Skip to main content
MathOverflow

Return to Question

Commonmark migration
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

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 $t$?

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 $t$?

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 $t$?

added 258 characters in body
Source Link
R B
R B
  • 618
  • 3
  • 18

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 $t$?

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 $p$.

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

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 $t$?

deleted 2 characters in body
Source Link
R B
R B
  • 618
  • 3
  • 18

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

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 $p$.

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

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 $p$.

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

edited body
Source Link
R B
R B
  • 618
  • 3
  • 18
Source Link
R B
R B
  • 618
  • 3
  • 18