# On the numerology of trigonometric polynomials

Hans-Christian Herbig  0000-0003-2676-3340<EMAIL_ADDRESS>Departamento de
Matemática Aplicada, Universidade Federal do Rio de Janeiro, Av. Athos da
Silveira Ramos 149, Centro de Tecnologia - Bloco C, CEP: 21941-909 - Rio de
Janeiro, Brazil  and  Mateus de Jesus Gonçalves
<EMAIL_ADDRESS>Departamento de Matemática Aplicada,
Universidade Federal do Rio de Janeiro, Av. Athos da Silveira Ramos 149,
Centro de Tecnologia - Bloco C, CEP: 21941-909 - Rio de Janeiro, Brazil

###### Abstract.

This note is about the observation that the various transition formulas
between bases of trigonometric polynomials can be expressed in terms binomial
coefficients. More specifically, we write the entries of the Chebyshev
matrices $T$ and $U$ in terms of binomial coefficients. We remark that the
inverses of certain submatrices of the Chebyshev matrix $T$ correspond to
classical power reduction formulas for trigonometric functions. Therefore the
entries of $T^{-1}$ are expressible in terms of binomial coefficients and
powers of two. The inverses of certain submatrices of the Chebyshev matrix $U$
appear in power reduction formulas involving the Catalan triangles. We could
not spot them in the literature. As a corollary the Catalan triangles are
interpreted as inverses of rather natural matrices of binomial coefficients.
We explain the matrix inversions in terms of Riordan arrays. We solve the
integral $\int_{0}^{2\pi}\cos^{2m}(x)\sin^{2n}(x)dx$ and deduce a
hypergeometric identity from the Fourier expansion of $\cos^{2m}(x)$ and
$\sin^{2n}(x)$. As a corollary we prove that super Catalan numbers are
integers. We emphasize that from the point of view of these base changes
between the trigonometric polynomials it is more natural to work with
$2\cos(x)$ and $2\sin(x)$ instead of $\cos(x)$ and $\sin(x)$, since the
transition matrices become invertible over the integers. We do a similar
analysis for the spread polynomials. This enables us to make a conjecture of
Goh and Wildberger more precise. In this exposition we give a mostly self-
contained account of the matter, the prerequisites just being four semesters
of calculus, linear algebra, elementary complex variables and the residue
theorem.

###### Key words and phrases:

Chebyshev polynomials, power reduction formulas, trigonometric polynomials,
Catalan triangle, pyramidal numbers, spread polynomials, Riordan arrays, super
Catalan numbers, golden ratio

###### 2020 Mathematics Subject Classification:

Primary 42A05 ; Secondary 05A10

###### Contents

  1. 1 Introduction
  2. 2 Chebyshev polynomials
  3. 3 Trigonometric polynomials
  4. 4 Chebyshev matrix and higher dimensional pyramidal numbers
  5. 5 Mnemonics
  6. 6 Catalan triangles and the Fuß-Catalan numbers
  7. 7 Power reduction formulas
  8. 8 Hypergeometric identities from matrix inversions
  9. 9 Interpretation in terms of the Riordan group
  10. 10 Hypergeometric identities from trigonometric intergrals
  11. 11 Spread polynomials
  12. 12 A conjecture of Goh and Wildberger

##  1\. Introduction

When teaching basic calculus and trigonometry the educator is faced with the
challenge that he or she cannot employ in a systematic way binomial
coefficients, as Pascal’s triangle is said to be too complicated for the
modern student. On the other hand, since the advent of the Wilf-Zeilberger
[PWZ96] method binomial coefficients are considered to be too shallow to be a
subject of serious research. In the end, nobody appears to be responsible for
this old subject, except the Online Encyclopaedia of Integer Sequences (OEIS)
[OEI23]. This invaluable resource contains no proofs however, but merely
references. To find proofs of classical formulas in the literature can be a
tedious task.

In this article we take the opportunity to discuss a number of trigonometric
formulas that arise when doing base changes between the various bases of
trigonometric polynomials. The formulas are all classical, even though the
ones involving Catalan triangles we were not able to spot in the literature.
We do not know of a reference where they are discussed in context and proofs
are provided. Moreover, we try to exhibit the appearance of binomial
coefficients as succinctly as possible and deduce the relevant simple
hypergeometric identities behind the inversions. Among other things, we point
out the close relation between the coefficients of the Chebyshev polynomials
and higher dimensional pyramidal numbers, a link we stumbled into while
contemplating on the OEIS. This is closely related to the observation that the
base change formulas simplify when working with $2\cos(x)$ and $2\sin(x)$
instead of $\cos(x)$ and $\sin(x)$ (but remember to keep the $1$ !). We do a
similar analysis for Norman Wildberger’s spread polynomials that play a
central role in his rational trigonometry [Wil05]. We express them in terms of
the even-dimensional pyramidal numbers. The most elegant point of view on our
matrix inversions is provided by the language of Riordan arrays [SSB+22],
relating generating functions and matrix multiplication. Since we got rid of
the annoying factors of four, we cleared the view to make a conjecture of Goh
and Wildberger on the factorization of spread polynomials more precise.

The paper is organized as follows. In Section 2 we review the definitions of
the Chebyshev polynomials. In Section 3 we recall the notion of a
trigonometric polynomial. In Section 4 we discover the higher pyramidal
numbers and binomial coefficients in the coeficients of the Chebyshev
polynomials. In Section 5 we propose an easy way to deduce quickly the first
few Chebyshev polynomials $T_{n}$ on a sheet of paper. In Section 6 (the only
section in this paper where we refer to other sources) we recall the
definition of the even and odd Catalan triangles. In Section 7 we provide
proofs of the well-known power reduction formulas and not-so-well-known
variants thereof. In Section 8 we relate the Chebyshev polynomials to the
power reduction formulas via matrix inversion. In Section 9 we provide another
proof of the matrix inversions using Riordan arrays. In Section 10 we deduce
further hypergeometric identities evaluating Fourier integrals and deduce the
integrality of the super Catalan numbers. In Section 11 we once again go
through the laundry list of the paper for the case of the spread polynomials.
In Section 12 elaborate on a conjecture of Goh and Wildberger and provide some
empirical evidence.

Acknowledgements. The paper arose from HCH teaching Calculus and Linear
Algebra. The more HCH dug into the material he came to the realize that it
should be known to Louis Shapiro in some form or another. The paper is part of
an undergraduate research project of MdJG. He is acknowledging financial
support of PROFAEX. We greatly benefited from discussions with Daniel Herden.
HCH wants to thank A.-K. Gallagher for disposing [PS98] on his bookshelf. HCH
thanks his administration at UFRJ for forcing him to teach beginner’s courses
for a decade, turning him into an _Elephant im Porzellanladen_ (He really
tried to teach them Eqn. (10.2), but they do not want to listen!).

Statements and Declarations. The authors declare that they have no conflicts
of interest.

##  2\. Chebyshev polynomials

For convenience of the reader we review here some basic facts about the
Chebyshev polynomials (see, e.g., [PS98, Exercise VI.2]). The Chebyshev
polynomials of the first kind $(T_{n}(x))_{n\geq 0}$ and of the second kind
$(U_{n}(x))_{n\geq 0}$ are defined by the recursions

$\displaystyle T_{0}(x)$ $\displaystyle=1,\ T_{1}(x)=x,$ (2.1) $\displaystyle
T_{n}(x)$ $\displaystyle=2xT_{n-1}(x)-T_{n-2}(x)\ \mbox{ for $n\geq 2$},$
$\displaystyle U_{0}(x)$ $\displaystyle=1,\ U_{1}(x)=2x,$ (2.2) $\displaystyle
U_{n}(x)$ $\displaystyle=2xU_{n-1}(x)-U_{n-2}(x)\ \mbox{ for $n\geq 2$}.$

######  Proposition 2.1.

(2.3) $\displaystyle\sum_{n=0}^{\infty}T_{n}(x)t^{n}$
$\displaystyle=\frac{1-tx}{1-2tx+t^{2}}$ (2.4)
$\displaystyle\sum_{n=0}^{\infty}U_{n}(x)t^{n}$
$\displaystyle=\frac{1}{1-2tx+t^{2}}$

###### Proof.

From

$\displaystyle\left(1-2tx+t^{2}\right)\sum_{n\geq 0}T_{n}(x)t^{n}$
$\displaystyle=\sum_{n\geq 0}T_{n}(x)t^{n}-\sum_{n\geq
0}T_{n}(x)2t^{n+1}x+\sum_{n\geq 0}T_{n}(x)t^{n+2}$ $\displaystyle=\sum_{n\geq
0}T_{n}(x)t^{n}-\sum_{m\geq 1}T_{m-1}(x)2t^{m}x+\sum_{m\geq 2}T_{m-2}(x)t^{m}$
$\displaystyle=1+tx-2tx+\sum_{m\geq
2}\left(T_{m}(x)-2xT_{m-1}(x)+T_{m-2}(x)\right)t^{m}=1-tx$

we deduce Eqn. (2.3); the proof of (2.4) is similar. ∎

######  Theorem 2.2.

For each integer $n\geq 0$ we have

$\displaystyle T_{n}(\cos(\theta))$ $\displaystyle=\cos(n\theta),\qquad
T_{n}(\sin(\theta))=(-1)^{\lfloor
n/2\rfloor}\begin{cases}\cos(n\theta)&\mbox{if}\ n\ is\ \mbox{even}\\\
\sin(n\theta)&\mbox{if}\ n\ is\ \mbox{odd}\end{cases},$ $\displaystyle
U_{n}(\cos(\theta))$
$\displaystyle=\frac{\sin((n+1)\theta)}{\sin(\theta)},\qquad
U_{n}(\sin(\theta))=(-1)^{\lfloor
n/2\rfloor}\begin{cases}\frac{\cos((n+1)\theta)}{\cos(\theta)}&n\
\text{even}\\\ \frac{\sin((n+1)\theta)}{\cos(\theta)}&n\
\text{odd}\end{cases}.$

###### Proof.

We give an inductive proof of $T_{m}(\cos(\theta))=\cos(m\theta)$ and
$\sin(\theta)U_{m}(\cos(\theta))=\sin(m\theta)$. The start of the induction is
obvious. Let us assume that these identities hold for $m=n-1$ and $m=n-2$. Let
us put for simplicity $U_{m}(x)=0=T_{m}(x)$ for $m<0$. From

$\displaystyle\sum_{n\geq 0}xT_{n-1}(x)+\sum_{n\geq
0}\left(x^{2}-1\right)U_{n-2}(x)t^{n}=\sum_{n\geq
0}xT_{n}(x)t^{n+1}+\sum_{n\geq 0}\left(x^{2}-1\right)U_{n}(x)t^{n+2}$
$\displaystyle=xt\frac{1-tx}{1-2tx+t^{2}}+t^{2}\left(x^{2}-1\right)\frac{1}{1-2tx+t^{2}}=\frac{xt-t^{2}}{1-2tx+t^{2}}=\frac{-1+2tx-t^{2}+1-xt}{1-2tx+t^{2}}$
$\displaystyle=\frac{1-tx}{1-2tx+t^{2}}-1=\sum_{n\geq 1}T_{n}(x)t^{n},$

we see that $xT_{n-1}(x)+\left(x^{2}-1\right)U_{n-2}(x)=T_{n}(x)$ for $n\geq
1$. Similarly, we observe that

$\displaystyle\sum_{n\geq 0}(T_{n}(x)t^{n}+xU_{n-1}(x))t^{n}=\sum_{n\geq
0}T_{n}(x)t^{n}+\sum_{n\geq
0}xU_{n}(x)t^{n+1}=\frac{1-tx}{1-2tx+t^{2}}+\frac{tx}{1-2tx+t^{2}}=\sum_{n\geq
0}U_{n}(x)t^{n},$

so that $T_{n}(x)+xU_{n-1}(x)=U_{n}(x)$ for $n\geq 0$. But, using the Addition
Theorem for cosine,

$\displaystyle\cos(n\theta)=\cos(\theta)\cos((n-1)\theta)-\sin(\theta)\sin((n-1)\theta)=\cos(\theta)T_{n-1}(\cos(\theta))-\sin^{2}(\theta)U_{n-2}(\cos(\theta)),$

the first relation proves $T_{n}(\cos(\theta))=\cos(n\theta)$. Moreover, the
second relation in combination with the Addition Theorem for sine

$\displaystyle\sin((n+1)\theta)=\cos(\theta)\sin(n\theta)+\sin(\theta)\cos(n\theta)=\sin(\theta)(\cos(\theta)U_{n-1}(\cos(\theta))+T_{n}(\cos(\theta)))$

shows $\sin(\theta)U_{n}(\cos(\theta))=\sin((n+1)\theta)$. ∎

##  3\. Trigonometric polynomials

A trigonometric polynomial of degree $\leq d$ in the variable $\theta$ is a
univariate real function of the form
$a_{0}+\sum_{n=1}^{d}a_{n}\cos(\theta)+\sum_{n=1}^{d}b_{n}\sin(n\theta)$ with
$a_{0},a_{i},b_{i}\in\mathbb{R}$ for $i=1,2,\dotsc,d$. The
$(2d+1)$-dimensional real vector space of trigonometric polynomials of degree
$\leq d$ will be denoted by $\operatorname{Trig}_{\leq d}$. By
$2\pi$-periodicity an $f\in\operatorname{Trig}_{\leq d}$ is uniquely
determined by its restriction to the interval $[-\pi,\pi]$. There is a inner
product111In Fourier analysis it is more convenient to work with the inner
product $2\langle f,g\rangle$.

