# Patterns of the zeros of the Faulhaber polynomials (modified)

Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying

$$S_{p}(n) = \sum_{k=1}^{n} k^p$$

for $n = 1, 2, 3, \cdots$. For example,

\begin{align*} S_0(x) &= x, \\ S_1(x) &= \frac{x(x+1)}{2}, \\ S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\ S_3(x) &= \frac{x^2 (x+1)^2}{4}. \end{align*}

In order to grasp some intuition on the partial decomposition of $1/S_p (x)$, I tried plotting the complex zeros of $S_p (x)$. The following graphics shows the distribution of the zeros of $S_{800}(x)$. (The precision of the calculated zeros $z_j$ of $S_{800}(z)$ above satisfy $|f(z_j)| \leq 10^{-300}$.)

It turns out that they exhibits a very neat, yet still a strange pattern as seen above.

So far I have never heard of the topic related to this pattern, and I want to know (out of curiosity) if there are some results concerning the pattern of zeros of $S_p(x)$.

Addendum. The previous fuzzy graphics turned out to be a result of numerical error. Now I removed those errors.

• A classic example of this type of problem appears as the Szego curve. Unfortunately I cannot recall any general references for this type of problem. Though the Faulhaber's are not Taylor polynomials, so maybe not as related as I might wish. [As an aside, this was also my first thought when I saw that $\sum(1+\cdots+n^p)^{-1}$ problem :) ]
– anon
Mar 1, 2013 at 1:18
• If you'd like, you can look at my MSc thesis for a survey of some of the results relating to the Szegő curve mentioned by anon. Mar 1, 2013 at 2:11
• Also I suspect that the more chaotic regions of your plot are due to numerical error. I'd be interested in seeing if the precision could be increased. Mar 1, 2013 at 2:13
• Dunno if this helps: There's a tenuous relation between Faulhaber Polynomial zeros and the zeros of the integrals of motion of a single solition solution to the Korteweg-de-Vries equation. The latter have been studied extensively so perhaps it could be connected to the zero distribution. See here: arxiv.org/pdf/math/0503175.pdf Mar 1, 2013 at 2:14
• Some of the discussion around this very similar question may be relevant.
– MJD
Mar 1, 2013 at 5:37

(This is just a comment, really.)

First it appears that the zeros are symmetric about the line $$x=-1/2$$, and indeed the polynomials

$$F_p(z) = S_p(z-1/2)$$

appear to have only even or only odd powers of $$z$$.

It seems that the zeros not on the real axis grow on the order of $$p/(2\pi)$$. Numerically they have the same limiting behavior as the zeros of the partial sums of the sine and cosine series (see this paper [PDF]) as well as the partial sums of the Bessel functions (see this preprint).

Below is a plot of the zeros of $$F_p\left(\frac{p}{2\pi}z\right)$$ for $$p=400$$, along with the modified Szegő curve

\begin{align} &\left\{z \in \mathbb{C} \,\colon \Im(z) \geq 0,\,\,\, |z| \leq 1, \,\,\,\text{and}\,\,\, \left|ze^{1+iz}\right| = 1 \right\} \\ &\qquad \cup \,\left\{z \in \mathbb{C} \,\colon \Im(z) \leq 0,\,\,\, |z| \leq 1, \,\,\,\text{and}\,\,\, \left|ze^{1-iz}\right| = 1 \right\} \\ &\qquad \cup \,\left\{x \in \mathbb{R} \,\colon -1/e \leq x \leq 1/e \right\} \end{align}

in blue. There is possibly a connection to the zeros of the Bernoulli polynomials $$B_p(x)$$ as a result of the fact that

$$S_p(z) = \frac{B_{p+1}(z+1) - B_{p+1}(0)}{p+1}.$$

You may wish to take a look at Karl Dilcher's memoir Zeros of Bernoulli, Generalized Bernoulli, and Euler Polynomials and this paper by John Mangual.

