# Extended Blumberg-Dieckmann Series

Robert Reynolds  Department of Mathematics and Statistics, York University,
Toronto, ON, Canada, M3J1P3<EMAIL_ADDRESS>
###### Abstract.

This paper introduces a set of finite summation formulas and utilize them to
establish various functional relationships involving the multivariable
Hurwitz-Lerch zeta function. Additionally, the paper examines several examples
of these functional relationships.

###### Key words and phrases:

Product, sum, special functions, contour integral, Catalan’s constant, Apéry’s
constant, definite integral

###### 2020 Mathematics Subject Classification:

Primary 30E20, 33-01, 33-03, 33-04

##  1\. Introduction

The finite sum of Hurwitz-Lerch zeta functions in the context of an infinite
array of line charges has been studied in the field of electrostatics and
mathematical physics [1, 2]. This specific topic falls under the broader area
of potential theory and the calculation of electrostatic potentials generated
by distributions of charges. The study of infinite arrays of line charges
involves analyzing the behavior of electric potentials and fields resulting
from an infinite sequence of evenly spaced line charges [3]. The Hurwitz-Lerch
zeta function, which is a generalization of the Riemann zeta function, appears
in the mathematical representation of these potentials. Researchers have
investigated various properties and applications of the finite sum of Hurwitz-
Lerch zeta functions in this context. They have explored convergence
properties, section (25.14) in [4], evaluated the sums for specific
parameters, derived analytical expressions or approximations, section (25.20)
in [4], and examined the behavior of the potentials as the number of charges
or their separation approaches infinity. These studies are important for
understanding the behavior of electrostatic fields in periodic systems, such
as crystal lattices or periodic structures in condensed matter physics [5].
They provide insights into the potential distribution and energy calculations
in these systems, aiding in the analysis and design of devices and materials.
Specific research papers and academic publications on the finite sum of Lerch
functions in the context of infinite arrays of line charges can be found by
searching scholarly databases or consulting textbooks and research papers in
the field of mathematical physics and electrostatics.

This paper introduces a table (1) of formulas similar to a table of integral
formulas [7]. The selection of formulas for this table was influenced by their
relevance to number theory [6]. A finite sum formula represents the sum of a
finite series, and the evaluation of such a sum is known as finite summation.
To achieve this, general sum formulas are developed and applied, providing
closed-form expressions that yield the exact or approximate sum depending on
the series type. The paper begins with the presentation of finite summation
theory, followed by the development of both general and special finite sum
formulas. Finally, the table itself is presented. In the context of studying
an infinite array of line charges [2], functions involving products of
trigonometric functions spaced at regular intervals arise, which are functions
of the complex variable $z=x+iy$. It has been demonstrated that these
functions can be simplified to a constant multiplied by a multiple angle
function. In this work, we apply the contour integral method from [8], to the
finite sum of the secant function given in [9, 2], resulting in

(1.1) $\frac{1}{2\pi
i}\int_{C}\sum_{j=0}^{2n}(-1)^{j}a^{w}w^{-k-1}\sec\left(\frac{\pi
j}{2n+1}+m+w\right)dw\\\ =\frac{1}{2\pi
i}\int_{C}(-1)^{n}(2n+1)a^{w}w^{-k-1}\sec((2n+1)(m+w))dw$

where $a,m,k\in\mathbb{C},Re(m+w)>0,n\in\mathbb{Z^{+}}$. Using equation (1.1)
the main Theorem to be derived and evaluated is given by

(1.2) $\sum_{j=0}^{2n}(-1)^{j}e^{i\left(\frac{\pi
j}{2n+1}+m\right)}\Phi\left(-e^{2i\left(\frac{\pi
j}{2n+1}+m\right)},-k,\frac{1}{2}(1-i\log(a))\right)\\\
=i^{-k}(-1)^{n}(2n+1)(i(2n+1))^{k}e^{im(2n+1)}\Phi\left(-e^{2im(2n+1)},-k,\frac{2n-i\log(a)+1}{4n+2}\right)$

where the variables $k,a,m$ are general complex numbers and $n$ is any
positive integer. This new expression is then used to derive special cases in
terms of trigonometric functions. The derivations follow the method used by us
in [8]. This method involves using a form of the generalized Cauchy’s integral
formula given by

(1.3) $\frac{y^{k}}{\Gamma(k+1)}=\frac{1}{2\pi
i}\int_{C}\frac{e^{wy}}{w^{k+1}}dw,$

where $y,w\in\mathbb{C}$ and $C$ is in general an open contour in the complex
plane where the bilinear concomitant [8] is equal to zero at the end points of
the contour. This method involves using a form of equation (1.3) then
multiplies both sides by a function, then takes the finite sum of both sides.
This yields a finite sum in terms of a contour integral. Then we multiply both
sides of equation (1.3) by another function and take the infinite sum of both
sides such that the contour integral of both equations are the same.

##  2\. The Hurwitz-Lerch zeta Function

We use equation (1.11.3) in [10] where $\Phi(z,s,v)$ is the Hurwitz-Lerch zeta
function which is a generalization of the Hurwitz zeta $\zeta(s,v)$ and
Polylogarithm function $\text{Li}_{n}(z)$. The Lerch function has a series
representation given by

(2.1) $\Phi(z,s,v)=\sum_{n=0}^{\infty}(v+n)^{-s}z^{n}$

where $|z|<1,v\neq 0,-1,-2,-3,..,$ and is continued analytically by its
integral representation given by

(2.2)
$\Phi(z,s,v)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{t^{s-1}e^{-vt}}{1-ze^{-t}}dt=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{t^{s-1}e^{-(v-1)t}}{e^{t}-z}dt$

where $Re(v)>0$, and either $|z|\leq 1,z\neq 1,Re(s)>0$, or $z=1,Re(s)>1$.

##  3\. Contour Integral Representation for the Finite Sum of the Hurwitz-
Lerch zeta Functions

In this section we derive the contour integral representations of the left-
hand side and right-hand side of equation (1.1) in terms of the Hurwtiz-Lerch
zeta and trigonometric functions.

###  3.1. Derivation of the left-hand side contour integral

We use the method in [8]. Using equation (1.3) we first replace
$\log(a)+i(2y+1)$ and multiply both sides by
$2(-1)^{y}e^{i(2y+1)\left(\frac{\pi j}{2n+1}+m\right)}$ then take the finite
and infinite sums over $j\in[0,2n]$ and $y\in[0,\infty)$ and simplify in terms
of the Hurwitz-Lerch zeta function to get

(3.1) $\sum_{j=0}^{2n}\frac{(-1)^{j}i^{k}2^{k+1}e^{i\left(\frac{\pi
j}{2n+1}+m\right)}}{k!}\Phi\left(-e^{2i\left(\frac{\pi
j}{2n+1}+m\right)},-k,\frac{1}{2}(1-i\log(a))\right)\\\ \ =\frac{1}{2\pi
i}\sum_{y=0}^{\infty}\sum_{j=0}^{2n}\int_{C}(-1)^{y}a^{w}e^{ib(2y+1)(m+w)}dw\\\
\\\ =\frac{1}{2\pi
i}\int_{C}\sum_{j=0}^{2n}\sum_{y=0}^{\infty}(-1)^{y}a^{w}e^{ib(2y+1)(m+w)}dw\\\
\\\ =\frac{1}{2\pi
i}\int_{C}\sum_{j=0}^{2n}(-1)^{j}a^{w}w^{-k-1}\sec\left(\frac{\pi
j}{2n+1}+m+w\right)dw$

from equation (1.232.2) in [11] where $Re(w+m)>0$ and $Im\left(m+w\right)>0$
in order for the sums to converge. We apply Tonelli’s theorem for multiple
sums, see page 177 in [12] as the summands are of bounded measure over the
space $\mathbb{C}\times[0,2n]\times[0,\infty)$.

###  3.2. Derivation of the right-hand side contour integral

We use the method in [8]. Using equation (1.3) we first replace
$\log(a)+i(2n+1)(2y+1)$ and multiply both sides by
$2(-1)^{y}e^{im(2n+1)(2y+1)}$ then take the finite and infinite sums over
$y\in[0,\infty)$ and simplify in terms of the Hurwitz-Lerch zeta function to
get

(3.2)
$\frac{2^{k+1}(-1)^{n}(2n+1)(i(2n+1))^{k}e^{im(2n+1)}}{k!}\Phi\left(-e^{2im(2n+1)},-k,\frac{2n-i\log(a)+1}{2(2n+1)}\right)\\\
\\\ =\frac{1}{2\pi
i}\sum_{y=0}^{\infty}\int_{C}2(-1)^{y}a^{w}e^{i(2n+1)(2y+1)(m+w)}dw\\\ \\\
=\frac{1}{2\pi
i}\int_{C}\sum_{y=0}^{\infty}2(-1)^{y}a^{w}e^{i(2n+1)(2y+1)(m+w)}dw\\\ \\\
=\frac{1}{2\pi i}\int_{C}(-1)^{n}(2n+1)a^{w}w^{-k-1}\sec((2n+1)(m+w))dw$

from equation (1.232.2) in [11] where $Re(w+m)>0$ and $Im\left(m+w\right)>0$
in order for the sums to converge. We apply Tonelli’s theorem for multiple
sums, see page 177 in [12] as the summands are of bounded measure over the
space $\mathbb{C}\times[0,\infty)$.

