Extremal Taylor polynomials

Suppose {f(z)=a_0+a_1z+a_2z^2+\cdots} is a holomorphic function in the unit disk {|z|<1} such that {|f|\le 1} in the disk. How large can its Taylor polynomial {T_n(z)=a_0+a_1z+\cdots +a_n z^n} be in the disk?

We should not expect {T_n} to be bounded by 1 as well. Indeed, the Möbius transformation {f(z)=(z+1/2)/(1+z/2)} has Taylor expansion {(z+1/2)(1-z/2+O(z^2)) = 1/2 + (3/4)z + O(z^2)}, so {T_1(1)=5/4} in this case. This turns out to be the worst case: in general {T_1} is bounded by 5/4 in the disk.

For the second-degree polynomial {T_2} the sharp bound is {89/64}, attained when {f(z) = (8z^2 + 4z + 3)/(3z^2 + 4z + 8)}; the image of the unit circle under the extremal {T_2} is shown below. Clearly, there is something nontrivial going on.

T2
Extremal T_2 attains 89/64 > 1.39

Edmund Landau established the sharp bound for {|T_n|} in his paper Abschätzung der Koeffizientensumme einer Potenzreihe, published in Archiv der Mathematik und Physik (3) 21 in 1913. Confusingly, there are two papers with the same title in the same issue of the journal: one on pages 42-50, the other on pages 250-255, and they appear in different volumes of Landau’s Collected Works. The sharp bound is in the second paper.

First steps

By rotation, it suffices to bound {|T_n(1)|}, which is {|a_0+\cdots +a_n|}. As is often done, we rescale {f} a bit so that it’s holomorphic in a slightly larger disk, enabling the use of the Cauchy integral formula on the unit circle {\mathbb T}. The Cauchy formula says {2\pi i a_k = \int_{\mathbb T} z^{-k-1} f(z) \,dz}. Hence

{\displaystyle 2\pi |T_n(1)| = \left| \int_{\mathbb T} z^{-n-1}(1+z+\dots+z^n) f(z) \,dz \right|}

It is natural to use {|f(z)|\le 1} now, which leads to

{\displaystyle 2\pi |T_n(1)| \le \int_{\mathbb T} |1+z+\dots+z^n|\, |dz| }

Here we can use the geometric sum formula and try to estimate the integral of {|(1-z^{n+1})/(1-z)|} on the unit circle. This is what Landau does in the first of two papers; the result is {O(\log n)} which is the correct rate of growth (this is essentially the Dirichlet kernel estimate from the theory of Fourier series). But there is a way to do better and get the sharp bound.

Key ideas

First idea: the factor {1+z+\dots+z^n} could be replaced by any polynomial {Q} as long as the coefficients of powers up to {n} stay the same. Higher powers contribute nothing to the integral that evaluates {T_n(1)}, but they might reduce the integral of {|Q|}.

Second idea: we should choose {Q} to be the square of some polynomial, {Q=P^2}, because {(2\pi)^{-1}\int_{\mathbb T} |P(z)|^2\, |dz|} can be computed exactly: it is just the sum of squares of the coefficients of {P}, by Parseval’s formula.

Implementation

Since {1+z+\dots+z^n} is the {n}-th degree Taylor polynomial of {(1-z)^{-1}}, it is natural to choose {P} to be the {n}-th degree Taylor polynomial of {(1-z)^{-1/2}}. Indeed, if {P_n(z) = (1-z)^{-1/2} + O(z^{n+1})}, then {P_n(z)^2 = (1-z)^{-1} + O(z^{n+1}) = 1+z+\dots+z^n + O(z^{n+1})} as desired (asymptotics as {z\to 0}). The binomial formula tells us that
{\displaystyle P_n(z)=\sum_{k=0}^n (-1)^k\binom{-1/2}{k}z^k }

The coefficient of {z^k} here can be written out as {(2k-1)!!/(2k)!!} or rewritten as {4^{-k}\binom{2k}{k}} which shows that in lowest terms, its denominator is a power of 2. To summarize, {|T_n(1)|} is bounded by the sum of squares of the coefficients of {P_n}. Such sums are referred to as the Landau constants,

{\displaystyle G_n = 1+ \left(\frac{1}{2}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4}\right)^2 + \cdots + \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 }

A number of asymptotic and non-asymptotic formulas have been derived for {G_n}, for example Brutman (1982) shows that {G_n - (1/\pi)\log(n+1)} is between 1 and 1.0663.

Sharpness

To demonstrate the sharpness of the bound {|T_n|\le G_n}, we want {|f|\equiv 1} and {P_n(z)^2f(z)/z^n\ge 0} on the unit circle. Both are arranged by taking {f(z) = z^n P_n(1/z) / P_n(z)} which is a Blaschke product of degree {n}. Note that the term {P_n(1/z)} can also be written as {\overline{P_n(1/\bar z)}}. Hence {P_n(z)^2f(z)/z^n = P_n(z) \overline{P_n(1/\bar z)}} which is simply {|P_n(z)|^2} when {|z|=1}. Equality holds in all the estimates above, so they are sharp.

Here are the images of the unit circle under extremal Taylor polynomials {T_5} and {T_{20}}.

T5
Extremal Taylor polynomial of 5th degree
T20
Extremal Taylor polynomial of 20th degree

These polynomials attain large values only on a short subarc of the circle; most of the time they oscillate at levels less than 1. Indeed, the mean value of {|T_n|^2} cannot exceed the mean of {|f|^2} which is at most 1. Here is the plot of the roots of extremal {T_n}:  they are nearly uniform around the circle, except for a gap near 1.

Troots10
Roots of extremail T_10
Troots20
Roots of extremal T_20

But we are not done…

Wait a moment. Does {f(z) = z^n P_n(1/z) / P_n(z)} define a holomorphic function in the unit disk? We are dividing by {P_n} here. Fortunately, {P_n} has no zeros in the unit disk, because its coefficients are positive and decreasing as the exponent {k} increases. Indeed, if {p(z)=c_0+c_1z+\cdots + c_nz^n} with {c_0>c_1>\dots>c_n > 0}, then {(1-z)p(z)} has constant term {c_0} and other coefficients {c_1-c_0}, {c_2-c_1}, … {c_n-c_{n-1}}, {-c_n}. Summing the absolute values of the coefficients of nonconstant terms we get {c_0}. So, when these coefficients are attached to {z^k} with {|z|<1}, the sum of nonconstant terms is strictly less than {c_0} in absolute value. This proves {P_n\ne 0} in the unit disk. Landau credits Adolf Hurwitz with this proof.

In fact, the zeros of {P_n} (Taylor polynomials of {(1-z)^{-1/2}}) lie just outside of the unit disk.

roots20
Zeros of P_20
roots50
Zeros of P_50

The zeros of the Blaschke products formed from {P_n} are the reciprocals of the zeros of  {P_n}, so they lie just inside the unit circle, much like the zeros of {T_n} (though they are different).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s