# A Curious Trigonometric Infinite Product in Context

Michael<EMAIL_ADDRESS>
Consulting Physicist, Geometrics Unlimited, Ltd

Box 1484, Deep River, Ont. Canada. K0J 1P0

Author’s manuscript, Orcid:0000-0002-7987-0820

revised

MSC classes: 40A20, 26-02, 26A09, 26E99, 33-02, 33B10, Keywords: Infinite
product, Viète’s Product, Accumulation point, Weierstrass’ Factor theorem,
trigonometric products

Abstract

By treating the multiple argument identity of the logarithm of the Gamma
function as a functional equation, we obtain a curious infinite product
representation of the $sinc$ function in terms of the cotangent function. This
result is believed to be new. It is then shown how to convert the infinite
product to a finite product, which turns out to be a simple telescoping of the
double angle $sin$ function. In general, this result unifies known infinite
product identities involving various trigonometric functions when the product
term index appears as an exponent. In one unusual case, what appears to be a
straightforward limit, suggests a counterexample to Weierstrass’ factor
theorem. A resolution is offered. An Appendix presents the general solution to
a simple functional equation. This work is motivated by its educational
interest.

##  1 Introduction and Motivation

Throughout, the symbols, $m,n,j,k,N$ are positive integers and $\gamma$ is the
Euler-Mascheroni constant; other symbols are real, although the extension to
general complex values is always available. In the course of a study of the
properties of $\eta(s)$, the alternating companion to Riemann’s function
$\zeta(s)$ defined [1, Eq. 2.3(1)] by

(1.1)
$\eta(s)\equiv\sum_{k=0}^{\infty}(-1)^{k}/(k+1)^{s}=(1-2^{1-s})\,\zeta(s),\hskip
15.0pt\Re(s)>0\,,$

I obtained the identity

(1.2)
$\overset{\infty}{\underset{j=1}{\sum}}\;\frac{\eta\\!\left(1+j\right)\left(-a\right)^{j}}{1+j}=\frac{1}{a}\ln\\!\left(\frac{\sqrt{\pi}\,\Gamma\left(\frac{a}{2}+1\right)}{\Gamma\left(\frac{a}{2}+\frac{1}{2}\right)2^{a}}\right),\hskip
10.0pt|a|<1$

corresponding to similar identities developed in [1, Section 3.4], in
particular the identity [1, Eq. 3.4(10)]

(1.3)
$\overset{\infty}{\underset{j=1}{\sum}}\;\frac{\zeta\\!\left(1+j\right)\left(-a\right)^{j}}{1+j}=-\frac{\ln\\!\left(\Gamma\\!\left(a+1\right)\right)}{a}-\gamma,\hskip
10.0pt|a|<1\,.$

To obtain (1.2), apply the second equality of (1.1) to yield

(1.4)
$\overset{\infty}{\underset{j=1}{\sum}}\;\frac{\zeta\\!\left(1+j\right)\left(-a\right)^{j}}{1+j}=\frac{1}{a}\ln\\!\left(\frac{\sqrt{\pi}\,\Gamma\left(\frac{a}{2}+1\right)}{\Gamma\left(\frac{a}{2}+\frac{1}{2}\right)2^{a}}\right)+\overset{\infty}{\underset{j=1}{\sum}}\;\frac{\zeta\\!\left(1+j\right)\left(-\frac{a}{2}\right)^{j}}{1+j}$

from which (1.2) can be easily verified by recourse to (1.3). In addition to
employing (1.4) to prove (1.2), cursory observation of (1.4) shows that it
satisfies a simple and easily solvable (see Appendix) functional equation of
the form

(1.5)
$f\\!\left(a\right)=g\\!\left(a\right)+x\,f\\!\left(\frac{a}{p}\right)\,,$

of which the application to $\eta(s)$ is discussed elsewhere [3]. However,
noticing that (1.3) and (1.4) also simply reproduce the half argument identity
of $\log(\Gamma(1+a))$ in the form

(1.6)
$\ln\\!\left(\Gamma\\!\left(1+a\right)\right)=\ln\\!\left(2\right)a-\frac{\ln\\!\left(\pi\right)}{2}+\ln\\!\left(\Gamma\\!\left(\frac{a}{2}+\frac{1}{2}\right)\right)+\ln\\!\left(\Gamma\\!\left(1+\frac{a}{2}\right)\right)\,,$

the question arises: “Will anything interesting be found if (1.6) is studied
as an instance of the functional equation (1.5)?” That is the motivation for
this work, which is intended to be an educational exercise in the merits of
curiosity.

In the following Section 2, that question is pursued, first in Subsection 2.1
where it is shown to yield interesting and curious infinite trigonometric
products and corresponding infinite summation identities, some of which are
thought to be new. In that same Subsection these identities are related to
known results by the operation of differentiation and the exceptional cases
are analysed. Curiosity then suggests that the new, curious identity be
generalized, and this is done in two different ways in Sections 2.2.1 and
2.2.2; in the latter Section, connections are also made with known identities,
thereby unifying known results listed elsewhere. A requirement that always
arises when studying what may be a new identity is to place it in context with
known material from the literature; this is pursued in Section 3 where a
closely-related, purported, and long-ago analysed, identity is reviewed.
Finally, curiosity leads one far afield by noticing that the limiting case of
two of the contextually related identities previously introduced ostensibly
appear to violate Weierstrass’ factor theorem. In Section 3.1 this issue is
considered and a resolution is offered.

##  2 $\log(\Gamma(a+1))$ treated as a functional equation

###  2.1 A curious identity

By applying the first corollary of Lemma (A.1) to (1.6)– see the Appendix– we
easily find the interesting identity

(2.1)
$\Gamma\\!\left(1+2\,a\right)\overset{\infty}{\underset{j=0}{\prod}}\;\frac{1}{2^{\,2\,a}}\left(\frac{\sqrt{\pi}\,\Gamma\left({a}/{2^{j}}+1\right)}{\Gamma\left({a}/{2^{j}}+\frac{1}{2}\right)}\right)^{2^{j}}={{\mathrm{e}}^{-2\,\gamma\,a}}\,,$

equivalent to the (convergent) sum

(2.2)
$\overset{\infty}{\underset{j=1}{\sum}}\;\left(2^{j}\,\ln\\!\left(\frac{\Gamma\\!\left({a}/{2^{j}}+1\right)}{\Gamma\\!\left({a}/{2^{j}}+{1}/{2}\right)}\right)+2^{j-1}\,\ln\\!\left(\pi\right)-2\,\ln\\!\left(2\right)a\right)+2\,\ln\\!\left(\Gamma\\!\left(a+1\right)\right)=-2\,\gamma\,a\,.$

Now add the identity (2.2) and its companion where $a\Rightarrow-a$, to obtain
the curious infinite product

(2.3)
$\overset{\infty}{\underset{j=1}{\prod}}\\!\left(\tan\\!\left({a}/{2^{j}}\right)\frac{2^{j}\,}{a}\right)^{2^{j-1}}=\frac{a}{\sin\\!\left(a\right)}\,,$

and, letting $a=ib$, its hyperbolic equivalent

(2.4)
$\overset{\infty}{\underset{j=1}{\prod}}\;\left(\frac{2^{j}\,\tanh\\!\left({b}/{2^{j}}\right)}{b}\right)^{2^{j-1}}=\frac{b}{\sinh\\!\left(b\right)}\,.$

Inverting also yields the following representation of the $\mathrm{sinc}(a)$
function, defined by $\mathrm{sinc}(a)\equiv\sin(a)/a$ :

(2.5)
$\mathrm{sinc}\\!\left(a\right)=\overset{\infty}{\underset{j=1}{\prod}}\;\left(\cot\\!\left({a}/{2^{j}}\right)\frac{a}{2^{j}}\right)^{2^{j-1}}\,.$

It is important to consider the exceptional cases where $a$ is in integral
multiple of $\pi$. There are two possibilities: $a$ is even or odd. In the
latter case, let $a=(2\,n-1)\,\pi,~{}n\geq 1$, in which case the only
divergent term in the product is labelled by the index $j=1$. So, by setting
$n\rightarrow n+\epsilon$, consider the limit

(2.6) $\underset{\mathit{\epsilon}\rightarrow
0}{\mathrm{lim}}\\!\frac{2\,\tan\\!\left(\frac{\left(2\,n+2\,\mathit{\epsilon}-1\right)\pi}{2}\right)}{\left(2\,n+2\,\mathit{\epsilon}-1\right)\pi}=-\frac{2}{\pi^{2}\left(2\,n-1\right)\mathit{\epsilon}}+O\\!\left(1\right)$

and compare to the corresponding divergent term on the right-hand side of
(2.3), to obtain

(2.7)
$\overset{\infty}{\underset{j=2}{\prod}}\;\left(\frac{\tan\\!\left(\frac{\left(2\,n-1\right)\pi}{2^{j}}\right)2^{j}}{\left(2\,n-1\right)\pi}\right)^{2^{j-1}}=\pi^{2}\left(1/2-n\right)^{2}\,.$

In the case that $a$ is even, recall that any even number can always be
written as the product of an integral power of $2$ and an odd number, so let
$a=2^{m}\,(2n-1)$, where $m\geq 0$ and $n\geq 1$. By setting $m\rightarrow
m+\epsilon$ and considering the limit as $\epsilon\rightarrow 0$, it is easy
to find that the first $m$ terms of the product each vanish to order
$(\epsilon\,\ln(2))^{2^{j-1}}$, and therefore

