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