• This is much more than a comment. It is a serious contribution to the discussion. Mar 1, 2013 at 5:21

As an addendum to Antonio Vargas's answer, let's prove that the roots of $S_p$ are indeed symmetrically distributed around $-1/2$, or in other words that if $r$ is a root, then so is $-1-r$. A somewhat more precise result is that $S_p(-1-x) = S_p(x)$ when $p$ is odd, and $S_p(-1-x) = -S_p(x)$ when $p$ is even (thus confirming that the polynomial $T(x) := S_p(x-1/2)$ is an odd function (has only odd powers of $x$) if $p$ is even, and is an even function (has only even powers of $x$) if $p$ is odd).

The key principle is that a polynomial is the zero polynomial iff it has infinitely many roots. For example, $f(x) = g(x)$ if $f(n) = g(n)$ for all $n \in \mathbb{N}$, and a polynomial $h$ is constant if $h(n) = h(0)$ for all $n \in \mathbb{N}$. If $h$ is a polynomial such that its first difference $(\Delta h)(x) := h(x) - h(x-1)$ satisfies $(\Delta h)(n) = 0$ for all $n \in \mathbb{N}$, then $h$ is constant with constant value $h(0)$.

Thus we have $(\Delta S_p)(x) = S_p(x) - S_p(x-1) = x^p$ and so $S_p(-x) - S_p(-1-x) = (-1)^p x^p$. For fixed $p$, put $g(x) := S_p(-1-x)$, so that $g(x-1) = S_p(-1-(x-1)) = S_p(-x)$. Thus

$$\Delta g(x) = g(x) - g(x-1) = S_p(-1-x) - S_p(-x) = (-1)^{p+1} x^p$$

which equals $\Delta S_p(x)$ if $p$ is odd, and $-\Delta S_p(x)$ if $p$ is even. Thus in the case where $p$ is odd, $\Delta(g-S_p) = 0$; applying the principle above, $g$ and $S_p$ differ by a constant: $S_p(-1-x) = S_p(x) + c$ for some constant $c$. For $x=0$, we note that $S_p(0) = 0$, and $S_p(0) - S_p(-1) = 0^p = 0$ so $S_p(-1) = 0$. It follows that $c = 0$, and we conclude $S_p(-1-x) = S_p(x)$ if $p$ is odd.

The case where $p$ is even is wholly similar. There we derive $\Delta(g + S_p) = 0$, so $g(x) + S_p(x) = c$ for some constant $c$, and we argue as before that $c = 0$, and so in this case $S_p(-1-x) = -S_p(x)$.

In addition to Antonio's comment/answer: Looking at the real roots (symmetrized by adding +1/2 (!)) of the 1,5,9,13,... polynomial we get the following list, where only the first three real roots are rational numbers. The rate of convergence to the half-integers is impressive... $$\small \begin{matrix} 0 & 1/2 & . & . & . & . & . & . \\ 0 & 1/2 & -1/2 & 0.763762615826 & -0.763762615826 & . & . & . \\ 0 & 1/2 & -1/2 & 0.949106003964 & -0.949106003964 & . & . & . \\ 0 & 1/2 & -1/2 & 0.999056597832 & -0.999056597832 & . & . & . \\ 0 & 1/2 & -1/2 & 0.999997848581 & -0.999997848581 & . & . & . \\ 0 & 1/2 & -1/2 & 0.999999998198 & -0.999999998198 & -1.50196566814 & 1.50196566814 & 1.74815179290 \\ 0 & 1/2 & -1/2 & 0.999999999999 & -0.999999999999 & -1.50001155318 & 1.50001155318 & 1.93305092402 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000003663 & 1.50000003663 & 1.99704558735 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000007 & 1.50000000007 & 1.99997147602 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 1.99999984071 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 1.99999999943 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 2.00000000000 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 2.00000000000 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 2.00000000000 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 2.00000000000 \\ 0 & 1/2 & -1/2 & 1.00000000000 & -1.00000000000 & -1.50000000000 & 1.50000000000 & 2.00000000000 \end{matrix}$$ The complex roots may be fit into that pattern perhaps by their absolute values, but this naive idea is not yet convincing to me

Another fascinating property of the zeros of the power sums is that for all $k\geq 3$, rational part of any nontrivial zero of $S_k(x)$ is always equal to $-1/2$. This is an analogue of the Riemann Hypothesis for the power sums! See also the paper by J. Singh (2009): http://www.worldscientific.com/doi/abs/10.1142/S179304210900189X