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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/