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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?
> 



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