Re: 2*m^z - m^z = ?
- To: mathgroup at smc.vnet.net
- Subject: [mg88139] Re: 2*m^z - m^z = ?
- From: "David Park" <djmpark at comcast.net>
- Date: Sat, 26 Apr 2008 03:42:55 -0400 (EDT)
- References: <fus8ca$81t$1@smc.vnet.net>
With the Presentations package you could use FactorOut. Needs["Presentations`Master`"] Table[FactorOut[m^z][2*m^z - m^z], {m, 1, 21}] {1, 2^z, 3^z, 4^z, 5^z, 6^z, 7^z, 8^z, 9^z, 10^z, 11^z, 12^z, 13^z, \ 14^z, 15^z, 16^z, 17^z, 18^z, 19^z, 20^z, 21^z} -- David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ "Alexey Popkov" <popkov at gmail.com> wrote in message news:fus8ca$81t$1 at smc.vnet.net... > Hello, > What do you think about this: > > Table[ > {2*m^z - m^z, > FullSimplify[2*m^z - m^z]}, > {m, 1, 21}] // TableForm > > The answer is very interesting (only odd numbers are treated well): > > 1 1 > -2^z + 2^(1 + z) 2^z > 3^z 3^z > 2^(1 + 2*z) - 4^z 4^z > 5^z 5^z > 2^(1 + z)*3^z - 6^z 2^(1 + z)*3^z - 6^z > 7^z 7^z > 2^(1 + 3*z) - 8^z 8^z > 9^z 9^z > 2^(1 + z)*5^z - 10^z 2^(1 + z)*5^z - 10^z > 11^z 11^z > 2^(1 + 2*z)*3^z - 12^z 2^(1 + 2*z)*3^z - 12^z > 13^z 13^z > 2^(1 + z)*7^z - 14^z 2^(1 + z)*7^z - 14^z > 15^z 15^z > 2^(1 + 4*z) - 16^z 16^z > 17^z 17^z > 2^(1 + z)*9^z - 18^z 2^(1 + z)*9^z - 18^z > 19^z 19^z > 2^(1 + 2*z)*5^z - 20^z 2^(1 + 2*z)*5^z - 20^z > 21^z 21^z > > Can anyone explain the reason for this behavior? >