something funny!
- To: mathgroup at smc.vnet.net
- Subject: [mg78696] something funny!
- From: dimitris <dimmechan at yahoo.com>
- Date: Sat, 7 Jul 2007 06:00:37 -0400 (EDT)
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