$\displaystyle\langle
f,g\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)g(\theta)d\theta$

making $\operatorname{Trig}_{\leq d}$ into an euclidean vector space. There is
an orthogonal decomposition $\operatorname{Trig}_{\leq
d}=\operatorname{Trig}_{\leq d}^{0}\oplus\operatorname{Trig}_{\leq d}^{1}$
into even and odd trigonometric polynomials $\operatorname{Trig}_{\leq
d}^{j}:=\left\\{f\in\operatorname{Trig}_{\leq
d}|f(-\theta)=(-1)^{j}f(\theta)\right\\}$, $j=0,1$. It is well-known that

$\displaystyle\begin{bmatrix}1&\sqrt{2}\cos(\theta)&\sqrt{2}\cos(2\theta)&\sqrt{2}\cos(3\theta)&\dotsc&\sqrt{2}\cos(d\theta)\end{bmatrix}$

forms an orthonormal basis for $\operatorname{Trig}_{\leq d}^{0}$ and

$\displaystyle\begin{bmatrix}\sqrt{2}\sin(\theta)&\sqrt{2}\sin(2\theta)&\sqrt{2}\sin(3\theta)&\dotsc&\sqrt{2}\sin(d\theta)\end{bmatrix}$

forms an orthonormal basis for $\operatorname{Trig}_{\leq d}^{1}$. In this
paper we find it convenient to view a basis as a frame, i.e., a row of vectors
with trivial kernel. We can understand the space of all trigonometric
polynomials as the direct limit $\operatorname{Trig}:=\cup_{d\geq
0}\operatorname{Trig}_{\leq d}$ with a decomposition
$\operatorname{Trig}=\operatorname{Trig}^{0}\oplus\operatorname{Trig}^{1}$,
$\operatorname{Trig}^{j}:=\cup_{d\geq 0}\operatorname{Trig}_{\leq d}^{j}$. It
follows that
$\begin{bmatrix}1&\sqrt{2}\cos(\theta)&\sqrt{2}\cos(2\theta)&\sqrt{2}\cos(3\theta)&\dotsc\end{bmatrix}$
forms an orthonormal basis for $\operatorname{Trig}^{0}$ and that
$\begin{bmatrix}\sqrt{2}\sin(\theta)&\sqrt{2}\sin(2\theta)&\sqrt{2}\sin(3\theta)&\dotsc\end{bmatrix}$
forms an orthonormal basis for $\operatorname{Trig}^{1}$.

There are a lot of other (non-orthogonal) bases for $\operatorname{Trig}^{j},\
j=0,1$. In this paper we are concerned, e.g., with the bases

$\displaystyle\begin{bmatrix}\cos^{0}(\theta)&\cos^{1}(\theta)&\cos^{2}(\theta)&\cos^{3}(\theta)&\dotsc\end{bmatrix},\
\begin{bmatrix}\frac{\sin^{1}(\theta)}{\sin(\theta)}&\frac{\sin^{2}(\theta)}{\sin(\theta)}&\frac{\sin^{3}(\theta)}{\sin(\theta)}&\frac{\sin^{4}(\theta)}{\sin(\theta)}&\dotsc\end{bmatrix}$

for $\operatorname{Trig}^{0}$ and

$\displaystyle\begin{bmatrix}\sin(x)&\sin^{2}(x)&\sin^{3}(x)&\dotsc\end{bmatrix}$

for $\operatorname{Trig}^{1}$. To verify that these form bases we refer the
reader to the classical literature (see, e.g., [PS98]). Alternatively, the
claim also follows from the calculations of Section 9.

The Chebyshev polynomials $T_{n}$ provide, for example, a base change between
the bases

$\displaystyle\begin{bmatrix}\cos^{0}(\theta)&\cos^{1}(\theta)&\cos^{2}(\theta)&\cos^{3}(\theta)&\dotsc\end{bmatrix}\mbox{
and
}\begin{bmatrix}1&\cos(\theta)&\cos(2\theta)&\cos(3\theta)&\dotsc\end{bmatrix},$

while the Chebyshev polynomials $U_{n}$ provide a base change between the
bases

$\displaystyle\begin{bmatrix}\frac{\sin^{1}(\theta)}{\sin(\theta)}&\frac{\sin^{2}(\theta)}{\sin(\theta)}&\frac{\sin^{3}(\theta)}{\sin(\theta)}&\frac{\sin^{4}(\theta)}{\sin(\theta)}&\dotsc\end{bmatrix}\mbox{
and
}\begin{bmatrix}1&\cos(\theta)&\cos(2\theta)&\cos(3\theta)&\dotsc\end{bmatrix}.$

In the interest of keeping the entries of the base change matrices as simple
as possible we advocate (see Section 8) for working with the basis
$\begin{bmatrix}1&2\cos(\theta)&2\cos(2\theta)&2\cos(3\theta)&\dotsc\end{bmatrix}$
instead of
$\begin{bmatrix}1&\cos(\theta)&\cos(2\theta)&\cos(3\theta)&\dotsc\end{bmatrix}$
and with
$\begin{bmatrix}2\sin(\theta)&2\sin^{2}(\theta)&2\sin^{3}(\theta)&\dotsc\end{bmatrix}$
instead of
$\begin{bmatrix}\sin(\theta)&\sin^{2}(\theta)&\sin^{3}(\theta)&\dotsc\end{bmatrix}$.
For the kabbalist the Chebyshev polynomials themselves appear to be a bit
convoluted, and are probably defined this way for historical reasons.

There is also a more modern variation of the subject. The so-called spread
polynomials, introduced by Norman Wildberger (see [Wil05]), mediate a base
change between
$\begin{bmatrix}\sin^{2}(\theta)&\sin^{4}(\theta)&\sin^{6}(\theta)&\dotsc\end{bmatrix}$
and
$\begin{bmatrix}\sin^{2}(\theta)&\sin^{2}(2\theta)&\sin^{2}(3\theta)&\dotsc\end{bmatrix}$.
Again, from the point of view of numerology it seems more natural to work with
$\begin{bmatrix}4\sin^{2}(\theta)&16\sin^{4}(\theta)&64\sin^{6}(\theta)&\dotsc\end{bmatrix}$
and
$\begin{bmatrix}4\sin^{2}(\theta)&4\sin^{2}(2\theta)&4\sin^{2}(3\theta)&\dotsc\end{bmatrix}$
instead. The details of the base change are to be addressed in Section 11.

##  4\. Chebyshev matrix and higher dimensional pyramidal numbers

We define the infinite matrix $T=(T_{mn})_{m,n\geq 0}$

(4.14) $\displaystyle\footnotesize
T:=\left[\begin{array}[]{ccccccccccccc}1&&-1&&1&&-1&&1&&-1&&\cdots\\\
&1&&-3&&5&&-7&&9&&-11&\\\ &&2&&-8&&18&&-32&&50&&\\\
&&&4&&-20&&56&&-120&&220&\\\ &&&&8&&-48&&160&&-400&&\\\
&&&&&16&&-112&&432&&-1232&\\\ &&&&&&32&&-256&&1120&&\\\
&&&&&&&64&&-567&&2816&\\\ &&&&&&&&128&&-1280&&\\\ &&&&&&&&&256&&-2816&\\\
&&&&&&&&&&512&&\\\ &&&&&&&&&&&1024&\\\
\cdots&&&&&&&&&&&&\cdots\end{array}\right],$

so that
$\begin{bmatrix}T_{0}(x)&T_{1}(x)&T_{2}(x)&\dots\end{bmatrix}=\begin{bmatrix}1&x&x^{2}&\dots\end{bmatrix}T$,
i.e., $T_{n}(x)=\sum_{m\geq 0}T_{mn}x^{m}$. We take the liberty to not
annotate zero entries.

Let us start our analysis by making an empirical observation. For $i\geq 1$ we
record in a table the sequence $\left(p_{j}^{[i]}\right)_{j\geq 0}$ of nonzero
entries of $T$ $i$th row multiplied by $(-1)^{j}2^{-i+1}$. The $0$th row does
not yet fit into the pattern.

$\displaystyle\begin{array}[]{ c|c c c c c c c|c }i\backslash
j&0&1&2&3&4&5&\cdots&\text{Entry in OEIS}\\\ \hline\cr 1\text{th
row}&1&3&5&7&9&11&\cdots&\text{A005408}\\\ 2\text{nd
row}&1&4&9&16&25&36&&\text{A000290}\\\ 3\text{rd
row}&1&5&14&30&55&91&&\text{A000330}\\\ 4\text{th
row}&1&6&20&50&105&196&&\text{A002415}\\\ 5\text{th
row}&1&7&27&77&182&378&&\text{A005585}\\\ \cdots&&&&&&&\cdots&\end{array}$

The number sequence $\left(p_{j}^{[i]}\right)_{j\geq 0}$ is usually referred
to as the _$i$ -dimensional pyramidal numbers_ (we are sloppy here about the
index shifts). Its generating function $p^{[i]}(t)=\sum_{j}p_{j}^{[i]}t^{j}$
is simply

$\displaystyle
p^{[i]}(t)=\frac{1+t}{(1-t)^{i+1}}=\frac{2-(1-t)}{(1-t)^{i+1}}=\frac{2}{(1-t)^{i+1}}-\frac{1}{(1-t)^{i}}=\sum_{j\geq
0}\left(2\binom{i+j}{j}-\binom{i+j-1}{j}\right)t^{j}.$

The matrix

(4.26) $\displaystyle\footnotesize
U=\left[\begin{array}[]{ccccccccccc}1&&-1&&1&&-1&&1&&\\\ &2&&-4&&6&&-8&&10&\\\
&&4&&-12&&24&&-40&&\\\ &&&8&&-32&&80&&-160&\\\ &&&&16&&-80&&240&&\\\
&&&&&32&&-192&&672&\\\ &&&&&&64&&-448&&\\\ &&&&&&&128&&-1024&\\\
&&&&&&&&256&&\\\ &&&&&&&&&512&\\\ &&&&&&&&&&\cdots\end{array}\right],$

on the other hand, is defined such that $U_{n}(x)=\sum_{m\geq 0}U_{mn}x^{m}$.
For $i\geq 0$ we record in a table the sequence of nonzero entries of the
$i$th row of $U$ multiplied by $(-1)^{j}2^{-i}$,

$\displaystyle\begin{array}[]{ c|c c c c c c c }i\backslash
j&0&1&2&3&4&5&\cdots\\\ \hline\cr 0\text{th row}&1&1&1&1&1&1&\cdots\\\
1\text{st row}&1&2&3&4&5&6&\\\ 2\text{nd row}&1&3&6&10&15&21&\\\ 3\text{rd
row}&1&4&10&20&35&56&\\\ 4\text{th row}&1&5&15&35&70&126&\\\
\cdots&&&&&&&\cdots\end{array},$

which apparently turns out to be the Pascal matrix.

We phrase our observation as follows (this type of formulas is known for quite
some time [Cle62]).

######  Theorem 4.1.

The Chebyshev polynomials are given by the formulas

(4.27) $\displaystyle
T_{2n}(x)=(-1)^{n}+\sum_{j=1}^{n}(-1)^{n+j}2^{2j-1}p_{n-j}^{[2j]}x^{2j},\ \ \
T_{2n+1}(x)=\sum_{j=0}^{n}(-1)^{n+j}2^{2j}p_{n-j}^{[2j+1]}x^{2j+1},$ (4.28)
$\displaystyle U_{2n}(x)=\sum_{j=0}^{n}(-1)^{n+j}2^{2j}\binom{n+j}{2j}x^{2j},\
\ \ U_{2n+1}(x)=\sum_{j=0}^{n}(-1)^{n+j}2^{2j+1}\binom{n+j+1}{2j+1}x^{2j+1},$

where $n\geq 0$ is an integer.

###### Proof.

The first step is to cleanse Eqn. (4.14) from its powers of $2$. The trick is
to define

(4.29) $\displaystyle P_{0}(z)$ $\displaystyle:=1,$ $\displaystyle P_{n}(z)$
$\displaystyle:=2T_{n}(z/2)\ \ \mbox{for $n>0$}$

to ensure that $P_{0}(1)=1$ and
$P_{n}(2\cos(\theta))=2T_{n}(\cos(\theta))=2\cos(n\theta)$ for $n>0$. We
easily deduce the bivariate generating function

(4.30) $\displaystyle\sum_{n\geq
0}P_{n}(z)t^{n}=\frac{2-zt}{1-zt+t^{2}}-1=\frac{1-t^{2}}{1-zt+t^{2}}.$

To cleanse the Taylor coefficients from the signs we make the substitution
$t\mapsto t/\sqrt{-1},\ z\mapsto\sqrt{-1}z$ to obtain222This is _almost_ the
generating function of the Lucas numbers $L(z)=\sum_{n\geq
0}L_{n}z^{n}=\frac{2-tz}{1-tz-t^{2}}$.

