Sequence question (analysis)

• Mar 16th 2013, 07:48 PM
director
Sequence question (analysis)
Hi guys,

I'm working on a complex analysis problem which states that I have
"a sequence {an} of distinct points in some region G such that an --> a in G as n --> infinity"

(then I'll need to do something with the sequences)

First question: Is {an} an infinite sequence? (In general, when I see this notation in analysis should I always assume an infinite sequence?)

I'm going to assume that it is an infinite sequence, then, I have a function represented by, lets say, f(an) = cos(an)

Since all sequences converge to the point a, is it true that f(an) = f(a) = cos(a)?
Do I have a constant function? That's what it looks like to me.

Thanks!

• Mar 16th 2013, 08:09 PM
Gusbob
Re: Sequence question (analysis)
Quote:

First question: Is {an} an infinite sequence? (In general, when I see this notation in analysis should I always assume an infinite sequence?)
In this case, yes. In general, I do think of this as an infinite sequence. There may be exceptions, however, and depends on context.

Quote:

Since all sequences converge to the point a, is it true that f(an) = f(a)?

Not in general. Consider the function $\displaystyle f(x)=1$ if $\displaystyle x\not= a$, $\displaystyle f(x)=0$ otherwise. The infinite sequence $\displaystyle a_n=\frac{1}{n}$ has limit $\displaystyle a=0$ and satisfies $\displaystyle f(a_n)=1$ for all $\displaystyle n$. But $\displaystyle f(0)=0$. You need stronger conditions (i.e. function is continuous) for this statement to be true. For your example, it would be so since $\displaystyle \cos(z)$ is analytic.
• Mar 16th 2013, 08:22 PM
director
Re: Sequence question (analysis)
Thanks Gusbob.

My function is, in fact, homomorphic on region G so I think that would make the previous statement true? Cos(zn) was just an example and it happened to be meet the criteria.