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Series Expansion of 1/(1-x)

Date: 07/30/98 at 20:49:50
From: Chyle Edic
Subject: Binomial expansion, roots

How do I get the roots of the following expression:

   (x^8)/(x^8 - exp(8*alpha))

I first tried to do the long division to simplify. Then I tried to find 
the roots of what was left in the bottom expression. But I am unclear 
how to find the roots if I can't complete the square or it is not in 
the form (ax + b)^n.

Date: 07/31/98 at 14:45:23
From: Doctor Pete
Subject: Re: Binomial expansion, roots

Try this technique. 

Let y = x^8. Then
   y/(y - exp(8*alpha))
is equivalent to your result. The roots satisfy
   y/(y - exp(8*alpha)) = 0
or y = 0. To be sure, when y = 0, the denominator is -exp(8*alpha), 
which is never 0 for finite alpha. Hence x = 0 is the only root. 
So I suspect perhaps I'm misunderstanding the problem.
- Doctor Pete, The Math Forum
Check out our web site!    

Date: 08/01/98 at 15:08:15
From: Chyle Edic
Subject: Re: Binomial expansion, roots

Dr Math:

I understand how to get the root of the expression. I guess what I 
wanted to ask is how do I expand this binomially. Isn't there an 
expression that goes as follows:

   (1 - x)^-1 = 1 + x^n     n = 1,2,3,4...

Thanks for your help in advance.


Date: 08/02/98 at 18:26:36
From: Doctor Pete
Subject: Re: binomial expansion, roots


The power series expansion of:

   (1 - x)^(-1)


   1 + x + x^2 + x^3 + x^4 + ....

and converges for all x such that -1 < x < 1. This power series 
expansion is usually considered as being derived from the theory of 
Taylor series (or more simply, from infinite geometric series), but 
not usually from the binomial theorem. However, the binomial theorem 
does have a generalization for negative and non-integer exponents and 
the above series can be obtained in this way. I won't go into this 
method now. If you would like more information on it, please write 

It is easiest to see the above series expansion as a geometric series 
with common ratio x. It follows that the product:

   (1 - x)(1 + x + x^2 + ... + x^n)

when expanded is: 

   1 + x + x^2 + ... + x^n
     - x - x^2 - ... - x^n - x^(n+1)
   1 - x^(n+1)

Therefore, taking the limit as n -> infinity, we see that the product
converges to 1, since x^(n+1) -> 0 as n -> infinity (but this is only 
true for -1 < x < 1, hence the convergence criterion). Thus, as n goes 
to infinity, we get:

   (1 - x)(1 + x + x^2 + ... + x^n) = 1

Hence dividing both sides out by 1 - x (since x is not equal to 1), 
results in the desired identity.

- Doctor Pete, The Math Forum
Check out our web site!   
Associated Topics:
High School Polynomials
High School Sequences, Series

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