(4.31)
$\displaystyle\frac{1-t^{2}}{1-zt+t^{2}}\mapsto\frac{1+t^{2}}{1-zt-t^{2}}.$

We extend our definition of the pyramidal numbers to include also
$\sum_{j}p_{j}^{[0]}t^{j}=(1+t)/(1-t)=1+2t+2t^{2}+2t^{3}+\dots$. Let
$K\subset\mathbb{C}^{\times}$ be compact and
$r=\operatorname{min}\left\\{||t|-|t|^{-1}|\ \mid\ t\in K\right\\}$. If
$|z|<r$ it follows that $|\frac{tz}{1-t^{2}}|<1\ $ for all $t\in K$. Then we
have for $t\in K$, $|z|<r$

$\displaystyle\sum_{m,l\geq 0}p_{l}^{[m]}z^{m}t^{m+2l}=\sum_{m\geq
0}\frac{1+t^{2}}{\left(1-t^{2}\right)^{m+1}}(tz)^{m}=\sum_{m\geq
0}\frac{1+t^{2}}{1-t^{2}}\left(\frac{tz}{1-t^{2}}\right)^{m}=\frac{1+t^{2}}{1-t^{2}}\frac{1}{1-\frac{tz}{1-t^{2}}}=\frac{1+t^{2}}{1-t^{2}-tz}.$

This holds everywhere outside of the poles, and obviously also at $t=0$. But
now
$\sum_{n}P_{n}(z)t^{n}=\frac{2-t^{2}}{1-zt+t^{2}}=\sum_{j}p_{j}^{[i]}t^{2j}(-z)^{i}$.
Similarly, we define

(4.32) $\displaystyle V_{n}(z)$ $\displaystyle:=U_{n}(z/2)\ \ \mbox{for $n\geq
0$},$

so that
$V_{n}(2\cos(\theta))=U_{n}(\cos(\theta))=\sin((n+1)\theta)/\sin(\theta)$ for
$n\geq 0$. Hence the bivariate generating function is

$\displaystyle\sum_{n\geq 0}V_{n}(z)t^{n}=\frac{1}{1-zt+t^{2}}.$

To get rid of the signs we make the substitution $t\mapsto t/\sqrt{-1},\
z\mapsto\sqrt{-1}z$ to obtain
$\frac{1}{1-zt+t^{2}}\mapsto\frac{1}{1-zt-t^{2}}$. The relation

$\displaystyle\sum_{m,l\geq 0}\binom{m+l}{l}z^{m}t^{m+2l}=\sum_{m\geq
0}\frac{(tz)^{m}}{\left(1-t^{2}\right)^{m+1}}=\sum_{m\geq
0}\frac{1}{1-t^{2}}\left(\frac{tz}{1-t^{2}}\right)^{m}=\frac{1}{1-t^{2}}\frac{1}{1-\frac{tz}{1-t^{2}}}=\frac{1}{1-t^{2}-tz},$

holds for $t\in K$ and $|z|<r$, hence everywhere outside the poles. ∎

For practical matters, let us mention that for $i\geq 0$ the coefficients of
$\sum_{j}p_{j}^{[i]}t^{j}$ are the difference pattern of the coefficients of
$\sum_{j}p_{j}^{[i+1]}t^{j}$. This is because the generating functions differ
by a factor $1-t$. We introduce the infinite matrix $P=(P_{mn})_{n,m\geq 0}$
so that $P_{n}(x)=\sum_{m\geq 0}P_{mn}x^{m}$ (cf. Eqn. (4.30)).

##  5\. Mnemonics

If you are stranded on a lonesome island and want to recall quickly the
Chebyshev polynomials $T_{n}(x)$ without entering some tedious recursions we
have a suggestion. Start with the row of odd numbers and write on top of it
its difference pattern. Then put on the bottom the row of the sequence whose
differences are the odd numbers, (i.e., the square numbers). Proceed the same
way with the resulting row and continue recursively. Introduce alternating
signs for the diagonals and fill the gaps with zeros. The result is $P$.

$\displaystyle\begin{array}[]{ c c c c c c c c c c }1&&2&&2&&2&&2&\cdots\\\
&1&&3&&5&&7&&\\\ &&1&&4&&9&&16&\cdots\\\ \end{array}\mapsto\begin{array}[]{ c
c c c c c c c c c}1&&2&&2&&2&&2&\cdots\\\ &1&&3&&5&&7&&\\\ &&1&&4&&9&&16&\\\
&&&1&&5&&14&&\\\ &&&&1&&6&&20&\\\ &&&&&1&&7&&\\\ &&&&&&1&&8&\\\ &&&&&&&1&&\\\
&&&&&&&&1&\\\ \cdots&&&&&&&&&\cdots\end{array}$ $\displaystyle\mapsto
P=\left[\begin{array}[]{ c c c c c c c c c c}1&0&-2&0&2&0&-2&0&2&\cdots\\\
0&1&0&-3&0&5&0&-7&0&\\\ 0&0&1&0&-4&0&9&0&-16&\\\ 0&0&0&1&0&-5&0&14&0&\\\
0&0&0&0&1&0&-6&0&20&\\\ 0&0&0&0&0&1&0&-7&0&\\\ 0&0&0&0&0&0&1&0&-8&\\\
0&0&0&0&0&0&0&1&0&\\\ 0&0&0&0&0&0&0&0&1&\\\ \cdots&&&&&&&&&\cdots\\\
\end{array}\right]$

To obtain $T$ turn all $2$s in the $0$th row into $1$s and for each $i\geq 1$
multiply the $i$th row with $2^{i}$.

There is another way to look at the coefficients of the Chebyshev polynomials
that we would like to mention. We plan to discuss it in a more general context
at another occasion. Namely, put

$\begin{Bmatrix}n\\\ k\end{Bmatrix}:=\begin{cases}T_{00}/2&\\\ T_{2k\
n}&\text{if}\ n>0\ \text{is even},\\\ T_{2k+1\ n}&\text{if}\ n\
\text{odd}.\end{cases}$

Then $\begin{Bmatrix}n\\\ k\end{Bmatrix}$ satisfies the recurrence
$\begin{Bmatrix}n\\\ k\end{Bmatrix}=a\begin{Bmatrix}n-1\\\
k\end{Bmatrix}+b\begin{Bmatrix}n\\\ k-1\end{Bmatrix}$ with $a=2,\ b=-1$ and
initial condition

$\begin{Bmatrix}n\\\ 0\end{Bmatrix}=2^{n-1},\ \begin{Bmatrix}0\\\
k\end{Bmatrix}=\begin{cases}1/2&\text{if}\ k=0,\\\
(-1)^{k}&\text{else}.\end{cases}$

##  6\. Catalan triangles and the Fuß-Catalan numbers

The _Catalan sequence_ $(C_{n})_{n\geq 0}$ can be defined by the Segner
recursion

$\displaystyle C_{0}$ $\displaystyle=1,$ $\displaystyle C_{n}$
$\displaystyle=\sum_{k=0}^{n-1}C_{k}C_{n-1-k}.$

In term of the generating function $C(x)=\sum_{n\geq 0}C_{n}x^{n}$ this can be
written as $xC=C^{2}-1$. On other words, if you have forgotten $C_{n}$ but
know $C_{0},C_{1},\dots,C_{n-1}$, let your computer expand
$(\sum_{k=0}^{n-1}C_{k}x^{k})^{2}$. The coefficient in front of $x^{n-1}$ is
$C_{n}$. By complementing the squares one can determine the two roots of
$C^{2}-xC-1$. This leads to the formula

(6.1) $\displaystyle C(x)=\frac{1-\sqrt{1-4x}}{2x},$

noting that only one root is a power series. From this one deduces
$C_{n}={1\over n+1}{2n\choose n}={2n\choose n}-{2n-1\choose n}$. The
literature on the Catalan sequence is extensive (see, e.g., [Sta12, Sta15,
GKP89, Kos09]). A closely related function is the generating function of the
_central binomial coefficients_

(6.2) $\displaystyle B(x)=\sum_{n\geq
0}\binom{2n}{n}x^{n}=\frac{C(x)}{2-C(x)}=\frac{1}{\sqrt{1-4x}}.$

The _Fuß-Catalan numbers_ are defined by

(6.3) $\displaystyle
F_{m}(p,r):=\frac{1}{mp+r}\binom{mp+r}{m}=\frac{r}{m(p-1)+r}\binom{mp+r-1}{m}=\frac{r}{m}\binom{mp+r-1}{m-1}$

for integers $m,p,r\geq 0$. They satisfy the important convolution property
$F_{m}(p,s+r)=\sum_{k=0}^{m}F_{k}(p,r)F_{m-k}(p,s)$; see [GKP89, Eqn. (5.36)]
or [Rio68]. From this it follows that

$\displaystyle\sum_{m\geq 0}F_{m}(p,r)x^{m}=\left(\sum_{m\geq
0}F_{m}(p,1)x^{m}\right)^{r},$

i.e., $\sum_{m\geq 0}F_{m}(p,r)x^{m}$ is the $r$-fold convolution of
$\sum_{m\geq 0}F_{m}(p,1)x^{m}$. By inspection, $\sum_{m\geq
0}F_{m}(1,1)x^{m}=(1-x)^{-1}$ and $\sum_{m\geq 0}F_{m}(2,1)x^{m}=C(x)$. The
series $\sum_{m\geq 0}F_{m}(p,1)x^{m}$ is also called the _generalized
binomial series_. Let us record the first $r$-fold convolutions of $C(x)$:

$\displaystyle C(x)$ $\displaystyle=1+\ \ x+\ 2x^{2}+\ \ 5x^{3}+\ \ 14x^{4}+\
\ \ 42x^{5}+\ 132x^{6}+\ \ 429x^{7}+\dotsc,$ $\displaystyle C^{2}(x)$
$\displaystyle=1+2x+\ 5x^{2}+\ 14x^{3}+\ \ 42x^{4}+\ \ 132x^{5}+\ 429x^{6}+\
1430x^{7}+\dotsc,$ $\displaystyle C^{3}(x)$ $\displaystyle=1+3x+9x^{2}+\ \
28x^{3}+\ \ 90x^{4}+\ \ 297x^{5}+1001x^{6}+\ 3432x^{7}+\dotsc,$ $\displaystyle
C^{4}(x)$ $\displaystyle=1+4x+14x^{2}+\ 48x^{3}+\ 165x^{4}+\
572x^{5}+2002x^{6}+\ 7072x^{7}+\dotsc,$ $\displaystyle C^{5}(x)$
$\displaystyle=1+5x+20x^{2}+\ 75x^{3}+\
275x^{4}+1001x^{5}+3640x^{6}+13260x^{7}+\dotsc,$ $\displaystyle C^{6}(x)$
$\displaystyle=1+6x+27x^{2}+110x^{3}+429x^{4}+1638x^{5}+6188x^{6}+23256x^{7}+\dotsc,$
$\displaystyle C^{7}(x)$
$\displaystyle=1+7x+35x^{2}+154x^{3}+637x^{4}+2548x^{5}+9996x^{6}+38760x^{7}+\dotsc,$
$\displaystyle C^{8}(x)$
$\displaystyle=1+8x+44x^{2}+208x^{3}+910x^{4}+3808x^{5}+15504x^{6}+62016x^{7}+\dotsc\dotsc$

The coefficients of the $2l$-fold convolutions $C^{2l}(x)$ for $l\geq 1$ have
been arranged by Louis Shapiro [Sha76] into an infinite lower triangular
matrix $B^{\operatorname{even}}$ that is usually referred to as the Catalan
triangle. We will call it here the even Catalan triangle since we also need
the _odd Catalan triangle_ $B^{\operatorname{odd}}$. The latter is formed by
arranging the $(2l-1)$-fold convolutions $C^{2l-1}(x)$ for $l\geq 1$ into an
infinite lower triangular matrix.

(6.4) $\displaystyle B^{\operatorname{even}}=\begin{bmatrix}1&&&&&&\cdots\\\
2&1&&&&&\\\ 5&4&1&&&&\\\ 14&14&6&1&&&\\\ 42&48&27&8&1&&\\\
132&165&110&44&10&1&\\\ \cdots&&&&&&\cdots\end{bmatrix},\ \ \ \ \
B^{\operatorname{odd}}=\begin{bmatrix}1&&&&&&\cdots\\\ 1&1&&&&&\\\
2&3&1&&&&\\\ 5&9&5&1&&&\\\ 14&28&20&7&1&&\\\ 42&90&75&35&9&1&\\\
\cdots&&&&&&\cdots\end{bmatrix}.$

The matrix entries are given by the formulas

(6.5) $\displaystyle B_{ij}^{\operatorname{even}}=\frac{j}{i}\binom{2i}{i-j},\
\ B_{ij}^{\operatorname{odd}}=\frac{2j+1}{2i+1}\binom{2i+1}{i-j},$

where $i,j\geq 1$. One of the authors met $B_{ij}^{\operatorname{even}}$
already on another occasion [CHI18]. We encounter the Catalan triangles in the
next section.

##  7\. Power reduction formulas

######  Theorem 7.1.

  1. (1)

$2^{2n-1}\cos^{2n}(\theta)=\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}\binom{2n}{n-k}\cos(2k\theta)$,

  2. (2)

$2^{2n}\cos^{2n+1}(\theta)=\sum_{k=0}^{n}\binom{2n+1}{n-k}\cos((2k+1)\theta)$,

  3. (3)

$2^{2n-1}\sin^{2n}(\theta)=\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}(-1)^{k}\binom{2n}{n-k}\cos(2k\theta)$,

  4. (4)

$2^{2n}\sin^{2n+1}(\theta)=\sum_{k=0}^{n}(-1)^{k}\binom{2n+1}{n-k}\sin((2k+1)\theta)$.

