Re: 2*m^z - m^z = ?
- To: mathgroup at smc.vnet.net
- Subject: [mg88149] Re: [mg88104] 2*m^z - m^z = ?
- From: Syd Geraghty <sydgeraghty at mac.com>
- Date: Sat, 26 Apr 2008 03:44:45 -0400 (EDT)
- References: <200804250926.FAA08036@smc.vnet.net>
Alexy,
Not so much an explanation (which I will leave to WR experts) but
another illustration that IF you are a "true believer" in Mathematica
you can find a way to get the result or form you desire with a little
patient exploration and trying.
Table[{2*m^z - m^z, Collect[FullSimplify[2*m^z - m^z], z, Modulus ->
2]}, {m,
1, 11}] // TableForm
( {
{1, 1},
{-2^z + 2^(z + 1), 2^z},
{3^z, 3^z},
{2^(2 z + 1) - 4^z, 4^z},
{5^z, 5^z},
{2^(z + 1) 3^z - 6^z, 6^z},
{7^z, 7^z},
{2^(3 z + 1) - 8^z, 8^z},
{9^z, 9^z},
{2^(z + 1) 5^z - 10^z, 10^z},
{11^z, 11^z}
} )
HTH ... Syd
Syd Geraghty B.Sc, M.Sc.
sydgeraghty at mac.com
My System
Mathematica 6.0.2.1 for Mac OS X x86 (64 - bit) (March 13, 2008)
MacOS X V 10.5.2
MacBook Pro 2.33 Ghz Intel Core 2 Duo 2GB RAM
On Apr 25, 2008, at 2:26 AM, Alexey Popkov wrote:
> 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?
>
- References:
- 2*m^z - m^z = ?
- From: Alexey Popkov <popkov@gmail.com>
- 2*m^z - m^z = ?