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*: "Jean-Marc 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*: owner-wri-mathgroup 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 non-zero 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.