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,
>
> 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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