Re: Re: something funny!
- To: mathgroup at smc.vnet.net
- Subject: [mg78783] Re: [mg78756] Re: something funny!
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Mon, 9 Jul 2007 01:42:02 -0400 (EDT)
- Reply-to: hanlonr at cox.net
(Integrate[1/(2*#1^z - #1^z), {z, 0, Infinity},
Assumptions -> {n > 1}] &)[n] /. n -> Range[2, 10]
{1/Log[2], 1/Log[3], 1/Log[4], 1/Log[5], 1/Log[6], 1/Log[7],
1/Log[8], 1/Log[9], 1/Log[10]}
Bob Hanlon
---- dimitris <dimmechan at yahoo.com> wrote:
>
> dimitris :
> > 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
>
> To see why I consider it funny:
>
> In[944]:=
> InputForm[({#1, Integrate[1/(2*#1^z - #1^z), {z, 0, Infinity}]} & ) /@
> Range[2, 10]]
>
> Out[944]//InputForm=
> {{2, Log[2]^(-1)}, {3, Log[3]^(-1)}, {4, Integrate[(2^(1 + 2*z) -
> 4^z)^(-1), {z, 0, Infinity}]}, {5, Log[5]^(-1)},
> {6, Integrate[(2^(1 + z)*3^z - 6^z)^(-1), {z, 0, Infinity}]}, {7,
> Log[7]^(-1)},
> {8, Integrate[(2^(1 + 3*z) - 8^z)^(-1), {z, 0, Infinity}]}, {9,
> Log[9]^(-1)},
> {10, Integrate[(2^(1 + z)*5^z - 10^z)^(-1), {z, 0, Infinity}]}}
>
> where I wait an output as below
>
> In[948]:=
> Integrate[1/(2*e^z - e^z), {z, 0, Infinity}, Assumptions -> e > 1]
> (% /. e -> #1 & )[Range[2, 10]]
>
> Out[948]=
> 1/Log[e]
> Out[949]=
> {1/Log[2], 1/Log[3], 1/Log[4], 1/Log[5], 1/Log[6], 1/Log[7], 1/Log[8],
> 1/Log[9], 1/Log[10]}
>
> Dimitris
>
>