Re: FullSimplify again ...
- To: mathgroup at smc.vnet.net
- Subject: [mg59159] Re: FullSimplify again ...
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sun, 31 Jul 2005 01:30:35 -0400 (EDT)
- Organization: The Open University, Milton Keynes, U.K.
- References: <dcf3d4$lgl$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Detlef Müller wrote:
> Hello.
> I happened to type the following lines:
>
>
>
> In[15]:= f=FullSimplify[Product[(5-i)/5,{i,1,n}]]
> Out[15]= 0
>
> In[17]:= Product[(5-i)/5,{i,1,n}]/.(n->4)
> Out[17]= 24/625
>
> Strange, isn't it?
>
> Version Number: 5.1.1.0
> Platform: X
>
> Greetings,
> Detlef
>
Hi Detlef,
The result might not look so strange when we investigate the behavior of
Mathematica functions such as *FullSimplify*. First, we notice that your
product involves special function:
In[1]:=
Product[(5 - i)/5, {i, 1, n}]
Out[1]=
(-(1/5))^n*Pochhammer[-4, n]
Therefore a function like *Simplify* will be unable to simplify it:
In[2]:=
Simplify[Product[(5 - i)/5, {i, 1, n}]]
Out[2]=
(-(1/5))^n*Pochhammer[-4, n]
However, *FullSimplify* uses an extensive set of replacement rules for
special functions and also may take in account that removing a finite
number of elements in a sequence does not change the asymptotic behavior
of the sequence. So *FullSimplify * returns zero as general value since
every terms of the sequence are equal to zero for n > 4.
In[3]:=
FullSimplify[Product[(5 - i)/5, {i, 1, n}]]
Out[3]=
0
In[4]:=
Product[(5 - i)/5, {i, 1, n}] /. n -> 1
Out[4]=
4/5
In[5]:=
Product[(5 - i)/5, {i, 1, n}] /. n -> 2
Out[5]=
12/25
In[6]:=
Product[(5 - i)/5, {i, 1, n}] /. n -> 3
Out[6]=
24/125
In[7]:=
Product[(5 - i)/5, {i, 1, n}] /. n -> 4
Out[7]=
24/625
In[8]:=
Product[(5 - i)/5, {i, 1, n}] /. n -> 5
Out[8]=
0
In[9]:=
Limit[%, n -> Infinity]
Out[9]=
0
Only the first four terms are not null. The behavior of *FullSimplify*
is coherent with the behavior of other Mathematica functions that are
designed to find general cases rather than specific ones. For instance,
although *Solve* complains that the solution involves inverse function,
*Reduce* just agree with the general result is always true (that is if
we exclude the first four terms in the sequence):
In[10]:=
Reduce[Product[(5 - i)/5, {i, 1, n}] == 0, n]
Out[10]=
True
In[11]:=
$Version
Out[11]=
"5.2 for Microsoft Windows (June 20, 2005)"
Hope this helps,
/J.M.
>