(2.8) $\underset{\epsilon\rightarrow
0}{\mathrm{lim}}\,\;\overset{m}{\underset{j=1}{\prod}}\;\left(\frac{\tan\\!\left(\frac{2^{m+\epsilon}\,(2n-1)\,\pi}{2^{j}}\right)2^{j}}{2^{m+\epsilon}\,(2n-1)\,\pi}\right)^{2^{j-1}}\approx\frac{\epsilon^{2^{m}}\,\ln\\!\left(2\right)^{2^{m}-1}}{\epsilon}\,.$

The next term of the product corresponding to $j=m+1$ is the only divergent
factor, being of order

(2.9) $\underset{\epsilon\rightarrow
0}{\mathrm{lim}}\;\left(\frac{\tan\\!\left(\frac{2^{m+\epsilon}\,(2n-1)\,\pi}{2^{m+1}}\right)2^{m+1}}{2^{m+\epsilon}\,(2n-1)\,\pi}\right)^{2^{m}}\approx\left(\frac{4}{\ln\\!\left(2\right)(2n-1)^{2}\,\pi^{2}\,\epsilon}\right)^{2^{m}}\,.$

Therefore the product of (2.8) and (2.9) diverges to order $\epsilon^{-1}$, as
does the right-hand side of (2.3) under the conditions specified. So, if $a$
is an integral multiple of $\pi$, we have the exceptional case

(2.10)
$\overset{\infty}{\underset{j=m+2}{\prod}}\;\left(\frac{\tan\\!\left(2^{m-j}\,(2n-1)\,\pi\right)}{2^{m-j}\,(2n-1)\,\pi}\right)^{2^{j-1}}=\left(\frac{(2n-1)\,\pi}{2}\right)^{2^{m+1}}$

reducing to (2.7) if $m=0$.

Remarks:

  * •

Differentiating (2.3) with respect to $a$ gives

(2.11)
$\overset{\infty}{\underset{j=1}{\sum}}\\!\left(2^{j-1}-\frac{a}{\sin\\!\left(a/2^{j-1}\right)}\right)=a\,\cot\\!\left(a\right)-1\,,$

an extension of the listed [2, Eq. 25.1.2] identity

(2.12)
$\overset{n}{\underset{j=0}{\sum}}\;\frac{1}{\sin\\!\left({x}/{2^{j}}\right)}=\cot\\!\left(\frac{x}{2^{n+1}}\right)-\cot\\!\left(x\right)\,.$

  * •

It is possible to reproduce (2.11) by subtracting the sum (2.12) from the
elementary identity

(2.13) $\overset{n}{\underset{j=0}{\sum}}\;2^{j}=2^{n+1}-1$

and evaluating the limit as $n\rightarrow\infty$.

  * •

Further, differentiation of (2.2) yields

(2.14)
$\overset{\infty}{\underset{j=1}{\sum}}\\!\left(2\,\ln\\!\left(2\right)-\psi\\!\left(1+\frac{a}{2^{j}}\right)+\psi\\!\left(\frac{a}{2^{j}}+\frac{1}{2}\right)\right)=2\,\psi\\!\left(a+1\right)+2\,\gamma\,,$

a result that is discussed elsewhere [3], which could also be obtained by
evaluating the $n\rightarrow\infty$ limit of [2, Eq. (55.8.4)].

  * •

A related result extracted from the classical literature [4, Eq. (1042)] is

(2.15)
$\underset{n\rightarrow\infty}{\mathrm{lim}}\\!\left(\overset{n}{\underset{j=1}{\prod}}\;\tan\\!\left(\frac{\pi\,j}{2\,n}\right)\right)^{\frac{1}{n}}=1\,.$

###  2.2 Generalizations

####  2.2.1 By recursion

It is possible to generalize (2.3) as follows

(2.16)
$\overset{\infty}{\underset{j=n+1}{\prod}}\;\left(\frac{2^{j-n}\,}{a}\tan\\!\left({2^{n-j}\,a}\right)\right)^{2^{j-1}}=\frac{a^{2^{n}}}{\sin^{2^{n}}\left(a\right)},\hskip
10.0ptn\geq 0,$

the proof being by induction.

Proof: Let $n\Rightarrow n+1$ and simultaneously let $a\Rightarrow a/2$ to
obtain

(2.17)
$\overset{\infty}{\underset{j=n+2}{\prod}}\;\left(\frac{2^{j-n}}{a}\,\tan\\!\left(2^{n-j}\,a\right)\right)^{2^{j-1}}=\frac{\left(\frac{a}{2}\right)^{2^{n+1}}}{\sin^{2^{n+1}}\left(\frac{a}{2}\right)}.$

Extend the product on the left-hand side and transfer the new factor to the
right-hand side, yielding

(2.18)
$\overset{\infty}{\underset{j=n+1}{\prod}}\;\left(\frac{2^{j-n}}{a}\,\tan\\!\left(2^{n-j}\,a\right)\right)^{2^{j-1}}=\frac{\left(\frac{a}{2}\right)^{2^{n+1}}}{\sin^{2^{n+1}}\left(\frac{a}{2}\right)}\left(\frac{2\,\tan\left(\frac{a}{2}\right)}{a}\right)^{2^{n}}\,.$

After some elementary simplification we find

(2.19)
$\overset{\infty}{\underset{j=n+1}{\prod}}\;\left(\frac{2^{j-n}}{a}\,\tan\\!\left(2^{n-j}\,a\right)\right)^{2^{j-1}}=a^{2^{n}}\left(\frac{1}{2\,\cos\\!\left(\frac{a}{2}\right)\sin\\!\left(\frac{a}{2}\right)}\right)^{2^{n}}$

reducing to (2.16) from the double angle identity

(2.20) $\displaystyle\sin(a)=2\cos(a/2)\sin(a/2).\hskip 50.0pt\mbox{ \bf{QED}
}$

Again generalizing, let $n\Rightarrow n_{1},a\Rightarrow a/2^{n_{1}}$ in
(2.16) and similarly $n\Rightarrow n_{2},a\Rightarrow a/2^{n_{2}}$ with
$n_{2}>n_{1}$. Divide the resulting identities and split the first product
from $n_{1}$ to $n_{2}$ and $n_{2}+1$ to $\infty$. After some simplification,
we have the finite product

(2.21)
$\overset{n_{2}}{\underset{j=n_{1}+1}{\prod}}\;\left(\frac{2^{j}}{a}\,\tan\\!\left(\frac{a}{2^{j}}\right)\right)^{2^{j-1}}=a^{N_{1}-N_{2}}\,2^{-N_{1}\,n_{1}+N_{2}\,n_{2}}\frac{\sin^{N_{2}}\\!\left(\frac{a}{N_{2}}\right)}{\sin^{N_{1}}\\!\left(\frac{a}{N_{1}}\right)}\,,$

where $N_{1}\equiv 2^{n_{1}}$ and $N_{2}\equiv 2^{n_{2}}$. Consider two
examples. Let $n_{1}=0,~{}n_{2}=4$ and $a\Rightarrow 2\,a$ giving

(2.22)
$\tan\\!\left(a\right)\tan^{2}\left(\frac{a}{2}\right)\tan^{4}\left(\frac{a}{4}\right)\tan^{8}\left(\frac{a}{8}\right)=2^{15}\frac{\sin^{16}\left(\frac{a}{8}\right)}{\sin\\!\left(2\,a\right)}\,.$

The case $n_{1}=n,~{}n_{2}=n+1$ yields the identity

(2.23)
$\tan\\!\left(\frac{a}{2^{n+1}}\right)=\frac{2\sin^{2}\left({a}/{2^{n+1}}\right)}{\sin\\!\left({a}/{2^{n}}\right)}\,,$

made obvious by setting $a\Rightarrow 2^{n}\,a$ and referring to (2.20). It is
easily seen that (2.22) is a simple recursive rendition of the double angle
formula (2.20) by rewriting it in the form

(2.24)
$\sin\\!\left(2\,a\right)\frac{\sin\\!\left(a\right)}{\cos\\!\left(a\right)}\tan^{2}\left(\frac{a}{2}\right)\tan^{4}\left(\frac{a}{4}\right)\tan^{8}\left(\frac{a}{8}\right)=2^{15}\sin^{16}\left(\frac{a}{8}\right)$

and applying (2.20) recursively.

