## Zeros of Taylor polynomials of (1+z)^p

This is post is related to Extremal Taylor polynomials where it was important to observe that the Taylor polynomials of the function ${(1+z)^{-1/2}}$ do not have zeros in the unit disk. Let’s see how far this generalizes.

The function ${f(z)=(1+z)^{-1}}$ has the rare property that all zeros of its Taylor polynomial have unit modulus. This is clear from

${\displaystyle T_n(z) = \sum_{k=0}^n (-z)^k = (1-(-z)^{n+1})/(1+z)}$.

In this and subsequent illustrations, the zeros of the first 50 Taylor polynomials are shown as blue dots, with the unit circle in red for reference.

When the exponent is less than -1, the zeros move inside the unit disk and begin forming nice patterns in there.

When the exponent is strictly between -1 and 1, the zeros are all outside of the unit disk. Some of them get quite large, forcing a change of scale in the image.

Why does this happen when the exponent approaches 1? The function ${1+z}$ is its own Taylor polynomial, and has the only zero at -1.  So, when ${p\approx 1}$, the Taylor polynomials are small perturbations of ${1+z}$. These perturbations of coefficients have to create additional zeros, but being small, they require a large value of ${z}$ to help them.

For a specific example, the quadratic Taylor polynomial of ${(1+z)^p}$ is ${1 + pz + p(p-1)z^2/2}$, with roots ${(1\pm \sqrt{(2-p)/p})/(1-p) }$. When ${p\approx 1}$, one of these roots is near ${-1}$ (as it has to be) and the other is large.

Finally, when ${p>1}$ and is not an integer, we get zeros on both sides of the unit circle. The majority of them are still outside. A prominent example of an interior zero is ${-1/p}$ produced by the first-degree polynomial ${1 + pz}$.

Another related post: Real zeros of sine Taylor polynomials.

## 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.

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 ${T_{20}}$.

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.

### 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.

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).

## Squarish polynomials

For some reason I wanted to construct polynomials approximating this piecewise constant function ${f}$:

Of course approximation cannot be uniform, since the function is not continuous. But it can be achieved in the sense of convergence of graphs in the Hausdorff metric: their limit should be the “graph” shown above, with the vertical line included. In concrete terms, this means for every ${\epsilon>0}$ there is ${N}$ such that for ${n\ge N}$ the polynomial ${p_n}$ satisfies

$\displaystyle |p_n-f|\le \epsilon\quad \text{ on }\ [0,2]\setminus [1-\epsilon,1+\epsilon]$

and also

$\displaystyle -\epsilon\le p_n \le 1+\epsilon\quad \text{ on }\ [1-\epsilon,1+\epsilon]$

How to get such ${p_n}$ explicitly? I started with the functions ${f_m(x) = \exp(-x^m)}$ when ${m}$ is large. The idea is that as ${m\rightarrow\infty}$, the limit of ${\exp(-x^m)}$ is what is wanted: ${1}$ when ${x<1}$, ${0}$ when ${x>1}$. Also, for each ${m}$ there is a Taylor polynomial ${T_{m,n}}$ that approximates ${f_m}$ uniformly on ${[0,2]}$. Since the Taylor series is alternating, it is not hard to find suitable ${n}$. Let’s shoot for ${\epsilon=0.01}$ in the Taylor remainder and see where this leads:

• Degree ${7}$ polynomial for ${\exp(-x)}$
• Degree ${26}$ polynomial for ${\exp(-x^2)}$
• Degree ${69}$ polynomial for ${\exp(-x^3)}$
• Degree ${180}$ polynomial for ${\exp(-x^4)}$
• Degree ${440}$ polynomial for ${\exp(-x^5)}$

The results are unimpressive, though:

To get within ${0.01}$ of the desired square-ness, we need ${\exp(-1.01^m)<0.01}$. This means ${m\ge 463}$. Then, to have the Taylor remainder bounded by ${0.01}$ at ${x=2}$, we need ${2^{463n}/n! < 0.01}$. Instead of messing with Stirling’s formula, just observe that ${2^{463n}/n!}$ does not even begin to decrease until ${n}$ exceeds ${2^{463}}$, which is more than ${10^{139}}$. That’s a … high degree polynomial. I would not try to ask a computer algebra system to plot it.

