Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$a_n\neq 0$, $b_n$ be complex null sequences, such that $\lim_{n}\frac{b_n}{a_n^k}=0\forall k\in\mathbb{N}$, suppose $f$ is analytic in domain $U$ which contains $0$ and all $a_n$, we need to show $f(a_n)=b_n=0\forall n$

well, by Identity Theroem of analytic function, zero set of $f$ has limit point in the domain of $f$ hence $f\equiv 0$ but how to show $f(a_n)=b_n=0\forall n$ ? Hint please. $|\frac{b_n}{a_n^k}|<\epsilon$ then?

share|improve this question
add comment

1 Answer

My interpretation of the question:

We have two sequences $(a_n), (b_n)$ which both converge to $0$. Also, $a_n\ne 0$ for all $n$, and for every $k$ we have $$ \lim_{n\to\infty }\frac{b_n}{a_n^k}\to 0 \tag{1}$$ Goal: if $f$ is analytic in a neighborhood of $0$ and $f(a_n)=b_n$ for all $n$, then $f\equiv0$ (and in particular, this means all $b_n$ had to be equal to $0$).

Idea of proof: suppose $f$ is not identically zero. Consider its Taylor expansion at $0$. Let $c_kz^k$ be its first nonzero term, that is, $f(z)=c_kz^k+\dots $. Obtain a constradiction with (1).

Spoiler included in case you get stuck again.

Further hint: since $f(z)/z^k\to c_k$ as $z\to 0$, it follows that $\lim_{n\to \infty} \frac{f(a_n)}{a_n^k} = c_k $, a contradiction.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.