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Topic: need algebraic identity
Replies: 1   Last Post: Oct 17, 1999 10:22 PM

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Chuck Cadman

Posts: 16
Registered: 12/12/04
Re: need algebraic identity
Posted: Oct 17, 1999 10:22 PM
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Larry Mintz wrote in message <7ue10b$f2u$1@cti15.citenet.net>...
>How can I factor x^n-y^m n<>m ?
>I would like it similar to the identity

b^n-a^n=(b-a)(b^(n-1)+b^(n-2)a+....+a^n-1) if it is
>possible.
>
>Basically I trying to do the following:
>|| a_n^n -a_m^m||<=|a_n-a_m| || ? ||
>
>where a={a_n} is a Cauchy sequence.



I don't think it's going to work. If a_n = 2^(n^(-1/2)), then a_n converges
to 1. However, a_n^n = 2^(n^(1/2)) diverges.

Chuck Cadman








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