GENERAL ⎜ ARTICLE
Some G Pólya Gems from Complex Analysis
Shobha Madan
George Pólya (1887–1985) was a brilliant Hungarian mathematician; and many of you may have
come across his famous (bestseller) book How to
Solve it. Elsewhere in this issue you will find a
more complete biography of Pólya. Here we will
talk about some of his work in complex analysis. Among Pólya’s contemporaries were mathematicians like Leopold (Lipót) Fejér, his thesis
advisor, Adolf Hurwitz (1859–1919), G H Hardy
(1877–1947), Gábor Szegö (1895–1985).
1. Introduction
From a first course in complex analysis, you know that
every analytic function has a power series valid in its
disc of convergence; this power series is unique, its coefficients being determined by the values of successive
derivatives at the centre of the disc of convergence. Now
if the function is entire, as soon as we know the values
of all the derivatives of the function at a single point,
we get a power series for the function valid everywhere.
But if the function f has some isolated singularities,
then it will require different power series to represent
the function in different regions; the coefficients for each
power series always come from values of the derivatives
at some point. This means then that every property
of the function is, at least implicitly, contained in the
set of its power series, i.e., in the sequence of its coefficients. Pólya was interested in finding the analytic
character of a function from these coefficients, and one
precise problem he worked on is discussed in Section 2.
This concerns finding where the singularities of the function are located from the knowledge of some properties
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Shobha Madan is a
Professor at the
Department of Mathematics and Statistics,
IIT Kanpur.
Her research interest is
in harmonic analysis.
Keywords
Analytic functions, power series, isolated singularities, poles
and zeros, meromorphic functions.
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GENERAL ⎜ ARTICLE
of the coefficients in the power series.
In Section 3, we will look at meromorphic functions and
show a surprising connection that Pólya found between
the location of the poles of a function and properties of
the set of zeros of successive derivatives of the function.
In this article, it will not be possible to give proofs of
any of Pólya’s theorems, but we will try to explain the
content of the results, and illustrate by doing explicit
computations on examples. As you will see, these examples are rather interesting.
2. On Gaps in Power Series
Consider a function defined by a power series
f(z) =
∞
an z n .
0
Let the radius of convergence of this series be 0 < R <
∞. We know that the above series then converges for
all z ∈ D(0, R) = {z : |z| < R} and that the series
diverges for all points in the set {z : |z| > R}. On the
circle, C(0, R) = {z : |z| = R}, the series may converge
at some points, not at others. The standard example
∞
1
=
z n , |z| < 1
1−z
0
is one where, although the power series diverges if |z| >
1, the expression on the left defines an analytic function
at all points except at z = 1, where it has a simple
pole. We say that this function, initially defined by a
power series in the disc of convergence has an analytic
continuation to an analytic function in a larger region (a
connected open set containing D(0, 1), namely, C \ {1}).
Next, let k be a fixed integer, and consider,
∞
1
=
z kn , |z| < 1.
k
1−z
0
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Now there are k simple poles on the unit circle at the
points e2πij/k , j = 0, 1, ..., k − 1. We notice immediately that the series has become somewhat sparse, that
the number of singularities of the function on the circle of convergence has increased, and the function has
an analytic continuation to a larger region, namely to
C \ {e2πij/k , j = 0, 1, ..., k − 1}. We will look at aspects
of Pólya’s work related to the following broad problem:
What, if any, is the
relation between the
coefficients of a power
series and the
singularities of the
function it represents ?
What, if any, is the relation between the coefficients of
a power series and the singularities of the function it
represents?
There must be a relation because after all, the power
series does completely determine a function, even in regions where the power series does not converge, since if
the function extends to an analytic function outside the
disc of convergence to some region, then it is uniquely
determined by the power series. (Remember that the
zeros of an analytic function are isolated.) In order to
really get into the spirit of what is to follow, we look at
some examples.
Example 2.1. Let
f(z) =
∞
z n!
0
be a power series, where many coefficients are zero. Its
radius of convergence is 1. (Why?) What can we say
about the possible extension of this function, i.e., its
analytic continuation to a region larger than the open
unit disc D(0, 1)? The question is really about what
happens at the boundary points. But it is easy to see
that if z = e2πip/q for some integers p, q = 0, then for
n ≥ q, the terms of the series are all equal to 1, and the
series diverges to ∞. So each such point is a singularity
of the function defined by this power series, and these
points are dense in C(0, 1); hence there is no way of
extending the function analytically beyond the unit disc,
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GENERAL ⎜ ARTICLE
since limr→1− f(re2πip/q ) = ∞.
