Re: 2*m^z - m^z = ?
- To: mathgroup at smc.vnet.net
- Subject: [mg88138] Re: 2*m^z - m^z = ?
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sat, 26 Apr 2008 03:42:44 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <fus8ca$81t$1@smc.vnet.net>
Alexey Popkov wrote:
> 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?
FWIW,
Though this might not answer your question, as a rule of thumb -- or
best practice -- one should always attempt to simplify symbolic
expressions before using them in any iterative or recursive construct.
FullSimplify[2*m^z - m^z]
m^z
Regards,
-- Jean-Marc