###### Proof.

The following proof of (1) we learnt from Daniel Herden.

$\displaystyle\sum_{k=1}^{n}\binom{2n}{n-k}\cos(2k\theta)$
$\displaystyle=\sum_{k=1}^{n}\binom{2n}{n-k}\frac{e^{2k\sqrt{-1}\theta}+e^{-2k\sqrt{-1}\theta}}{2}=e^{-2n\sqrt{-1}\theta}\sum_{k=1}^{n}\binom{2n}{n-k}\frac{e^{2(n+k)\sqrt{-1}\theta}+e^{2(n-k)\sqrt{-1}\theta}}{2}$
$\displaystyle=\frac{e^{-2n\sqrt{-1}\theta}}{2}\left(\sum_{k=1}^{n}\binom{2n}{n-k}e^{2(n-k)\sqrt{-1}\theta}+\sum_{k=1}^{n}\binom{2n}{n-k}e^{2(n+k)\sqrt{-1}\theta}\right)$
$\displaystyle=\frac{e^{-2n\sqrt{-1}\theta}}{2}\left(\sum_{k=1}^{n}\binom{2n}{n-k}e^{2(n-k)\sqrt{-1}\theta}+\sum_{k=1}^{n}\binom{2n}{n+k}e^{2(n+k)\sqrt{-1}\theta}\right)$
$\displaystyle=\frac{e^{-2n\sqrt{-1}\theta}}{2}\left(-\binom{2n}{n}e^{2n\sqrt{-1}\theta}+\sum_{j}\binom{2n}{j}e^{2j\sqrt{-1}\theta}\right)=\frac{e^{-2n\sqrt{-1}\theta}}{2}\left(1+e^{2\sqrt{-1}\theta}\right)^{2n}-\frac{1}{2}\binom{2n}{n}$
$\displaystyle=\frac{1}{2}\left(e^{\sqrt{-1}\theta}+e^{-\sqrt{-1}\theta}\right)^{2n}-\frac{1}{2}\binom{2n}{n}=2^{2n-1}\cos^{2n}(\theta)-\frac{1}{2}\binom{2n}{n}.$

By shifting the argument we deduce (3):

$\displaystyle 2^{2n-1}\sin^{2n}(\theta)$
$\displaystyle=2^{2n-1}\cos^{2n}\left(\theta-\frac{\pi}{2}\right)=\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}\binom{2n}{n-k}\cos\left(2k\left(\theta-\frac{\pi}{2}\right)\right)$
$\displaystyle=\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}(-1)^{k}\binom{2n}{n-k}\cos(2k\theta).$

To show (2) along the lines of Herden’s proof we calculate

$\displaystyle\sum_{k=0}^{n}\binom{2n+1}{n-k}\cos((2k+1)\theta)$
$\displaystyle=\sum_{k=0}^{n}\binom{2n+1}{n-k}\frac{e^{(2k+1)\sqrt{-1}\theta}+e^{-(2k+1)\sqrt{-1}\theta}}{2}$
$\displaystyle=\frac{e^{-(2n+1)\sqrt{-1}\theta}}{2}\left(\sum_{k=0}^{n}\binom{2n+1}{n-k}e^{(2(n+k)+2)\sqrt{-1}\theta}+\sum_{k=0}^{n}\binom{2n+1}{n-k}e^{2(n-k)\sqrt{-1}\theta}\right)$
$\displaystyle=\frac{e^{-(2n+1)\sqrt{-1}\theta}}{2}\left(\sum_{j=n+1}^{2n+1}\binom{2n+1}{j}e^{2j\sqrt{-1}\theta}+\sum_{j=0}^{n}\binom{2n+1}{j}e^{2j\sqrt{-1}\theta}\right)$
$\displaystyle=\frac{e^{-(2n+1)\sqrt{-1}\theta}}{2}\sum_{j=0}^{2n+1}\binom{2n+1}{j}e^{2j\sqrt{-1}\theta}=\frac{e^{(-2n-1)\sqrt{-1}\theta}}{2}\left(1+e^{2\sqrt{-1}\theta}\right)^{2n+1}$
$\displaystyle=\frac{1}{2}\left(e^{\sqrt{-1}\theta}+e^{-\sqrt{-1}\theta}\right)^{2n+1}=2^{2n}\cos^{2n+1}(\theta).$

By shifting the argument we deduce (4):

$\displaystyle 2^{2n}\sin^{2n+1}(\theta)$
$\displaystyle=2^{2n}\cos^{2n+1}\left(\theta-\frac{\pi}{2}\right)=\sum_{k=0}^{n}\binom{2n+1}{n-k}\cos\left((2k+1)\left(\theta-\frac{\pi}{2}\right)\right)$
$\displaystyle=\sum_{k=0}^{n}\binom{2n+1}{n-k}\cos\left((2k+1)\theta-k\pi-\frac{\pi}{2}\right)=\sum_{k=0}^{n}(-1)^{k}\binom{2n+1}{n-k}\sin((2k+1)\theta).$

∎

The following banal calculation has a pretty matrix inversion as a consequence
(see Eqn. (8.6)––(8.9)).

######  Corollary 7.2.

  1. (1)

$2^{2n}\cos^{2n}(\theta)=\sum_{k=0}^{n}B_{nk}^{\operatorname{odd}}\
\frac{\sin((2k+1)\theta)}{\sin(\theta)}$,

  2. (2)

$2^{2n-1}\cos^{2n-1}(\theta)=\sum_{k=1}^{n}B_{nk}^{\operatorname{even}}\
\frac{\sin(2k\theta)}{\sin(\theta)}$.

###### Proof.

We derive (1) of Theorem 7.1:

$\displaystyle-2^{2n-1}2n\sin(\theta)\cos^{2n-1}(\theta)$
$\displaystyle={d\over
d\theta}\left(2^{2n-1}\cos^{2n}(\theta)\right)=-\sum_{k=1}^{n}\binom{2n}{n-k}2k\sin(2k\theta)$
$\displaystyle\Longrightarrow 2^{2n-1}\cos^{2n-1}(\theta)$
$\displaystyle=\sum_{k=1}^{n}\frac{k}{n}\binom{2n}{n-k}\frac{\sin(2k\theta)}{\sin(\theta)},$

which shows (2). Similarly, taking the derivative of (2) of Theorem 7.1 we
obtain:

$\displaystyle-2^{2n}(2n+1)\sin(\theta)\cos^{2n}(\theta)$
$\displaystyle={d\over
d\theta}\left(2^{2n}\cos^{2n+1}(\theta)\right)=-\sum_{k=0}^{n}\binom{2n+1}{n-k}(2k+1)\sin((2k+1)\theta)$
$\displaystyle\Longrightarrow 2^{2n}\cos^{2n}(\theta)$
$\displaystyle=\sum_{k=0}^{n}\frac{2k+1}{2n+1}\binom{2n+1}{n-k}\frac{\sin((2k+1)\theta)}{\sin(\theta)},$

proving (1). ∎

##  8\. Hypergeometric identities from matrix inversions

In order to distill the relevant hypergeometric identities we choose to get
rid of the superficial powers of $2$. We put

(8.1) $\displaystyle\kappa(\theta):=2\cos(\theta),\
\sigma(\theta):=2\sin(\theta),\
\nu_{m}(\theta):=\frac{\sin((m+1)\theta)}{\sin(\theta)}$

and work with the following bases

$\displaystyle\begin{bmatrix}1&\kappa(\theta)&\kappa(2\theta)&\kappa(3\theta)&\dotsc\end{bmatrix},$
$\displaystyle\begin{bmatrix}1&\nu_{1}(\theta)&\nu_{2}(\theta)&\nu_{3}(\theta)&\dotsc\end{bmatrix}$

for the even trigonometric polynomials and

$\displaystyle\begin{bmatrix}\sigma(\theta)&\sigma(2\theta)&\sigma(3\theta)&\dotsc\end{bmatrix}$

for the odd trigonometric polynomials.

We are prepared to understand that $T_{2n}(\cos(\theta))=\cos(2n\theta)$ and
Theorem 7.1 (1) can be interpreted as the following mutually inverse linear
systems

(8.2)
$\begin{bmatrix}1&\kappa(2\theta)&\kappa(4\theta)&\kappa(6\theta)&\kappa(8\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}\kappa^{0}(\theta)&\kappa^{2}(\theta)&\kappa^{4}(\theta)&\kappa^{6}(\theta)&\kappa^{8}(\theta)&\cdots\end{bmatrix}\begin{bmatrix}p_{0}^{[0]}&-p_{1}^{[0]}&p_{2}^{[0]}&-p_{3}^{[0]}&p_{4}^{[0]}&\cdots\\\
&p_{0}^{[2]}&-p_{1}^{[2]}&p_{2}^{[2]}&-p_{3}^{[2]}&\\\
&&p_{0}^{[4]}&-p_{1}^{[4]}&p_{2}^{[4]}&\\\ &&&p_{0}^{[6]}&-p_{1}^{[6]}&\\\
&&&&p_{0}^{[8]}&\\\ \cdots&&&&&\cdots\end{bmatrix},$ (8.3)
$\begin{bmatrix}\kappa^{0}(\theta)&\kappa^{2}(\theta)&\kappa^{4}(\theta)&\kappa^{6}(\theta)&\kappa^{8}(\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}1&\kappa(2\theta)&\kappa(4\theta)&\kappa(6\theta)&\kappa(8\theta)&\cdots\end{bmatrix}\begin{bmatrix}\binom{0}{0}&\binom{2}{1}&\binom{4}{2}&\binom{6}{3}&\binom{8}{4}&\cdots\\\
&\binom{2}{0}&\binom{4}{1}&\binom{6}{2}&\binom{8}{3}&\\\
&&\binom{4}{0}&\binom{6}{1}&\binom{8}{2}&\\\ &&&\binom{6}{0}&\binom{8}{1}&\\\
&&&&\binom{8}{0}&\\\ \cdots&&&&&\cdots\end{bmatrix}.$

The equivalence of the relation $T_{2n}(\sin(\theta))=(-1)^{n}\cos(2n\theta)$
and Theorem 7.1 (3) boils down to essentially the same matrix inversion.

Similary, $T_{2n+1}(\cos(\theta))=\cos((2n+1)\theta)$ can be compared with
Theorem 7.1 (2):

(8.4)
$\begin{bmatrix}\kappa(\theta)&\kappa(3\theta)&\kappa(5\theta)&\kappa(7\theta)&\kappa(9\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}\kappa^{1}(\theta)&\kappa^{3}(\theta)&\kappa^{5}(\theta)&\kappa^{7}(\theta)&\kappa^{9}(\theta)&\cdots\end{bmatrix}\begin{bmatrix}p_{0}^{[1]}&-p_{1}^{[1]}&p_{2}^{[1]}&-p_{3}^{[1]}&p_{4}^{[1]}&\cdots\\\
&p_{0}^{[3]}&-p_{1}^{[3]}&p_{2}^{[3]}&-p_{3}^{[3]}&\\\
&&p_{0}^{[5]}&-p_{1}^{[5]}&p_{2}^{[5]}&\\\ &&&p_{0}^{[7]}&-p_{1}^{[7]}&\\\
&&&&p_{0}^{[9]}&\\\ \cdots&&&&&\cdots\end{bmatrix},$ (8.5)
$\begin{bmatrix}\kappa^{1}(\theta)&\kappa^{3}(\theta)&\kappa^{5}(\theta)&\kappa^{7}(\theta)&\kappa^{9}(\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}\kappa(\theta)&\kappa(3\theta)&\kappa(5\theta)&\kappa(7\theta)&\kappa(9\theta)&\cdots\end{bmatrix}\begin{bmatrix}\binom{1}{0}&\binom{3}{1}&\binom{5}{2}&\binom{7}{3}&\binom{9}{4}&\cdots\\\
&\binom{3}{0}&\binom{5}{1}&\binom{7}{2}&\binom{9}{3}&\\\
&&\binom{5}{0}&\binom{7}{1}&\binom{9}{2}&\\\ &&&\binom{7}{0}&\binom{9}{1}&\\\
&&&&\binom{9}{0}&\\\ \cdots&&&&&\cdots\end{bmatrix}.$

Essentially the same matrix inversion shows up when comparing
$T_{2n+1}(\sin(\theta))=(-1)^{n}\sin((2n+1)\theta)$ and Theorem 7.1 (4).

