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I have an equation

$$ p_0 + \frac{p_1}{1 + x} + \frac{p_2}{(1 + x)^2} + \frac{p_3}{(1 + x)^3} + \dots + \frac{p_n}{(1 + x)^n} = 0 $$

I want a formula for $x$, may I know if there is any way of finding a formula for $x$.

Also all $ p_0, p_1, \ldots, p_n$ are independent variables.

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This is an expression, where is the equation? – Aang Jan 24 '13 at 10:25
Thanks, now you can see my equation. – Anshul Shukla Jan 24 '13 at 10:28
what do u mean by $p0,p1,\cdots,pn$ are independent variables? Which set does they belong ? – Aang Jan 24 '13 at 10:30
p0,p1...,pn. they all are random variables.they are not depend on each other. – Anshul Shukla Jan 24 '13 at 10:32
This is not a GP at all. :| – hjpotter92 Jan 24 '13 at 10:41
up vote 0 down vote accepted

$$ p_0 + \frac{p_1}{1 + x} + \frac{p_2}{(1 + x)^2} + \frac{p_2}{(1 + x)^3} + \dots + \frac{p_n}{(1 + x)^n} = 0 $$ $$\implies Var(p_0 + \frac{p_1}{1 + x} + \frac{p_2}{(1 + x)^2} + \frac{p_2}{(1 + x)^3} + \dots + \frac{p_n}{(1 + x)^n} )=0$$ $$\implies Var(p_0)+Var(\frac{p_1}{1 + x})+\cdots+Var(\frac{p_n}{(1 + x)^n})=0$$ $$Var(\frac{p_i}{(1 + x)^i})=0\forall i\in \{0,1,...,n\}$$

$$\implies \frac{p_i}{(1 + x)^i}$$ is constant

Thus, $$X=k_i(p_i)^{\frac{1}{i}}-1$$

EDIT: As you can see $X=k_i(p_i)^{\frac{1}{i}}-1=k_j(p_j)^{\frac{1}{j}}-1$ which implies $p_i,p_j$ are dependent.

Thus, there is no value of $x$ that can satisfy this equation for independent random variables $p_0,p_1,...,p_n$

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Please elaborate your answer, What is the value of ki, and it not two times is like p0,p1,p2, – Anshul Shukla Jan 24 '13 at 10:58

A simple case, perhaps... $$ P_0+\frac{P_1}{1+X} = 0, $$ where $P_0, P_1$ are independent random variables. Solving: $$ X = -1-\frac{P_0}{P_1} $$ so that we see $X$ is also a random variable. NOT just a constant.

Is this what the OP means? So then the case $n=2$ involves solving a quadratic equation?

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Perhaps the $p_{i}$ are arbitrary constants? In this case, assume WLOG $p_{n} \ne 0$, set for simplicity $y = 1 + x$, multiply by $y^n$ to get a polynomial equation in $y$ of degree $n$.

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