1

I started thinking about this after this MathSE thread.

Take a sequence of Taylor polynomials $f_n$ that converge to $f$. Does $f_n$ always have a growing number or roots in $\mathbb{C}$ which grow in magnitude unboundedly as $n\rightarrow\infty$?

Intuitively, it seems to me that every Taylor series should have at least one "root at infinity" (not that that is an actual rigourous concept here).

So the main question is this:
Let $A_n$ be the set of roots of Taylor polynomial $f_n$. Does there exist a disc around the origin $D_n$ whose radius grows unboundedly in $n$, such that the number of roots outside of $D_n$ grows unboundedly (i.e. #|$A_n\cap D_n^c|\overset{n\rightarrow\infty}{\longrightarrow}\infty$)?

  • Does the set of roots of $f_n$ with large magnitude always tend to infinite cardinality?
  • Is there an $f$ for which the roots of $f_n$ have a global bound, i.e. all $|z_n|<R$ for any $n$?
  • Does the answer depend on the radius of convergence of the Taylor series of $f$?

Intuitively, I feel like there will always be some sequence of roots $z_n$ such that $|z_n|\rightarrow\infty$ as $n\rightarrow\infty$ (at least when the radius of convergence is infinity). And that there may or may not be sequences of roots that tend to finite limits. But my argument for that is just hand-wavy using polynomials with infinite coefficients.

For example $\sin(z)$ has infinitely many finite roots but its Taylor polynomials also appear to have a growing number of roots with magnitudes that grow unboundedly. Of course, $e^z$ only has the latter.

I've never studied complex analysis beyond the bits and pieces I've picked up from other courses. Maybe there are standard results that answer my question.

jdods
  • 6,248
  • For the example of $\sin x$ and its Taylor series truncations, note that the roots "interlace" with the roots of its derivative $\cos x$. So an ad hoc argument on that basis makes it clear that the roots of the truncations converge to roots of the analytic function. – hardmath May 05 '16 at 00:08
  • Aside: I suspect you may want to hear about the great Picard theorem for meromorphic functions; in particular, in the special case where the essential singularity is taken to be $\infty$. (note, however, that $0$ and $\infty$ can be the two excluded values; e.g. as is the case for $\exp(z)$) –  Jul 22 '17 at 18:00
  • Thanks for the hint, I don't really know anything about complex analysis. So it sounds like the answer to what I'm asking is unknown. I expected someone here would know a counterexample, or at least have some quick intuition on whether it is true or not. – jdods Jul 22 '17 at 23:55

0 Answers0