####  2.2.2 Triplication and Onwards

Since (2.1) is essentially based on the duplication formula (1.6), consider
the triplication analogue

(2.25)
$\ln\\!\left(\Gamma\\!\left(a+1\right)\right)=\ln\\!\left(\frac{3^{a+\frac{1}{2}}\,\Gamma\\!\left(\frac{a}{3}+\frac{1}{3}\right)\Gamma\\!\left(\frac{a}{3}+\frac{2}{3}\right)}{2\,\pi}\right)+\ln\\!\left(\Gamma\\!\left(\frac{a}{3}+1\right)\right)$

a special case of the well-known general identity [5, Eq. 5.5.6]

(2.26)
$\Gamma\\!\left(n\,a\right)=\left(2\,\pi\right)^{\frac{1}{2}-\frac{n}{2}}\,n^{na-\frac{1}{2}}\,\overset{n-1}{\underset{k=0}{\prod}}\;\Gamma\\!\left(a+\frac{k}{n}\right)\,.$

Apply Corollary 1 from the Appendix as before to find

(2.27)
$\ln\\!\left(\Gamma\\!\left(a+1\right)\right)=\overset{\infty}{\underset{j=0}{\sum}}\;\left(\frac{\ln\\!\left(3\right)}{2}+\ln\\!\left(3^{{}^{\frac{3\,a}{3^{1+j}}}}\right)+\ln\\!\left(\Gamma\\!\left(\frac{a}{3^{1+j}}+\frac{1}{3}\right)\right)+\ln\\!\left(\Gamma\\!\left(\frac{a}{3^{1+j}}+\frac{2}{3}\right)\right)-\ln\\!\left(2\,\pi\right)\right)\,,$

the analogue of (2.2). Adding (2.25) and its partner with $a\Rightarrow-a$
then yields

(2.28)
$\overset{\infty}{\underset{j=0}{\prod}}\;\frac{1+2\,\cos\\!\left({2\,a}/{3^{1+j}}\right)}{3}=\frac{\sin\\!\left(a\right)}{a}\,,$

equivalent to

(2.29)
$\overset{\infty}{\underset{j=1}{\prod}}\\!\left(1-\frac{4\sin^{2}\left({a}/{3^{j}}\right)}{3}\right)=\frac{\sin\\!\left(a\right)}{a}\,,$

a known result [2, Eq. 91.8.14] . Similarly, by differentiating (2.27) we find

(2.30)
$\overset{\infty}{\underset{j=0}{\sum}}\;\left(\frac{3\,\ln\\!\left(3\right)}{3^{1+j}}+\frac{\psi\\!\left({a}/{3^{1+j}}+\frac{1}{3}\right)}{3^{1+j}}+\frac{\psi\\!\left({a}/{3^{1+j}}+\frac{2}{3}\right)}{3^{1+j}}\right)=\psi\\!\left(a+1\right)\,,$

the analogue of (2.14). Remark: Hansen [2] sources (2.29) by the initials
“CJ’, which label does not appear in the Bibliography.

Continuing, and similarly, the quadruplication and quintuplication cases
produce

(2.31)
$\overset{\infty}{\underset{j=0}{\prod}}\;\left(\frac{\cos\\!\left({a}/{4^{1+j}}\right)}{2}+\frac{\cos\\!\left({3\,a}/{4^{1+j}}\right)}{2}\right)=\frac{\sin\\!\left(a\right)}{a}$

and

(2.32)
$\overset{\infty}{\underset{j=0}{\prod}}\;\frac{1+2\,\cos\\!\left({2\,a}/{5^{1+j}}\right)+2\,\cos\\!\left({4\,a}/{5^{1+j}}\right)}{5}=\frac{\sin\\!\left(a\right)}{a}\,,$

the general even and odd cases being respectively

(2.33)
$\overset{\infty}{\underset{j=0}{\prod}}\;\frac{1}{N}\;\overset{N}{\underset{n=1}{\sum}}\;{\cos\\!\left(\frac{\left(2\,n-1\right)a}{\left(2\,N\right)^{1+j}}\right)}{}=\frac{\sin\\!\left(a\right)}{a}\,,$
(2.34)
$\overset{\infty}{\underset{j=0}{\prod}}\;\left(\frac{1}{2\,N+1}+\frac{2}{2\,N+1}\;\overset{N}{\underset{n=1}{\sum}}\;\cos\\!\left(\frac{2\,n\,a}{\left(2\,N+1\right)^{1+j}}\right)\right)=\frac{\sin\\!\left(a\right)}{a}\,.$

In (2.33), let $N=1$ to find Euler’s well-known infinite product formula [2,
Eq. 91.8.2]

(2.35)
$\overset{\infty}{\underset{j=0}{\prod}}\;\cos\\!\left({a}/{2^{1+j}}\right)=\frac{\sin\\!\left(a\right)}{a}\,.$

Both (2.33) and (2.34) have been obtained elsewhere through the use of both
Chebyshev polynomials and Fourier transforms (see Nishimura [6] where the
words $odd$ and $even$ are erroneously scrambled).

Remarks

  * •

Although the identity (2.35) as cited requires $|a|<1$, the derivation given
here removes that limitation.

  * •

Additionally, (2.35) might be compared to the equally well-known [2, Eq.
91.8.3, unsourced] and occasionally re-cited (e.g. [7]) identity

(2.36)
$\overset{k-1}{\underset{j=0}{\prod}}\;\cos\\!\left(2^{j}\,a\right)=\,\frac{\sin\\!\left(2^{k}\,a\right)}{2^{k}\,\sin\\!\left(a\right)}\,.$

  * •

We also find from the classical literature [4, Eq. (1050)] the related
identity

(2.37)
$\overset{2\,k-1}{\underset{j=1}{\prod}}\;\cos\\!\left(\frac{\pi\,j}{k}\right)=\frac{\left(-1\right)^{k}-1}{2^{2\,k-1}}\,.$

Superficially, the equality of the left-hand sides of (2.31) and (2.32)
appears to be surprising; likewise the corresponding sides of (2.33) and
(2.34). However, from [2, Eq. 17.1.1] we have

(2.38)
$\overset{N}{\underset{n=1}{\sum}}\\!\cos\\!\left(\frac{\left(2\,n-1\right)a}{\left(2\,N\right)^{1+j}}\right)=\frac{\sin\\!\left({a}/{\left(2\,N\right)^{j}}\right)}{2\,\sin\\!\left({a}/{\left(2\,N\right)^{1+j}}\right)}$

and with some simplification help from Maple [8]

(2.39)
$\overset{N}{\underset{n=1}{\sum}}\;\cos\\!\left(\frac{2\,n\,a}{\left(2\,N+1\right)^{1+j}}\right)=\frac{\sin\\!\left(a/\left(2\,N+1\right)^{j}\right)}{2\,\sin\\!\left(a/\left(2\,N+1\right)^{1+j}\right)}-\frac{1}{2}\,,$

so that, after the application of (2.39) to (2.34), we have

(2.40)
${\color[rgb]{0,0,1}\left[\frac{{\sin\\!\left(a\right)}\;}{{\color[rgb]{0,0,1}\sin\\!\left(\frac{a}{(2\,N+1)}\right)\left(2\,N+1\right)}}\right]}{\color[rgb]{1,0,0}\left[\frac{{\color[rgb]{1,0,0}\sin\\!\left(\frac{a}{2\,N\,+1}\right)}\;}{{\color[rgb]{1,0,0}\sin\\!\left(\frac{a}{(2\,N+1)^{2}}\right)\left(2\,N+1\right)}}\right]}{\underset{j=2}{\prod}}{\color[rgb]{0,0,0}\frac{\sin\left(a/\left(2\,N\,+1\right)^{j}\right)}{~{}\sin\left(a/\left(2\,N+1\right)^{1+j}\right)\left(2\,N+1\right)}}=\frac{\sin\\!\left(a\right)}{a}$

where the first two terms of the product have been extracted (and coloured) to
demonstrate that the product partially telescopes (e.g. the denominator of the
j=0 factor (blue) cancels the corresponding factor in the numerator of the j=1
factor (red), each contributing a factor $2N+1$ in the denominator. The
limiting term(s) of the product then approach unity because

(2.41)
$\underset{j\rightarrow\infty}{\mathrm{lim}}\;\frac{1}{\sin\\!\left(a/\left(2\,N+1\right)^{1+j}\right)}=\frac{\left(2\,N+1\right)^{j+1}}{a}+O\\!\left(\frac{1}{\left(2\,N+1\right)^{j}}\right)\,,$

cancelling the accumulating multiplicative factor, thereby demonstrating the
veracity of (2.34), and similarly (2.33).

##  3 Some Perspectives

Infinite products have significance rooted in the early history of analysis
exemplified by Euler’s 1734 product formula [5, Eq. 4.22.1] for the sin
function

(3.1)
$\frac{\sin\\!\left(a\right)}{a}=\overset{\infty}{\underset{j=1}{\prod}}\;\left(1-\frac{a^{2}}{\pi^{2}\,j^{2}}\right)\,.$

In the realm of infinite products involving trigonometric functions, we have
Euler’s infinite product formula (2.35). Setting $a=\pi/2$ in (2.35) yields

(3.2)
$\frac{2}{\pi}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{2+\sqrt{2}}}{2}\times\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\times\dots\,$

as a special case, originally obtained recursively by François Viète in 1593
and recently generalized by both Moreno and Garcia [9] and Levin [10], the
latter of whose approach was based upon an an analysis of the functional
equation

(3.3) $F\\!\left(a\,z\right)=g\\!\left(F\\!\left(z\right)\right)$

with simple conditions on $F(z)~{}\mathrm{and}~{}g(z)$.

