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

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 $g(n)\triangleq t(n,0.6)$?

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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 $p$.

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

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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 $t$?

Answer 1

Too long for comment, here is partial answer for $n=5$ per R B's request.

According to Maple:

n:=5:q:=(1-p)*sum(t^i,i=0..n)-p*sum( (1-t)^i,i=0 .. n):so:=solve(q,t):

so[1] and so[2] are complex and this is messy for latex, so:

convert(so[3],string);convert(so[4],string);convert(so[5],string);

"1/6*(-4116*p^2-28+1428*p+2744*p^3+12*(-43218*p^3+21903*p^2+21609*p^4-294*p\
    +9)^(1/2))^(1/3)-6*(49/9*p+2/9-49/9*p^2)/(-4116*p^2-28+1428*p+2744*p^3+\
    12*(-43218*p^3+21903*p^2+21609*p^4-294*p+9)^(1/2))^(1/3)-2/3+7/3*p"

"-1/12*(-4116*p^2-28+1428*p+2744*p^3+12*(-43218*p^3+21903*p^2+21609*p^4-294\
    *p+9)^(1/2))^(1/3)+3*(49/9*p+2/9-49/9*p^2)/(-4116*p^2-28+1428*p+2744*p^\
    3+12*(-43218*p^3+21903*p^2+21609*p^4-294*p+9)^(1/2))^(1/3)-2/3+7/3*p+1/\
    2*I*3^(1/2)*(1/6*(-4116*p^2-28+1428*p+2744*p^3+12*(-43218*p^3+21903*p^2\
    +21609*p^4-294*p+9)^(1/2))^(1/3)+6*(49/9*p+2/9-49/9*p^2)/(-4116*p^2-28+\
    1428*p+2744*p^3+12*(-43218*p^3+21903*p^2+21609*p^4-294*p+9)^(1/2))^(1/3\
    ))"

"-1/12*(-4116*p^2-28+1428*p+2744*p^3+12*(-43218*p^3+21903*p^2+21609*p^4-294\
    *p+9)^(1/2))^(1/3)+3*(49/9*p+2/9-49/9*p^2)/(-4116*p^2-28+1428*p+2744*p^\
    3+12*(-43218*p^3+21903*p^2+21609*p^4-294*p+9)^(1/2))^(1/3)-2/3+7/3*p-1/\
    2*I*3^(1/2)*(1/6*(-4116*p^2-28+1428*p+2744*p^3+12*(-43218*p^3+21903*p^2\
    +21609*p^4-294*p+9)^(1/2))^(1/3)+6*(49/9*p+2/9-49/9*p^2)/(-4116*p^2-28+\
    1428*p+2744*p^3+12*(-43218*p^3+21903*p^2+21609*p^4-294*p+9)^(1/2))^(1/3\
    ))"

Experimentally so[3] is real and in the requested range.

Answer 2

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 $n \ge 1$.