##  4\. A functional equation for the Hurwitz-Lerch zeta function and its
evaluation

In this section we will derive and evaluate formulae involving the sum and
product of the Hurwitz-Lerch zeta function in terms other Special functions
and fundamaental constants.

######  Theorem 4.1.

For all $k,a,m\in\mathbb{C}$ then,

(4.1) $\sum_{j=0}^{2n}(-1)^{j}e^{i\left(\frac{\pi
j}{2n+1}+m\right)}\Phi\left(-e^{2i\left(\frac{\pi
j}{2n+1}+m\right)},-k,\frac{1}{2}(1-i\log(a))\right)\\\
=i^{-k}(-1)^{n}(2n+1)(i(2n+1))^{k}e^{im(2n+1)}\Phi\left(-e^{2im(2n+1)},-k,\frac{2n-i\log(a)+1}{4n+2}\right)$

###### Proof.

With respect to equation (1.1) and observing the addition of the right-hand
sides of relation (3.1), and relation (3.2) are identical; hence, the left-
hand sides of the same are identical too. Simplifying with the Gamma function
yields the desired conclusion. ∎

######  Example 4.2.

An alternate form: A functional equation for the Hurwitz-Lerch zeta function.

(4.2)
$\sum_{j=0}^{2n}(-1)^{j\left(\frac{1}{2n+1}+1\right)}\Phi\left(-e^{i\left(\frac{2\pi
j}{2n+1}+m\right)},-k,a\right)\\\
=(-1)^{n}(2n+1)^{k+1}e^{imn}\Phi\left(-e^{im(2n+1)},-k,\frac{a+n}{2n+1}\right)$

###### Proof.

Use equation (4.1) and rewrite using $a=e^{ai}$ and simplify. ∎

######  Example 4.3.

The Degenerate Case.

(4.3) $\sum_{j=0}^{2n}(-1)^{j}\sec\left(\frac{\pi
j}{2n+1}+m\right)=(-1)^{n}(2n+1)\sec(2mn+m)$

###### Proof.

Use equation (4.1) and set $k=0$ and simplify using entry (2) in Table below
(64:12:7) in [13]. ∎

######  Example 4.4.

Hurwitz-Lerch zeta functional equation.

(4.4) $\Phi(z,s,a)\\\
=3^{1-s}z\Phi\left(z^{3},s,\frac{a+1}{3}\right)-(-1)^{2/3}\Phi\left(-\sqrt[3]{-1}z,s,a\right)+\sqrt[3]{-1}\Phi\left((-1)^{2/3}z,s,a\right)$

###### Proof.

Use equation (4.1) and set $n=1,m=-\frac{1}{2}i\log(-z),a=e^{i(2a-1)},k=-s$
and simplify. ∎

######  Example 4.5.

Hurwitz-Lerch zeta functional equation.

(4.5) $\Phi(-z,s,a)\\\
=5^{1-s}z^{2}\Phi\left(-z^{5},s,\frac{a+2}{5}\right)+(-1)^{3/5}\Phi\left(\sqrt[5]{-1}z,s,a\right)+\sqrt[5]{-1}\Phi\left(-(-1)^{2/5}z,s,a\right)\\\
-(-1)^{2/5}\left((-1)^{2/5}\Phi\left((-1)^{3/5}z,s,a\right)+\Phi\left(-(-1)^{4/5}z,s,a\right)\right)$

###### Proof.

Use equation (4.1) and set $n=2,m=\frac{1}{2}i\log(z),a=e^{i(2a-1)},k=-s$ and
simplify. ∎

######  Example 4.6.

Special case of the first partial derivative of the Hurwitz-Lerch zeta
function in terms of the log-gamma function.

(4.6)
$\sqrt[3]{-1}\Phi^{\prime}\left((-1)^{2/3},0,a\right)-(-1)^{2/3}\Phi^{\prime}\left(-\sqrt[3]{-1},0,a\right)=\log\left(\frac{2\pi
3^{\frac{1}{2}-a}\Gamma(a)}{\Gamma\left(\frac{a+1}{3}\right)^{3}}\right)$

###### Proof.

Use equation (4.4) set $z=1$ and take the first partial derivative with
respect to $s$ and set $s=0$ and simplify using equation (25.11.18) in [4]. ∎

######  Example 4.7.

Special case of the first partial derivative of the Hurwitz-Lerch zeta
function in terms of the log-gamma function.

(4.7)
$\sqrt[3]{-1}\Phi^{\prime}\left(-(-1)^{2/3},0,a\right)-(-1)^{2/3}\Phi^{\prime}\left(\sqrt[3]{-1},0,a\right)\\\
=3\log\left(\frac{\sqrt{3}\sqrt[3]{\frac{\Gamma\left(\frac{a-2}{2}\right)}{\Gamma\left(\frac{a-1}{2}\right)}}\Gamma\left(\frac{a+1}{6}\right)}{\left((a-2)\sqrt{a-1}\right)^{2/3}\Gamma\left(\frac{a-2}{6}\right)}\right)+\log(2)$

###### Proof.

Use equation (4.4) set $z=-1$ and take the first partial derivative with
respect to $s$ and set $s=0$ and simplify using equation (25.11.18) in [4]. ∎

######  Example 4.8.

Finite product of the cosine and sine functions in terms of the ratio of the
sine and tangent functions.

(4.8)
$\prod_{j=0}^{2n}\left(\frac{\cos\left(\frac{1}{2}\left(m+\frac{2j\pi}{1+2n}+r\right)\right)+\sin\left(\frac{m-r}{2}\right)}{\cos\left(\frac{m+2mn+2j\pi+r+2nr}{2+4n}\right)-\sin\left(\frac{m-r}{2}\right)}\right)^{(-1)^{j}}\\\
=\left(-\frac{\sin\left(\frac{1}{4}(m(2+4n)+\pi)\right)}{\tan\left(\frac{1}{4}(\pi+2r+4nr)\right)\sin\left(m\left(\frac{1}{2}+n\right)-\frac{\pi}{4}\right)}\right)^{(-1)^{n}}$

###### Proof.

Use equation (4.1) and form a second equation by replacing $m\to r$ take the
difference of both these equations then set $k=-1,a=1$, take the exponential
function of both sides and simplify using entry (3) of Section (64:12) in
[13]. ∎

######  Example 4.9.

The product of the ratio of trigonometric and exponential of trigonometric
functions.

(4.9)
$\prod_{j=0}^{2n}\exp\left(-2(-1)^{j}\sec\left(\frac{j\pi}{1+2n}+x\right)\sec\left(\frac{j\pi}{1+2n}+\frac{x}{b}\right)\right.\\\
\left.\sin\left(\frac{(-1+b)x}{2b}\right)\sin\left(\frac{1}{2}\left(\frac{2j\pi}{1+2n}+x+\frac{x}{b}\right)\right)\right)\\\
\left(\frac{\left(-1+\sin\left(\frac{j\pi}{1+2n}+x\right)\right)\left(1+\sin\left(\frac{j\pi}{1+2n}+\frac{x}{b}\right)\right)}{\left(1+\sin\left(\frac{j\pi}{1+2n}+x\right)\right)\left(-1+\sin\left(\frac{j\pi}{1+2n}+\frac{x}{b}\right)\right)}\right)^{\frac{(-1)^{j}}{2}}\\\
=\exp\left(\frac{2(-1)^{n}(1+2n)\left(\cos(x+2nx)-\cos\left(\frac{x+2nx}{b}\right)\right)}{\cos\left(\frac{(-1+b)(x+2nx)}{b}\right)+\cos\left(\frac{(1+b)(x+2nx)}{b}\right)}\right)\\\
\left(\frac{(-1+\sin(x+2nx))\left(1+\sin\left(\frac{x+2nx}{b}\right)\right)}{(1+\sin(x+2nx))\left(-1+\sin\left(\frac{x+2nx}{b}\right)\right)}\right)^{\frac{(-1)^{n}}{2}}$

###### Proof.

Use equation (4.1) and set $k=1,a=e,m=x$ and simplify using the method in
section (8.1) in [14]. ∎

######  Example 4.10.

Finite product of the ratio of the sine function.