Relevant to the identity reported here, in 1876, Dobinski [11, Eq. (72)]
proposed the identity

(3.4)
$\overset{\infty}{\underset{j=1}{\prod}}\;\left(\tan\left(2^{j}\,a\right)\right)^{2^{-j}}=4\sin^{2}\left(a\right)\,.$

Written otherwise, we have

(3.5)
$\tan\\!\left(a\right)\times\left({\tan}\left(2\,a\right)\right)^{\frac{1}{2}}\times\left(\tan\left(4\,a\right)\right)^{\frac{1}{4}}\times\left(\tan\left(8\,a\right)\right)^{\frac{1}{8}}\times\dots=4\sin^{2}\left(a\right)$

to be compared to (2.23) and (2.24). In 1947, the identity (3.4) was
questioned by Agnew and Walker [12]. By reproducing Dobinski’s derivation,
Agnew and Walker claim that (3.4) is true iff

(3.6)
$\underset{j\rightarrow\infty}{\mathrm{lim}}\\!\left(\sin\left(2^{1+j}\,a\right)\right)^{2^{-j}}=1\,;$

therefore, (3.4) is true only for special values of $a\neq(2n+1)\pi/2^{k}$. To
reinforce their scepticism, it is worth noting that for certain real values of
$a$, some of the product factors are complex while the right-hand side is
real. So, (3.4) thus requires that, if for some value $j=j_{+}$
$\tan(2^{j_{+}}a)>0$, then $2^{j}a$ must lie in a region where
$\tan(2^{j}a)>0$ for all $j\neq j_{+}$. As might be expected, this imposes a
severe restriction on allowable values of $a$, and omits the possibility that
the imaginary part of all the (complex) product factors corresponding to
$\tan(2^{j_{+}}a)<0$ somehow vanishes. Unfortunately, Agnew and Walker do not
identify a concrete example to demonstrate the limited veracity of (3.4) and
numerical experimentation by myself has uncovered no examples.

###  3.1 A digression

The juxtaposition of the proof of (3.4) in proximity with (2.36) leads the
curious analyst to consider the question of the evaluation of (2.36) when its
right-hand side vanishes, in the light of “Weierstrass’ Factor-Theorem”, [13,
page 9, Theorem 1] that states “A convergent product has the value zero if and
only if, one of its factors vanishes”. Which of the factors of (2.36) vanishes
in this eventuality? Consider the candidate case $a=2^{n}\pi$. After
evaluating the appropriate limit on the right-hand side, we find

(3.7)
$\overset{k-1}{\underset{j=0}{\prod}}\;\cos\\!\left(2^{j+n}\,\pi\right)=1+\left(-\frac{2^{2\,k}}{6}+\frac{1}{6}\right)\left(a-2^{n}\,\pi\right)^{2}+O\\!\left(\left(a-2^{n}\,\pi\right)^{4}\right)$

and there is no issue, since each of the factors comprising the product are
equal to unity. In the case $a=\pi/2^{m}$, that is

(3.8)
$\overset{k-1}{\underset{j=0}{\prod}}\;\cos\\!\left(\frac{2^{j}\,\pi}{2^{m}}\right)=\frac{\sin\\!\left({2^{k-m}\,\pi}\right)}{2^{k}\,\sin\\!\left({2^{-m}\pi}\right)}$

only if $k\geq m$ does the right-hand side vanish, and accordingly, only the
single left-hand side factor indexed by $j=m-1$ vanishes, and both (2.36) and
Weierstrass’ Factor-Theorem are satisfied. Now consider (2.36) in the case
that $k\rightarrow\infty$, that is

(3.9)
$\underset{k\rightarrow\infty}{\mathrm{lim}}\;\overset{k-1}{\underset{j=0}{\prod}}\;\cos\\!\left(2^{j}\,a\right)=\underset{k\rightarrow\infty}{\mathrm{lim}}\;\frac{\sin\\!\left(2^{k}\,a\right)}{2^{k}\,\sin\\!\left(a\right)}\rightarrow
0\,.$

Unlike the previous two cases, (3.9) is apparently analytically (and, for
large values of $k$, numerically) valid for all values of $a$. Consequently,
consistency of the left-hand side requires the existence of a solution to

(3.10) $2^{j-1}a/N=\pi$

for general values of $a$, if $N$ is an odd integer. That (3.10) is clearly
impossible (e.g. $a$ is rational) suggests that (3.9) provides a
counterexample to Weierstrass’ factor theorem – an unlikely prospect. Is there
a resolution?

After due consideration, one must conclude that equating the right-hand side
limit of (3.9) to zero is unjustified (indicated by the use of the arrow),
because the limit $k\rightarrow\infty$ yields an accumulation, rather than a
limit, point, courtesy of the numerator factor that varies discontinuously for
consecutive values of $k$ in a neighbourhood of zero acting as a limit.
Similarly, the factors comprising the left-hand side of (3.9) span an infinite
number of possibilities, any one of which can approach arbitrarily close to,
but not equal to, zero, thus emulating an accumulation point and the infinite
product does not converge, although it comes arbitrarily close to zero.
However, let us now rewrite (3.9) as follows

(3.11)
$\frac{\sin\\!\left(a\right)\overset{k-1}{\underset{j=0}{\prod}}\;\cos\\!\left(2^{j}\,a\right)}{\sin\\!\left(2^{k}\,a\right)}=\frac{1}{2^{k}}\,$

an identity that is also analytically and numerically valid. Again,
considering the limit $k\rightarrow\infty$ gives

(3.12)
$\underset{k\rightarrow\infty}{\mathrm{lim}}\;\frac{\sin\\!\left(a\right)\overset{k-1}{\underset{j=0}{\prod}}\;\cos\\!\left(2^{j}\,a\right)}{\sin\\!\left(2^{k}\,a\right)}=\underset{k\rightarrow\infty}{\mathrm{lim}}\;\frac{1}{2^{k}}\,=0,$

and this time there is no ambiguity about the limit of the right-hand side.
However, as seen above, since the infinite product term on the left-hand side
must be non-convergent, Weierstrass’ factor theorem does not apply and we have
an example of a simple equality whose left and right-hand side limits differ,
since neither of the other two left-hand side factors will lead to a zero or
negate the arbitrariness of the product term; if the product were to be
convergent then we would have a counterexample to Weierstrass’ factor theorem.
Since $k$ is not a continuous variable, I am unaware that this example
violates any theorems of analysis.

In addition to the above, note that (2.37) exhibits similar behaviour when
$k\rightarrow\infty$, although in that case, the resolution is obvious since
the right-hand side clearly alternates in value between even and odd values of
$k$. However, (2.37) does exhibit another curiosity: when $k$ is even, the
left-hand side product possesses two vanishing factors indexed by $j=k/2$ and
$j=3k/2$, whereas the right-hand side possess a simple zero. Again, since $k$
is not a continuous variable, the disagreement in the order of the zero (if
one can talk about the order of a zero attached to a discrete variable) does
not violate Weierstrass’ theorem as written. For a collection of other
interesting analytic anomalies, see [14].

##  4 Summary

What began as a simple derivation of what appears to be a new, curious
identity, led to a study of some related and well-known identities. This in
turn led to a unification of heretofore disparate listed results for a family
of infinite product identities characterized by arguments containing the index
variable appearing as an exponent. Further investigation of related identities
selected from the literature led to some interesting examples of unexpected
analytic behaviour.

##  Appendix A Appendix: Lemma

######  Lemma A.1.

Consider the functional equation

(A.1) $f\\!\left(a\right)=g\\!\left(a\right)+x\,f\\!\left(\frac{a}{p}\right)$

Then, if $|x|<1$, $f(0)$ is finite, $1<p\in\Re$ and if the series converges

(A.2)
$f\\!\left(a\right)=\overset{\infty}{\underset{j=0}{\sum}}\;x^{j}\,g\\!\left(\frac{a}{p^{j}}\right)$

Proof: Let $a:=a/p$ in (A.1) and substitute into itself, to yield

(A.3)
$f\\!\left(a\right)=g\\!\left(a\right)+g\\!\left(\frac{a}{p}\right)\,x+f\\!\left(\frac{a}{\,p^{2}}\right)\,x^{2}\,.$

Let $a:=a/p^{2}$ in (A.1), substitute into (A.3) and repeat recursively
altogether $N$ times to obtain

(A.4)
$f\\!\left(a\right)=f\\!\left(\frac{a}{p^{N}}\right)x^{N}+\overset{N-1}{\underset{j=0}{\sum}}\;x^{j}\,g\\!\left(\frac{a}{p^{j}}\right)\,.$

According to the suppositions, if $N\rightarrow\infty$, the first term in
(A.4) vanishes, leaving (A.2). QED

Corollary 1: If $x=1$ and $f(0)=0$, (A.2) also follows.

Corollary 2: If $x>1$,
$\underset{x\rightarrow\infty}{\mathrm{lim}}\;f\left(x\right)$ is finite, and
if the sum converges, by reversing (A.1) to read

(A.5)
$f\\!\left(\frac{a}{p}\right)=-\frac{g\\!\left(a\right)}{x}+\frac{f\\!\left(a\right)}{x}$

we have the solution

(A.6)
$f\\!\left(a\right)=-\overset{\infty}{\underset{j=1}{\sum}}\;\frac{g\\!\left(p^{j}\,a\right)}{x^{j}}\,.$

## References

  * [1] H.M. Srivastava and Junesang Choi.  Zeta and q-Zeta Functions and Associated Series and Integrals.  Elsevier, 32 Jamestown Rd.,London, NW1 7BY, first edition, 2012.
  * [2] Eldon R Hansen.  A Table of Series and Products.  Prentice-Hall Inc., Englewood Cliffs, N.J., 1975.
  * [3] M. Milgram.  in preparation.
  * [4] L.B.W. Jolley.  Summation of Series, Second Revised Edition.  Dover, New York, 1961.
  * [5] F. W. J. Olver, D. W. Lozier, R. F. Boisvert, and C. W. Clark, editors.  NIST Handbook of Mathematical Functions.  Cambridge University Press, New York, NY, 2010.  Print companion to [15].
  * [6] R. Nishimura.  A generalization of Viète’s infinite product and new mean iterations.  The Australian Journal of Mathematical Analysis and Applications, 13(1), 2016.
  * [7] July 2022.  https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Infinite_product_formulae.
  * [8] Maplesoft, a division of Waterloo Maple Inc., version 2021.  Maple.
  * [9] Moreno S.G. and Garcia E.M.  New infinite products of cosines and Viète-like formulae.  Mathematics Magazine, 86(1):15–25, February 2013.
  * [10] Aaron Levin.  A new class of infinite products generalizing Viète’s product formula for $\pi$.  The Ramanujan Journal, 10, 2005.  https://doi.org/10.1007/s11139-005-4852-z.
  * [11] E.W. Weisstein.  Infinite product, mathworld – a wolfram web resource., June 2022.  Retrieved from https://mathworld.wolfram.com/InfiniteProduct.html, Eq. (72).
  * [12] Agnew R.P. and Walker R.J.  A trigonometric infinite product.  The American Mathematical Monthly, 54(4), 1947.
  * [13] Konrad Knopp.  Theory of Functions, Part II.  Dover, 1947.  translated by Frederick Bagemihl from the Fourth German Edition.
  * [14] Gelbaum Bernard R. and Olmsted John M.H.  Counterexamples in Analysis.  Holden-Day Inc, San Francisco, 1964.
  * [15] NIST Digital Library of Mathematical Functions.  http://dlmf.nist.gov/, Release 1.0.9 of 2014-08-29.


