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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

Hi,

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! http://mathforum.org/dr.math/
```

```
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.

Chyle
```

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

Hi,

The power series expansion of:

(1 - x)^(-1)

is:

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
back.

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! http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials
High School Sequences, Series

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