Finally, $U_{2n}(\cos(\theta))=\nu_{2n}(\theta)$ is the inverse linear system
to Corollary 7.2 (1)

(8.6)
$\begin{bmatrix}1&\kappa^{2}(\theta)&\kappa^{4}(\theta)&\kappa^{6}(\theta)&\kappa^{8}(\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}1&\nu_{2}(\theta)&\nu_{4}(\theta)&\nu_{6}(\theta)&\nu_{8}(\theta)&\cdots\end{bmatrix}\begin{bmatrix}1&1&2&5&14&\cdots\\\
&1&3&9&28&\\\ &&1&5&20&\\\ &&&1&7&\\\ &&&&1&\\\
\cdots&&&&&\cdots\end{bmatrix},$ (8.7)
$\begin{bmatrix}1&\nu_{2}(\theta)&\nu_{4}(\theta)&\nu_{6}(\theta)&\nu_{8}(\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}1&\kappa^{2}(\theta)&\kappa^{4}(\theta)&\kappa^{6}(\theta)&\kappa^{8}(\theta)&\cdots\end{bmatrix}\begin{bmatrix}\binom{0}{0}&-\binom{1}{1}&\binom{2}{2}&-\binom{3}{3}&\binom{4}{4}&\cdots\\\
&\binom{2}{0}&-\binom{3}{1}&\binom{4}{2}&-\binom{5}{3}&\\\
&&\binom{4}{0}&-\binom{5}{1}&\binom{6}{2}&\\\
&&&\binom{6}{0}&-\binom{7}{1}&\\\ &&&&\binom{8}{0}&\\\
\cdots&&&&&\cdots\end{bmatrix},$

and $U_{2n+1}(\cos(\theta))=\nu_{2n+1}(\theta)$ is the inverse linear system
to Corollary 7.2 (2)

(8.8)
$\begin{bmatrix}\kappa&\kappa^{3}(\theta)&\kappa^{5}(\theta)&\kappa^{7}(\theta)&\kappa^{9}(\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}\nu_{1}&\nu_{3}(\theta)&\nu_{5}(\theta)&\nu_{7}(\theta)&\nu_{9}(\theta)&\cdots\end{bmatrix}\begin{bmatrix}1&2&5&14&42&\cdots\\\
&1&4&14&48&\\\ &&1&6&27&\\\ &&&1&8&\\\ &&&&1&\\\
\cdots&&&&&\cdots\end{bmatrix},$ (8.9)
$\begin{bmatrix}\nu_{1}(\theta)&\nu_{3}(\theta)&\nu_{5}(\theta)&\nu_{7}(\theta)&\nu_{9}(\theta)&\cdots\end{bmatrix}\\\
=\begin{bmatrix}\kappa(\theta)&\kappa^{3}(\theta)&\kappa^{5}(\theta)&\kappa^{7}(\theta)&\kappa^{9}(\theta)&\cdots\end{bmatrix}\begin{bmatrix}\binom{1}{0}&-\binom{2}{1}&\binom{3}{2}&-\binom{4}{3}&\binom{5}{4}&\cdots\\\
&\binom{3}{0}&-\binom{4}{1}&\binom{5}{2}&-\binom{6}{3}&\\\
&&\binom{5}{0}&-\binom{6}{1}&\binom{7}{2}&\\\
&&&\binom{7}{0}&-\binom{8}{1}&\\\ &&&&\binom{9}{0}&\\\
\cdots&&&&&\cdots\end{bmatrix}.$

The transposed matrices of Eqn. (8.7) and (8.9) are the odd and the even
Catalan triangles $B^{\operatorname{odd}}$ and $B^{\operatorname{even}}$. The
latter two matrix inversions were already observed by Louis Shapiro in
[Sha01], and we gave here a trigonometric proof. In fact, all the matrices of
this section are transposes of Riordan arrays. This can be used to explain all
the inversions of this section; we do this in the next Section.

##  9\. Interpretation in terms of the Riordan group

The Chebyshev matrix $T$ itself cannot be interpreted as the transpose of a
Riordan array, but the matrizes $P,B^{\operatorname{even}}$ and
$B^{\operatorname{odd}}$ can. This means that the matrix inversions of the
previous section can be understood as inversions in the so-called Riordan
group. The calculations are based on Lagrange inversion and turn out to be
surprisingly simple. Some of the statements of this section can be found in
some form in the PhD thesis of Aoife Hennessy [Hen11].

Let us review some basic ideas of the theory of Riordan arrays. The proofs of
the facts reviewed here are actually quite accessible; we refer the reader to
the monograph [SSB+22]. We write $\operatorname{\mathbb{R}}\llbracket
x\rrbracket$ for the formal power series with real coefficients. By
$\operatorname{\mathbb{R}}\llbracket x\rrbracket^{\times}$ we mean the
multiplicatively invertible formal power series.

The composition map $\operatorname{\operatorname{\mathbb{R}}}\llbracket
x\rrbracket\times x\operatorname{\operatorname{\mathbb{R}}}\llbracket
x\rrbracket\rightarrow\operatorname{\operatorname{\mathbb{R}}}\llbracket
x\rrbracket$ that sends $g(x)=\sum_{n\geq
0}g_{n}x^{n}\in\operatorname{\operatorname{\mathbb{R}}}\llbracket x\rrbracket$
and $f(x)=\sum_{m\geq 1}f_{m}x^{m}\in x\operatorname{\mathbb{R}}\llbracket
x\rrbracket$ to $\sum_{n\geq 0}g_{n}\left(\sum_{m\geq 1}f_{m}x^{m}\right)^{n}$
is denoted by $(g,f)\mapsto g\circ f$. The space
$x\operatorname{\mathbb{R}}\llbracket x\rrbracket^{\times}$ is comprised of
compositionally invertible formal power series. I.e., if $f\in
x\operatorname{\mathbb{R}}\llbracket x\rrbracket^{\times}$ then the map
$\operatorname{\operatorname{\mathbb{R}}}\llbracket
x\rrbracket\rightarrow\operatorname{\operatorname{\mathbb{R}}}\llbracket
x\rrbracket,\ g\mapsto g\circ f$ is invertible. The inversion is provided by
the Langrange inversion formula: if $f(x)=\frac{x}{\phi(x)}$ then the
coefficients of the compositional inverse $\overline{f}$ are given by the
formula (see, e.g., [Ges16])

(9.1)
$\displaystyle\left[y^{n}\right]\overline{f}\left(y^{n}\right)=\frac{1}{n}\left[x^{n-1}\right](\phi(x))^{n}.$

######  Definition 9.1.

An infinite lower triangular matrix $A=[a_{nk}]_{n,k\geq
0}\in\operatorname{\mathbb{R}}^{\infty\times\infty}$ is called a Riordan array
if there exist $(g,f)\in\operatorname{\mathbb{R}}\llbracket x\rrbracket\times
x\operatorname{\mathbb{R}}\llbracket x\rrbracket^{\times}$ such that
$a_{nk}=\left[x^{n}\right]g(x)f(x)^{k}$. It is tradition to write, abusing
notation, $A=(g,f)$, so that $(g,f)$ can be interpreted as a matrix or as a
tuple. If $g\in\mathbb{R}\llbracket x\rrbracket^{\times}$ then $(g,f)$ is
called a proper Riordan array.

It turns out that $\operatorname{\mathbb{R}}\llbracket
x\rrbracket^{\times}\times x\operatorname{\mathbb{R}}\llbracket
x\rrbracket^{\times}$ with the composition
$(g_{1},f_{1})*(g_{2},f_{2})=(g_{1}(g_{2}\circ f_{1}),f_{2}\circ f_{1})$ forms
a group, the so-called Riordan group, with identity $(1,x)$ and inverse
$(g,f)^{-1}=\left(\frac{1}{g\circ\overline{f}},\overline{f}\right)$.

The Fundamental Theorem of Riordan Arrays links the Riordan group to matrix
multiplication. Let us denote by
$\vec{h}=\begin{bmatrix}h_{0}&h_{1}&h_{2}&\dotsc\end{bmatrix}^{T}$ the
infinite column vector associated to the formal power series $h(x)=\sum_{n\geq
0}h_{n}x^{n}\in\operatorname{\mathbb{R}}\llbracket x\rrbracket$. Then the
theorem says that $(g,f)\vec{h}=\overrightarrow{g(h\circ f)}$. As a
consequence one deduces the relation
$(g_{1},f_{1})(g_{2},f_{2})=\begin{bmatrix}(g_{1},f_{1})\overrightarrow{g_{2}}&(g_{1},f_{1})\overrightarrow{g_{2}f_{2}}&(g_{1},f_{1})\overrightarrow{g_{2}f_{2}^{2}}&\dotsc\end{bmatrix}=(g_{1},f_{1})*(g_{2},f_{2})$.

######  Lemma 9.2.

Assume for $i=1,2$ the following inversions in the Riordan group
$(g_{i},f)^{-1}=(G_{i},F)$. Then $(g_{1}g_{2},f)^{-1}=(G_{1}G_{2},F)$ follows.

###### Proof.

The assumption means $(g_{i},f)*(G_{i},F)=(1,x)$. It follows that

$(g_{1}g_{2},f)*(G_{1}G_{2},F)=(g_{1}g_{2}(G_{1}G_{2}\circ F),f\circ
F)=(g_{1}(G_{1}\circ F)g_{2}(G_{2}\circ F),f\circ F)=(1,x).$

∎

The key observation we want to make now (it seems to be combinatorialist’s
folklore) is that

$\displaystyle B^{\operatorname{odd}}=\left(C,xC^{2}\right),\ \ \
B^{\operatorname{even}}=\left(C^{2},xC^{2}\right),$
$\displaystyle\left(p^{[0]}(-x),x\frac{p^{[2]}(-x)}{p^{[0]}(-x)}\right)=\left(\frac{1-x}{1+x},\frac{x}{(1+x)^{2}}\right),\
\ \ \
\left(p^{[1]}(-x),x\frac{p^{[3]}(-x)}{p^{[1]}(-x)}\right)=\left(\frac{1-x}{(1+x)^{2}},\frac{x}{(1+x)^{2}}\right).$

are Riordan arrays. This permits to interprete Equations (8.2)–(8.9) in
Section 8 as inversions in the Riordan group.

######  Theorem 9.3.

Let $C$ and $B$ the generating functions of the Catalan numbers and of the
central binomial coefficients, respectively. Then

  1. (1)

$\left(C,xC^{2}\right)^{-1}=\left(\frac{1}{1+x},\frac{x}{(1+x)^{2}}\right)$,

  2. (2)

$\left(C^{2},xC^{2}\right)^{-1}=\left(\frac{1}{(1+x)^{2}},\frac{x}{(1+x)^{2}}\right)$,

  3. (3)

$\left(B,xC^{2}\right)^{-1}=\left(\frac{1-x}{1+x},\frac{x}{(1+x)^{2}}\right)$,

  4. (4)

$\left(BC,xC^{2}\right)^{-1}=\left(\frac{1-x}{(1+x)^{2}},\frac{x}{(1+x)^{2}}\right)$.

###### Proof.

First we show $\overline{xC^{2}}(x)=\frac{x}{(1+x)^{2}}$ using formula (9.1).
Namely, for $n\geq 1$ we have

$\displaystyle\left[y^{n}\right]\overline{\frac{x}{(1+x)^{2}}}(y)=\frac{1}{n}\left[x^{n-1}\right](1+x)^{2n}=\frac{1}{n}\left[x^{n-1}\right]\sum_{j\geq
0}\binom{2n}{j}x^{j}=\frac{1}{n}\binom{2n}{n-1}=\left[x^{n}\right]xC^{2}.$

To prove (1) it remains to check that $C\circ\overline{xC^{2}}=1+x$. We do
this by comparing $C=C\circ\overline{xC^{2}}\circ\left(xC^{2}\right)$ with
$(1+x)\circ\left(xC^{2}\right)=\left(1+xC^{2}\right)=1+C-1=C$, and remember
that $g\mapsto g\circ\left(xC^{2}\right)$ is invertible.

To prove (3) we observe that
$\left(\frac{1-xC^{2}}{1+xC^{2}}\right)^{-1}=\frac{C}{2-C}=B$ is the
generating function for the central binomial coefficients.

Now (2) and (4) follow from (1) and (3) by Lemma 9.2. ∎

######  Theorem 9.4.

Let $n,m$ be integers with $m\geq 0$. Then $\sum_{j\geq
0}\binom{2j+n}{j-m}x^{j}=BC^{n}(C-1)^{m}$.

###### Proof.

We generalize here a trick from [Pap99, Exercise 8.6]. Note that for a simple
closed curve around $0\in\mathbb{C}$ we have for $N,M\in\mathbb{Z}$ that
$\binom{N}{M}=\frac{1}{2\pi\sqrt{-1}}\int_{\gamma}\frac{(1+z)^{N}}{z^{M}}\frac{dz}{z}$.
For $x\in\mathbb{C}$ with $|x|<\operatorname{dist(0,\gamma})$ we can swap
summation and integral in the following calculation:

$\displaystyle\sum_{j\geq 0}\binom{2j+n}{j-m}x^{j}$ $\displaystyle=\sum_{j\geq
0}\frac{1}{2\pi\sqrt{-1}}\int_{\gamma}\frac{(1+z)^{n+2j}}{z^{j-m}}x^{j}\frac{dz}{z}$
$\displaystyle=\sum_{j\geq
0}\frac{1}{2\pi\sqrt{-1}}\int_{\gamma}z^{m}(1+z)^{n}\left(\frac{(1+z)^{2}}{z}x\right)^{j}\frac{dz}{z}=\frac{1}{2\pi\sqrt{-1}}\int_{\gamma}z^{m}(1+z)^{n}\frac{1}{1-\frac{(1+z)^{2}}{z}x}\frac{dz}{z}$
$\displaystyle=\frac{1}{2\pi\sqrt{-1}}\frac{1}{x}\int_{\gamma}\frac{z^{m}(1+z)^{n}}{\frac{z}{x}-(1+z)^{2}}dz=-\frac{1}{2\pi\sqrt{-1}}\frac{1}{x}\int_{\gamma}\frac{z^{m}(1+z)^{n}}{(1+z)^{2}-\frac{z}{x}}dz.$

We substitute $1+z=w,\ xw^{r}=(w-1)^{d}$ and write

$\displaystyle\sum_{j\geq
0}\binom{n+2j}{k+j}x^{j}=-\frac{1}{2\pi\sqrt{-1}}\frac{1}{x}\int_{\gamma+1}\frac{w^{n}(w-1)^{m}}{w^{2}-\frac{w-1}{x}}dw.$

Factoring the quadratic polynomial in the denominator

$\displaystyle w^{2}-\frac{w-1}{x}$
$\displaystyle=\left(w-\frac{1}{2x}\right)^{2}-\frac{1-4x}{4x^{2}}$
$\displaystyle=\left(w-\frac{1-\sqrt{1-4x}}{2x}\right)\left(w-\frac{1+\sqrt{1-4x}}{2x}\right)=(w-C(x))\left(w-\frac{1}{x}+C(x)\right)$

and observing that $C(x)-\frac{1}{x}+C(x)=\frac{-1}{xB(x)}$ we can apply now
the Residue Theorem and deduce the claim. In fact the second pole is outside
$\gamma+1$ for $|x|$ small enough. ∎

######  Corollary 9.5.

$\left(B,xC^{2}\right)=\left[\binom{2n}{n-m}\right]_{n,m\geq 0},\
\left(BC,xC^{2}\right)=\left[\binom{2n+1}{n-m}\right]_{n,m\geq 0}$.