$\mathcal{E}_{ij}:=\bigg{\\{}\exists v\in
A_{ij}:1-\frac{\|Xv\|_{2}}{\sqrt{n}}\geq 2[g_{1}(\|v\|_{2})+g_{2}(\|\Gamma
v\|_{1})]\bigg{\\}}$

as well as the event

$\mathcal{E}:=\bigg{\\{}\exists
v\in\mathbb{R}^{p}:\|\Sigma^{1/2}v\|_{2}=1\text{ and
}1-\frac{\|Xv\|_{2}}{\sqrt{n}}\geq 2\left[g_{1}(\|v\|_{2})+g_{2}(\|\Gamma
v\|_{1})\right]\bigg{\\}}$

Note that
$\mathcal{E}=\cup_{i=1}^{\infty}\cup_{j=1}^{\infty}\mathcal{E}_{ij}$. Our goal
is to prove that $\mathbb{P}(\mathcal{E})\leq c_{1}\exp(-nc_{2})$, from which
the lemma follows.

If we have $v\in A_{ij}$ such that $1-\frac{\|Xv\|_{2}}{\sqrt{n}}\geq
2[g_{1}(\|v\|_{2})+g_{2}(\|\Gamma v\|_{1})]$ holds, then by definition of
$A_{ij}$,

$1-\frac{\|Xv\|_{2}}{\sqrt{n}}\geq
2(2^{i-1}\mu+2^{j-1}\mu)=2^{i}\mu+2^{j}\mu=g_{1}(g_{1}^{-1}(2^{i}\mu))+g_{2}(g_{2}^{-1}(2^{j}\mu))$

Again by definition of $A_{ij}$, $g_{1}(\|v\|_{2})\leq 2^{i}\mu$ and
$g_{2}(\|\Gamma v\|_{1})\leq 2^{j}\mu$, and so

$\|v\|_{2}\leq g_{1}^{-1}(2^{i}\mu)\qquad\text{and}\qquad\|\Gamma v\|_{1}\leq
g_{2}^{-1}(2^{j}\mu)$

Therefore, we must have

$\displaystyle\mathbb{P}(\mathcal{E}_{ij})$ $\displaystyle\leq
2\exp\left(-\frac{n}{18}[g_{1}(g_{1}^{-1}(2^{i}\mu))+g_{2}(g_{2}^{-1}(2^{j}\mu))]^{2}\right)$
$\displaystyle=2\exp\left(-\frac{n}{18}(2^{i}+2^{j})^{2}\mu^{2}\right)$
$\displaystyle\leq
2\exp\left(-\frac{n}{18}2^{2i}\mu^{2}\right)\exp\left(-\frac{n}{18}2^{2j}\mu^{2}\right)$

Hence,

$\displaystyle\mathbb{P}(\mathcal{E})$ $\displaystyle\leq
2\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\exp\left(-\frac{n}{18}2^{2i}\mu^{2}\right)\exp\left(-\frac{n}{18}2^{2j}\mu^{2}\right)$
$\displaystyle=2\left(\sum_{i=1}^{\infty}\exp\left(-\frac{n}{18}2^{2i}\mu^{2}\right)\right)^{2}$
$\displaystyle\leq
2\left(\sum_{i=1}^{\infty}\exp\left(-\frac{ni}{18}\mu^{2}\right)\right)^{2}$
$\displaystyle=2\left(\frac{\exp\left(-\frac{n}{18}\mu^{2}\right)}{1-\exp\left(-\frac{n}{18}\mu^{2}\right)}\right)^{2}\leq
c_{1}\exp(-nc_{2})$

∎

######  Corollary A.1.

Under the settings of Lemma A.2,

$\frac{\|\tilde{X}v\|_{2}^{2}}{n}\geq
v^{T}\left(\frac{1}{64}\Sigma+\lambda_{2}L\right)v-\frac{72\gamma_{\text{max}}(\Sigma)n_{c}}{n}\|v\|_{2}^{2}-576\rho(\Gamma)^{2}\frac{\gamma_{\text{max}}(\Sigma)\log
p}{n}\|\Gamma v\|_{1}^{2}$

holds for all $v\in\mathbb{R}^{p}$ with probability at least
$1-c_{1}\exp(-nc_{2})$

Proof We argue in a manner similar to the proof of Theorem 7.16 in Wainwright
(2019). For any real numbers $a,b,c$ such that $c\geq\max(a-b,0)$, we claim
that $c^{2}\geq(1-\delta)^{2}a^{2}-\frac{b^{2}}{\delta^{2}}$ for any
$\delta\in(0,1)$. This is because if $b\geq a\delta$, then
$(1-\delta)^{2}a^{2}-\frac{b^{2}}{\delta^{2}}\leq a^{2}[(1-\delta)^{2}-1]\leq
0\leq c$, and if $b<a\delta$, then since $c\geq a-b$, we have $c\geq
a-a\delta=(1-\delta)a$.

Letting $c=\frac{\|Xv\|_{2}}{\sqrt{n}}$, $a=\frac{1}{4}\|\Sigma^{1/2}v\|_{2}$,
$b=3\sqrt{\frac{\gamma_{\text{max}}(\Sigma)n_{c}}{n}}\|v\|_{2}+6\sqrt{2}\rho(\Gamma)\sqrt{\frac{\gamma_{\text{max}}(\Sigma)\log
p}{n}}\|\Gamma v\|_{1}$ and $\delta=\frac{1}{2}$, we obtain for all
$v\in\mathbb{R}^{p}$ with probability at least $1-c_{1}\exp(-nc_{2})$:

$\displaystyle\frac{\|Xv\|_{2}^{2}}{n}$
$\displaystyle\geq\frac{1}{64}\|\Sigma^{1/2}v\|_{2}^{2}-36\left(\sqrt{\frac{\gamma_{\text{max}}(\Sigma)n_{c}}{n}}\|v\|_{2}+2\sqrt{2}\rho(\Gamma)\sqrt{\frac{\gamma_{\text{max}}(\Sigma)\log
p}{n}}\|\Gamma v\|_{1}\right)^{2}$
$\displaystyle\geq\frac{1}{64}\|\Sigma^{1/2}v\|_{2}^{2}-72\frac{\gamma_{\text{max}}(\Sigma)n_{c}}{n}\|v\|_{2}^{2}-576\rho(\Gamma)^{2}\frac{\gamma_{\text{max}}(\Sigma)\log
p}{n}\|\Gamma v\|_{1}^{2}$

By adding $\lambda_{2}v^{T}Lv$ to both sides, we obtain what we need to prove.
∎

######  Lemma A.3 (High-probability bound on $\|\Pi X^{T}\epsilon\|_{2}$).

For any $\delta>0$,

$\|\Pi
X^{T}\epsilon\|_{2}\leq\sqrt{2\sigma^{2}n\gamma_{\text{max}}(\Sigma)}(\sqrt{n_{c}}+\delta)$

with probability at least $1-e^{-n/8}-e^{-\delta^{2}/2}$.

Proof We make use of the fact that $X$ and $\epsilon$ are independent. Note
that $X\Pi$ has i.i.d. $N(0,\Pi\Sigma\Pi)$ rows, which we denote by
$\tilde{x}_{1},\cdots,\tilde{x}_{n}$. Then

$\|\Pi
X^{T}\epsilon\|_{2}=\left\|\sum_{i=1}^{n}\epsilon_{i}\tilde{x}_{i}\right\|_{2}=\|\epsilon\|_{2}\left\|\frac{1}{\|\epsilon\|_{2}}\sum_{i=1}^{n}\epsilon_{i}\tilde{x}_{i}\right\|_{2}$

which has the same distribution as $\|\epsilon\|_{2}\|\tilde{x}\|_{2}$, where
$\tilde{x}\sim N(0,\Pi\Sigma\Pi)$ is independent of $\epsilon$. Since
$\gamma_{\text{max}}(\Sigma)\Pi-\Pi\Sigma\Pi$ is positive semi-definite, by
Lemma A.1, $\|\tilde{x}\|_{2}$ is stochastically dominated by
$\sqrt{\gamma_{\text{max}}(\Sigma)}\|\Pi h\|_{2}$ (where $h\sim N(0,I_{p})$),
which in turn has the same distribution as
$\sqrt{\gamma_{\text{max}}(\Sigma)}\|h^{\prime}\|_{2}$ where $h^{\prime}\sim
N(0,I_{n_{c}})$.

