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Re: What we from 1^Infinity, Infinity^0, and similar stuff

<ted.ersek at> a écrit dans le message de news: 
dogjvd$pno$1 at
|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 (

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 
( example,


In[1]:= Limit[z, z -> Infinity]
Out[1]= Infinity


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,


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