2
$\begingroup$

I have an entire function which is not a polynomial. Is there a way to use the Casorati-Weierstrass theorem to prove there exists a point $z_0$ such that every coefficient of the Taylor series at $z_0$ is not zero?

$\endgroup$
5
$\begingroup$

The set of points where the $n$-th Taylor coefficient is zero is the set $D_n = \{w \in \mathbb{C}\,:\,f^{(n)}(w) = 0\}$. This is a closed discrete set, hence it is countable (because if it were not discrete the identity theorem would imply that $f^{(n)} \equiv 0$, hence $f$ would be a polynomial of degree at most $n-1$). Thus, the set of points where at least one Taylor coefficient is zero is the countable set $D = \bigcup_{n=0}^\infty D_n$. Since $\mathbb{C}$ is uncountable, $\mathbb{C} \smallsetminus D$ is non-empty.

$\endgroup$
  • 1
    $\begingroup$ At the moment I don't see an immediate way to answer your question using the Casorati-Weierstrass theorem. $\endgroup$ – t.b. Apr 15 '12 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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