By an application of a concentration inequality for Lipschitz functions of
Gaussian vectors, we have for any $\delta>0$ (see Example 2.28 of Wainwright
(2019)):

$\mathbb{P}(\|h^{\prime}\|_{2}\geq\sqrt{n_{c}}+\delta)\leq e^{-\delta^{2}/2}$

and we also have $\|\epsilon\|_{2}\leq\sigma\sqrt{2n}$ with probability at
least $1-e^{-n/8}$ (see Example 2.11 of Wainwright (2019)). Combining all the
pieces yields the result. ∎

######  Lemma A.4 (Choice of $\lambda_{1}$).

With probability at least $1-\frac{2}{m}-e^{-n/8}$, we have

$\|(\Gamma^{\dagger})^{T}X^{T}\epsilon\|_{\infty}\leq
4\sigma\rho(\Gamma)\sqrt{\gamma_{\text{max}}(\Sigma)n\log p}$

and hence $\lambda_{1}$ should be chosen such that $\lambda_{1}\geq
32\sigma\rho(\Gamma)\sqrt{\frac{\gamma_{\text{max}}(\Sigma)\log p}{n}}$

Proof Recall that the columns of $\Gamma^{\dagger}\in\mathbb{R}^{p\times m}$
are denoted as $s_{1},\cdots,s_{m}$, and let the rows of $X$ be
$x_{1},\cdots,x_{n}$, which by assumption are i.i.d. $N(0,\Sigma)$ vectors.

For any $t>0$:

$\displaystyle\mathbb{P}(\|(\Gamma^{\dagger})^{T}X^{T}\epsilon\|_{\infty}\geq
t)$
$\displaystyle=\mathbb{P}\left(\max_{j\in[m]}\bigg{|}\bigg{\langle}s_{j},\sum_{i=1}^{n}\epsilon_{i}x_{i}\bigg{\rangle}\bigg{|}\geq
t\right)$
$\displaystyle=\mathbb{P}\left(\|\epsilon\|_{2}\max_{j\in[m]}\bigg{|}\bigg{\langle}s_{j},\frac{1}{\|\epsilon\|_{2}}\sum_{i=1}^{n}\epsilon_{i}x_{i}\bigg{\rangle}\bigg{|}\geq
t\right)$
$\displaystyle\leq\mathbb{P}\left(\sqrt{2n}\sigma\max_{j\in[m]}\bigg{|}\bigg{\langle}s_{j},\frac{1}{\|\epsilon\|_{2}}\sum_{i=1}^{n}\epsilon_{i}x_{i}\bigg{\rangle}\bigg{|}\geq
t\right)+\mathbb{P}(\|\epsilon\|_{2}>\sigma\sqrt{2n})$

Using the same trick as in Lemma A.3,
$x:=\frac{1}{\|\epsilon\|_{2}}\sum_{i=1}^{n}\epsilon_{i}x_{i}\sim N(0,\Sigma)$
independent of $\epsilon$. Also, we note again that
$\mathbb{P}(\|\epsilon\|_{2}>\sigma\sqrt{2n})\leq e^{-n/8}$. Hence,
$\mathbb{P}(\|(\Gamma^{\dagger})^{T}X^{T}\epsilon\|_{\infty}\geq t)$ is
bounded above by

$\mathbb{P}\left(\sqrt{2n}\sigma\max_{j\in[m]}|\langle s_{j},x\rangle|\geq
t\right)+e^{-n/8}\leq\mathbb{P}\left(\sqrt{2n\gamma_{\text{max}}(\Sigma)}\sigma\max_{j\in[m]}|\langle
s_{j},g\rangle|\geq t\right)+e^{-n/8}$

where $g\sim N(0,I_{p})$ and we used Lemma A.1 in the last inequality. Since
$\\{\langle s_{j},g\rangle:j\in[m]\\}$ are Normal variables with variance at
most $\rho(\Gamma)^{2}$, by applying the union bound, the expression above can
be bounded above by

$2\exp\left(-\frac{t^{2}}{4\gamma_{\text{max}}(\Sigma)n\sigma^{2}\rho(\Gamma)^{2}}+\log
m\right)+e^{-n/8}$

If $t$ is chosen such that
$t^{2}=8\log(m)\gamma_{\text{max}}(\Sigma)n\sigma^{2}\rho(\Gamma)^{2}$, we can
conclude that

$\|(\Gamma^{\dagger})^{T}X^{T}\epsilon\|_{\infty}\leq
2\sqrt{2}\sigma\rho(\Gamma)\sqrt{\gamma_{\text{max}}(\Sigma)n\log m}\leq
4\sigma\rho(\Gamma)\sqrt{\gamma_{\text{max}}(\Sigma)n\log p}$

with probability at least $1-\frac{2}{m}-e^{-n/8}$, where we used $m\leq
p^{2}$. ∎

######  Lemma A.5 (Lemma 3 of Hütter and Rigollet (2016)).

If $\Gamma$ is the incidence matrix of a graph $G=(V,E)$ with maximum degree
$d$ and $\emptyset\neq S\subseteq E$, then

$k_{S}^{-2}\leq 4\min(d,|S|)$

######  Lemma A.6 (Lower bound in Lemma 2.2).

If $\Gamma$ is the incidence matrix of the 2D grid, then $\rho(\Gamma)\gtrsim
1$.

Proof Let $N:=\sqrt{p}$. In the proof of Proposition 4 of Hütter and Rigollet
(2016), it was shown that $\Gamma^{\dagger}$ has $2N(N-1)$ columns
$((s_{i,j}^{(1)})_{\begin{subarray}{c}i\in[N-1]\\\
j\in[N]\end{subarray}},((s_{i,j}^{(2)})_{\begin{subarray}{c}j\in[N-1]\\\
i\in[N]\end{subarray}})$, each of which has column norm such that

$\|s_{i,j}^{(\diamond)}\|_{2}^{2}=\sum_{k=0}^{N-1}\sum_{l=1}^{N-1}\frac{1}{(\lambda_{k}+\lambda_{l})^{2}}\langle
v_{l},d_{i}\rangle^{2}\langle v_{k},e_{j}\rangle^{2}$

where $\diamond\in\\{1,2\\}$, $\lambda_{k}=2-2\cos\frac{k\pi}{N}$ for $0\leq
k\leq N-1$ (these are the eigenvalues of the Laplacian of the one-dimensional
chain graph with $N$ vertices), $d_{i}$ is the $i^{\text{th}}$ column of
$D_{1}^{T}$ where $D_{1}$ is the incidence matrix of the chain graph with $N$
vertices, $e_{1},\dots,e_{n}$ are the canonical basis vectors of
$\mathbb{R}^{N}$, and $v_{k}\in\mathbb{R}^{N}$ ($0\leq k\leq N-1$) are the
orthonormal eigenvectors of the Laplacian of the one-dimensional chain graph
with $N$ vertices:

$(v_{0})_{j}=\frac{1}{\sqrt{N}}$
$(v_{k})_{j}=\sqrt{\frac{2}{N}}\cos\left(\frac{(j+1/2)k\pi}{N}\right)\text{
for }0\leq j\leq N-1,1\leq k\leq N-1$

Since $\rho(\Gamma)$ is defined as the maximum column norm of
$\Gamma^{\dagger}$, we can bound it below by the column norm of
$s_{i,j}^{(\diamond)}$ where $i=j=\diamond=1$. We have:

$\|s_{1,1}^{(1)}\|_{2}^{2}=\sum_{k=0}^{N-1}\sum_{l=1}^{N-1}\frac{1}{\left(4-2\cos\frac{k\pi}{N}-2\cos\frac{l\pi}{N}\right)^{2}}\langle
v_{l},d_{1}\rangle^{2}\langle v_{k},e_{1}\rangle^{2}$

Using the inequality $2-2\cos(x)\leq x^{2}$, we have
$4-2\cos\frac{k\pi}{N}-2\cos\frac{l\pi}{N}\leq\frac{\pi^{2}}{N^{2}}(k^{2}+l^{2})$.
Furthermore, note that

$\displaystyle\langle v_{l},d_{1}\rangle^{2}$
$\displaystyle=\frac{2}{N}\left(\cos\frac{(5/2)l\pi}{N}-\cos\frac{(3/2)l\pi}{N}\right)^{2}$
$\displaystyle=\frac{2l^{2}\pi^{2}}{N^{3}}\sin^{2}(x^{\prime})$

for some $x^{\prime}\in\left[\frac{(3/2)l\pi}{N},\frac{(5/2)l\pi}{N}\right]$,
by the mean value theorem. Given the inequality $\sin(x)\geq x/2$ for
$x\in[0,\pi/2]$, we can conclude that
$\sin^{2}(x^{\prime})\geq(x^{\prime})^{2}/4\geq\frac{9}{16}\frac{l^{2}\pi^{2}}{N^{2}}$
if we assume $l\leq\frac{N}{5}$, and so

$\langle v_{l},d_{1}\rangle^{2}\geq\frac{9\pi^{4}}{8}\frac{l^{4}}{N^{5}}$

if $l\leq N/5$. Moreover, $\langle v_{k},e_{1}\rangle^{2}=\frac{1}{N}$ if
$k=0$, and if we assume $k\leq\frac{2}{3\pi}N$, then
$1-\frac{9}{8}\frac{k^{2}\pi^{2}}{N^{2}}\geq\frac{1}{2}$ and an application of
$1-\cos(x)\leq\frac{x^{2}}{2}$ gives

