# Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)

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 $t$?

• Probably not. Most irreducible polynomials of degree $n>4$ have a non-solvable Galois group. Of course it would help if you defined more precisely what you mean by "closed form". Solvability in radicals is equivalent to a solvable Galois group for the polynomial. – GH from MO Nov 17 '14 at 14:50
• I think I found closed form for n in {3,5}. – joro Nov 17 '14 at 15:35
• @joro - I have a closed form for $t(3,p)$, would love to hear about $t(5,p)$ ! – R B Nov 17 '14 at 15:36
• The Galois group of the equation for $p=3/5$ and $n=7$ is $S_7$, which is not soluble. So don't expect a solution by radicals for $n=7$. – Chris Wuthrich Nov 17 '14 at 17:26

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.

• For $n=5$ you're solving the cubic $${t}^{3}-7\,p{t}^{2}+2\,{t}^{2}+7\,pt+2\,t-7\,p+1$$ – Robert Israel Nov 17 '14 at 18:16
• @RobertIsrael for n=5 I think I am solving higher degree than cubic, which might be reducible. – joro Nov 17 '14 at 18:26
• Yes, the point is that it factors, and the cubic factor is the one that contains $p$, so that's what you're actually solving. – Robert Israel Nov 17 '14 at 18:35

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 $n \ge 1$.