Yatir Halevi

Posted on Tuesday, 04 March, 2003 - 07:35 pm:

Does anybody have an example of when

1^{f (x)}

doesn't tend to 1 as

f (x) \to \infty

x \to \infty

?
(There might be a simple example, I haven't given it much thought yet...)

Yatir

Dan Goodman

Posted on Tuesday, 04 March, 2003 - 07:42 pm:

1^x =1 for all x...

Colin Prue

Posted on Tuesday, 04 March, 2003 - 08:16 pm:

perhaps this is cheating, but:

$1^{3 / 2} = (e^{2 π i})^{3 / 2} = - 1$

Colin Prue

Posted on Tuesday, 04 March, 2003 - 08:17 pm:

...therefore if you are careful how you define things you can have a divergent sequence (1,-1,1,-1,1...etc)

David Loeffler

Posted on Wednesday, 05 March, 2003 - 12:30 am:

Yes. It's cheating. You are trying to tell me that 1 = -1. From this I can validly deduce that I am the Pope.

David

Hauke Worpel

Posted on Wednesday, 05 March, 2003 - 08:36 am:

How is it cheating, your Holiness? ;) 1^3/2 =(1³ )^1/2

1³ =1 all the time, no disagreement there, but what is 1^1/2 ? Either 1 or -1. Is this a semantic quibble or a genuine mathematical point?

David Loeffler

Posted on Wednesday, 05 March, 2003 - 09:37 am:

It is not the case for general complex numbers that (a^b )^c = a^bc ; the values of the latter will be a subset, in some cases a proper subset, of the values of the former.

-----

Yatir, can I check exactly what question you were asking? Did you really mean to ask,

"Do there exist real functions a(x) and b(x) such that a(x) -> 1 but a(x)^b(x) -> infinity as x -> infinity?"

In this case the answer is yes, and there is a fairly easy counterexample.

David

Yatir Halevi

Posted on Wednesday, 05 March, 2003 - 05:46 pm:

David, I know that a(x) exists. But a lot of things in mathematics are very un-intuitional, and I wanted to know if maybe here lies one of them.

Yatir