Re: something funny!
- To: mathgroup at smc.vnet.net
- Subject: [mg78757] Re: [mg78696] something funny!
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 8 Jul 2007 06:19:21 -0400 (EDT)
- Reply-to: hanlonr at cox.net
$Version
5.2 for Mac OS X (June 20, 2005)
(1/(2*#1^z - #1^z)& )[n]/.n->Range[10]
{1, 2^(-z), 3^(-z), 4^(-z), 5^(-z),
6^(-z), 7^(-z), 8^(-z), 9^(-z),
10^(-z)}
Same result for v6.
Bob Hanlon
---- dimitris <dimmechan at yahoo.com> wrote:
> The version is 5.2.
>
> Say
>
> In[125]:=
> o = 1/(2*e^z - e^z)
>
> Out[125]=
> e^(-z)
>
> Then
>
> In[126]:=
> o = (1/(2*e^z - e^z) /. e -> #1 & ) /@ Range[2, 10]
>
> Out[126]=
> {2^(-z), 3^(-z), 4^(-z), 5^(-z), 6^(-z), 7^(-z), 8^(-z), 9^(-z), 10^(-
> z)}
>
> However,
>
> In[128]:=
> FullSimplify[(1/(2*#1^z - #1^z) & ) /@ Range[10]]
>
> Out[128]=
> {1, 2^(-z), 3^(-z), 1/(2^(1 + 2*z) - 4^z), 5^(-z), 1/(2^(1 + z)*3^z -
> 6^z), 7^(-z), 1/(2^(1 + 3*z) - 8^z), 9^(-z),
> 1/(2^(1 + z)*5^z - 10^z)}
>
> No matter what I tried I could not simplify the expressions with even
> base of z above.
> Any ideas?
>
> My query becomes bigger considering that as I was informed even in
> version
> 6 we get
>
> FullSimplify[(1/(2*#1^z - #1^z) & ) /@ Range[10]]
>
> {1, 2^(-z), 3^(-z), 4^(-z), 5^(-z), 1/(2^(1 + z)*3^z - 6^z),
> 7^(-z), 8^(-z), 9^(-z), 1/(2^(1 + z)*5^z - 10^z)}
>
> What it is so exotic I can't figure out. It is so trivial!
>
> In[140]:=
> FullSimplify[1/(2^(1 + 2*z) - 4^z) == 1/(2*4^z - 4^z)]
>
> Out[140]=
> True
>
> In another CAS I took
>
> convert("(1/(2*#1^z - #1^z) & ) /@ Range[10]",FromMma);value(%);
>
> 1
> map(unapply(----, _Z1), [seq(i, i = 1 .. 10)])
> z
> _Z1
>
>
> 1 1 1 1 1 1 1 1 1
> [1, ----, ----, ----, ----, ----, ----, ----, ----, ---]
> z z z z z z z z z
> 2 3 4 5 6 7 8 9 10
>
>
> Dimitris
>
>