some time.

Could you point me in the right direction? (a name for such function would be enough)

When you say $g(j)$, do you mean $g(x_j)$, where $x_j$ are the terms of some partition of $(0,1)$? Or do you mean the actual integers $1,\dots,i$ (none of which are contained in $(0,1)$? — John Gowers, Sep 19 '13 at 20:48
@Donkey_2009 No, I mean exactly the natural numbers... I'm editing the question to make it more clear — Barranka, Sep 19 '13 at 20:49
When you talk about the limit as $k\to 1$, what topology are you taking on the natural numbers? — John Gowers, Sep 19 '13 at 20:56
OK, that's right. But then $\lim_{k\to1}g(j)=g(j)$, since $g(j)$ doesn't depend on $k$ at all. — John Gowers, Sep 19 '13 at 20:58
duh, right! Forgot to write that $k$ is a (deterministic) parameter of the function — Barranka, Sep 19 '13 at 20:59

score 2 · Accepted Answer · answered Sep 19 '13 at 20:54

2

You just need to find $g(0)$, for then $g(i)=k^ig(0)$ You have a geometric series, so $$\sum_{j=0}^{i}g\left(j\right)=\sum_{j=0}^{i}g(0)k^j=g(0)\frac {1-k^{i+1}}{1-k}=1\\g(0)=\frac {1-k}{1-k^{i+1}}$$

answered Sep 19 '13 at 20:54

Ross Millikan

374,822

Thank you!!! Do you know is this function has some special name (other than geometric series)? (I know I found it in a book some years ago, but I can't remember the name... or the book) – Barranka Sep 19 '13 at 21:18
I don't know of one – Ross Millikan Sep 19 '13 at 21:31

A function that creates a partition of values such that the sum is 1

1 Answers1