$\displaystyle\langle
v_{k},e_{1}\rangle^{2}=\frac{2}{N}\left[\cos\left(\frac{3k\pi}{2N}\right)\right]^{2}$
$\displaystyle=\frac{2}{N}\left[1-\left(1-\cos\frac{3k\pi}{2N}\right)\right]^{2}$
$\displaystyle\geq\frac{2}{N}\left[1-\frac{9k^{2}\pi^{2}}{8N^{2}}\right]^{2}\geq\frac{1}{2N}$

Hence, if $k\leq\frac{2}{3\pi}N$, we have $\langle
v_{k},e_{1}\rangle^{2}\geq\frac{1}{2N}$. Let
$c=\min\left(\frac{2}{3\pi},\frac{1}{5}\right)=\frac{1}{5}$. Now,

$\displaystyle\|s_{1,1}^{(1)}\|_{2}^{2}$ $\displaystyle\geq\sum_{k=0}^{\lfloor
cN\rfloor}\sum_{l=1}^{\lfloor
cN\rfloor}\frac{1}{\left(4-2\cos\frac{k\pi}{N}-2\cos\frac{l\pi}{N}\right)^{2}}\langle
v_{l},d_{1}\rangle^{2}\langle v_{k},e_{1}\rangle^{2}$
$\displaystyle\geq\sum_{k=0}^{\lfloor cN\rfloor}\sum_{l=1}^{\lfloor
cN\rfloor}\left(\frac{N^{4}}{\pi^{4}(k^{2}+l^{2})^{2}}\right)\left(\frac{9\pi^{4}}{8}\frac{l^{4}}{N^{5}}\right)\left(\frac{1}{2N}\right)$
$\displaystyle=\frac{9}{16N^{2}}\sum_{l=1}^{\lfloor
cN\rfloor}\sum_{k=0}^{\lfloor cN\rfloor}\frac{l^{4}}{(k^{2}+l^{2})^{2}}$

Since $\frac{1}{(k^{2}+l^{2})^{2}}$ is a decreasing function of $k$,

$\displaystyle\|s_{1,1}^{(1)}\|_{2}^{2}$
$\displaystyle\geq\frac{9}{16N^{2}}\sum_{l=1}^{\lfloor
cN\rfloor}l^{4}\int_{0}^{cN}\frac{1}{(x^{2}+l^{2})^{2}}dx$
$\displaystyle=\frac{9}{16N^{2}}\sum_{l=1}^{\lfloor
cN\rfloor}l^{4}\frac{(l^{2}+c^{2}N^{2})\arctan(cN/l)+cNl}{2l^{3}(l^{2}+c^{2}N^{2})}$
$\displaystyle=\frac{9}{32N^{2}}\sum_{l=1}^{\lfloor
cN\rfloor}l\frac{(l^{2}+c^{2}N^{2})\arctan(cN/l)+cNl}{l^{2}+c^{2}N^{2}}$
$\displaystyle\geq\frac{9}{32N^{2}}\sum_{l=1}^{\lfloor cN\rfloor}l\arctan(c)$
$\displaystyle=\frac{9\arctan(c)}{32N^{2}}\frac{\lfloor cN\rfloor(\lfloor
cN\rfloor+1)}{2}\gtrsim 1$

∎

######  Lemma A.7 (Lower bound in Lemma 2.3).

If $\Gamma$ is the incidence matrix of the r-dimensional grid for $r\geq 3$,
then $\rho(\Gamma)\geq c(r)$, where $c(r)$ is a constant depending only on
$r$.

Proof Note that the $\gtrsim$ sign is used in this proof to omit constant
multipliers that may depend on $r$. Similarly to the previous lemma, it is
sufficient to lower bound

$\|s_{\bm{1}}^{(1)}\|_{2}^{2}=\sum_{l=1}^{N-1}\sum_{k_{1}=0}^{N-1}\dots\sum_{k_{r-1}=0}^{N-1}\frac{\langle
v_{l},d_{1}\rangle^{2}\prod_{j=1}^{r-1}\langle
v_{k_{j}},e_{1}\rangle^{2}}{(\lambda_{l}+\sum_{j=1}^{r-1}\lambda_{k_{j}})^{2}}$

where $d_{1},e_{1}$ as well as $\lambda_{0},\dots\lambda_{N-1}$ and
$v_{0},\dots,v_{N-1}$ are defined in relation to the chain graph with $N$
vertices as in the previous lemma.

By applying the inequality $2-2\cos(x)\leq x^{2}$, we have

$\left(\lambda_{l}+\sum_{j=1}^{r-1}\lambda_{k_{j}}\right)^{2}\leq\left(2-2\cos\frac{l\pi}{N}+\sum_{j=1}^{r-1}\left(2-2\cos\frac{k_{j}\pi}{N}\right)\right)^{2}\leq\frac{\pi^{4}}{N^{4}}\left(l^{2}+\sum_{j=1}^{r-1}k_{j}^{2}\right)^{2}$

Also, $k\leq\frac{2}{3\pi}N$ implies $\langle
v_{k},e_{1}\rangle^{2}\geq\frac{1}{2N}$, and $l\leq N/5$ implies $\langle
v_{l},d_{1}\rangle^{2}\geq\frac{9\pi^{4}}{8}\frac{l^{4}}{N^{5}}$. Hence, if we
define $c=\min(\frac{2}{3\pi},\frac{1}{5})=\frac{1}{5}$ as in the previous
lemma,

$\displaystyle\|s_{\bm{1}}^{(1)}\|_{2}^{2}$
$\displaystyle\gtrsim\frac{1}{N^{r}}\sum_{l=1}^{\lfloor
cN\rfloor}\sum_{k_{1}=0}^{\lfloor cN\rfloor}\dots\sum_{k_{r-1}=0}^{\lfloor
cN\rfloor}\frac{l^{4}}{(l^{2}+\sum_{j=1}^{r-1}k_{j}^{2})^{2}}$
$\displaystyle\geq\frac{1}{N^{r}}\sum_{l=1}^{\lfloor cN\rfloor}\int_{0\leq
x_{j}\leq cN,j=1,\dots,r-1}\frac{l^{4}}{(l^{2}+\|x\|_{2}^{2})^{2}}dx$
$\displaystyle\geq\frac{1}{N^{r}}\sum_{l=1}^{\lfloor
cN\rfloor}\int_{\|x\|_{2}\leq cN}\frac{l^{4}}{(l^{2}+\|x\|_{2}^{2})^{2}}dx$
$\displaystyle=\frac{1}{N^{r}}\sum_{l=1}^{\lfloor
cN\rfloor}\int_{0}^{cN}\int_{S_{r-2}}\frac{l^{4}R^{r-2}}{(l^{2}+R^{2})^{2}}d\sigma_{r-2}(u)dR$

where we changed to polar coordinates in the last equality; here, $S_{r-2}$ is
the unit sphere in $\mathbb{R}^{r-1}$, and $\sigma_{r-2}$ is a measure on
$S_{r-2}$ such that, if $A\subseteq S_{r-2}$ is a Borel set and $\tilde{A}$ is
the set of all points $ru$ with $0<r<1$ and $u\in A$, then
$\sigma_{r-2}(A)=(r-1)m_{r-1}(\tilde{A})$, where $m_{r-1}$ is the Lebesgue
measure on $\mathbb{R}^{r-1}$ (see Exercise 6, Chapter 8 of Rudin (1974)). We
continue:

$\displaystyle\|s_{\bm{1}}^{(1)}\|_{2}^{2}$
$\displaystyle\gtrsim\frac{1}{N^{r}}\sum_{l=1}^{\lfloor
cN\rfloor}\int_{0}^{cN}\frac{l^{4}R^{r-2}}{(l^{2}+R^{2})^{2}}dR$
$\displaystyle\geq\frac{1}{N^{r}}\sum_{l=1}^{\lfloor
cN\rfloor}\int_{cN/2}^{cN}\frac{l^{4}R^{r-2}}{(l^{2}+R^{2})^{2}}dR$
$\displaystyle\gtrsim\frac{1}{N^{3}}\sum_{l=1}^{\lfloor
cN\rfloor}\int_{cN/2}^{cN}\frac{l^{4}R}{(l^{2}+R^{2})^{2}}dR$

where we used the fact that $r\geq 3$. Note that
$\int_{a}^{b}\frac{R}{(l^{2}+R^{2})^{2}}dR=\frac{b^{2}-a^{2}}{2(b^{2}+l^{2})(a^{2}+l^{2})}$
and hence

$\displaystyle\|s_{\bm{1}}^{(1)}\|_{2}^{2}$
$\displaystyle\gtrsim\frac{1}{N^{3}}\sum_{l=1}^{\lfloor
cN\rfloor}\frac{l^{4}N^{2}}{(l^{2}+N^{2})^{2}}\geq\frac{1}{N}\int_{0}^{N/10}\frac{l^{2}}{(l^{2}+N^{2})^{2}}dl$

where we used the fact that $\frac{l^{2}}{(l^{2}+N^{2})^{2}}$ is increasing in
$l$. Since

$\int_{0}^{N/10}\frac{l^{2}}{(l^{2}+N^{2})^{2}}dl=\frac{151-1515\arctan(1/10)}{1010}N$

and $151-1515\arctan(1/10)>0$, the proof is complete. ∎

##  Appendix B The interior point method on the dual objective

For the special case where the design matrix is the identity and
$\lambda_{2}=0$, Kim et al. (2009) applies the interior point method on the
dual objective. Similarly, we can apply interior point method to solve our
more general dual objective