###### Proof.

According to the Theorem 9.4 we have

$\displaystyle\sum_{n\geq 0}\binom{2n}{n-m}x^{n}=(C-1)^{m}B=x^{m}BC^{2m},\ \
\sum_{n\geq 0}\binom{2n+1}{n-m}x^{n}=C(C-1)^{m}B=x^{m}BC^{2m+1}.$

∎

######  Corollary 9.6.

For $|x|<1$ we have

  1. (1)

$\frac{1-x^{2}}{(1+x)^{2}-4\cos^{2}(\theta)x}=1+2\sum_{n\geq
1}\cos(2n\theta)x^{n}$,

  2. (2)

$\frac{2(1-x)\cos(\theta)}{(1+x)^{2}-4x\cos^{2}(\theta)}=\sum_{n\geq
0}\cos((2n+1)\theta)x^{n}$,

  3. (3)

$\frac{1+x}{(1+x)^{2}-4\cos^{2}(\theta)}=\sum_{n\geq
0}\frac{\sin((2n+1)\theta)}{\sin(\theta)}x^{n}$,

  4. (4)

$\frac{2\cos(\theta)}{(1+x)^{2}-4x\cos^{2}(\theta)}=\sum_{n\geq
0}\frac{\sin(2(n+1)\theta)}{\sin(\theta)}x^{n}$.

###### Proof.

We put $\kappa_{\theta}^{\operatorname{even}}(x):=\sum_{n\geq
0}\left(\kappa^{2}(\theta)x\right)^{n}=\frac{1}{1-\kappa^{2}(\theta)x}$ and
note that the right hand side of Equation (8.2) is

$\left(\frac{1-x}{1+x},\frac{x}{(1+x)^{2}}\right)\overrightarrow{\kappa_{\theta}^{\operatorname{even}}}=\overrightarrow{\frac{1-x}{1+x}\frac{1}{1-\kappa^{2}(\theta)\frac{x}{(1+x)^{2}}}}=\overrightarrow{\frac{\left(1-x^{2}\right)}{(1+x)^{2}-x\kappa^{2}(\theta)}}.$

This shows $\frac{1-x^{2}}{(1+x)^{2}-\kappa^{2}(\theta)x}=1+\sum_{n\geq
1}\kappa(2n\theta)x^{n}$, which is equivalent to (1). Similarly, we put
$\kappa_{\theta}^{\operatorname{odd}}(x):=\sum_{n\geq
0}\kappa(\theta)\left(\kappa^{2}(\theta)x\right)^{n}=\frac{\kappa(\theta)}{1-\kappa^{2}(\theta)x}$
and express the right hand side of Equation (8.4) as

$\left(\frac{1-x}{(1+x)^{2}},\frac{x}{(1+x)^{2}}\right)\overrightarrow{\kappa_{\theta}^{\operatorname{odd}}}=\overrightarrow{\frac{1-x}{(1+x)^{2}}\frac{\kappa(\theta)}{1-\kappa^{2}(\theta)\frac{x}{(1+x)^{2}}}}=\overrightarrow{\frac{(1-x)\kappa(\theta)}{(1+x)^{2}-x\kappa^{2}(\theta)}}.$

This shows
$\frac{(1-x)\kappa(\theta)}{(1+x)^{2}-\kappa^{2}(\theta)x}=\sum_{n\geq
0}\kappa((2n+1)\theta)x^{n}$, which is equivalent to (2). Writing the right
hand side of Equation 8.7

$\left(\frac{1}{1+x},\frac{x}{(1+x)^{2}}\right)\overrightarrow{\kappa_{\theta}^{\operatorname{even}}}=\overrightarrow{\frac{1}{1+x}\frac{1}{1-\kappa^{2}(\theta)\frac{x}{(1+x)^{2}}}}=\overrightarrow{\frac{1+x}{(1+x)^{2}-x\kappa^{2}(\theta)}}$

we prove $\frac{1+x}{(1+x)^{2}-x\kappa^{2}(\theta)}=\sum_{n\geq
0}\nu(2n\theta)x^{n}$, and hence (3). Finally, writing the right hand side of
Equation (8.9) as

$\left(\frac{1}{(1+x)^{2}},\frac{x}{(1+x)^{2}}\right)\overrightarrow{\kappa_{\theta}^{\operatorname{odd}}}=\overrightarrow{\frac{1}{(1+x)^{2}}\frac{\kappa(\theta)}{1-\kappa^{2}(\theta)\frac{x}{(1+x)^{2}}}}=\overrightarrow{\frac{\kappa(\theta)}{(1+x)^{2}-x\kappa^{2}(\theta)}}$

we deduce $\frac{\kappa(\theta)}{(1+x)^{2}-x\kappa^{2}(\theta)}=\sum_{n\geq
0}\nu((2n+1)\theta)x^{n}$, which is equivalent to (4). ∎

##  10\. Hypergeometric identities from trigonometric intergrals

This section is not essential for the logic of the paper, but gives an example
how the matrix inversions can be used. Our first aim is to deduce further
hypergeometric identities by inspecting the trigonometric integral
$\int_{0}^{2\pi}\cos^{n}(\theta)\sin^{m}(\theta)d\theta$. This could be
evaluated using the beta function, but we prefer here an more elementary
treatment. When $n\geq 1,\ m\geq 0$ we perform the partial integration

$\displaystyle\ I_{n,m}=\int_{0}^{2\pi}\cos^{n}(\theta)\sin^{m}(\theta)\
d\theta=\left[\frac{\cos^{n-1}(\theta)\sin^{m+1}(\theta)}{m+1}\right]_{0}^{2\pi}\
+\int_{0}^{2\pi}(n-1)\ \sin(\theta)\ \cos^{n-2}(\theta)\frac{\
\sin^{m+1}(\theta)}{m+1}\ d\theta.$

We deduce $(m+1)I_{n,m}=(n-1)\ I_{n-2,m+2}$. From the substitution
$s=\theta-\pi/2$ we find

$\displaystyle
I_{n,m}=\int_{0}^{2\pi}\cos^{n}(\theta)\sin^{m}(\theta)d\theta=\int_{-\pi/2}^{3\pi/2}\left(-\sin(s)^{n}\right)\cos^{m}(s)ds=(-1)^{n}I_{m,n}.$

Evidently, $I_{n,1}=\int_{0}^{2\pi}\cos^{n}(\theta)\sin(\theta)\
d\theta=\left[-\frac{\cos^{n+1}(t)}{n+1}\right]_{0}^{2\pi}=0$. Moreover, from
the power reduction formulas (see Theorem 7.1) we obtain

(10.1) $\displaystyle I_{n,0}=\begin{cases}0&\text{if}\ n\ \text{odd}\\\ 2\pi\
2^{-n}\binom{n}{n/2}&\text{if}\ n\ \text{even}\end{cases}.$

Hence $I_{n,m}\neq 0$ if and only if $n$ are $m$ even. By induction we obtain
for $k\geq l\geq 0$:

$\displaystyle I_{2k,2l}=\left(\prod_{m=1}^{l}\ \
\frac{(2m+1)}{(2k+2m-1)}\right)\ I_{2(k+l),0}.$

But $\prod_{m=1}^{l}(2k+2m-1)=\frac{(2k+2l-1)!!}{(2k-1)!!}\ $and hence
$I_{2k,2l}=\frac{(2k-1)!!}{(2k+2l-1)!!\ \ }(2l-1)!!\ I_{2(k+l),0}$. From

$\displaystyle\frac{(2k-1)!!}{(2k+2l-1)!!}(2l-1)!!=\frac{2^{k+l-1}(k+l-1)!}{{(2k+2l-1)!}}\frac{(2k-1)!}{2^{k-1}(k-1)!}\
\frac{(2l-1)!}{2^{l-1}(l-1)!}$
$\displaystyle=2\frac{\binom{k+l-1}{k-1}l!}{{\binom{2k+2l-1}{2k-1}}{(2l)!}}\frac{(2l-1)!}{(l-1)!}=\frac{\binom{k+l-1}{k-1}}{\binom{2k+2l-1}{2k-1}}=\frac{\binom{k+l-1}{l}}{\binom{2k+2l-1}{2l}}$

Remembering $I_{2k,2l}=I_{2l,2k}$, we finally deduce that for $k,l\geq 0$

$\displaystyle
I_{2k,2l}=\frac{\binom{k+l-1}{l}}{\binom{2k+2l-1}{2l}}I_{2(k+l),0}.$

Combining this with (10.1) we have proven the following.

######  Theorem 10.1.

For integers $k,l\geq 0$ we have
$\int_{0}^{2\pi}\cos^{2k}(\theta)\sin^{2l}(\theta)\ d\theta=2\pi\
2^{-2(k+l)}\frac{\binom{k+l-1}{l}\binom{2(k+l)}{k+l}}{\binom{2(k+l)-1}{2l}}$.

We evaluate

$\displaystyle 2\pi=\int_{0}^{2\pi}\left(\cos^{2}(\theta)\
+\sin^{2}(\theta)\right)^{m}\ d\theta=\sum_{{\begin{subarray}{c}k,l\geq 0\\\
k+l=m\end{subarray}}}\binom{k+l}{l}\int_{0}^{2\pi}\cos^{2k}(\theta)\ \
\sin^{2l}(\theta)\ d\theta$
$\displaystyle=2\pi\sum_{\begin{subarray}{c}k,l\geq 0\\\
k+l=m\end{subarray}}\binom{k+l}{l}\
2^{-2(k+l)}\frac{\binom{k+l-1}{l}\binom{2(k+l)}{k+l}}{\binom{2k+2l-1}{2l}}$

and find as a corollary

(10.2)
$\displaystyle\frac{2^{2m}}{\binom{2m}{m}}=\sum_{l=0}^{m}\frac{\binom{m-1}{l}\binom{m}{l}}{\binom{2m-1}{2l}}.$

The rest of this section we use our base change of the previous section to
show that the infinite symmetric matrix $M=[M_{kl}]_{k,l\geq 0}$,

$\displaystyle
M_{kl}:=\frac{1}{2\pi}\int_{-\pi}^{\pi}\kappa^{2k}(\theta)\sigma^{2l}(\theta)\
d\theta=\frac{\binom{k+l-1}{l}\binom{2(k+l)}{k+l}}{\binom{2(k+l)-1}{2l}}.$

has integer entries. On the way we deduce another hypergeometric identity
(namely Eqn. (10.3) below). For the kabbalist333In the words of Fields
medalist Edward Witten, $M$ stands for Magical, Mystery or Membrane. the
matrix $M$ looks actually a bit curious

$\displaystyle M=\begin{bmatrix}1&2&6&20&70&252&924&\cdots\\\
2&2&4&10&28&84&264&\\\ 6&4&6&12&28&72&198&\\\ 20&10&12&20&40&90&220&\\\
70&28&28&40&70&140&308&\\\ 252&84&72&90&140&252&504&\\\
924&264&198&220&308&504&924&\\\ \cdots&&&&&&&\cdots\end{bmatrix}.$

The first row is comprised of the central binomial coefficients and the second
row of doubled Catalan numbers. The lower triangular part of $M$ appears in
the OEIS as entry https://oeis.org/A182411. Let us mention other empirical
observations about the entries of $M$. It appears that
$M_{kl}=\frac{(2k)!(2l)!}{k!l!(k+l)!}$ and the textbook [UH39] asks in
Exercise 4. on p.103 to show that this is an integer. Nowadays these numbers
are called _super Catalan numbers_ (see [Ges92, Bor08])444In [LW22] the
rational numbers in Theorem 10.1 have been named _circular super Catalan
numbers_.. That our $M_{kl}$ are the super Catalan numbers can be easily
verified

$\displaystyle\frac{(2k+2l)!\ }{(k+l)!^{2}}\frac{\ (2l)!\ (2k-1)!\ !}{\
(2k+2l-1)!\ }\frac{\ (k+l-1)!}{\ \ (k-1)!l!}=\frac{2(k+l)\ (2l)!\ (2k)!\
}{(k+l)!(k+l)\ (k-1)!l!2k}=\frac{\ (2l)!\ (2k)!\ }{(k+l)!\ (k-1)!l!k}.$

Moreover, according to computer algebra the matrix $M$ admits the neat LU
factorization

(10.3) $\displaystyle M=L\operatorname{diag}(1,-2,2,-2,2,\dots)L^{T},$

where $L$ is the lower triangular matrix
$\left[\binom{2i}{i-j}\right]_{i,j\geq 0}$. This entails that the upper left
$n\times n$-block $M^{(n)}$ of $M$ has determinant $(-1)^{\lfloor
n/2\rfloor}2^{n-1}$ and that $M$ is integer. Once again, the number sequence
$1,2,2,2,\dots$ pops up.

