Re: correction to my last post
- To: mathgroup at smc.vnet.net
- Subject: [mg36021] Re: correction to my last post
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Mon, 12 Aug 2002 03:34:39 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
In terms of Mathematica this can be expressed as the difference between the following: In[1]:= Limit[Limit[(1 + y/n)^m, n -> Infinity], m -> Infinity] Out[1]= 1 In[2]:= Limit[(1 + y/Abs[x])^Abs[x], x -> Infinity] Out[2]= E^(I*Im[y] + Re[y]) The latter can of course be made equal to any number, real or complex except 0. So the whole original confusion relates to the meaning of 1^Infinity. If one interpreted it in the first sense (or simply as Limit[1^x,x->Infinity]) than the answer would have been 1. However, Mathematica adopts the more general approach, considering an expression involving Infinity to be Indeterminate unless all ways of representing it as a limit lead to the same answer. Of course one should not take this too strictly, for Mathematica gives: In[3]:= Infinity! Out[3]= Infinity although a pretty good case can be made for Sqrt[2Pi], see <http://functions.wolfram.com/10.01.06.0013> :) Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Monday, August 12, 2002, at 04:39 AM, Jonathan Rockmann wrote: > MathGroup, > > I don't know why all of my symbols disappeared when I copied and pasted > to my last post. My last post makes no sense. I bet you now I'm going > to get a flood of comments about this. Sorry. > > It should say: > > A lot of people think that Lim n->00 (1+(1/n))^n =1. Their reasoning > is this: "When n->00, then 1+(1/n)->1. Now compute lim n->00 1^n =1". > Of course, this reasoning is too simple. You have to deal with both of > the n's in the expression (1+(1/n))^n at the same time -- i.e., they > both go to infinity simutaneously: you can't figure that one goes to > infinity and then the other goes to infinity. And in fact, if you let > the other one go to infinity first, you'd get a different answer: lim > n->00 (1+.0000001)^n=00. So evidently the answer lies somewhere > between 1 and 00. That doesn't tell us much; my point here is that > easy methods do not work on this problem. The correct answer is a > number that is near 2.718281828... i.e. the constant "e". There's no > way you could get that by an easy method. > > > Jonathan > mtheory at msn.com > > > > ----- Original Message ----- > From: Jonathan Rockmann To: mathgroup at smc.vnet.net > Sent: Sunday, August 11, 2002 6:12 AM > Subject: [mg36021] [mg35978] Re: [mg35940] Re: One to the power Infinity > > As a follow up to David's example: > > Some students seem to think that lim n->0 (1+(1/n))n = 1. Their > reasoning > is this: "When n->0, then 1+(1/n) -> 1. Now compute lim n->0 1 n = 1." > This reasoning is just too simplistic. You have to deal with both of > the n's > in the expression (1+(1/n))n at the same time -- i.e., the > y both go to infinity simultaneously; you can't figure that one goes to > infinity > and then the other goes to infinity. In fact, if you let the other one > go > to infinity first, you'd get a different answer: lim n->0 > (1+0.0000001)n = > 0. So evidently the answer lies somewhere between 1 and > 0. Easy methods do not work on this problem. > > The correct answer is a number that is near 2.718. (It's an important > constant, > known to mathematicians as "e" = aprrox. 2.718281828...) There's no way > you > could get that by an easy method. > > Jonathan Rockmann > mtheory at msn.com > > ----- Original Message ----- > From: David W. Cantrell To: mathgroup at smc.vnet.net > Subject: [mg36021] [mg35978] [mg35940] Re: One to the power Infinity > > Matthias.Bode at oppenheim.de wrote: > > 0^\[Infinity] => 0, as expected; > > 0.9^\[Infinity] => 0, as expected; > > 2^\[Infinity] => Infinity, as expected; > > 1^\[Infinity] => Indeterminate, unexpected. Naively expected: 1. > > > > For which reason(s) is 1^\[Infinity] defined as Indeterminate? > > Presumably because, as a limit form, 1^Infinity is indeterminate. > In other words, if f(x) approaches 1 and g(x) increases without bound, > f(x)^g(x) need not approach 1. Consider the limit of (1+1/x)^x as x > increases without bound, for example. > > David > >