       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:

>
> 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

```

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