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MathGroup Archive 2005

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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.

> 


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