(4.10) $\prod_{j=0}^{2n}\left(\frac{2}{\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{4}\right)+1}-1\right)^{(-1)^{j}}\left(\frac{\left(\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)+1\right)^{3}\left(\sin\left(\frac{\pi
j}{2n+1}+x\right)-1\right)}{\left(\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)-1\right)^{3}\left(\sin\left(\frac{\pi
j}{2n+1}+x\right)+1\right)}\right)^{\frac{(-1)^{j}}{2}}\\\
=\left(\frac{2}{\sin\left(\frac{1}{4}(2nx+x)\right)+1}-1\right)^{(-1)^{n}}\left(\frac{\left(\sin\left(\left(n+\frac{1}{2}\right)x\right)+1\right)^{3}(\sin(2nx+x)-1)}{\left(\sin\left(\left(n+\frac{1}{2}\right)x\right)-1\right)^{3}(\sin(2nx+x)+1)}\right)^{\frac{(-1)^{n}}{2}}$

###### Proof.

Use equation (4.1) and set $k=1,a=1,m=x$ and simplify using the method in
section (8.1) in [14]. ∎

######  Example 4.11.

The finite product of the ratio of trigonometric and exponential of
trigonometric functions.

(4.11) $\prod_{j=0}^{2n}\left(\frac{\left(\sin\left(\frac{\pi
j}{2n+1}-x\right)-1\right)\left(\sin\left(\frac{\pi
j}{2n+1}-\frac{x}{2}\right)+1\right)}{\left(\sin\left(\frac{\pi
j}{2n+1}-x\right)+1\right)\left(\sin\left(\frac{\pi
j}{2n+1}-\frac{x}{2}\right)-1\right)}\right)^{\frac{1}{2}i\pi(-1)^{j}}\\\
\left(\frac{\left(\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)+1\right)\left(\sin\left(\frac{\pi
j}{2n+1}+x\right)-1\right)}{\left(\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)-1\right)\left(\sin\left(\frac{\pi
j}{2n+1}+x\right)+1\right)}\right)^{\frac{1}{2}i\pi(-1)^{j}}\\\
\exp\left((-1)^{j}\left(-\sec\left(\frac{\pi
j}{2n+1}-x\right)\right.\right.\\\ \left.\left.+\sec\left(\frac{\pi
j}{2n+1}-\frac{x}{2}\right)+\sec\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)-\sec\left(\frac{\pi
j}{2n+1}+x\right)\right)\right)\\\
=\left(\frac{\left(\sin\left(\left(n+\frac{1}{2}\right)x\right)+1\right)(\sin(2nx+x)-1)}{\left(\sin\left(\left(n+\frac{1}{2}\right)x\right)-1\right)(\sin(2nx+x)+1)}\right)^{\frac{1}{2}i\pi(-1)^{n}}\\\
\left(\frac{\left(\sin\left(\left(n+\frac{1}{2}\right)x\right)-1\right)(\sin(2nx+x)+1)}{\left(\sin\left(\left(n+\frac{1}{2}\right)x\right)+1\right)(\sin(2nx+x)-1)}\right)^{\frac{1}{2}i\pi(-1)^{n}}\\\
\exp\left(2(-1)^{n}(2n+1)\left(\cos\left(\left(n+\frac{1}{2}\right)x\right)-1\right)\right.\\\
\left.\left(\sec\left(\left(n+\frac{1}{2}\right)x\right)+2\right)\sec(2nx+x)\right)$

###### Proof.

Use equation (4.1) and set $k=1,a=-1,m=x$ and simplify using the method in
section (8.1) in [14]. ∎

######  Example 4.12.

Finite product involving the ratio of the Gamma function.

(4.12) $\prod_{j=0}^{2n}\exp\left(\frac{(-1)^{j-n}e^{\frac{i\pi
j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},0,x\right)}{2n+1}\right)=\frac{1}{\sqrt{2(1+2n)}}\frac{\Gamma\left(\frac{n+x}{2+4n}\right)}{\Gamma\left(\frac{1+3n+x}{2+4n}\right)}$

###### Proof.

Use equation (4.1) and set $m=0$ then take the first partial derivative with
respect to $k$ and set $k=0,a=e^{ai}$ and simplify using equation (25.11.18)
in [4]. Then take the exponential function of both sides and simplify. ∎

######  Example 4.13.

The first partial derivative of the Hurwitz-Lerch zeta function in terms of
Euler’s constant $\gamma$ and $\pi$.

(4.13) $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi
j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi j}{2n+1}},1,\frac{1}{2}\right)\\\
=\frac{1}{4}\pi(-1)^{n}\left(-2\log(i(2n+1))+i\pi+2\gamma+\log\left(\frac{64\pi^{6}}{\Gamma\left(\frac{1}{4}\right)^{8}}\right)\right)$

###### Proof.

Use equation (4.1) and take the first partial derivative with respect to $k$
and set $m=0,k=-1,a=1$ and simplify. ∎

######  Example 4.14.

The first partial derivative of the Hurwtiz-Lerch zeta function in terms of
Catalan’s constant $K$ and $\pi$.

(4.14) $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi
j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},-1,\frac{1}{2}\right)=\frac{K}{\pi}(-1)^{n}(2n+1)^{2}$

###### Proof.

Use equation (4.1) and take the first partial derivative with respect to $k$
and set $m=0,k=1,a=1$ and simplify using equation (10) in [15] and equation
(2.2.1.2.7) in [16] ∎

######  Example 4.15.

The first partial derivative of the Hurwtiz-Lerch zeta function in terms of
the Glaisher-Kinkelin constant $A$.

(4.15) $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi
j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},-1,n+1\right)=\frac{1}{8}(-1)^{n}(2n+1)^{2}\log\left(\frac{A^{24}}{4\
2^{2/3}e^{2}(2n+1)^{2}}\right)$

###### Proof.

Use equation (4.1) and take the first partial derivative with respect to $k$
and set $m=0,k=1,a=1$ and simplify using equation (8) in [15] ∎

######  Example 4.16.

The first partial derivative of the Hurwtiz-Lerch zeta function in terms of
zeta(3) is Apéry’s constant $\zeta(3)$.

(4.16) $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi
j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},-2,n+1\right)=\frac{7(-1)^{n}(2n+1)^{3}}{4\pi^{2}}\zeta(3)$

###### Proof.

Use equation (4.1) and take the first partial derivative with respect to $k$
and set $m=0,k=1,a=1$ and simplify using equation (9) in [15] ∎

######  Example 4.17.

Special case of the Hurwitz-Lerch zeta function in terms of the Polylogarithm
function.

(4.17) $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi
j}{2n+1}}\Phi\left(-e^{2i\left(\frac{\pi j}{2n+1}+m\right)},k,n+1\right)\\\
=-(-1)^{n}(2n+1)^{1-k}e^{-2im(n+1)}\text{Li}_{k}\left(-e^{2im(2n+1)}\right)$

###### Proof.

Use equation (4.1) and set $k\to-k,a=e^{i(2n+1)}$ and simplify using equation
(25.14.3) in [4]. ∎

######  Example 4.18.

The first partial derivative of the Polylogarithm function in terms of $\pi$.

(4.18)
$Li_{0}^{\prime}\left(\sqrt[3]{-1}\right)+Li_{0}^{\prime}\left(-(-1)^{2/3}\right)=\log\left(\frac{6\sqrt{3}}{\pi}\right)$

###### Proof.

Use equation (4.17) and set $n=1,m=0$, then take the first partial derivative
with respect to $k$ and set $k=0$ and simplify using equations (25.11.2),
(25.14.2) and (25.14.3) in [4]. ∎

######  Example 4.19.

The first partial derivative of the Polylogarithm function in terms of
Glaisher’s constant $A$.

(4.19)
$Li_{-1}^{\prime}\left(\sqrt[3]{-1}\right)+Li_{-1}^{\prime}\left(-(-1)^{2/3}\right)=\log\left(\frac{36\
2^{2/3}\sqrt[4]{3}e^{2}}{A^{24}}\right)$

###### Proof.

Use equation (4.17) and set $n=1,m=0$, then take the first partial derivative
with respect to $k$ and set $k=1$ and simplify using equations (25.11.2),
(25.14.2) and (25.14.3) in [4] and equation (8) in [15]. ∎

######  Example 4.20.

The first partial derivative of the Polylogarithm function in terms of
$\log(2)\log(3)$.