It turns out that with our preparations we can easily find an explanation for
Eqn. (10.3). By the results of Section 8 we have

$\displaystyle\begin{bmatrix}\kappa^{0}(\theta)&\kappa^{2}(\theta)&\kappa^{4}(\theta)&\kappa^{6}(\theta)&\cdots\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1&\kappa(\theta)&\kappa(2\theta)&\kappa(4\theta)&\cdots\end{bmatrix}L^{T},$
$\displaystyle\begin{bmatrix}\sigma^{0}(\theta)&\sigma^{2}(\theta)&\sigma^{4}(\theta)&\sigma^{6}(\theta)&\cdots\end{bmatrix}$
$\displaystyle=\begin{bmatrix}1&\kappa(\theta)&\kappa(2\theta)&\kappa(4\theta)&\cdots\end{bmatrix}\operatorname{diag}(1,-1,1,-1,\dots)L^{T}.$

Transposing and integrating we deduce

$\displaystyle M$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}\begin{bmatrix}\kappa^{0}(\theta)\\\
\kappa^{2}(\theta)\\\ \kappa^{4}(\theta)\\\ \kappa^{6}(\theta)\\\
\cdots\end{bmatrix}\begin{bmatrix}\sigma^{0}(\theta)&\sigma^{2}(\theta)&\sigma^{4}(\theta)&\sigma^{6}(\theta)&\cdots\end{bmatrix}d\theta$
$\displaystyle=L\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}\begin{bmatrix}1\\\
\kappa(2\theta)\\\ \kappa(4\theta)\\\ \kappa(6\theta)\\\
\cdots\end{bmatrix}\begin{bmatrix}1&\kappa(2\theta)&\kappa(4\theta)&\kappa(6\theta)&\cdots\end{bmatrix}d\theta\right)\operatorname{diag}(1,-1,1,-1,\dotsc)L^{T}$
$\displaystyle=L\operatorname{\operatorname{diag}(1,2,2,2,\dotsc)diag}(1,-1,1,-1,\dotsc)L^{T}.$

##  11\. Spread polynomials

The _spread polynomials_ $(S_{n}(x))_{n\geq 0}$ (see [Wil05]) play a central
role in Wildberger’s _Rational Trigonometry_ , an approach to plane geometry
free of transcendental expressions. They encode rotations in the plane and are
defined by the following recursion:

$\displaystyle S_{0}(x)$ $\displaystyle=0,\ S_{1}(x)=x,$ $\displaystyle
S_{n}(x)$ $\displaystyle=2(1-2x)S_{n-1}(x)-S_{n-2}(x)+2x.$

######  Proposition 11.1.

For every integer $n\geq 0$ we have $S_{n}(x)=\frac{1-T_{n}(1-2x)}{2}$.

###### Proof.

We define

$S_{n}(x):=\frac{1-T_{n}(1-2x)}{2}\ \ \ \ \ (*)_{n}.$

Obviously, $S_{0}(x)=0,\ S_{1}(x)=x$. Assuming $(*)_{n}$ we want to deduce the
recurrence for $S_{n}(x)$ from the recurrence for $T_{n}(x)$. We observe that
$(*)_{n}$ is equivalent to

$T_{n}(1-2x)=1-2S_{n}(x)\ \ \ \ \ (**)_{n}.$

We have

$T_{n}(1-2x)=2(1-2x)T_{n-1}(1-2x)-T_{n-2}(1-2x)$

and hence

$S_{n}(x)=\frac{1-2(1-2x)T_{n-1}(1-2x)+T_{n-2}(1-2x)}{2}.$

From $(**)_{n-1}$ and $(**)_{n-2}$ we now obtain

$\displaystyle S_{n}(x)$
$\displaystyle=\frac{1}{2}(1-2(1-2x)(1-2S_{n-1}(x))+1-2S_{n-2}(x))$
$\displaystyle=\frac{1}{2}(4x+2(1-2x)2S_{n-1}(x)-2S_{n-2}(x))=2x+2(1-2x)S_{n-1}(x)-S_{n-2}(x).$

∎

######  Corollary 11.2 (M. Hirschhorn).

The bivariate generating function of the spread polynomials is given by

(11.1) $\displaystyle S(x,t)=\sum_{n\geq
0}S_{n}(x)t^{n}=\frac{tx(1+t)}{(1-t)\left(1-2t+t^{2}+4tx\right)}.$

###### Proof.

Using formula (2.3) we deduce

$\displaystyle\sum_{n\geq 0}S_{n}(x)t^{n}$ $\displaystyle=\sum_{n\geq
0}\frac{1-T_{n}(1-2x)}{2}t^{n}=\frac{1}{2}\sum_{n\geq
0}t^{n}-\frac{1}{2}\sum_{n\geq 0}T_{n}(1-2x)t^{n}$
$\displaystyle=\frac{1}{2}\sum_{n\geq
0}t^{n}-\frac{1}{2}\frac{1-t(1-2x)}{1-2t(1-2x)+t^{2}}=\frac{1}{2}\left(\frac{1}{1-t}-\frac{1-t(1-2x)}{1-2t(1-2x)+t^{2}}\right)\
$ $\displaystyle=\frac{1}{2}\
\frac{1-2t(1-2x)+t^{2}-(1-t(1-2x))(1-t)}{(1-t)(1-2t(1-2x)+t^{2}}$
$\displaystyle=\frac{1}{2}\
\frac{1-2t+4tx+t^{2}-\left(1-t-t+t^{2}+2tx-2t^{2}x\right)}{(1-t)\left(1-2t(1-2x)+t^{2}\right)}$
$\displaystyle=\frac{1}{2}\
\frac{2tx+2t^{2}x}{(1-t)\left(1-2t(1-2x)+t^{2}\right)}=\frac{tx(1+t)}{(1-t)\left(1-2t+t^{2}+4tx\right)}.$

∎

######  Corollary 11.3.

For each integer $n\geq 0$ we have
$S_{n}(\sin^{2}(\theta))=\sin^{2}(n\theta)$.

###### Proof.

From the Addition Theorem $\cos(2z)=1-2\sin^{2}(z)$ and Proposition 11.1 we
deduce

$S_{n}\left(\sin^{2}(\theta)\right)=\frac{1-T_{n}\left(1-2\sin^{2}(\theta)\right)}{2}=\frac{1-T_{n}(\cos(2\theta))}{2}=\frac{1-\cos(2n\theta)}{2}=\sin^{2}(n\theta).$

∎

The _spread matrix_ $S=(S_{mn})_{m,n\geq 1}$ is then defined such that

$\displaystyle\begin{bmatrix}S_{1}(x)&S_{2}(x)&S_{3}(x)&\dotsc\end{bmatrix}=\begin{bmatrix}x&x^{2}&x^{3}&\dotsc\end{bmatrix}S.$

######  Corollary 11.4.

For each integer $n\geq 1$ we have
$2^{2n-2}\sin^{2n}(\theta)=\sum_{k=1}^{n}(-1)^{k-1}\binom{2n}{n-k}\sin^{2}(k\theta)$.

###### Proof.

We use Theorem 7.1(3) to deduce from Proposition 11.1

$\displaystyle 2^{2n-1}\sin^{2n}(\theta)$
$\displaystyle=\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}(-1)^{k}\binom{2n}{n-k}\left(1-2\sin^{2}(k\theta)\right)$
$\displaystyle=\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}(-1)^{k}\binom{2n}{n-k}-2\sum_{k=1}^{n}(-1)^{k}\binom{2n}{n-k}\sin^{2}(k\theta)).$

The first two terms cancel each other. In fact, we have

$\displaystyle\frac{1}{2}\binom{2n}{n}+\sum_{k=1}^{n}(-1)^{k}\binom{2n}{n-k}$
$\displaystyle=\frac{1}{2}\binom{2n}{n}+\frac{1}{2}\sum_{j=1}^{n}(-1)^{n-j}\binom{2n}{j}+\frac{1}{2}\sum_{j=2n+1}^{2n}(-1)^{n-j}\binom{2n}{2n-j}$
$\displaystyle=\frac{1}{2}(-1)^{n}\left((-1)^{n}\binom{2n}{n}+\sum_{j=1}^{n}(-1)^{j}\binom{2n}{j}+\sum_{j=2n+1}^{2n}(-1)^{2n-j}\binom{2n}{2n-j}\right)$
$\displaystyle=\frac{1}{2}(-1)^{n}\sum_{j=1}^{2n}(-1)^{j}\binom{2n}{j}=\frac{1}{2}(-1)^{n}(1-1)^{2n}=0,$

since $n\geq 1$. ∎

Let $\operatorname{Wild}$ be the subspace of the space
$\operatorname{Trig}^{0}$ consisting of functions that vanish at $\theta=0$.
Put $\shuffle(\theta):=4\sin^{2}(\theta)$ and consider the two alternative
bases for $\operatorname{Wild}$:

(11.2)
$\displaystyle\begin{bmatrix}\shuffle(\theta)&\shuffle(2\theta)&\shuffle(3\theta)&\shuffle(4\theta)&\shuffle(5\theta)&\dotsc\end{bmatrix},$
(11.3)
$\displaystyle\begin{bmatrix}\shuffle(\theta)&\shuffle^{2}(\theta)&\shuffle^{3}(\theta)&\shuffle^{4}(\theta)&\shuffle^{5}(\theta)&\dotsc\end{bmatrix}.$

The idea is now to get rid of the powers of $4$ in the coefficients of $S_{n}$
by putting $Z_{n}(x):=4S_{n}\left(\frac{x}{4}\right)$. We refer to the
polynomials $Z_{n}(x)$ as the _zpread polynomials_. It follows that

(11.4) $\displaystyle Z_{n}(\shuffle(\theta))=\shuffle(n\theta),$

and we will see in a moment that $Z_{n}(x)\in\mathbb{Z}[x]$. We introduce the
infinite _zpread matrix_ $(Z_{mn})_{m,n\geq 1}$ such that

$\displaystyle\begin{bmatrix}Z_{1}(x)&Z_{2}(x)&Z_{3}(x)&\dots\end{bmatrix}=\begin{bmatrix}x&x^{2}&x^{3}&\dots\end{bmatrix}Z.$

It turns out $Z$ is a matrix of even pyramidal numbers of dimension $\geq 2$.

######  Theorem 11.5.

For $m,n\geq 1$ we have $Z_{mn}=(-1)^{m+1}p_{n-m}^{[2m]}$.

###### Proof.

Let $k\geq 0$ be an integer. From the generating function of the pyramidal
numbers we deduce

$\sum_{n\geq k}p_{n-k}^{[2m]}t^{n}=\sum_{n\geq
0}p_{n}^{[2m]}t^{n+k}=t^{k}\frac{1+t}{(1-t)^{2m+1}},$

and hence obtain

$\sum_{n\geq m}(-1)^{m+1}p_{n-m}^{[2m]}t^{n}=t^{m}\frac{1+t}{(1-t)^{2m+1}}.$

For the bivariate generating function this leads us to

(11.5) $\displaystyle\sum_{n,m\geq
1}(-1)^{m+1}p_{n-m}^{[2m]}x^{m}t^{n}=-\sum_{m\geq
1}\frac{(-x)^{m}t^{m}}{(1-t)^{2m}}\frac{1+t}{1-t}=\frac{1+t}{1-t}\frac{\frac{xt}{(1-t)^{2}}}{1+\frac{xt}{(1-t)^{2}}}=\frac{1+t}{1-t}\frac{xt}{(1-t)^{2}+xt},$

where we assume $-1<x,t<1$ and by default $p_{l}^{[2m]}=0$ for $l<0$. The last
expression in Equation (11.5) coincides with $4S\left(\frac{x}{4},t\right)$. ∎

The mutual inverse linear systems corresponding to Corollary 11.4 turn out to
be

(11.6)
$\begin{bmatrix}\shuffle(\theta)&\shuffle^{2}(\theta)&\shuffle^{3}(\theta)&\shuffle^{4}(\theta)&\shuffle^{5}(\theta)&\dotsc\end{bmatrix}\\\
=\begin{bmatrix}\shuffle(\theta)&\shuffle(2\theta)&\shuffle(3\theta)&\shuffle(4\theta)&\shuffle(5\theta)&\dotsc\end{bmatrix}\begin{bmatrix}1&4&15&56&210&\cdots\\\
&-1&-6&-28&-120&\\\ &&1&8&45&\\\ &&&-1&-10&\\\ &&&&1&\\\
\cdots&&&&&\cdots\end{bmatrix},$ (11.7)
$\begin{bmatrix}\shuffle(\theta)&\shuffle(2\theta)&\shuffle(3\theta)&\shuffle(4\theta)&\shuffle(5\theta)&\dotsc\end{bmatrix}\\\
=\begin{bmatrix}\shuffle(\theta)&\shuffle^{2}(\theta)&\shuffle^{3}(\theta)&\shuffle^{4}(\theta)&\shuffle^{5}(\theta)&\dotsc\end{bmatrix}\begin{bmatrix}1&4&9&16&25&\cdots\\\
&-1&-6&-20&-50&\\\ &&1&8&35&\\\ &&&-1&-10&\\\ &&&&1&\\\ \cdots&&&&&\cdots\\\
\end{bmatrix},$

the latter matrix being $Z$, whose entries are relatively small. The former
equation can be rephrased as

(11.8)
$\displaystyle\shuffle(\theta)^{n}=\sum_{k=1}^{n}(-1)^{k-1}\binom{2n}{n-k}\shuffle\\!(k\theta).$

The zpread matrix $Z$ can also be understood as the transpose of a Riordan
array.

