construct a power series with infinitely many zeros in the complex plane, bounded coefficients??? 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 $F(z)=\sum_{n=0}^\infty c_nz^n$?
Thanks!
 A: 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

plot(x=.99,.9993,R(x))    
for(i=11,17,print(solve(x=1-3/(2^(i-1)),1-3/2^i,R(x))))

