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# Self-repellent branching random walk

Anton Bovier  A. Bovier, Institute for Applied Mathematics, University of
Bonn, Endenicher Allee 60, 53115 Bonn, Germany<EMAIL_ADDRESS>,  Lisa
Hartung  L. Hartung, Institut für Mathematik, Johannes Gutenberg-Universität
Mainz, Staudingerweg 9, 55128 Mainz, Germany<EMAIL_ADDRESS>and  Frank
den Hollander  F. den Hollander, Mathematical Institute, Leiden University,
Einsteinweg 55, 2333 CC Leiden, The Netherlands<EMAIL_ADDRESS>
###### Abstract.

We consider a discrete-time binary branching random walk with independent
standard normal increments subject to a penalty ${\beta}$ for every pair of
particles that get within distance $\epsilon$ of each other at any time. We
give a precise description of the most likely configurations of the particles
under this law for $N$ large and ${\beta},\epsilon$ fixed. Particles spread
out over a distance $2^{2N/3}$, essentially in finite time, and subsequently
arrange themselves so that at time $2N/3$ they cover a grid of width
$\epsilon$ with one particle per site. After time $2N/3$, the bulk of the
particles and their descendants do not move anymore, while the particles in a
boundary layer of width $2^{N/3}$ form a “staircase” to the particles in the
bulk. At time $N$, each site in the boundary layer is occupied by $2^{N/3}$
particles.

###### Key words and phrases:

Branching random walk; Excluded volume; Optimal strategy.

###### 2000 Mathematics Subject Classification:

60J80, 60G70, 82B44

AB was supported by the Deutsche Forschungsgemeinschaft (DFG, German Research
Foundation) under Germany’s Excellence Strategy - GZ 2047/1, Projekt-ID
390685813, and through Project-ID 211504053 - SFB 1060. LH was supported by
the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) through
Project-ID 233630050 -TRR 146, Project-ID 443891315 within SPP 2265, and
Project-ID 446173099. FdH was supported by the Netherlands Organisation for
Scientific Research (NWO) through Gravitation Grant NETWORKS 024.002.003 and
by the Alexander von Humboldt Foundation. We thank Stefan Müller for
interesting discussions on Gamma-convergence that inspired the analysis of the
fine structure of optimal profiles.

##  1\. Introduction and main results

###  1.1. Background

Both branching Brownian motion (BBM) and branching random walk (BRW) can be
seen as elementary models for the evolution of a population of individuals
that are subject to birth, death, and motion in space [21, 3]. Of primary
interest in these models is the evolution of the spatial distribution of the
population over time. Over the past 40 years, BBM and BRW have been
investigated from the point of view of extreme-value theory [10, 18, 12, 13,
5, 6, 7, 4, 14, 8].

As models for population dynamics, both BBM and BRW are unrealistic as they
lead to uncontrolled exponential growth of the population size and therefore
to unsustainably high densities of the population. For instance, the
population size of binary BRW at time $n$ grows like $2^{n}$, while the
population spreads out over a volume of order $n$. Several variants have been
proposed where, according to some selection rule, offspring is selected in
such a way that the total population size remains controlled [11, 20, 15, 19],
or competitive interactions between particles are added to suppress the growth
[16, 17, 2, 1].

In [9], a BBM is considered that is penalised when any two particles get close
to each other at any time. It turns out that the optimal strategy for this
process is to delay branching until a late random time, to guarantee that the
total number of particles generated remains finite. In short, in this model
the optimal strategy is to do nothing for a long time, and branch and diffuse
only shortly before the time horizon. In the present paper, we look at a
similar process, but in a setting where branching cannot be avoided.
Specifically, we consider a BRW where all particles duplicate at each unit of
time, jump according to independent Gaussian increments, and receive a penalty
for every pair of particles that get close to each other at any time.

Sections 1.2 and 1.3 define the model and introduce notation. Section 1.4
defines a subclass of optimal configurations called admissible. Sections 1.5
and 1.6 formulate our main theorems and provide the underlying heuristics.
Section 1.7 gives the outline of the remainder of the paper.

###  1.2. Model

Let

$X=\left((X_{i}(n))_{i=1}^{2^{n}}\right)_{n\in{\mathbb{N}}}$ (1.1)

denote the particles of a discrete-time binary BRW whose increments are i.i.d.
standard normal random variables. Write ${\mathbb{P}}$ to denote the law of
$X$. For $N\in{\mathbb{N}}$ and $\epsilon>0$, define

$J_{N,\epsilon}(X)=\sum_{n=1}^{N}\sum_{{1\leq i,j\leq 2^{n}}\atop{i\neq
j}}\mathbbm{1}_{|X_{i}(n)-X_{j}(n)|<\epsilon},$ (1.2)

which is the total time up to time $N$ that two particles are within distance
$\epsilon$ of each other. This quantity can be viewed as the _total collision
local time_ on length scale $\epsilon$ of all the particles up to time $N$. We
are interested in the law of $X$ under the tilted probability measure
$P_{N,{\beta},\epsilon}$ given by

$P_{N,{\beta},\epsilon}(\cdot)=\frac{1}{Z_{N,{\beta},\epsilon}}\,{\mathbb{E}}\left[\mathbbm{1}_{X\in\,\cdot\,}\,{\mathrm{e}}^{-{\beta}J_{N,\epsilon}(X)}\right]$
(1.3)

with

$Z_{N,{\beta},\epsilon}={\mathbb{E}}\left[{\mathrm{e}}^{-{\beta}J_{N,\epsilon}(X)}\right]$
(1.4)

the normalising partition function and ${\beta}>0$ the _inverse temperature_ ,
i.e., all collisions of all particles up to time $N$ are penalised by a factor
${\mathrm{e}}^{-{\beta}}$. To avoid unnecessary complications, we assume
throughout this paper that ${\beta}>\epsilon^{2}/2$ (the reason will become
clear shortly). Note that the birth of particles is _not_ penalised: the
branching is deterministic and binary.

