TITLE: Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)
QUESTION [5 upvotes]: We are given $n\in \mathbb N^+$ and $p\in[\frac{1}{2},\frac{n+1}{n+2}]$.
Our goal is to find $t\in[0,1]$ such that
$$(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i$$

Is there a closed-form solution $t(n,p)$?
How about a close formula for some non-trivial $p$, e.g. for $g(n)\triangleq t(n,0.6)$?


A few observations:

This is equivalent, for $t\neq0,1$, to:
$$(1-p)t(1-t^{n+1})=p(1-t)(1-(1-t)^{n+1})$$

$\forall n:t(n,\frac{1}{2})=\frac{1}{2}$

$\forall p:t(1,p)=3p-1$

$\forall n:t(n,\frac{n+1}{n+2})=1$

$\forall n:t(n,p)$ is monotonically increasing in $p$.

$\forall p:t(2,p)=\frac{2 p + 1-\sqrt{-3+28 p-28 p^2}}{4 p - 2}$



If this is not possible, is it possible to bound it with simple function? e.g. I think I showed $$p\leq t(n,p)\leq \frac{(n+2)p - 1}{n}$$
Which works great for large $n$, but not so much for small values.

Can we give tighter bound for $t$?

REPLY [2 votes]: You might find this useful.
Write the equation as
$$ p = \dfrac{t (t^{n+1}-1)}{(1-t)^{n+2} + t^{n+2}-1} = \dfrac{(1-s)((1-s)^{n+1}-1)}{s^{n+2}+(1-s)^{n+2}-1}$$
where $s = 1-t$.  If $n \ge 3$, I get
$$ p = \dfrac{n+1}{n+2} - \dfrac{(n+1)}{2(n+2)} s - \dfrac{(n^2+4n+3)}{12(n+2)} s^2 + \ldots $$
so we should have $$ t > \dfrac{2(n+2)p}{n+1} - 1 $$
for $p$ near $(n+1)/(n+2)$.  Indeed this appears to hold for $1/2 \le p \le 1$
and all $n \ge 1$.