(4.20)
$Li_{1}^{\prime}\left(\sqrt[3]{-1}\right)+Li_{1}^{\prime}\left(1,-(-1)^{2/3}\right)=\log(2)\log(3)$

###### Proof.

Use equation (4.17) and set $n=1,m=0$, then take the first partial derivative
with respect to $k$ and take the limit as $k\to-1$ and simplify using
equations (25.11.2), (25.14.2) and (25.14.3) in [4] and equation (1) in [15]
and the book by Lewin [16]. ∎

######  Example 4.21.

The first partial derivative of the Hurwitz-Lerch zeta function in terms of
the log-gamma function.

(4.21)
$(-1)^{2/3}\Phi^{\prime}\left(\sqrt[3]{-1},0,a\right)-\sqrt[3]{-1}\Phi^{\prime}\left(-(-1)^{2/3},0,a\right)=\log\left(\frac{12\sqrt{3}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{a+4}{6}\right)^{3}}{\Gamma\left(\frac{a}{2}\right)\Gamma\left(\frac{a+1}{6}\right)^{3}}\right)$

###### Proof.

Use equation (4.1) and set $m=-\frac{\pi}{2n+1},n=1,a=e^{ai}$ and simplify
using equation (25.11.18) in [4]. Then take the first partial derivative with
respect to $k$ and set $k=0$ and simplify. ∎

######  Example 4.22.

The Hurwitz-Lerch zeta function in terms of the digamma function.

(4.22)
$\left(1-i\sqrt{3}\right)\Phi\left(\sqrt[3]{-1},1,a\right)+\left(1+i\sqrt{3}\right)\Phi\left(-(-1)^{2/3},1,a\right)\\\
=-\psi^{(0)}\left(\frac{a}{2}\right)-\psi^{(0)}\left(\frac{a+1}{6}\right)+\psi^{(0)}\left(\frac{a+1}{2}\right)+\psi^{(0)}\left(\frac{a+4}{6}\right)$

###### Proof.

Use equation (4.1) and set $k=-1,m=-\frac{\pi}{2n+1},n=1,a=e^{ai}$ and
simplify using equation (25.11.18) in [4]. ∎

######  Example 4.23.

The finite product of the ratio of trigonometric and exponential of
trigonometric functions.

(4.23) $\prod_{j=0}^{2n}\left(\frac{\left(\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)+1\right)\left(\sin\left(\frac{\pi
j}{2n+1}+x\right)-1\right)}{\left(\sin\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)-1\right)\left(\sin\left(\frac{\pi
j}{2n+1}+x\right)+1\right)}\right)^{(-1)^{j}(2n+1)}\\\
\left(\cos\left(2e^{i\pi j}\left(\sec\left(\frac{\pi
j}{2n+1}+\frac{x}{2}\right)-\sec\left(\frac{\pi
j}{2n+1}+x\right)\right)\right)\right.\\\ \left.-i\sin\left(2e^{i\pi
j}\left(\sec\left(\frac{\pi j}{2n+1}+\frac{x}{2}\right)-\sec\left(\frac{\pi
j}{2n+1}+x\right)\right)\right)\right)\\\
=2^{-2(-1)^{n}(2n+1)}\left(i\sin\left(4e^{i\pi
n}(2n+1)\sin^{2}\left(\frac{1}{4}(2nx+x)\right)\left(\sec\left(\left(n+\frac{1}{2}\right)x\right)+2\right)\right.\right.\\\
\left.\left.\sec(2nx+x)\right)+\cos\left(4e^{i\pi
n}(2n+1)\sin^{2}\left(\frac{1}{4}(2nx+x)\right)\right.\right.\\\
\left.\left.\left(\sec\left(\left(n+\frac{1}{2}\right)x\right)+2\right)\sec(2nx+x)\right)\right)\left(\csc\left(\frac{1}{4}(4nx+2x+\pi)\right)\right.\\\
\left.\sec\left(\frac{1}{4}(2nx+x+\pi)\right)\left(\cos\left(\frac{3}{4}(2nx+x)\right)-\sin\left(\frac{1}{4}(2nx+x)\right)\right)\right)^{2(-1)^{n}(2n+1)}$

###### Proof.

Use equation (4.1) and set $k=1,a=e^{i(2n+1)},m=x$ and simplify using the
method in section (8.1) in [14]. ∎

######  Example 4.24.

The finite sum of the ratio of trigonometric functions.

(4.24)
$\sum_{j=0}^{2n}\frac{e^{ij\left(1+\frac{1}{1+2n}\right)\pi}}{\left(\cos(2m)+\cos\left(\frac{2j\pi}{1+2n}\right)\right)^{2}}\left((1+n)\left(\cos(2m)+\cos\left(\frac{2j\pi}{1+2n}\right)\right)-i\sin\left(\frac{2j\pi}{1+2n}\right)\right)\\\
=\frac{(-1)^{n}(1+2n)^{2}\sin(2m(1+n))}{2\left(\sin(2m)\cos^{2}(m(1+2)n)\right)}$

###### Proof.

Use equation (4.17) and form a second equation by replacing $m$ by $-m$ and
taking their difference. Next set $k=-1$ and simplify. ∎

######  Example 4.25.

Finite summation formulae with the first generalized Stieltjes constant at
complex argument.

(4.25)
$\sum_{j=0}^{2n}(-1)^{j\left(\frac{1}{2n+1}+1\right)}\Phi^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},1,a\right)\\\
=\frac{1}{2}(-1)^{n}\left(-\gamma_{1}\left(\frac{a+n}{4n+2}\right)+\gamma_{1}\left(\frac{a+3n+1}{4n+2}\right)\right.\\\
\left.+\log(4n+2)\left(\psi^{(0)}\left(\frac{a+n}{4n+2}\right)-\psi^{(0)}\left(\frac{a+3n+1}{4n+2}\right)\right)\right)$

###### Proof.

Use equation (4.1) and set $m=0$ and simplify using equation (25.11.18) in
[4]. Next apply l’Hopital’s rule as $k\to-1$ and simplify using equation (3)
in [17]. ∎

######  Example 4.26.

Finite summation formulae with the first and second generalized Stieltjes
constant at complex argument.

(4.26)
$\sum_{j=0}^{2n}(-1)^{j\left(\frac{1}{2n+1}+1\right)}\Phi^{\prime\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},1,a\right)\\\
=\frac{1}{2}(-1)^{n}\left(\gamma_{2}\left(\frac{a+n}{4n+2}\right)-\gamma_{2}\left(\frac{a+3n+1}{4n+2}\right)\right.\\\
\left.+2\log(4n+2)\left(\gamma_{1}\left(\frac{a+n}{4n+2}\right)-\gamma_{1}\left(\frac{a+3n+1}{4n+2}\right)\right)\right.\\\
\left.-\left(\left(\log(2n+1)\log(8n+4)+\log^{2}(2)\right)\left(\psi^{(0)}\left(\frac{a+n}{4n+2}\right)-\psi^{(0)}\left(\frac{a+3n+1}{4n+2}\right)\right)\right)\right)$

###### Proof.

Use equation (4.1) and set $m=0$ and simplify using equation (25.11.18) in
[4]. Next take the second partial derivative with respect to $k$ and set $k=1$
and simplify using equation (3) in [17]. ∎

######  Example 4.27.

Finite product of the exponential of the Polylogarithm function in terms of
the ratio of gamma functions.

(4.27) $\prod_{j=0}^{2n}\exp\left(\frac{(-1)^{j-n}e^{-\frac{i\pi
j}{2n+1}}Li_{0}^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}}\right)}{2n+1}\right)=\sqrt{4n+2}\left(\frac{\Gamma\left(\frac{3n+2}{4n+2}\right)}{\Gamma\left(\frac{n+1}{4n+2}\right)}\right)$

###### Proof.

Use equation (4.1) and set $a=e^{i},m=0$ and simplify. Next take the first
partial derivative with respect to $k$ and set $k=0$ and simplify using
equation (25.11.8) in [4]. Then take the exponential function of both sides
and simplify. ∎

######  Example 4.28.

Finite sum of the logarithm function in terms of the digamma function.

(4.28) $\sum_{j=0}^{2n}(-1)^{j}e^{-\frac{i\pi
j}{2n+1}}\log\left(1+e^{\frac{2i\pi j}{2n+1}}\right)\\\
=-\frac{1}{2}(-1)^{n}\left(\psi^{(0)}\left(\frac{n+1}{4n+2}\right)-\psi^{(0)}\left(\frac{3n+2}{4n+2}\right)\right)$

###### Proof.

Use equation (4.1) and set $a=e^{i},m=0$ and simplify. Next apply l’Hopital’s
rule as $k\to-1$ and simplify. ∎

######  Example 4.29.

Finite sum of the Polylogarithm function in terms of the Hurwitz zeta
function.

