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Re: need algebraic identity
Posted:
Oct 17, 1999 10:22 PM


Larry Mintz wrote in message <7ue10b$f2u$1@cti15.citenet.net>... >How can I factor x^ny^m n<>m ? >I would like it similar to the identity b^na^n=(ba)(b^(n1)+b^(n2)a+....+a^n1) if it is >possible. > >Basically I trying to do the following: > a_n^n a_m^m<=a_na_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



