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I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they ask

Which of the following series converge to a rational function in some domain? $$\sum_{k=0}^\infty \frac{1}{k!+k^2+k}z^{k^2+2k}\\\sum_{k=0}^\infty \frac{2^k}{(1+z^2)^k}\\\sum_{k=0}^\infty\frac{1}{k! (z-k)^k}$$

I had no problem with the second and third ones. The second one converges to $\frac{1}{1-\frac{2}{1+z^2}}$ given that $\left|\frac{2}{1+z^2}\right|<1$, and the third one has an infinite number of singularities so it can't be a rational function.

I don't know how to verify that in the first series. Is there any general method to see if some power series converge to a rational function, or in this particular case, a way to see if this series does?

Edit: Now I am not that convinced about my argument for the third series (Since the series might only converge in a bounded domain from which the number of singularities would be finite). Is there anything wrong with it or any way to formalize it further? I also just found the same argument here.

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    $\begingroup$ For the first, try to prove that it is entire (then in order to be rational, it must be a polynomial). Re third, your logic looks sound (by analytical continuation, if it coincides with the rational on a good enough domain, it must coincide with it pretty much everywhere). $\endgroup$ – user58697 Jul 16 '18 at 2:20
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    $\begingroup$ @user58697 How can the first one converge in $|z|>1?$ The $k$th term doesn't go to $0$ does it? The logarithm of the modulus of the numerator is $(k^2+2k)\log|z|$ and that of the denominator is $\approx k\log k$. And if I'm right about that, then it's not rational because on $|z|<1$ it's analytic and bounded by $e$. $\endgroup$ – saulspatz Jul 16 '18 at 2:30
  • $\begingroup$ PMI by using induction $\endgroup$ – Amruth A Jul 16 '18 at 9:34
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The first series converges uniformly to some $f$ in $\overline {\mathbb D},$ and diverges for all other $z.$ Thus $f$ is analytic in $\mathbb D.$ Suppose $f=R$ in some domain $U,$ where $R$ is a rational function. Then $U\subset \mathbb D.$ Let $P$ be the set of poles of $R.$ Then $f,R$ are both analytic in $\mathbb D\setminus P,$ which is also a domain. By the identity principle, $f=R$ in $\mathbb D\setminus P.$ Since $f$ has no singularities in $D(0,1),$ neither does $R.$ Therefore $f=R$ in $\mathbb D.$ Now $f$ extends continuously to $\overline {\mathbb D},$ hence so does $R.$ Therefore $R$ has no poles in $\overline {\mathbb D}.$ It follows that $R$ is analytic in some $D(0,r), r>1.$ Thus $f$ has an analytic extension to $D(0,r).$ Therefore the power series defining $f$ converges in $D(0,r),$ contradiction.

Added later For the third series: Let $\Omega =\mathbb C\setminus \{0,1,2,\dots\}.$ Note that $\Omega$ is an open connected set. The given series converges uniformly on compact subsets of $\Omega$ to an analytic function $f,$ and $f$ has a pole at each point of $\{0,1,2,\dots\}.$ (Please ask if you have questions on this.) Let $R$ be a rational function, and let $P$ be the set of poles of $R.$ Then both $f,R$ are analytic on $\Omega \setminus P,$ which is also an open connected set. Suppose $f=R$ on an open subset of $\Omega \setminus P.$ Then by the identity principle, $f=R$ everywhere in $\Omega \setminus P.$ It follows that $R$ has a pole at each point of $\{0,1,2,\dots\}.$ This is a contradiction, since $R$ has at most a finite number of poles.

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  • $\begingroup$ I was thinking about how to formalize my thoughts in the third series and that's just what I thought. I like how the first case is developed. Thanks. $\endgroup$ – David Molano Jul 17 '18 at 5:23
  • $\begingroup$ I was thinking about this but why does the third series converge to an analytic function $f$ (on compact sets) if the series are not power series? $\endgroup$ – David Molano Jul 23 '18 at 21:29
  • $\begingroup$ If a sequence or series of analytic functions converges uniformly on compact subsets of an open set $U,$ then the limit function is analytic in $U.$ Are you familiar with this result? $\endgroup$ – zhw. Jul 23 '18 at 22:01
  • $\begingroup$ Oh. That's it. I was overthinking it. $\endgroup$ – David Molano Jul 23 '18 at 22:21
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Your logic is fine in the second case.


For the third case, recall the following:

If $f$ and $g$ are analytic in an open, connected set $D$ and are equal on a set having an accumulation point in $D$, then $f = g$ throughout $D$.

This result quickly gives that analytic continuations are unique. We can likewise show that meromorphic continuations are unique.

To this end, suppose that $m_1$ and $m_2$ are two meromorphic extensions of $f$. Then $m_1$ and $m_2$ must each have a (at most) countable sets of poles $P_1$ and $P_2$ respectively. Then $D \equiv \mathbb{C}\setminus(P_1 \cup P_2)$ will be connected and open since poles of meromorphic functions are isolated; as such, the restrictions of $m_1$ and $m_2$ to $D$ will be analytic functions equal on the open, connected set $D$ and are thus equal on $D$. To finish the argument, I quote Daniel Fischer:

$m_1$ and $m_2$ are the same function (at least outside the union of the singular sets). The Laurent expansion around an isolated singularity is determined by the values of the function on an arbitrarily small punctured neighborhood. That means $m_1$ and $m_2$ have the same Laurent expansion around each $z \in S_1 \cup S_2$. In other words, a meromorphic continuation is unique modulo gratuitous introduction of removable singularities.

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    $\begingroup$ No, $|z|^{2k+3}/(k+1)\to \infty$ if $|z|>1.$ The series in the first case converges iff $|z|\le 1.$ $\endgroup$ – zhw. Jul 16 '18 at 4:31
  • $\begingroup$ @zhw. You are correct - I will remove it. I wrote that off handedly at the end without thinking. Fortunately you took the time to correct the mistake in your post $\endgroup$ – Brevan Ellefsen Jul 16 '18 at 4:46
  • $\begingroup$ Yes, I forgot about using the identity theorem in these cases. Thanks. $\endgroup$ – David Molano Jul 17 '18 at 5:24

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