(4.29) $\sum_{j=0}^{2n}-1)^{j}e^{-\frac{i\pi
j}{2n+1}}\text{Li}_{-k}\left(-e^{\frac{2ij\pi}{2n+1}}\right)\\\
=i\left(\frac{i}{2}\right)^{-k}(-1)^{n}(i(2n+1))^{k+1}\left(\zeta\left(-k,\frac{n+1}{4n+2}\right)-\zeta\left(-k,\frac{3n+2}{4n+2}\right)\right)$

###### Proof.

Use equation (4.1) and set $a=e^{i},m=0$ and simplify using equation (25.14.2)
in [4]. ∎

######  Example 4.30.

Finite trigonometric Euler form.

(4.30)
$\prod_{j=0}^{2n}\left(1-\frac{2}{1+\sec\left(\frac{j\pi}{1+2n}\right)\sin(m)}\right)^{(-1)^{j}}=-\left(\frac{\sin\left(\frac{m}{2}+mn-\frac{\pi}{4}\right)}{\sin\left(\frac{m}{2}+mn+\frac{\pi}{4}\right)}\right)^{2(-1)^{n}}$

###### Proof.

Use equation (4.1) and form a second equation by replacing $m\to-m$ take the
difference of both these equations then set $k=-1,a=1$, take the exponential
function of both sides and simplify using entry (3) of Section (64:12) in
[13]. Similar forms are given the the work by Chamberland [18]. ∎

######  Example 4.31.

Finite sum in terms of a definite integral.

(4.31) $\prod_{j=0}^{2n}\exp\left(\frac{(-1)^{j-n}e^{\frac{i\pi
j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi
j}{2n+1}},0,x\right)}{2n+1}\right)=\frac{1}{\sqrt{2\pi}\sqrt{2n+1}}\int_{0}^{\infty}\frac{z^{\frac{n+x}{4n+2}-1}}{(z+1)^{\frac{3n+x+1}{4n+2}}}dz$

###### Proof.

Use equation (4.12) and equation (2) in [19], compare the right-hand sides and
simplify to achieve the stated result. ∎

A short table of formulae
---
Feature | Formula
Log-gamma | $\sqrt[3]{-1}\Phi^{\prime}\left((-1)^{2/3},0,a\right)-(-1)^{2/3}\Phi^{\prime}\left(-\sqrt[3]{-1},0,a\right)=\log\left(\frac{2\pi 3^{\frac{1}{2}-a}\Gamma(a)}{\Gamma\left(\frac{a+1}{3}\right)^{3}}\right)$
Catalan’s constant, $K$ | $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi j}{2n+1}},-1,\frac{1}{2}\right)=\frac{K}{\pi}(-1)^{n}(2n+1)^{2}$
Glaisher’s constant, $A$ | $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi j}{2n+1}},-1,n+1\right)=\frac{1}{8}(-1)^{n}(2n+1)^{2}\log\left(\frac{A^{24}}{4\ 2^{2/3}e^{2}(2n+1)^{2}}\right)$
Apéry’s constant, $\zeta(3)$ | $\sum_{j=0}^{2n}(-1)^{j}e^{\frac{i\pi j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi j}{2n+1}},-2,n+1\right)=\frac{7(-1)^{n}(2n+1)^{3}}{4\pi^{2}}\zeta(3)$
Logarithm | $Li_{0}^{\prime}\left(\sqrt[3]{-1}\right)+Li_{0}^{\prime}\left(-(-1)^{2/3}\right)=\log\left(\frac{6\sqrt{3}}{\pi}\right)$
Glaisher’ constant, $A$ | $Li_{-1}^{\prime}\left(\sqrt[3]{-1}\right)+Li_{-1}^{\prime}\left(-(-1)^{2/3}\right)=\log\left(\frac{36\ 2^{2/3}\sqrt[4]{3}e^{2}}{A^{24}}\right)$
Product Logarithm | $Li_{1}^{\prime}\left(\sqrt[3]{-1}\right)+Li_{1}^{\prime}\left(1,-(-1)^{2/3}\right)=\log(2)\log(3)$
Log-gamma | $(-1)^{2/3}\Phi^{\prime}\left(\sqrt[3]{-1},0,a\right)-\sqrt[3]{-1}\Phi^{\prime}\left(-(-1)^{2/3},0,a\right)=\log\left(\frac{12\sqrt{3}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{a+4}{6}\right)^{3}}{\Gamma\left(\frac{a}{2}\right)\Gamma\left(\frac{a+1}{6}\right)^{3}}\right)$
Gamma function | $\prod_{j=0}^{2n}\exp\left(\frac{(-1)^{j-n}e^{-\frac{i\pi j}{2n+1}}Li_{0}^{\prime}\left(-e^{\frac{2i\pi j}{2n+1}}\right)}{2n+1}\right)=\sqrt{4n+2}\left(\frac{\Gamma\left(\frac{3n+2}{4n+2}\right)}{\Gamma\left(\frac{n+1}{4n+2}\right)}\right)$
Euler product form | $\prod_{j=0}^{2n}\left(1-\frac{2}{1+\sec\left(\frac{j\pi}{1+2n}\right)\sin(m)}\right)^{(-1)^{j}}=-\left(\frac{\sin\left(\frac{m}{2}+mn-\frac{\pi}{4}\right)}{\sin\left(\frac{m}{2}+mn+\frac{\pi}{4}\right)}\right)^{2(-1)^{n}}$
Definite integral | $\prod_{j=0}^{2n}\exp\left(\frac{(-1)^{j-n}e^{\frac{i\pi j}{2n+1}}\Phi^{\prime}\left(-e^{\frac{2i\pi j}{2n+1}},0,x\right)}{2n+1}\right)=\frac{1}{\sqrt{2\pi}\sqrt{2n+1}}\int_{0}^{\infty}\frac{z^{\frac{n+x}{4n+2}-1}}{(z+1)^{\frac{3n+x+1}{4n+2}}}dz$

Table 1. Table of functional equations, sum and product formulae.

##  5\. Conclusion

This work offers a method that allows for the derivation of finite and
infinite sums and product identities in terms of the the Hurwitz-Lerch zeta
function and other Special functions. A summary table of some interesting
formulae are summarized in Table (1). For parameters in the derivations
comprising real, imaginary, and complex values, the results were numerically
confirmed using Wolfram’s Mathematica software.

## References

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  * [3] Vigren, Erik, and Andreas Dieckmann. 2020. _”Simple Solutions of Lattice Sums for Electric Fields Due to Infinitely Many Parallel Line Charges”_ , Symmetry 12, no. 6: 1040. https://doi.org/10.3390/sym12061040
  * [4] Olver, F.W.J.; Lozier, D.W.; Boisvert, R.F.; Clark, C.W. (Eds.) _NIST Digital Library of Mathematical Functions_ ; U.S. Department of Commerce, National Institute of Standards and Technology: Washington, DC, USA; Cambridge University Press: Cambridge, UK, 2010; With 1 CD-ROM (Windows, Macintosh and UNIX). MR 2723248 (2012a:33001).
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  * [6] T. M. Apostol (1976) _Introduction to Analytic Number Theory_. Springer-Verlag, New York.
  * [7] Bromwich, Thomas John I’Anson. & MacRobert, T. M. 1926, _An introduction to the theory of infinite series by T.J. I’A. Bromwich_ Macmillan London
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  * [10] Erdéyli, A.; Magnus, W.; Oberhettinger, F.; Tricomi, F.G. _Higher Transcendental Functions_ ; McGraw-Hill Book Company, Inc.: New York, NY, USA; Toronto, ON, Canada; London, UK, 1953; Volume I.
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  * [13] Oldham, K.B.; Myland, J.C.; Spanier, J. _An Atlas of Functions: With Equator, the Atlas Function Calculator_ , 2nd ed.; Springer: New York, NY, USA, 2009.
  * [14] Reynolds, R., & Stauffer, A. (2022). , _A Note on the Infinite Sum of the Lerch function_. European Journal of Pure and Applied Mathematics, 15(1), 158–168. https://doi.org/10.29020/nybg.ejpam.v15i1.4137
  * [15] Weisstein, Eric W. ”Lerch Transcendent.” From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/LerchTranscendent.html
  * [16] L. Lewin (1981) _Polylogarithms and Associated Functions_. North-Holland Publishing Co., New York.
  * [17] I.V. Blagouchine, _A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations_ , J. Number Theory, 148 (2015), pp. 537-592
  * [18] M. Chamberland (2012), _Finite Trigonometric Product and Sum Identities_ , Fibonacci Quarterly 50(3), 217-221.
  * [19] A. Erdélyi, F. G. Tricomi _”The asymptotic expansion of a ratio of gamma functions.,”_ , Pacific Journal of Mathematics, Pacific J. Math. 1(1), 133-142, (1951)