Example 2.2. Consider
f(z) =
∞
n
z2 .
0
The sequence of coefficients looks like
{0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, ..., }.
(Notice how the gaps between non-zero coefficients keep
increasing, as in the previous example!) Once again,
the radius of convergence equals 1, and z = 1 is clearly
a singular point. Now, for every positive integer k, we
have
∞
∞
k+n
2k
2k 2n
(z ) =
z2
f(z ) =
0
so that
2k
f(z ) = f(z) −
0
k−1
n
z2 .
0
2πip
It follows that each point of the form e 2k , with p odd, is
a singular point for f. Such points form a dense subset
k
of the unit circle, and again limr→1− f(re2πip/2 ) = ∞,
so no analytic continuation of f is possible.
These two examples have three things in common:
1. The power series is very sparse, in the sense that
there are huge gaps between nonzero coefficients.
2. The function blows up on the boundary on a dense
subset of the boundary of the region of definition.
3. The function cannot be analytically continued to
any larger domain.
The reader may begin to get an idea that the fact that
the power series blows up is the reason for the circle of
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convergence being the natural boundary of the function
represented by it. But consider the following example:
Example 2.3. We modify Example 2.1 and let
∞
z n!
.
f(z) =
n2
0
Now the power series has radius of convergence 1, and
converges absolutely for all points on the unit circle, and
so defines a continuous function on the closed unit disc.
The question is whether this function has an analytic
continuation to a larger region or not.
This problem was considered by the French mathematician Jacques Hadamard (1865–1963) in his thesis as
early as in 1892. Hadamard defined the existence of
adequate gaps in a quantitative way as follows:
DEFINITION.
n
A power series ∞
0 an z is called a Hadamard lacunary
series if there exists a sequence n0 , n1 , ...nk , ... such that
nk+1 /nk ≥ λ > 1,
and for which aj = 0, ∀ nk < j < nk+1 , and ank = 0.
All the examples above are Hadamard lacunary series.
Theorem 2.1 (Hadamard’s
Theorem). Any
∞ Gap
nk
Hadamard lacunary series 0 ank z with radius of convergence R > 0 does not have an analytic continuation
to any region larger than D(0, R). Moreover, every point
of the boundary C(0, R) is a singular point.
The author finds this result rather strange because on
the one hand, with the large gaps, the series converges
quite fast in D(0, R) and on the other, has singularities
everywhere on the boundary.
In 1899, E Fabry proved that the gap condition can be
greatly improved to get the same result as in Hadamard’s
theorem, which then becomes a special case.
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First a definition: Given a sequence {nk }∞
0 of positive
integers, let
k
ρ = lim inf .
k→∞ nk
Then ρ is called the lower asymptotic density of the
sequence, and is also a measure of lacunarity. We see
easily that every Hadamard lacunary series has ρ = 0.
Theorem 2.2 (Fabry). Suppose {nk } is an increasing
sequence of positive integers with lower asymptotic
den∞
nk
sity ρ = 0, and the power series f(z) = 0 ank z has
radius of convergence R > 0, then f cannot be analytically continued from D(0, R).
At this point a natural question arises: Can the density condition be weakened further to get the same result as in Fabry’s theorem. Pólya said, “No”; and he
completely solved the problem by proving a converse of
Fabry’s theorem.
Theorem 2.3. Suppose {nk } is an increasing sequence
of positive integers. Suppose that every power series
ank z nk with a finite radius of convergence R > 0, cannot be continued analytically to any larger region than
D(0, R), then limk→∞ nkk = 0.
So Pólya’s big contribution is to say that Fabry’s theorem is the best possible.
Pólya's big
contribution is to
say that Fabry's
theorem is the best
possible.
