0
$\begingroup$

Let $a, a_n \in \mathbb{C}$ ($n \ge 2$). Suppose we are given the Laurent series

$$ \sum_{n=2}^{\infty} a_n(z - a)^{-n}.$$

I know that we can calculate the set on which this expression converges using Hadamard's formula. That is, we have that the series converges for $|z - a| > \frac{1}{R}$, where $\frac{1}{R} = \limsup \sqrt[n]{|a_n|}$. In case $\limsup \sqrt[n]{|a_n|} = 0$, the series converges for $|z-a|> 0$. And in case $\limsup \sqrt[n]{|a_n|} = \infty$, the series is nowhere convergent.

To continue, suppose that we formally anti-differentiate our series to obtain

$$ \sum_{n=1}^{\infty} \frac{-a_{n+1}}{n}(z - a)^{-n}.$$

My question is, how does the radius of convergence of this new series relate to the old one? Maybe the new radius of convergence is potentially larger? I see that the question boils down to just being smart about Hadamard's formula. We need to compare $\limsup \sqrt[n]{|a_n|}$ with $\limsup \sqrt[n]{\frac{|a_{n+1}|}{n}}$, but this seems like a tricky bit of analysis. It's already clear to me that $\limsup \sqrt[n]{\frac{|a_{n+1}|}{n}} \le \limsup \sqrt[n]{|a_{n+1}|}$, so my hope is that I can somehow compare $\limsup \sqrt[n]{|a_{n+1}|}$ with $\limsup \sqrt[n]{|a_n|}$.

Hints or solutions are greatly appreciated.

$\endgroup$
1
$\begingroup$

Hint: $\lim_{n\to\infty} n^{1/n} = 1$, and $$ |a_{n+1}|^{1/n} = |a_{n+1}|^{1/(n+1)}\cdot |a_{n+1}|^{1/(n(n+1))} $$

$\endgroup$
1
$\begingroup$

Ok, I think with the hint from @mrf I am able to answer my own question. First, I was able to establish the following lemma--and I think it's worth trying to prove this for yourself:

Suppose that $a_n \ge 0, b_n > 0$ with $b_n \to \alpha > 0$. Then

$$ \limsup \frac{a_n}{b_n} = \frac{\limsup a_n}{\alpha}.$$

With this lemma in hand, the following equality is proved, and we proceed to establish the inequality:
$$ \limsup \frac{|a_{n+1}|^{\frac{1}{n}}}{n^{\frac{1}{n}}} = \limsup |a_{n+1}|^{\frac{1}{n}} \le \limsup |a_n|^{\frac{1}{n}}.$$

Hence the radius of convergence of the anti-derivative is at least as large as that of the original Laurent series.

Notice that if $ \limsup |a_n|^{\frac{1}{n}} = \infty $ there is nothing to show. Hence we may assume that $ \limsup |a_n|^{\frac{1}{n}} = M < \infty $. Hence each $ \displaystyle \sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{m+1}}\} = M_n < \infty$. And this in turn says that, for each $n$, we have:

$$\sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{m(m+1)}}\}\le \sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{n(m+1)}}\} = M_n^{\frac{1}{n}} \le \sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{m+1}}\} = M_n < \infty.$$

Now, using the hint given by @mrf, we get

$$0 \le \sup_{m \ge n}\{|a_{m+1}|^{\frac{1}{m}}\} \le M_n \cdot M_n^{\frac{1}{n}} \to M.$$

I think this does the trick. Let me know if I've messed something up.

$\endgroup$
  • $\begingroup$ Looks good to me. $\endgroup$ – mrf May 14 '13 at 10:34
0
$\begingroup$

Derivatives of Taylor series and Laurent series keep the same radius, simply because they always converge where they exist in their respect Ball or Annulus. That is, derivatives do not introduce new poles.

Thus antidifferentiating a Taylor series or Laurent series cannot yield a new radius, because if you differentiate the antiderivative you should have the same radius as your antiderivative.

$\endgroup$

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.