Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
This is an expression, where is the equation? –  Aang Jan 24 '13 at 10:25
    
Thanks, now you can see my equation. –  TechDon Jan 24 '13 at 10:28
1  
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. –  TechDon Jan 24 '13 at 10:32
    
This is not a GP at all. :| –  hjpotter92 Jan 24 '13 at 10:41

3 Answers 3

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 $p_0,p_1,...,p_n$

share|improve this answer
    
Please elaborate your answer, What is the value of ki, and it not two times p2...it is like p0,p1,p2,p3..pn. –  TechDon 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?

share|improve this answer

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 $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.