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Before ending this section we will mention another of
Pólya’s insightful results; that there are many (in fact,
uncountably many) series, not necessarily lacunary,
which have their circle of convergence as their natural
boundary. In fact, as soon as we have a power series
with a finite radius of convergence, then by changing
signs of the coefficients suitably, singularities can appear
in a big way. A Hurwitz and G Pólya wrote a paper [3],
where the first part is authored by Hurwitz giving one
proof, and the second part is authored by Pólya giving
a different proof of the following result:
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GENERAL ⎜ ARTICLE
n
Theorem 2.4 (Hurwitz–Pólya). Let ∞
0 an z be a
power series with a finite radius of convergence R > 0.
Then the set of functions of the form
∞
ǫn an z n , ǫn ∈ {−1, 1} ,
0
which have no analytic continuation outside the disc
D(0, R), is uncountable.
Observe that the theorem does not say that the set of
functions of the above form which can be analytically
continued beyond D(0, R) is countable!
3. Zeros of Derivatives
Pólya thought a lot about what happens to the zeros
of successive derivatives of analytic functions. In this
section, we will discuss a particularly elegant result for
meromorphic functions on the plane. Recall that a function f : C → C is called meromorphic if all its singularities are poles (hence these are isolated points). It will
be useful to begin with simple examples.
Pólya thought a lot
about what
happens to the
zeros of
successive
derivatives of
analytic functions.
Example 3.1. Consider the simplest example of a meromorphic function,
1
f(z) = ,
z
with a single simple pole. Observe that f does not have
any zero in C, and a simple zero at the point at infinity.
As we take successive derivatives of f, namely f (n) , we
find that no new poles, nor new zeros appear, although
the order of the pole at z = 0 and of the zero at z = ∞
change.
Next, consider any rational function with a single pole,
i.e., a function such as
1
+ P (z),
zk
where P (z) is any polynomial, and k ≥ 1 an integer.
Now f and the first few derivatives (as many as the
f(z) =
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degree of P (z)) have finitely many zeros, and for large
n, f (n) has one pole at z = 0 and a zero at the point at
infinity.
We now consider the next simplest example of a function
with two poles and we begin to see that two different
poles begin to interact to produce new zeros of successive
derivatives.
Example 3.2. Let,
f(z) =
1
1
− .
z −1 z
Clearly f has no zeros, except the point at infinity. For
the derivatives,
f (n−1) (z) = (−1)n−1 (n − 1)!
z n − (z − 1)n
z n (z − 1)n
we see that the set of zeros
Zf (n−1) = {z : z n = (z − 1)n }
are the solutions of the equation
z = (z − 1)e2πij/n
with j = 0, 1, ..., n − 1, i.e., the n points given by
zj =
1
1 − e−2πij/n
, j = 0, 1, ..., n − 1.
All these points lie on the line {Re(z) = 1/2} (points
equidistant from z = 0 and z = 1). Further, the reader
can compute the imaginary part of the zj ’s and then
easily verify that ∪∞
0 Zf (n) is a dense subset of this line.
One can check easily that if we had chosen the poles to
lie at points a, b ∈ C, instead of at 0, 1, the zeros of
the derivatives would lie on the line consisting of points
equidistant from a and b. But what will happen if we
change the residues at the poles? Consider
f(z) =
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a
− ,
z−1 z
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where a ∈ C, |a| =
1. We see that the zeros of f (n−1)
are given by
The poles of f act
as repulsive
centres for the
zeros of its
Zf (n−1) = {z : az n = (z − 1)n }
successive
derivatives.
and will not lie on the line Rez = 1/2 unless |a| = 1.
But as n → ∞, the zeros will condense towards this
⁀ ltimately it is only the
line, since |a|1/n → 1, so that u
location of the poles that had any influence.
There is an observation that can be made at this stage:
The poles of f act as repulsive centres for the zeros of
its successive derivatives.
(*)
Let us try and check this out with another example:
Example 3.3. We will investigate a simple case of three
simple poles. Let
f(z) =
1
1
1
+ +
.
z −i z z −1
Then
f
(n−1)
n−1
(z) = (−1)
(n−1)!
1
1
1
+ n+
n
(z − i)
z
(z − 1)n
.
We are interested in the zeros of these functions, and
hence the zeros of the polynomials,
Pn (z) = z n (z − 1)n + (z − i)n (z − 1)n + z n (z − i)n .
We will not find the zeros explicitly, but will try to find
their location. Let us write a1 = i, a2 = 0, a3 = 1.
