Hi all.

I want to construct a power series $F(z)=\sum_{n=0}^\infty c_nz^n$ centered at zero and with finite radius of convergence in the complex plane, and which has infinitely many zeros (in its disk of convergence) but is nonconstant (and hence nonpolynomial, i.e. has infinitely many nonzero coefficients $c_n$).

Unfortunately I have a strict bound on the coefficients. I am limited to those $c_n$ such that $\sum_{n=0}^\infty n|c_n|<\infty$.

Is this possible?

A harder problem is this: Suppose instead I am choosing the $c_n$'s to satisfy $|c_n-c_{n-1}|< n^{-3}$ for all $n$. Can I still get infinitely many zeros inside some finite radius for $F(z)=\sum_{n=0}^\infty c_nz^n$?


  • 3
    $\begingroup$ Consider an infinite product $(1 + m_1 z)(1 + m_2 z)(1 + m_3 z) \cdots$, with $1 > m_1 > m_2 > m_3 > \ldots$. If I'm not mistaken, the $n$th coefficient of the corresponding power series is bounded by $|m_1 m_2 \cdots m_n|$, and you can make this product arbitrarily small by choosing small $m_\bullet$. Right? $\endgroup$
    – Vectornaut
    Commented Jun 20, 2013 at 18:22
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    $\begingroup$ Vectornaut is correct. You can also choose something as simple as $F(z)=sin(z)$ which has infinitely many zeros and $|c_n| \leq 1/n!$ are bounded. If you want infinitely many zeros in the unit disc (and radius of convergence 1) you can choose $F(z)=\sin(1/(1-z))$. It is easy to see that in this case $|c_n| \leq e$. $\endgroup$ Commented Jun 20, 2013 at 18:27
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    $\begingroup$ My comment about $\sin(1/(1-z))$ is not quite correct since even if $c_n$ are bounded, $\sum_n n|c_n|=\infty$, but then you can choose $F(z)=\sin(1/(1-z/2))$ instead. The radius of convergence will now be 2 instead, but the number of zeros in the disc of absolute convergence will be infinite. I wonder if the question is stated correctly because of such trivial constructions. Do you want to exclude entire functions, or do you want maybe to assume that the radius of convergence is one or is there something else that you want to assume? $\endgroup$ Commented Jun 20, 2013 at 19:14
  • $\begingroup$ Why not just use $e^z-1=\sum_{n=1}^\infty z^n/n!$. It has zeros at every $2\pi i k$ with $k\in\mathbb{Z}$, and the coefficients certainly satisfy your convergence condition since $\sum_{n=1}^\infty n/n!=e$. So for your second questions, it's possible with $a_n=n!b_n$ provided $\sum b_n$ converges. Possibly you meant to specify that the function has a finite radius of convergence, rather than an infinite radius of convergence as is true for $e^z-1$. $\endgroup$ Commented Jun 20, 2013 at 21:10
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    $\begingroup$ $c_n - c_{n-1}$ is the $z^n$ coefficient of $(1-z) \phantom. f(z)$, so you might as well work with $(1-z) \phantom. f(z)$ and impose the condition $|c_n| < n^{-3}$. $\endgroup$ Commented Jun 21, 2013 at 2:06

1 Answer 1


It's been noted already in the comments that the problem is still too easy as stated, because one can easily find functions $f(z) = \sum_{n=0}^\infty c_n z^n$ such as $f(z) = \sin(1 / (r-z))$, with infinitely many zeros inside the circle of convergence $|z|<r$, and as long as $r>1$ the coefficients $c_n$ decay exponentially, and thus satisfy constraints such as $\sum_n n|c_n| < \infty$ or $\left|c_n-c_{n-1}\right| \ll n^{-3}$ with plenty of room to spare.

[It's apparently not been noted that the "nonpolynomial" condition is automatic unless $f$ is identically zero, since a nonzero polynomial has finitely many roots anywhere in ${\bf C}$.]

However, the problem becomes somewhat more interesting if we require (as Johan Andersson suggested, and as the proposer may have intended) that $f$ have infinitely many roots in the unit disc $|z|<1$, which is the largest disc on which the power series is guaranteed to converge by the constraints on $c_n$.

Still, even with that interpretation there are power series whose coefficients that satisfy either condition, or even $|c_n - c_{n-1}| \ll n^{-k}$ for any constant $k$, and nevertheless have infinitely many real zeros in $\left|z\right|<1$. I'll deal with $\left|c_n - c_{n-1}\right| \ll n^{-3}$, though the same technique applies for all $k$. Since $c_n - c_{n-1}$ is the $z^n$ coefficient of $(1-z) \phantom. f(z)$, it is enough to construct $g(z) = \sum_{n=0}^\infty a_n z^n$ with $|a_n| \ll n^{-3}$ and infinitely many real zeros, and then set $f(z) = g(z) / (1-z)$.

Define $g(z) = S(z) - P(z)$, where $$ S(z) = z - \frac{z^2}{2^3} + \frac{z^4}{4^3} - \frac{z^8}{8^3} + \frac{z^{16}}{16^3} - + \cdots = \sum_{m=0}^\infty (-1)^m \frac{z^{2^m}}{2^{3m}}, $$ and $P(z)$ is the cubic polynomial $$ P(z) = \frac89 - \frac45(1-z) - \frac1{15} (1-z)^2 - \frac1{60}(1-z)^3. $$ Clearly $S$ satisfies the functional equation $S(z) = z - \frac18 S(z^2)$. The cubic $P$ was chosen to satisfy the same equation within $O((1-z)^4)$: we have $$ P(z) = z - \frac18 P(z^2) + (1-z)^4 \frac{(1+z)(3+z)}{480}. $$ Hence $g(z) = -\frac18 g(z^2) + O((1-z)^4)$. It follows that as $z \rightarrow 1$ the ratio $f(z) / (\log(1/z))^3$ approaches some function $\phi(z)$ that satisfies $\phi(z^2) = -\phi(z)$ exactly. Unless this function is identically zero, it must have at least one zero in each interval $(z,z^{1/2}]$, and thus an infinite sequence of zeros approaching $1$. Now there are several ways to show that $\phi$ is not identically zero, but possibly the simplest is numerical computation, which shows $\phi$ oscillating with amplitude about $2.2 \cdot 10^{-5}$. The first zero of $f(z)$ appears near $z=0.998131$, then $0.998891$, $0.999493$, $0.999736$, $0.999871$, $0.999935$, $0.999967$, etc. with each zero about twice as close to $1$ as the previous one.

Here's some gp code that gives a rough plot showing $f$ first crossing zero after not quite reaching it around $0.9945$ (so it's 99.44% certain to be negative?...), and then locates the first seven zeros exhibited above.

f(z) = suminf(n=0,(-1)^n*z^2^n/8^n)
P(z) = 8/9 - (1-z)*4/5 - (1-z)^2/15 - (1-z)^3/60
R(z) = (f(z)-P(z)) / log(1/z)^3


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