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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



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