Find a function that satisfies the following property I am finding it hard to find such a function $v:$ { 1,2,...,n } $\rightarrow \mathbb{R}$ which satisfies the following property:
$$v(k) - v(q) > 2^{n-1}(v(k) - v({q+1}))$$
where $$v(1) < v(2) < ... < v(q) < v({q+1}) < ... < v(k) \text{ and } n\geq  k \neq {q+1}.$$
In words, the difference between 2 points (first maximal and second some other point $q$) is strictly greater than the $2^{n-1}$ times the difference of the maximal point and point $q+1$.
 A: It's easier to start from $n$ and work your way backwards. Start by assigning the value 0 to $n$. There is no restriction on $v(n-1)$, so let it be $-1$.
The inequality inequalities for $k=n$ and $q<n-1$ become
$$
\begin{align}
v(n) - v(q) &> 2^{n-1}(v(n)-v(q+1))\\
v(q) &< 2^{n-1}v(q+1).
\end{align}
$$
Inspired by this fact you can take any value $c>2^{n-1}$ and let $v(n-i)=-c^{i-1}$.
Note that this satisfies the condition that $v$ is always increasing.
To prove this works for all other inequalities take $k<n$:
$$
\begin{align}
v(k)-v(q) &= -c^{n-k-1} + c^{n-q-1}\\
&=c(c^{n-q-2} - c^{n-k-2})\\
&>2^{n-1}(c^{n-(q+1)-2}-c^{n-(k+1)-1})\\
&=2^{n-1}(-v(q+1)+v(k+1))\\
&>2^{n-1}(v(k)-v(q+1))
\end{align}
$$
Of course not every sequence $v$ that satisfies these equations is of this form, but these sequences all satisfy the equations.
For $n=3$ you can choose sequence $(-5, -1, 0)$.
For $n=4$ you can choose sequence $(-81, -9, -1, 0)$.
For $n=5$ you can choose sequence $(-4913, -289, -17, 0)$, etc.
