Given $q \in (0,1)$ and $n \in \mathbb{N}$, define the points $\{ 0, q^n, q^{n-1}, \dots, q^2, q, 1 \}$. What is the largest interval between two adjacent points?
So far I've got the following. The interval lengths are $q^{k-1}-q^k$ for all intervals, except for the leftmost, which has length $q^n$. The largest interval is never a middle interval (if $n>1$), since $q^{k-2}-q^{k-1} < q^{k-1}-q^k$ by a factor of $q$. Thus the largest interval is either the leftmost or the rightmost interval, with length $q^n$ or $1-q$, respectively. But this depends on the choice of $q$ and $n$, so to come up with an answer, I'm trying to determine when $1-q > q^n$, but here I am stuck (no luck on Wolfram etc). The best answer I have right now is $max(1-q,q^n)$.
Is my work so far correct? How can I get any further? Thanks!
You can shorten your argument a little by noting that $q^{k-1}-q^k=q^{k-1}(1-q)$; since the $q^{k-1}$ factor decreases with increasing $k$, this difference is clearly largest when $k$ is minimal, i.e., when $k=1$ and the difference is $1-q$. This of course immediately yields your $\max\{1-q,q^n\}$. In the case $n=1$ this boils down to $\max\{1-q,q\}$, which clearly cannot be improved, though it can of course be expressed differently, e.g., as
$$\begin{cases} q,&\text{if }q\ge\frac12\\ 1-q,&\text{otherwise}\;. \end{cases}$$
It seems unlikely that the cases with $n>1$ are any better behaved, so I doubt that you can get rid of the maximum.