# Machine Learning Resistant Amorphous Silicon Physically Unclonable Functions
(PUFs)

Velat Kilic1  Neil Macfarlane1  Jasper Stroud1  Samuel Metais1  Milad
Alemohammad1  A. Brinton Cooper1  Amy C. Foster1  Mark A. Foster1  1Department
of Electrical and Computer Engineering, Johns Hopkins University, Baltimore,
U.S.A<EMAIL_ADDRESS>
###### Abstract

We investigate usage of nonlinear wave chaotic amorphous silicon (a-Si)
cavities as physically unclonable functions (PUF). Machine learning attacks on
integrated electronic PUFs have been demonstrated to be very effective at
modeling PUF behavior. Such attacks on integrated a-Si photonic PUFs are
investigated through application of algorithms including linear regression,
k-nearest neighbor, decision tree ensembles (random forests and gradient
boosted trees), and deep neural networks (DNNs). We found that DNNs performed
the best among all the algorithms studied but still failed to completely break
the a-Si PUF security which we quantify through a private information metric.
Furthermore, machine learning resistance of a-Si PUFs were found to be
directly related to the strength of their nonlinear response.

††journal: oe††articletype: Research Article

##  1 Introduction

The pervasive introduction of interconnected devices into our lives has
generated increased concerns for privacy and has made cryptographic systems
more important than ever before. Many crypto protocols rely heavily on the
security of keys that are stored in device memory and are susceptible to
malware attacks. Physically unclonable functions (PUFs) have been proposed as
an alternative [1] whose response to external stimuli (challenge) is
determined by their microscopic structure, which is difficult to clone. PUF
operation consists of two phases: enrollment and deployment. During the
enrollment process, the manufacturer creates a challenge response pair (CRP)
library by probing the device with unique binary challenges and
measuring/generating the corresponding digitized response. The CRP data set is
then stored for the deployment phase where the PUF device can be authenticated
by probing it with a subset of challenges in the CRP data set and comparing
the responses. To be a strong security primitive a PUF must exhibit behavior
that is i) deterministic, ii) unpredictable, and iii) unique. Additionally,
iv) the device should be unclonable in that its behavior cannot be replicated
by a copy of the device. An unclonable but deterministic pseudo-random
function generator makes a desirable PUF.

Since their inception, PUFs built on various platforms have been proposed [2,
3, 4, 5]; however most, if not all, electronic PUFs have been shown to be
susceptible to modelling and machine learning (ML) attacks [6] and therefore
do not satisfy one of the four requirements of a strong security primitive. In
contrast, bulk optical scattering PUFs have remained resistant to machine
learning attacks due to the sheer size of their challenge-response space [7].
However, such bulk optical scattering PUFs are large, unreliable, and
difficult to integrate into conventional electronics. ML resistance of
scattering PUFs rely on high dimensional challenge-response space which is
physically upper bounded by the number of optical modes and therefore device
size. To solve this trade-off between security and practicality, a novel PUF
architecture has been recently demonstrated which is known as a silicon
photonic PUF [3, 8, 9, 10]. Silicon photonic PUFs are devices fabricated using
conventional CMOS-compatible planar semiconductor fabrication but leverage
nonlinear optical wave physics for their challenge-response behavior. Thus
silicon photonic PUFs strike a balance between the practicality of electronic
PUFs and the security offered by optical PUFs. To this end nonlinear
crystalline silicon (c-Si) photonic PUFs have been recently shown to be
resistant to modeling attacks based on deep neural networks (DNNs) and the
degree of resistance was shown to be correlated with the strength of the
optical nonlinearity [11].

In this work, we introduce PUFs based on hydrogenated amorphous silicon
(a-Si:H) which has an effective nonlinearity that is an order of magnitude
larger than c-Si [12]. Additionally, high-speed parametric amplification has
been shown in a-Si:H waveguides, showing a better nonlinear figure of merit
(defined as the ratio of the Kerr coefficient ($n_{2}$) to the two-photon
absorption coefficient ($\beta_{TPA}$) time wavelength) than c-Si at C-band
wavelengths [13, 14]. This is caused both by the higher Kerr coefficient of
a-Si:H, but also the lower TPA from a-Si:H having a larger band gap energy.
Therefore, a-Si:H photonic PUFs have the potential to be even more resistant
to modelling attacks due to increased nonlinearity and better integration.
Also, we further study the machine learning resistance of nonlinear a-Si:H
photonic PUFs through the application of a suite of learning strategies
including linear regression, k-nearest neighbor, decision trees, and DNNs. We
find that the PUF’s resistance to machine learning attacks is tied to its
degree of optical nonlinearity and quantify this resistance in terms of the
amount of private information.

Figure 1: a) Scanning electron microscope (SEM) image of an amorphous silicon
ray chaotic cavity used as a PUF. b) The device is probed by patterning a
chirped pulse with the binary challenges and the output time series signal is
measured in the Hadamard basis. c) Finite difference time domain (FDTD)
simulation of a mode in the PUF device.

##  2 A Nonlinear Integrated a-Si:H Photonic PUF

In contrast to bulky scattering optical PUFs, we have designed an integrated
a-Si:H photonic PUF as shown in Fig. 1a which has two optical single mode
wave-guides and a cavity. Single mode waveguide ensures stable and repeatable
coupling of light in (and out of) the device using telecommunications
commercial grade fiber optics, which proves more practical than using a large
scattering medium. The geometry of the cavity is a clipped disc, which is
known to be ray-chaotic. The input optical signal reverberates around the
cavity: the behaviour is then similar to the one of an optical scattering PUF,
where reverberation is used to limit its size. The surface roughness of the
boundaries, that comes from the fabrication variation help randomize the
cavity behaviour, which makes cloning the PUF intrinsically challenging as
that would require near atomic level precision in both measurement and
fabrication. Furthermore, integrated photonic devices are very stable in time,
with consistent behaviour over multiple days[3]. Yet, the limited size of the
device limits its degrees of freedom (size of the independent pair-response
challenges space), and so we make use of the large $\chi^{(3)}$ nonlinearity
of a-Si:H [15] inside the chaotic cavity to add another layer of complexity on
the intrinsic function of this object, which should further protect it against
machine learning attacks.

Ray chaotic design is necessary to protect against nearest neighbor attacks.
PUF response needs to decorrelate fast even for a small change in the
challenge (i.e large Lyapunov exponent). Otherwise, a large measurement
database can be used to find a similar enough challenge-response pair. Ray
chaoticity along with Kerr nonlinearity provides protection against these
attacks because a change in the challenge leads to a change in the intensity
which modulates the cavity index through $\chi^{(3)}$. A ray chaotic cavity
can then translate even small changes in refractive index to large variations
in the cavity response.

###  2.1 Fabrication

The a-Si:H devices are fabricated using mature CMOS fabrication techniques.
Fabrication begins with a 4-inch silicon wafer with 3000-nm of thermal oxide
on the surface. First, 250-nm of a-Si:H is deposited using plasma-enhanced
chemical vapor deposition (PECVD) at 350°C with 200-SCCM of silane gas, and
800 SCCM of argon gas. Then 150-nm of $\mathrm{SiO_{2}}$ is deposited also
using PECVD to act as a hard mask for etching. The devices are patterned using
electron-beam lithography (EBL). The devices are then etched using reactive
ion etching (RIE) and inductively coupled plasma etching (ICP). Finally, about
1000-nm of $\mathrm{SiO_{2}}$ is deposited using PECVD to act as an upper
cladding and optically isolate the devices from air.

Figure 2: Ultrafast optical setup for creating the CRP (challenge-response
pair) library. Pulses from the MLL (mode locked laser) are stretched using
dispersive fiber. Pulses are then patterned by an MZM (Mach-Zehnder modulator)
modulated by a PPG (pulse pattern generator) triggered by the 8th harmonic of
the MLL. The pulses are then compressed with equal but opposite amounts of
dispersion. They are then fed through an EDFA (Erbium doped fiber amplifier),
PC (polarization controller), PUF (physically unclonable function), EDFA
again, and finally a programmable spectral filter.

###  2.2 Experiment

Authentication in PUF systems is often done by a process known as challenge-
response verification. Upon manufacturing, each device is fully characterized
by applying a sequence of binary input keys (i.e. challenges). The challenge-
response library (CRL) is then constructed as a library of all unique, and
unclonable responses to individual challenges. Authentication is carried out
by comparing the PUF response to a known challenge, against the response
stored in the library.

