Function $f = (f_1,\dots,f_n)$ s.t. $[f_1(x),\dots,f_n(x)]$ goes from $[1,0,\dots,0]$ to $[0,\dots,0,1]$ in a smooth way

Question

Question

I would like to have a function: $$ f: \mathbb{R} \rightarrow \left\{x \in \mathbb{R}^n \mid \sum_{i=1}^n x_i = 1\right\}:x \mapsto (f_1(x),\dots,f_n(x)) $$ s.t. $f(0) = (1,0,\dots,0)$ and $f(1) = (0,\dots,0,1)$. This is of course very easy to find but I would like to have these such that the conversion happens in a smooth way. What I mean by that is that I would like to have the following extra properties:

• $f_1(1/2) = \dots = f_n(1/2) = 1/n$
• $f_1(x)$ is descending
• $f_n(x)$ is ascending
• If we define the function $g_x(k) := f_k(x)$ and we draw the graphs of $g_x(k)$ for $x$ going from $0$ to $1$ this should look like a bit like a wave (idea further explained in next two points)
• On $]0,1/2[$ we have $f_1(x) > \dots > f_n(x)$ and on $]1/2,1[$ we have $f_1(x) < \dots < f_n(x)$

Example method

For example for $n = 2$ we can define: $$ f(x) := (x, 1-x) $$ now for $n = 3$ we can write: $$ f(x) := (x,f_2(x),1-x) $$ but then we have a problem as we need to have $f_2(x) = 0$, therefore we need to let $f_1(x)$ descend to $1/3$ on $[0,1/2]$ and $f_3(x)$ ascend to $1/3$ on $[0,1/2]$ we define: $$ f_1(x) :=\begin{cases} (1 - \frac{4}{3} \cdot x) & \mbox{ if } x \in [0,1/2]\\ \frac{2}{3}(1 - x) & \mbox{ if } x \in [1/2,1] \end{cases} $$ and $$ f_3(x) :=\begin{cases} \frac{2}{3} \cdot x & \mbox{ if } x \in [0,1/2]\\ \frac{4}{3} x - \frac{1}{3} & \mbox{ if } x \in [1/2,1] \end{cases} $$ then we automatically find for $f_2(x)$: $$ f_2(x) = \begin{cases} \frac{2}{3} x & \mbox{ if } x \in [0,1/2]\\ \frac{2}{3}(1 - x) & \mbox{ if } x \in [1/2,1] \end{cases} $$

Now we can continue for $n = 4$, the logical way to continue would be to define $f_1(x)$ to be $(1 - \frac{3}{2} x)$ on $[0,1/2]$ and $f_4(x) := \frac{1}{2} x$ on $[0,\frac{1}{2}]$. We could then use $f_2 = f_3$ and solve the equation $f_1 + 2 f_2 + f_4 = 1$ on $[0,\frac{1}{2}]$. But this does not satisfy the last property and doesn't look enough like a wave to me.

Reason For My Question

The Reason I ask this is because I would like to have a sequence of Markov Chains which goes from perfect correlation (i.e. transition matrix the unity matrix) to some other Markov Chain which is strongly negatively correlated.

Answer 1

$$f_k(x)={q^{k-1}\over\sum\limits_{i=0}^{n-1}q^i},\text{ where }q={x\over1-x}$$ would do the job.