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Re: What we from 1^Infinity, Infinity^0, and similar stuff
 To: mathgroup at smc.vnet.net
 Subject: [mg63325] Re: What we from 1^Infinity, Infinity^0, and similar stuff
 From: "JeanMarc Gulliet" <jeanmarc.gulliet at gmail.com>
 Date: Sat, 24 Dec 2005 07:18:51 0500 (EST)
 Organization: The Open University, Milton Keynes, U.K.
 References: <dogjvd$pno$1@smc.vnet.net>
 Sender: ownerwrimathgroup at wolfram.com
<ted.ersek at tqci.net> a écrit dans le message de news:
dogjvd$pno$1 at smc.vnet.net...
I am using Mathematica 4.1, and version 5 may work different in this case.

 It seems I can compute (1^z) where (z) has any numeric value and
 Mathematica returns the Integer 1. I can also compute (z^0) where (z) is
 any non zero value and Mathematica returns the Integer 1. Hence I think
 the following should return {1,1,1,1,1,1,1}. Can someone explain why that
 would be wrong?
Hi Ted,
From a mathematical point of view, I am afraid that your hunch is totally
wrong since you are treating infinity as a number that own a definite
numeric quantity, perhaps extremely very enormously large but still finite!
Infinity is NOT a number (http://mathworld.wolfram.com/Infinity.html).
Expression such as 1^z makes sense (and have a definite meaning) only for
nonzero value of z. Since infinity is NOT a number, z cannot be equal to
it. However, z can approaches infinity and the expression 1^z as z
approaches infinity makes sense. Indeed, we deal with limiting value:
In[1]:= Limit[1^z, z > Infinity]
Out[1]= 1
Therefore, what is the meaning of an expression such that 1^infinity?
Mathematically, this is the expression of an indeterminate form that results
from the computation of the limits of two functions of the form f(x)^g(x)
and the result is generally not equal to 1
(http://mathworld.wolfram.com/Indeterminate.html).For example,
although
In[1]:= Limit[z, z > Infinity]
Out[1]= Infinity
and
In[2]:= Limit[1/z + 1, z > Infinity]
Out[2]= 1
concluding that (1/z + 1)^z approaches 1 as z approaches infinity is an
erroneous conclusion
In[3]:= Limit[(1/z + 1)^z, z > Infinity]
Out[3]= E
So, redefining the behavior of Mathematica could be a dangerous stand for
the unforeseeable side effects that might occur in different kind of
computations yielding unexpected ? weird ? results in doing some integration
or differentiations to name a few.
Best regards,
/J.M.
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