Bernstein polynomials turn out to work better. On the interval ${[0,2]}$ they are given by

$\displaystyle p_n(x) = 2^{-n} \sum_{k=0}^n f(2k/n) \binom{n}{k} x^k (2-x)^{n-k}$

To avoid dealing with ${f(1)}$, it is better to use odd degrees. For comparison, I used the same or smaller degrees as above: ${7, 25, 69, 179, 439}$.

Looks good. But I don’t know of a way to estimate the degree of Bernstein polynomial required to obtain Hausdorff distance less than a given ${\epsilon}$ (say, ${0.01}$) from the square function.

## Real zeros of sine Taylor polynomials

The more terms of Taylor series ${\displaystyle \sin x = x-\frac{x^3}{3!}+ \frac{x^5}{5!}- \cdots }$ we use, the more resemblance we see between the Taylor polynomial and the sine function itself. The first-degree polynomial matches one zero of the sine, and gets the slope right. The third-degree polynomial has three zeros in about the right places.

The fifth-degree polynomial will of course have … wait a moment.

Since all four critical points are in the window, there are no real zeros outside of our view. Adding the fifth-degree term not only fails to increase the number of zeros to five, it even drops it back to the level of ${T_1(x)=x}$. How odd.

Since the sine Taylor series converges uniformly on bounded intervals, for every ${ A }$ there exists ${ n }$ such that ${\max_{[-A,A]} |\sin x-T_n(x)|<1 }$. Then ${ T_n }$ will have the same sign as ${ \sin x }$ at the maxima and minima of the latter. Consequently, it will have about ${ 2A/\pi }$ zeros on the interval ${[-A,A] }$. Indeed, the intermediate value theorem guarantees that many; and the fact that ${T_n'(x) \approx \cos x }$ on ${ [-A,A]}$ will not allow for extraneous zeros within this interval.

Using the Taylor remainder estimate and Stirling's approximation, we find ${A\approx (n!)^{1/n} \approx n/e }$. Therefore, ${ T_n }$ will have about ${ 2n/(\pi e) }$ real zeros at about the right places. What happens when ${|x| }$ is too large for Taylor remainder estimate to be effective, we can't tell.

Let's just count the zeros, then. Sage online makes it very easy:

sineroots = [[2*n-1,len(sin(x).taylor(x,0,2*n-1).roots(ring=RR))] for n in range(1,51)]
scatter_plot(sineroots)

The up-and-down pattern in the number of zeros makes for a neat scatter plot. How close is this data to the predicted number ${ 2n/(\pi e) }$? Pretty close.

scatter_plot(sineroots,facecolor='#eeee66') + plot(2*n/(pi*e),(n,1,100))

The slope of the blue line is ${ 2/(\pi e) \approx 0.2342 }$; the (ir)rationality of this number is unknown. Thus, just under a quarter of the zeros of ${ T_n }$ are expected to be real when ${ n }$ is large.

The actual number of real zeros tends to exceed the prediction (by only a few) because some Taylor polynomials have real zeros in the region where they no longer follow the function. For example, ${ T_{11} }$ does this:

Richard S. Varga and Amos J. Carpenter wrote a series of papers titled Zeros of the partial sums of ${ \cos z }$ and ${\sin z }$ in which they classify real zeros into Hurwitz (which follow the corresponding trigonometric function) and spurious. They give the precise count of the Hurwitz zeros: ${1+2\lfloor n/(\pi e)\rfloor }$ for the sine and ${2\lfloor n/(\pi e)+1/2\rfloor }$ for the cosine. The total number of real roots does not appear to admit such an explicit formula. It is the sequence A012264 in the OEIS.