$\hat{\beta}:=\arg\min_{\beta\in\mathbb{R}^{p}}\frac{1}{2}\|\tilde{Y}-\tilde{X}\beta\|_{2}^{2}+\lambda_{1}\|\Gamma\beta\|_{1}$
(63)

We will specify the update directions, update step size, and measure of
suboptimality. Let $\bm{1}_{m}$ denote the vector in $\mathbb{R}^{m}$ with all
entries equal to 1. The dual problem only has inequality constraints
$f_{1}(u)=u-\lambda_{1}\bm{1}_{m}\leq 0$ and
$f_{2}(u)=-u-\lambda_{1}\bm{1}_{m}\leq 0$. Let $\mu_{1},\mu_{2}$ be the dual
variables corresponding to $f_{1},f_{2}$. We apply the standard Newton’s
updates on the perturbed KKT conditions (by parameter $t$) of this dual
problem. That is, the directions of the updates $(\Delta
u,\Delta\mu_{1},\Delta\mu_{2})$ are solutions of the following linear system:

The residuals are:

$\begin{split}r_{t}(u,\mu_{1},\mu_{2})=\begin{bmatrix}\check{\Gamma}\check{\Gamma}^{T}u-\check{\Gamma}\check{Y}+\mu_{1}-\mu_{2}\\\
-\mathrm{diag}(\mu_{1})f_{1}(u)-\frac{1}{t}\bm{1}_{m}\\\
-\mathrm{diag}(\mu_{2})f_{2}(u)-\frac{1}{t}\bm{1}_{m}\end{bmatrix}\end{split}$
(64)

By Newton’s method, we need to solve:

$\nabla r_{t}(u,\mu_{1},\mu_{2})\begin{bmatrix}\Delta u\\\ \Delta\mu_{1}\\\
\Delta\mu_{2}\end{bmatrix}=-r_{t}(u,\mu_{1},\mu_{2})$ (65)

That simplifies to 3 linear equations below, where divisions between vectors
are element-wise:

$\left[\check{\Gamma}\check{\Gamma}^{T}-\mathrm{diag}(\mu_{1}/f_{1}(u))-\mathrm{diag}(\mu_{2}/f_{2}(u))\right]\Delta
u=-\left[\check{\Gamma}\check{\Gamma}^{T}u-\check{\Gamma}\check{Y}-\frac{\bm{1}_{m}}{tf_{1}(u)}+\frac{\bm{1}_{m}}{tf_{2}(u)}\right]$
(66)

$\Delta\mu_{1}=-\left[\mathrm{diag}(\mu_{1}/f_{1}(u))\Delta
u+\mu_{1}+\frac{\bm{1}_{m}}{tf_{1}(u)}\right]$ (67)

$\Delta\mu_{2}=-\left[-\mathrm{diag}(\mu_{2}/f_{2}(u))\Delta
u+\mu_{2}+\frac{\bm{1}_{m}}{tf_{2}(u)}\right]$ (68)

The step size for each update are computed in a standard way as Section 11.7.3
of Boyd et al. (2004). We apply standard backtracking line search to find the
step size $s$ for the updates. Choose parameters $\alpha,\gamma\in(0,1)$ for
backtracking. Denote the updates as $(u^{+},\mu_{1}^{+},\mu_{2}^{+})$. For
example, $u^{+}=u+s\Delta u$. To ensure the updates to be feasible, we first
make sure that $\mu_{1}^{+},\mu_{2}^{+}\geq 0$. That is, we set
$s_{\max}=\min\\{1,\min\\{-\mu_{1i}/\Delta\mu_{1i}\>|\>\Delta\mu_{1i}<0\\},\min\\{-\mu_{2i}/\Delta\mu_{2i}\>|\>\Delta\mu_{2i}<0\\}\\}$.
Next, continuously set $s=\gamma s$ until $f_{1}(u^{+}),f_{2}(u^{+})<0$.
Finally, set $s=\gamma s$ until
$\|r_{t}(u^{+},\mu_{1}^{+},\mu_{2}^{+})\|_{2}\leq(1-\alpha
s)\|r_{t}(u,\mu_{1},\mu_{2})\|_{2}$.

As a standard measure of suboptimality, the surrogate duality gap (see Section
11.7.2 of Boyd et al. (2004) for details) at the $k^{\text{th}}$ iteration is:

$\eta^{(k)}=-f_{1}(u^{(k)})^{T}\mu^{(k)}_{1}-f_{2}(u^{(k)})^{T}\mu^{(k)}_{2}$
(69)

And the residual at the $k^{\text{th}}$ iteration is:

$r^{(k)}=r_{t}(u^{(k)},\mu_{1}^{(k)},\mu_{2}^{(k)})$ (70)

Our interior point algorithm is presented below.

Input: $\lambda_{1},\lambda_{2},\Gamma,Y,X$, tolerance $\epsilon$

Output: $\hat{\beta}$ as defined in (63)

1 Initialize $u^{(0)}=0$, $\mu_{1}^{(0)},\mu_{2}^{(0)}>0$, $\tau>1$

2 while  _$r^{(k)} >\epsilon$ or $\eta^{(k)}>\epsilon$_ do

3 Set $t=2\tau m/\eta^{(k)}$

4 Compute update direction $(\Delta u,\Delta\mu_{1},\Delta\mu_{2})$ as in
(66), (67), (68)

5 Determine step size $s$ by using $\alpha,\gamma$ backtracking line search

6 Update:  $\displaystyle u^{(k+1)}$ $\displaystyle=u^{(k)}+s\Delta u$
$\displaystyle\mu_{1}^{(k+1)}$ $\displaystyle=\mu_{1}^{(k)}+s\Delta\mu_{1}$
$\displaystyle\mu_{2}^{(k+1)}$ $\displaystyle=\mu_{2}^{(k)}+s\Delta\mu_{2}$

7Compute
$\hat{\beta}\leftarrow\tilde{X}^{\dagger}(\check{Y}-\check{\Gamma}^{T}u)$

8 Return $\hat{\beta}$

Algorithm 2 Interior point method on the dual objective

##  Appendix C Additional details on data processing

Chicago Crime Data. As per the main text, statistics on the number of crimes
per community between 2001 and 2022 are available on the city’s data portal.
For the purpose of our analysis, we consider the data between 2004 and 2022,
since by preliminary inspection of the data, the first years of collection
seem to have more missing data (see Figure 15(a)). We define the monthly crime
rates as the number of crimes per 100,000 inhabitants. The latter are computed
from the raw crime data by aggregating crime counts over neighborhoods and
dividing by neighborhood population estimates found at the following  link.
These crime rates are usually modeled by Poisson distributions Osgood (2000),
which we transform here into a normal distribution through the use of an
Anscombe transform. Examples of the resulting estimates are displayed on
Figure 15(b). We note that the crime rates vary substantially over the years
and across the communities, and are also subject to significant seasonal
effects.

(a) Temporal evolution of the Anscombe transformed crime rate (per 100,000) as
a function of time across 6 specific neighborhoods.

(b) Crime rate across neighborhoods, coloured by month of the year to
highlight seasonal effects.

Figure 15: Anscombe transform of the number of crimes per month per 100,000
inhabitants in a few neighborhoods of Chicago. Note the seasonal effect in the
crime rate and the consistent drop across neighborhoods during the coldest
months of the year.

COVID Data. We consider the problem of predicting the number of COVID-19 cases
14 days in advance for a given county in California. As decribed in the main
text, this could be an interesting use case for local public health decisions,
such as for instance, trying to plan 2 weeks in advance appropriate resources
at a local clinic. To this end, we used the New York Times-curated COVID
database. The NYT COVID data provides a description of the total number of
cases across all US counties, from January 2020 to October 2022 (time of
writing). For the purpose of our analysis, we focus more specifically on
analyzing new cases in the 25 densest California counties using data from June
$1^{\text{st}}$, 2020 to July $1^{\text{st}}$, 2021. This time window was
selected to provide more consistency in the epidemics dynamics: by June 2020,
all counties in California had non zero daily incidence data. On the other
hand, restricting the analysis to before July 2021 allows selecting a more
cohesive window of time where the epidemic propagation was not dominated by
(other unobserved) covariates, such as the advent of new contagious strains of
the virus (Delta in Summer 2021, and subsequently Omicron in Winter 2022). We
pre-process the data and make it amenable to data analysis through the
following steps:

  1. 1.

Conversion of cumulative case counts to incidence data

  2. 2.

Correction of aberrations and smoothing: we fix data aberrations (e.g.
negative incidences, due to small errors in the reporting) by imposing the
lower bound on the number of new cases to be 0. We further transform the
incidence data using a seven-day rolling average so as to get rid of known
spurious phenomena (e.g. the “weekend effect”, by which the number of new
cases is lower over the weekend but typically followed by a spike on the
following Monday).

  3. 3.

Anscombe transform. We apply a variance stabilizing transform to transform
incidence data (here modeled as a Poisson process, as per Agosto and Giudici
(2020); Bu et al. (2021); Cori et al. (2013); Toharudin et al. (2020)):
$\tilde{x}\leftarrow 2\sqrt{x+\frac{3}{8}}$.

Figure 16: $R^{2}$ for the simple autoregressive model of Equation 49 on the
seven different folds (see main text). Note that most models have $R^{2}$ of
over 0.8, thus indicating the validity of the model.