We now turn to the construction of an entire function with prescribed zeros
Theorem (Weierstrass). Given any sequence
of complex numbers with
as
, there exists an entire function
that vanishes at all
and nowhere else. Any other such entire function is of the form
, where
is entire.
To begin the proof, note first that if and
are two entire functions that vanish at all
and nowhere else, then
has removable singularities at all the points an. Hence
is entire and vanishes nowhere, so that there exists an entire function
with
. Therefore
and the last statement of the theorem is verified.
Hence we are left with the task of constructing a function that vanishes at all the points of the sequence and nowhere else. A naive guess, suggested by the product formula for
, is the product
.
The problem is that this product converges only for suitable sequences , so we correct this by inserting exponential factors. These factors will make the product converge without adding new zeros.
For each integer we define canonical factors by
and
,
for . The integer
is called the degree of the canonical factor.
Lemma. If , then
for some
.
Suppose that we are given a zero of order m at the origin, and that are all non-zero. Then we define the Weierstrass product by
.
We claim that this function has the required properties; that is, is entire with a zero of order
at the origin, zeros at each point of the sequence
, and
vanishes nowhere else.
Fix , and suppose that
belongs to the disc
. We shall prove that
has all the desired properties in this disc, and since
is arbitrary, this will prove the theorem.
We can consider two types of factors in the formula defining , with the choice depending on whether
or
. There are only finitely many terms of the first kind (since
), and we see that the finite product vanishes at all
with
. If
, we have
, hence the previous lemma implies
.
Note that by the above remark, does not depend on
. Therefore, the product
defines a holomorphic function when , and does not vanish in that disc. This shows that the function
has the desired properties, and the proof of Weierstrass’s theorem is complete.
Question (Harvard’s QE). Given a sequence of complex numbers such that
, does there exist an entire function
with
?
Proof. By the Weierstrass theorem, we can construct an entire function with simple zeros
Then we define
where is chosen such that when
,
.
Because , for any
, there exists
such that
when
. Hence
holds for all when
. In other words,
converges uniformly for all , so that
in analytic in
. Since
can be arbitrarily large,
is an entire function. It is easy to see that
while for ,
,
which implies that is an entire function satisfying the required condition.
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