######  Theorem 11.6.

The transpose of the zpread matrix $Z$ is the Riordan array
$Z^{T}=\left(\frac{1+x}{(1-x)^{3}},\frac{-x}{(1-x)^{2}}\right)$ with inverse
$(Z^{T})^{-1}=\left(BC^{2},-xC^{2}\right)=\left(C^{\prime},-xC^{2}\right)$.

###### Proof.

From Lemma 9.2 and Theorem 9.3 item (2) and (3) we deduce

$\left(\frac{1-x}{(1+x)^{3}},\frac{x}{(1+x)^{2}}\right)*\left(BC^{2},xC^{2}\right)=(1,x).$

To adjust the sign in front of $x$ we argue as follows. Let $\rho\in
x\mathbb{R}\llbracket x\rrbracket^{\times}$ and consider the formal pullback
$\rho^{*}:\mathbb{R}\llbracket x\rrbracket\rightarrow\mathbb{R}\llbracket
x\rrbracket,\ f\mapsto f\circ\rho$, which is an automorphism of the algebra
$(\mathbb{R}\llbracket x\rrbracket,\cdot)$. Let us denote the product of two
Riordan arrays $(g_{1},f_{1})$ and $(g_{2},f_{2})$ as
$(g_{1},f_{1})*(g_{2},f_{2})=(g_{1}(g_{2}\circ f_{1}),f_{2}\circ
f_{1})=:(G,F)$. It then follows that

$\left(\rho^{*}g_{1},\rho^{*}f_{1}\right)*(g_{2},f_{2})=((g_{1}(g_{2}\circ
f_{1}))\circ\rho,(f_{2}\circ
f_{1})\circ\rho)=\left(\rho^{*}G,\rho^{*}F\right).$

In the special case, when $\rho(x)=-x$ we deduce

$\left(\frac{1+x}{(1-x)^{3}},\frac{-x}{(1-x)^{2}}\right)*\left(BC^{2},xC^{2}\right)=(1,-x).$

But $(1,-x)$ is an idempotent, i.e., $(1,-x)^{*2}=(1,x)$. Hence we obtain

$(1,x)=\left(\frac{1+x}{(1-x)^{3}},\frac{-x}{(1-x)^{2}}\right)*\left(BC^{2},xC^{2}\right)*(1,-x)=\left(\frac{1+x}{(1-x)^{3}},\frac{-x}{(1-x)^{2}}\right)*\left(BC^{2},-xC^{2}\right).$

The verification of the well-known fact $BC^{2}=C^{\prime}$ we leave to the
reader. ∎

We refer to the following formula as the _spreadometric series_ since it
specializes to $(1-x^{2})^{-1}$ when $\theta=\pi/2$.

######  Corollary 11.7.

If $|x|<1$ then
$\frac{1+x}{1-x}\frac{\sin^{2}(\theta)}{(1-x)^{2}+4x\sin^{2}(\theta)}=\sum_{n\geq
0}\sin^{2}((n+1)\theta)x^{n}.$

###### Proof.

We use the language of Riordan arrays. Let us put
$\shuffle_{\theta}(x):=\sum_{n\geq
0}\shuffle(\theta)(\shuffle(\theta)x)^{n}=\frac{\shuffle(\theta)}{1-\shuffle(\theta)x}.$
The right hand side of Equation (11.7) can be understood as

$Z^{T}\overrightarrow{\shuffle_{\theta}}=\left(\frac{1+x}{(1-x)^{3}},-\frac{x}{(1-x)^{2}}\right)\overrightarrow{\shuffle_{\theta}}=\overrightarrow{\frac{1+x}{(1-x)^{3}}\frac{\shuffle(\theta)}{1+\shuffle(\theta)\frac{x}{(1-x)^{2}}}}=\overrightarrow{\frac{1+x}{1-x}\frac{\shuffle(\theta)}{(1-x)^{2}+x\shuffle(\theta)}}.$

Hence Equation (11.7) means

$\frac{1+x}{1-x}\frac{\shuffle(\theta)}{(1-x)^{2}+x\shuffle\\!(\theta)}=\sum_{n\geq
0}\shuffle((n+1)\theta)x^{n}.$

The claim follows by writing this in terms of $\sin^{2}(\theta)$. ∎

To memorize $S_{n}(x)$ one can proceed as the follows. Start with the square
numbers and put below it the sequence whose differences are made by the square
numbers. Underneath that sequence put the sequence whose differences are made
by the latter. Continue recursively.

(11.22) $\displaystyle\begin{array}[]{ c c c c c c c c c c c c c
c}1&&4&&9&&16&&25&&36&&49&\cdots\\\ &1&&5&&14&&30&&55&&91&&\\\
&&1&&6&&20&&50&&105&&196&\\\ &&&1&&7&&27&&77&&182&&\\\
&&&&1&&8&&35&&112&&294&\\\ &&&&&1&&9&&44&&156&&\\\ &&&&&&1&&10&&54&&210&\\\
&&&&&&&1&&11&&63&&\\\ &&&&&&&&1&&12&&77&\\\ &&&&&&&&&1&&13&&\\\
&&&&&&&&&&1&&14&\\\ &&&&&&&&&&&1&&\\\ &&&&&&&&&&&&1&\cdots\end{array}$

In the resulting pattern take every second row and arrange them into a
triangle. The spread matrix $S$ is then made by multiplying the $m$th row with
$(-4)^{m-1}$, keeping in mind that the first row carries the row index $m=1$:

$\displaystyle\begin{array}[]{ c c c c c c c c }1&4&9&16&25&36&49&\cdots\\\
&1&6&20&50&105&196&\\\ &&1&8&35&112&294&\\\ &&&1&10&54&210&\\\ &&&&1&12&77&\\\
&&&&&1&14&\\\ &&&&&&1&\\\ &&&&&&&\cdots\end{array}\mapsto
S=\left[\begin{array}[]{ c c c c c c c c }1&4&9&16&25&36&49&\cdots\\\
&-4&-24&-80&-120&-410&-784&\\\ &&16&128&560&1792&4704&\\\
&&&-64&-640&-3456&-13440&\\\ &&&&256&3072&19712&\\\ &&&&&-1024&-14336&\\\
&&&&&&4096&\\\ \cdots&&&&&&&\cdots\end{array}\right].$

To obtain $Z$ one has to multiply the $m$th row with $(-1)^{m-1}$ instead. The
transpose of the spread matrix is the Riordan array
$S^{T}=\left(\frac{1+x}{(1-x)^{3}},\frac{-4x}{(1-x)^{2}}\right)$, whose
inverse $\left(BC^{2},-xC^{2}/4\right)$ does of course not have integer
entries.

##  12\. A conjecture of Goh and Wildberger

Goh and Wildberger [GW09, Wil22] formulated an interesting conjecture
concerning the factorizations of the spread polynomials over the ring
$\mathbb{Z}[x]$. We take here the liberty to formulate it in terms of the
zpread polynomials and speculate a bit about numbers appearing in the
coefficients of the factors. Using the zpread polynomials those become less
obscure. As spread polynomials are encoding rotations in the plane555That is,
they come straight from the almighty. we find these unexpected phenomena
intriguing.

######  Conjecture 12.1.

There are $\Phi_{d}(x)\in\mathbb{Z}[x]$, $d\geq 1$, with
$\deg\Phi_{d}(x)=\phi(d)$ such that for each $n\geq 1$ we have
$Z_{n}(x)=\prod_{d|n}\Phi_{d}(x)$. For $d\geq 3$ the polynomial
$\Phi_{d}(x)=\psi_{d}(x)^{2}$ is the square of an irreducible polynomial
$\psi_{d}(x)=\mathbb{Z}[x]$ with constant term $\psi_{d}(0)>0$. This constant
term evaluates to $\psi_{d}(0)=\prod_{d|n}\left(\frac{n}{d}\right)^{\mu(d)}$
(see entry A014963 in [OEI23].) Here $\phi(d)$ denotes the Euler totient
function and $\mu(d)$ the Möbius $\mu$-function. If $p\geq 5$ is a prime then
$\psi_{p}(1)=(-1)^{\phi(p)/2}$.

In order to convince the reader we provide some empirical evidence:

$\displaystyle\begin{array}[]{ c|c c c c c c c c }d&1&2&3&4&5&6&7&8\\\
\hline\cr\Phi_{d}(x)&x&4-x&(3-x)^{2}&(2-x)^{2}&\left(5-5x+x^{2}\right)^{2}&(1-x)^{2}&\left(7-14x+7x^{2}-x^{3}\right)^{2}&\left(2-4x+x^{2}\right)^{2}\end{array}$
$\displaystyle\begin{array}[]{ c c c c }9&10&11&12\\\
\hline\cr\left(3-9x+6x^{2}-x^{3}\right)^{2}&\
\left(1-3x+x^{2}\right)^{2}&\left(11-55x+77x^{2}-44x^{3}+11x^{4}-x^{5}\right)^{2}&\left(1-4x+x^{2}\right)^{2}\end{array}$
$\displaystyle\begin{array}[]{ c c c }13&14&15\\\ \hline\cr\
\left(13-91x+182x^{2}-156x^{3}+65x^{4}-13x^{5}+x^{6}\right)^{2}&\left(1-6x+5x^{2}-x^{3}\right)^{2}&\left(1-8x+14x^{2}-7x^{3}+x^{4}\right)^{2}\end{array}$
$\displaystyle\begin{array}[]{ c c }16&17\\\
\hline\cr\left(2-16x+20x^{3}-8x^{3}+x^{4}\right)^{2}&\left(17-204x+714x^{2}-1122x^{3}+935x^{4}-442x^{5}+119x-17x^{7}+x^{8}\right)^{2}\end{array}.$

It comes as a surprise that the coefficients of $\psi_{d}(x)$ seem to almost
coincide with certain columns of the array (11.22). Up to shifts, there are
merely differences in the linear and/or constant coefficients. The column
index in the array relevant for $\psi_{d}(x)$ seems to be the coefficient in
front of $-x^{\phi(d)/2-1}$. From the empirical data we guess it is
$\phi(d)-\mu(d)$ (see entry A053139 in [OEI23]). We would like to mention that
it seems to be the case that for each prime $p>2$ we have
$\Phi_{p}(x)=\Phi_{2p}(2-x)$. Furthermore, the evaluations of the
$\Phi_{n}(x)$ at the integers $x=0,1,2,3,4$ look interesting. One may ask
which of those values $\Phi_{n}(x)$ appear for infinitely many $n$.

Clearly, for each $n\geq 1$ we have that $Z_{n}$ maps the square $[0,4]^{2}$
into itself and all the zeros of $Z_{n}(x)$ have to be in $[0,4]$. The
question for what $x\in[0,4]$ there are distinct $n,m$ such that
$Z_{n}(x)=Z_{m}(x)$ deserves further attention. With regards to this let us
mention some empirical observations.

######  Conjecture 12.2.

Let $n\geq 1$ be an integer and $\varphi=\left(1+\sqrt{5}\right)/2$ be the
golden ratio. Then

  1. (1)

$2=Z_{n}(2)$ if and only if $n\equiv 1\bmod 2$,

  2. (2)

$3=Z_{n}(3)$ if and only if $n\equiv 1,2\bmod 3$,

  3. (3)

$2+\varphi=Z_{n}(2+\varphi)$ if and only if $n\equiv 1,4\bmod 5$,

  4. (4)

$3-\varphi=Z_{n}(2+\varphi)$ if and only if $n\equiv 2,3\bmod 5$.

Johann Cigler [Cig10] has noted that

$\displaystyle
S_{2n}\\!\left(x^{2}\right)=\left(1-x^{2}\right)U_{2n-1}(x)^{2},\ \
S_{2n+1}\\!\left(x^{2}\right)=T_{2n+1}(x)^{2},$

so that

$\displaystyle Z_{2n}\\!\left(x^{2}\right)$
$\displaystyle=4S_{2n}\left((x/2)^{2}\right)=\left(4-x^{2}\right)U_{2n-1}(x/2)^{2},$
$\displaystyle Z_{2n+1}\\!\left(x^{2}\right)$
$\displaystyle=4S_{2n+1}\left((x/2)^{2}\right)=4T_{2n+1}(x/2)^{2}.$

Hence all the claims above correspond to statements about the Chebyshev
polynomials, and those have been addressed in the literature. For example, the
zeros of the Chebyshev polynomials correspond obviously the zeros of
$\cos(n\theta)$ and $\sin(n\theta)$, respectively. Hence the main task is to
understand the combinatorics of how the linear factors are distributed to form
the $\Phi_{d}$.

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