In order to achieve an extremely large challenge-response space, we use a
spectrally encoded ultra-fast optical scheme shown in Fig 2. The ultra-fast
encoder module comprises of a dispersive fiber which performs the frequency-
to-time mapping of mode-locked laser (MLL) pulses. Dispersed pulses enter a
high speed electro-optic modulator, driven by a pulse pattern generator (PPG).
The PPG is synchronized to the laser pulses and each pulse is encoded with a
known sequence of 336 pseudo-random bits (i.e. challenges). After patterning,
the laser pulses are temporally compressed using a complimentary dispersion
compensating fiber. Compressed pulses are further amplified using an Erbium
Doped Fiber Amplifier (EDFA) in order to increase the peak power required for
non-linear optical interactions inside the silicon PUF device. Inside the PUF
device, each encoded laser pulse experiences a unique, complex spectro-
temporal modulation due to a combination of linear and nonlinear optical
interactions. It is important to note that these interactions will vary based
on the choice of the binary challenge sequence. These pulses are then detected
using a photodetector and the pulse peaks are digitized with 14 bit
resolution. By using the orthogonal Hadamard spectral filter at the output of
the PUF device we effectively extract 127$\times$14 bits of key data for each
binary challenge sequence.

We use fractional Hamming distance (FHD) to compare the responses from two
challenges

$FHD(a,b)=\frac{1}{L}\sum_{i=1}^{L}a_{i}\oplus b_{i}$ (1)

where $a$ and $b$ are bit strings (i.e. response sequence) of length $L$ and
$\oplus$ is the xor operation. Calculating the FHD results in values between
$0$ (similar) and $0.5$ (dissimilar). Ideally, true PUF responses would
achieve an FHD of $0$ and the clone would achieve $0.5$ with very small
variance, although in practice, this is rarely the case due to environmental
factors (i.e temperature, humidity, etc.) and noise.

In the following section, we investigate the machine learning resistance of
our nonlinear silicon photonic PUF device. Modelling attacks are performed on
the the Hadamard transform of the analog time series signal. In other words,
we assume snooping of an overt channel (in an application, the PUF security
can be further strengthened by techniques such as histogram equalization). For
comparison however digitized responses are used.

Figure 3: a) Average fractional Hamming distance (FHD) between PUF responses
as well as linear regression guess as a function of the bit level (from the
most significant bit to the least significant bit). Different colors
correspond to different pulse energies (blue: 5pJ, green: 1.5 pJ, red: 0.5
pJ). Circles correspond to FHD between PUF responses probed twice for the same
challenges and is a measure of the repeatability of the measurement. Triangles
correspond to FHD between ground truth PUF response and the linear regression
guess. LIN stands for linear model. b) Histogram of FHDs for the 5 pJ pulse
energy for a 20kbit key length both for the PUF and the linear model
prediction. Figure 4: Average fractional Hamming distance as a function of
training data set size for bit levels 1-4. Training set size beyond a few
thousand samples provides diminishing returns.

##  3 Machine Learning Attacks

###  3.1 Linear Regression

Optical response of the PUF can be expanded into linear and nonlinear parts
which are large and perturbative respectively. Linear component is easy to
learn by using linear regression and the number of measurements needed is
small (see Section 4). Here we investigate how much of the PUF response is
linear and can be learned by solving the following linear regression problem:

$\hat{W}_{lin}=\operatorname*{arg\,min}_{W}\lvert
W\mathbf{x}-\mathbf{y}\rvert^{2}_{2}$ (2)

where $\mathbf{x}$ and $\mathbf{y}$ are data matrices of the challenges and
the corresponding responses. A vector of ones is appended to the challenges to
compute the bias term. Training and test sets consist of 180,000 and 20,000
challenge response pairs respectively. Optimization problem is solved by using
Scikit-learn, a machine learning toolbox[16]. Training is performed on the
analog information and the results are then digitized (14 bits). Finally, the
fractional Hamming distance (FHD) between the prediction and the ground truth
is calculated. Ground truth is generated by probing the PUF twice with the
same challenge and averaging the responses. Fig. 3a shows the results for the
linear regression for the same device at three different pulse energy levels
(0.5, 1.5 and 5 pJ). As expected, the most significant bit which corresponds
to roughly guessing the total power level and the linear contribution can be
learned very well. However, following bits can be readily distinguished and as
shown in 3b, a decision boundary can be very clearly drawn. This supports our
claim that integrated PUF security is derived from the perturbative nonlinear
response.

Ideally, FHD between PUF responses would be very small and the FHD between PUF
responses and the prediction would be 0.5 corresponding to a random guess. As
shown in Fig. 3, increasing the pulse energy increases the gap between the
prediction and the ground truth value which is achieved through two mechanisms
in this case: i) increased pulse energy leads to more nonlinear interactions
and ii) signal to noise ratio increases which makes the responses more
repeatable. Therefore, the largest gap is seen between the blue curves(higher
energy) and the smallest between the red curves(lower energy).

Although, the linear response can be learned with the given dataset, an
attacker often will not have access to such a large CRP data set. In order to
test the effect of data set size, we randomly sampled a fraction of the set to
train a linear model. As shown in Fig. 4, increased CRP set size leads to
better prediction but with diminishing returns after a few thousand samples.
This number is determined by the number of modes in the optical device,
receiver/transmitter degrees of freedom as well as SNR which we investigate in
section 4.

###  3.2 k-Nearest Neighbor (kNN)

k-Nearest neighbor is a lazy learning algorithm that keeps all the training
data without any generalization. Test data is compared to all the samples in
the training set and the output of the k-closest samples are used in making
the final prediction. Since kNN does not require any modelling, it might be
suitable for predicting highly nonlinear data given a large training set.

However, as shown in Fig. 5 and 10, kNN algorithm performed worse than the
linear regression model. The training set had 180k samples and the test set
had 20k samples. Scikit-learn implementation is used with Euclidean distance
as the similarity metric and the best performance was achieved for $k=100$
[16]. Performance of the kNN algorithm in general depends on the neighbors
exhibiting similar behavior and we observe that it performs the worst for data
corresponding to high energy pulses which shows nonlinear response is key to
ML resistance. Physically, as the pulse energy is varied, the refractive index
profile of the PUF changes due to Kerr nonlinearity. Ray chaotic cavity then
maps small changes of the refractive index to large changes in device
response.

Figure 5: Average FHD at 3 different power levels as a function of the bit
level. Triangles correspond to kNN predictions and circles correspond to PUF
ground truth (displayed for comparison). b) FHD histogram at each bit level
for a key length of 20kbits.

Due to non-flat laser pulse profile, not all bits in the challenge vectors
have the same importance. In fact, matrix weights from the linear regression
model obtain larger values for the challenge bits near the pulse energy
maximum. Therefore, we implemented a hybrid linear regression - kNN algorithm
that assigns larger weights to these bits in the challenge vectors. However,
the hybrid model still performs worse than a simple linear regression.

Figure 6: Average FHD at 3 different power levels as a function of the bit
level. Triangles correspond to random forest (FOR) predictions. b) FHD
histogram at each bit level for a key length of 20kbits. Figure 7: Average FHD
at 3 different power levels as a function of the bit level. Triangles
correspond to gradient boosted tree (GBT) predictions. b) FHD histogram at
each bit level for a key length of 20kbits.

###  3.3 Decision Tree Models

Decision trees combine multiple levels of if-else statements to make a
prediction. Decision trees are often used in ensembles such as random forests
since they are prone to over-fitting. Getting a consensus vote from multiple
weak predictors improves accuracy and reduces variance. We used Scikit-learn
implementation of random forests with 100 decision trees per forest. Results
from this algorithm are shown in Fig. 6 which performed worse than linear
regression. We also used gradient boosting which constructs trees one at a
time to compensate weakness of the existing trees in the ensemble as opposed
to random forests which randomly constructs them. For that purpose, we use a
state of the art algorithm XGBoost [17] which performed significantly better
as shown in 7; better than linear regression but worse than DNNs as shown in
Fig. 10. Implementation details of XGBoost are beyond the scope of this work.

###  3.4 Deep Neural Networks (DNNs)

With the availability of powerful GPUs, deep neural networks have gained
popularity in the past years and have outperformed other machine learning
algorithms across many tasks from computer vision to natural language
processing and decision making [18]. DNNs can be mathematically described as
linear transformations followed by nonlinearities where parameters of the
network can be optimized using gradient descent. In this work, we use Pytorch,
an open source software platform for building neural networks to implement the
DNN shown in Fig. 8. All the layers except for the output use the rectified
linear unit (ReLU) as the nonlinear activation function. Training, validation
and test sets consisted of 179k, 1k and 20k samples respectively. The
validation set is used to optimize the hyperparameters of the network such as
number of hidden layers and their size, optimizer type, and learning rate
using a hyperparameter optimizer Optuna [19]. A batch size of 200 is used and
training is performed for 200 epochs. Adam is used as the optimizer [20] with
a learning rate of $4.99\times 10^{-4}$.