In Figure 1, the complex plane is shown as the union of
three half lines rj = rk starting at the point 12 (1+i), and
the three open regions marked as I, II and III. We will see
that each point in these open regions cannot be a zero
of f (n) for n large enough. Consider first z ∈ III, then
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Figure 1.
with rj = |z − aj |, j = 1, 2, 3, we have r3 < min(r1, r2 )
in this region. Suppose first that r3 < r1 ≤ r2 , then
|Pn (z)| = |z n (z − 1)n + (z − i)n (z − 1)n + z n (z − i)n |
≥ |z n (z − i)n | − |z n (z − 1)n | − |(z − i)n (z − 1)n |
= r1n r2n − r2n r3n − r1n r3n
n n
r1
r1
n n
= r2 r3
−
−1
r3
r2
n
r1
n n
−2 >0
≥ r2 r3
r3
if n is large enough, since rr13 > 1. We can do a similar
estimate if r3 < r2 ≤ r1, and so it follows that for all
z ∈ III, there is an n such that f (n) (z) = 0. The same
happens for the regions I and II.
The three examples show that indeed (*) holds. Now
let us be more precise about this. We begin with two
definitions.
DEFINITION.
Let f be a meromorphic function and a0 ∈ C a pole of
f. The domain of a0 is defined as the set
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Dom(a0) = {z ∈ C : |z−a0| < |z−a| ∀ pole a = a0 of f} .
In Example 3.2 discussed above, the domain of the pole
at z = 0 is the half plane to the left of the line {z : Rez =
1/2} and the domain of z = 1 is the right half plane. The
regions I, II and III in Example 3.3 are respectively the
domains of the poles a1, a2, a3. In general, the domain
of a pole is an open convex set whose boundary consists
of polygonal lines.
Next, Pólya defined a set, called the ‘final set’ D of a
function f : C → C consisting of accumulation points of
the set of all zeros of all derivatives of f. In other words,
we can say that z ∈
/ D if there exists an r > 0 such that
(n)
no derivative f vanishes on the set {w : 0 < |w − z| ≤
r}.
Now we are ready to state Pólya’s result, proved in [4],
where he shows a very interesting connection between
the location of the poles of f and the final set D. See
also [5].
Pólya shows a very
interesting
connection
between the
location of the
poles of f and the
final set D.
The final set D of a meromorphic function f contains
no point of the domain of any of its poles, but contains
all points which are on the common boundary of two or
more poles.
It is remarkable that the final set D here is completely
determined by the location of the poles of f (and does
not even depend on the order of the poles, nor on their
residues). Further, observe that the result does not say
anything about the location of the zeros of successive
derivatives of f, and in fact these may lie inside the
domain of a pole. In Example 3.2, all the zeros of all
the derivatives do lie on the common boundary of the
domains of the two poles, i.e., the line Re z = 1/2, if
the two residues have equal size, and if not, the set of
zeros still condenses towards that line. In Example 3.3,
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we have not found the zeros of the derivatives explicitly;
it was enough to show that each point of the domain of
any pole is not a zero of f (n) for all large enough n.
Our final example is rather interesting.
Example 3.4.
1
.
1 − ez
There are infinitely many poles of f, namely the set:
f(z) =
{2πin; n ∈ Z} .
Pólya’s result stated above tells us that the final set D
of f is given as a union of lines,
D = ∪∞
−∞ {z : Im z = π(2n + 1)}
and each line is the common boundary of the domain of
two consecutive poles. It turns out that in this case, all
the zeros of all the derivatives also lie in D, which is more
than what Pólya’s result says. We will show this below.
For this we first need to find the nth derivatives. We can
write our function as a composition of two functions as
follows:
f(z) = F (φ(z)) ,
1
, and φ(z) = ez . Now it would be
where F (z) = 1−z
nice to have a Leibniz formula for the nth derivative of
the composition of two functions (the reader may want
to find such a formula), but in our case, the task is
somewhat simpler, since φ(n) (z) = φ(z) for all n. First,
let us compute the nth derivative of a function of the
form F (ez ), where F is a differentiable function. Taking
the first derivative, we get,
d
F (ez ) = F ′(ez )ez .
dz
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Taking subsequent derivatives we will get an expression
of the form
n
n
d
z
F (e ) =
Skn F (k)(ez )ekz ,
dz
1
where the coefficients satisfy
n
S11 = 1, Snn = 1, Skn+1 = Sk−1
+ kSkn
as is easy to verify. We put
Skn = 0, unless 1 ≤ k ≤ n.