Figure 8: Deep neural network with rectified linear unit (ReLU) activation
function in the hidden layers. Input is a 336 bit long challenge and the
output is an analog signal of length 127. Figure 9: Average FHD at 3 different
power levels as a function of the bit level. Triangles correspond to the deep
neural network (DNN) predictions. b) FHD histogram at each bit level for a key
length of 20kbits.

As shown in Fig. 9 and 10, DNN provided the best predictions. However, as seen
in Fig. 9b a decision boundary can still be drawn accurately. Indeed, this
improvement comes from the ability of the neural network to model some of the
device non-linearity. In order to better study our PUF’s security, we
investigate its private information content in the following section.

Figure 10: Average FHD comparison for all models at different pulse energies
a) 0.5 pJ, b) 1.5 pJ, c) 5.0 pJ. ML attack performance remains the same across
different pulses energies with DNN performing the best and kNN the worst.

###  3.5 Private Information

PUFs can be used to securely transmit $H$ bits of information as determined by
entropy of the FHD distribution. In the presence of an attack, some of that
information is revealed through modelling and thus private information is
reduced. Therefore, we introduce a private information metric which is a
measure of the amount of information that can be transmitted securely in the
presence of an attack. We define private information as a symmetric, averaged
Kullback–Leibler (KL) divergence which is also known as Jeffrey divergence:

$PV(p,q)=\frac{1}{2}\left(KL(p,q)+KL(q,p)\right)$ (3)

where

$KL(p,q)=-\int^{\infty}_{-\infty}p\left(x\right)ln\left(\frac{q(x)}{p(x)}\right)$
(4)

$p(x)$ and $q(x)$ are probability density functions (in this case that of PUF
and model FHD distributions for 20kbit keys). We assume FHD histograms are
generated through a Gaussian process so the private information metric can be
further simplified:

$PV(p,q)=\frac{1}{2}\frac{\left(\sigma_{1}^{2}-\sigma_{2}^{2}\right)^{2}+\left(\mu_{1}-\mu_{2}\right)^{2}(\sigma_{1}^{2}+\sigma_{2}^{2})}{2\sigma_{1}^{2}\sigma_{2}^{2}}$
(5)

where $\sigma_{1,2}$ and $\mu_{1,2}$ are the standard deviations and means of
$p(x)$ and $q(x)$ respectively. Private information in this form is in natural
unit of information (nats) but can be converted to bits: 1 bit =
$\frac{1}{ln(2)}$ nat. In the limit
$\sigma_{1}\approx\sigma_{2}\approx\sigma$, eq. 5 simplifies to
$\frac{\Delta\mu^{2}}{2\sigma^{2}}$ which makes intuitive sense since private
information is maximal for narrow distributions and large separation of the
means between the model and the real PUF histograms.

Figure 11: Private information measured in bits as a function of the bit level
for different pulse energies in the presence of a DNN attack for 20kbit keys.
Bit level 4 retains the largest amount of private information striking a
balance between ML resistance and SNR trade-off.

In Fig. 11, we plot private information in the presence of a DNN attack as a
function of the bit level for different pulse energies. Although, the same
plot can be generated for all other attacks, DNN is picked since it performs
the best as shown in 10 and sets an upper limit on the private information. As
expected, total private information increases as the pulse energy increases
due to increased nonlinearity. Further, it is observed that the bit levels 2-7
have the largest amount of private information as they balance enhanced
security from nonlinear effects at the corresponding precision level and
random noise.

##  4 Linear Collapse Problem for PUFs Based on Linear Optical Devices

Traditional electronic PUFs scale in complexity through the addition of
discrete unit cells of circuit elements. In contrast linear optical PUFs
leverage optical wave physics which scale in complexity with the number of
optical modes. Small private information content of the first few bit levels
(as shown in Fig. 11) in our a-Si PUFs is due to predictability of the linear
contribution to the response. In this section, we provide theoretical bounds
on PUFs based on linear optical devices and show how nonlinearity can improve
device security.

It has been noted in the literature that large optical scattering PUFs contain
many degrees of freedom, which implies a large CRP set [1]; however, this is
not always true in practice due to limited degrees of freedom in the source
and the receiver, which we call “linear collapse.” Even though the number of
modes in the linear optical device might be very large, the degrees of freedom
provided by the transmitter and/or the receiver significantly reduces the
number of measurements needed to model the PUF for the purposes of the
transmitter and the receiver. Let us define basis functions (modes) for each
component: $\\{\lvert t_{i}\rangle\\}$, $\\{\lvert c_{j}\rangle\\}$, and
$\\{\lvert r_{k}\rangle\\}$ for the transmitter, the channel and the receiver
respectively. Transmitter basis functions $\\{\lvert t_{i}\rangle\\}$ encode
the challenge information (i.e digital micromirror device (DMD) [7]). Light
emanating from the transmitter then interacts with the linear optical device
(i.e scatterer) whose modes are $\\{\lvert c_{j}\rangle\\}$ and is finally
detected by the receiver (i.e camera). Suppose the transmitter has $P_{1}$
degrees of freedom (i.e at most the number of pixels of a DMD), the optical
device has $M$ degrees of freedom (i.e number of modes) and the receiver has
$P_{2}$ degrees of freedom (i.e at most the number of pixels of a camera). Let
us define the following operators $\hat{T}$, $\hat{C}$, and $\hat{R}$:

$\displaystyle\hat{T}_{M*P_{1}}$
$\displaystyle=\sum^{P_{1}}_{i=1}\sum^{M}_{j=1}T_{ij}\lvert
c_{j}\rangle\langle t_{i}\rvert$ (6) $\displaystyle\hat{C}_{M*M}$
$\displaystyle=\sum^{M}_{i=1}\sum^{M}_{j=1}C_{ij}\lvert c_{j}\rangle\langle
c_{i}\rvert$ (7) $\displaystyle\hat{R}_{P_{2}*M}$
$\displaystyle=\sum^{P_{2}}_{i=1}\sum^{M}_{j=1}R_{ij}\lvert
r_{j}\rangle\langle c_{i}\rvert$ (8)

where $\hat{T}$ maps transmitter modes to the channel modes, $\hat{C}$ maps
the channel modes at the input to the channel modes at the output, and
$\hat{R}$ maps the output channel modes to the receiver modes. Final mapping
from transmitter to the receiver is then
$\hat{D}_{P_{2}*P_{1}}~{}=~{}\hat{R}_{P_{2}*M}\hat{C}_{M*M}\hat{T}_{M*P_{1}}$.
Usually $P_{1}<P_{2}$ since PUFs are very similar to stream ciphers.

The PUF response for the purposes of the transmitter and the receiver can be
modeled using $P_{1}$ independent measurements (i.e camera shots) even for a
device with large number of modes. Of course, noise would increase the number
of required measurements, whereas prior information on any part of the system
(i.e sparsity, correlated pixels) would significantly reduce that number. In
fact, optical scattering PUFs are being broken on a regular basis in
computational imaging, and often with far fewer measurements than required by
the Nyquist limit using compressed sensing [21, 22, 23, 24].

The end result is that linear optical scattering PUFs must leverage large
scattering volumes and large and complex actuators and sensors with abundant
degrees of freedom to achieve sufficient behavioral complexity to ensure a
resistance to machine learning attacks. While this is possible in bulk
geometries, this requirement prevents the application of linear optical
scattering physics for integrated photonic geometries. Thus in integrated
photonic geometries nonlinearity is crucial.

##  5 Conclusion

In conclusion, we have presented a CMOS compatible amorphous silicon photonic
PUF architecture that uses nonlinear wave optics to achieve complex behavior.
Complex behavior of the device can be modified by changing probe pulse energy.
A large challenge response pair data set is used to implement machine learning
attacks through linear regression, k-nearest neighbor, boosted decision trees,
and deep neural networks (DNNs). Private information after a DNN attack is
quantified since it provides an upper bound. DNN achieved the best performance
however failed to break the PUF completely leaving as much as $8000$ bits of
private information per bit level for $20$ kbit key length. Lastly,
theoretical bounds for PUFs based on linear optical devices are presented.
Future study will focus on a deeper level of integration with on-chip laser
source and challenge generation as well as optical to digital read-out. Yet,
even in its current form, compatibility of the PUF with telecommunications
devices holds potential for secure communications.

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