The notation of using Skn for the coefficients is because
these numbers are well known as Stirling numbers of the
second kind; and in fact, Skn stands for the number of
distinct partitions of a set of n elements into k classes.
With this we can now write the expression for the nth
derivative of the given function f as follows
d
dz
n
z −1
(1 − e )
n
ekz
=
(1 − ez )k+1
1
k
n
1 n
ez
=
S k!
1 − ez 1 k
(1 − ez )
1
Pn (ω) ,
1 − ez
=
where we have put ω =
given by
Skn k!
ez
(1−ez )
P (ω) =
n
and the polynomial Pn is
Skn k!ω k .
1
Clearly we need to get information on the zeros of these
polynomials for all n ≥ 1. For this we use induction.
We follow Pólya and use the notation
ank = k!Skn .
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(In a footnote in the paper [8], Pólya remarks that in his
search for a proof, the turning point was the introduction of this notation!) Then the ank ’s satisfy the following
recurrence relation,
= k(ank−1 + ank )
an+1
k
and so
Pn+1 (ω) =
=
n+1
1
n+1
k
an+1
k ω
k(ank−1 + ank )ω k
1
= ω
n+1
((k − 1)ank−1 ω k−1 + ank−1 ω k−1 )
1
+
n
kank ω k−1
1
n
= ω ω
=
kank ω k−1 + Pn (ω) + Pn′ (ω)
1
′
ω[Pn (ω)(ω
= ω
d
dω
+ 1) + Pn (ω)]
[(ω + 1)Pn (ω)]
We can now prove
Lemma 3.1. For every n ≥ 1, all zeros of the polynomial
Pn are real and simple, and are contained in the interval
(−1, 0].
Proof. We prove by induction on n. We know that the
degree of the polynomial Pn is n. For n = 1, P1 (ω) = ω
and the only zero is at ω = 0. We assume the result for
an integer n, and we may simply restrict the polynomials
to the real line, since all zeros will be found to be real.
In view of the formula derived above for Pn+1 , note first
that the polynomial (ω + 1)Pn (ω) has n + 1 simple zeros
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in [−1, 0] by the induction hypothesis; and apart from
the two simple zeros at the end-points of this interval,
the other (n − 1) simple zeros lie in its interior and are
distinct. Therefore by Rolle’s theorem, we see that there
are n distinct zeros of Pn+1 in (−1, 0), and along with
the zero at ω = 0, we have determined all the zeros of
Pn+1 , as required.
The author would have liked to read the proofs of these
and other remarkable results due to Pólya, but feels limited by the fact that a large body of Pólya’s papers are
written in Hungarian, Russian and German. It needs to
be said that the same limitation determined the choice
of the two themes in this article.
Suggested Reading
[1]
R P Boas, Selected Topics from Pólya's work in Complex
Analysis, Math. Mag., Vol.60, pp.271–274, 1987.
[2]
J Hadamard, Essai sur l'étude des fonctione données par leur
développement de Taylor, J. Math. Pur. Appl., Vol.8, pp.102–186,
[3]
A Hurwitz and G Pólya, Zwei Beweise eines von Herrn Fatou
vermuteten Satzes, Acta Math., Vol.40, pp.179–183, 1917.
[4]
G Pólya, Über die Nullstellen sukzessiver Derivierten, Math
Z., Vol.12, pp.36–60, 1922.
[5]
G Pólya, On the zeros of the derivatives of a function and its
1892.
analytic character, Bull. Amer. Math. Soc., Vol.49, pp.178–191,
1943.
Address for Correspondence
[6]
G Pólya, Sur les séries entiéres lacunaires non-prologeables.
C.R. Acad. Sci. Paris., Vol.208, pp.709–711, 1939.
Department of Mathematics
[7]
G Pólya and G Szegö, Problems and Theorems in Analysis,
Springer-Verlag, 1972–75.
[8]
Shobha Madan
Indian Institute of Technology
Kanpur 208 016, India.
Email: [email protected]
G Pólya, On the zeros of successive derivatives, an example,
J. D'Analyse Mathématique, Vol.30, pp.452–455, 1976.
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