It is well known that under the law ${\mathbb{P}}$ the spread of a _typical_
particle configuration at time $n$ is of order $n$. Since there are $2^{n}$
particles at time $n$, the total collision local time in a typical particle
configuration at time $n$ is of order $\epsilon\,2^{2n}/n$. When summed over
$n$ this gives a value of $J_{N,\epsilon}$ that is of order
$\epsilon\,2^{2N}/N$.

The situation is _very different_ under the law $P_{N,{\beta},\epsilon}$. Our
main theorems below give a precise description of the most likely
configurations of the particles under the law $P_{N,{\beta},\epsilon}$ for $N$
large and ${\beta},\epsilon$ fixed. Before stating these theorems we need to
introduce some notation.

###  1.3. Notation

We denote the infinite binary tree by ${\mathbb{T}}$ and the finite binary
subtree with $n$ generations by ${\mathbb{T}}^{(n)}$. We label the leaves of
${\mathbb{T}}$ by an infinite multi-index
${\boldsymbol{x}}=(x_{1}x_{2}\dots)\in\\{0,1\\}^{\mathbb{N}}$, and write
${\boldsymbol{x}}^{(n)}$ for the ancestor of ${\boldsymbol{x}}$ in the $n$-th
generation. The root of ${\mathbb{T}}$ is denoted by $\emptyset$, and
${\mathbb{T}}$ is endowed with the natural lexicographic order. Let
$h\colon\,{\mathbb{T}}\rightarrow{\mathbb{R}}$ denote the collection of
positions of all the particles in the BRW, i.e., $h({\boldsymbol{x}}^{(n)})$
is the position of the ancestor of ${\boldsymbol{x}}$ in the $n$-th
generation. Then the probability density of a configuration of particles in
the first $N$ generations is given by

$p(h;{\mathbb{T}}^{(N)})=\exp\left(-S_{\mathrm{spr}}(h,{\mathbb{T}}^{(N)})\right)$
(1.5)

with

$S_{\mathrm{spr}}(h,{\mathbb{T}}^{(N)})=\sum_{n=0}^{N-1}W_{n}(h),$ (1.6)

where

$W_{n}(h)=\tfrac{1}{2}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\left(\left(h({\boldsymbol{z}}^{(n)}0)-h({\boldsymbol{z}}^{(n)})\right)^{2}+\left(h({\boldsymbol{z}}^{(n)}1)-h({\boldsymbol{z}}^{(n)})\right)^{2}\right).$
(1.7)

The penalty function reads

$J_{N,\epsilon}(h,{\mathbb{T}}^{(N)})=\sum_{n=1}^{N}I_{n,\epsilon}(h),$ (1.8)

where

$I_{n,\epsilon}(h)=\sum_{{{\boldsymbol{z}}^{(n)},{\boldsymbol{y}}^{(n)}\in{\mathbb{T}}^{(n)}}\atop{{\boldsymbol{z}}^{(n)}\neq{\boldsymbol{y}}^{(n)}}}\mathbbm{1}_{|h({\boldsymbol{z}}^{(n)})-h({\boldsymbol{y}}^{(n)})|<\epsilon}.$
(1.9)

Thus, for given $N\in{\mathbb{N}}$, the optimal strategy will be given by the
configurations minimising the functional

$S(h,{\mathbb{T}}^{(N)})=S_{\mathrm{spr}}(h,{\mathbb{T}}^{(N)})+{\beta}J_{N,\epsilon}(h,{\mathbb{T}}^{(N)}).$
(1.10)

###  1.4. Admissible configurations

We collect some a priori properties of possible minimising configurations. We
begin by showing that $h({\boldsymbol{z}}^{(N)})$ takes values only in
$\frac{1}{2}+\epsilon\,{\mathbb{Z}}$ (so that particles are either at the same
location or at a distance at least $\epsilon$ from each other). Let $h$ be any
configuration such that in generation $N$ at some site $i\epsilon$ there are
at least two particles more than at one of the neighbouring sites $j\epsilon$
with $j=i\pm 1$. From $h$ we can obtain a configuration $h^{\prime}$ by moving
one of the particles at site $i\epsilon$ to one of the neighbouring sites
$j\epsilon$ without changing anything else. This move entails a cost in the
spread $W_{N-1}$ of at most $\epsilon^{2}/2$ and produces a gain in the
interaction $I_{N}$ of at least $2{\beta}$. Thus, if ${\beta}>\epsilon^{2}/2$,
then any minimising configuration $h$ must satisfy the condition that the
difference in occupation numbers of any two neighbouring sites in
$\epsilon\,{\mathbb{Z}}$ is at most one. We call this the set of ‘smooth
configurations’. We want to find the configurations within this set that
minimise the functional in (1.10).

Next, consider configurations that increase linearly over length
$\epsilon\,r$, subsequently stay constant over length $\epsilon d$, and
afterwards decrease linearly over length $\epsilon\,r$ again, for some
$r\in{\mathbb{N}}$. The total number of particles in such configurations is
$r(r+1)+dr$, which must be equal to $2^{N}$, i.e.,

$d=(2^{N}/r)-(r+1).$ (1.11)

We denote the set of such configurations by ${\mathcal{H}}_{r,d,N}$, and call
them _admissible configurations_.

###  1.5. Main theorems

The following theorems identify the _optimal profiles_ for $N$ large and will
be proved in Section 4. Recall that we assume that ${\beta}>\epsilon^{2}/2$.

######  Theorem 1.1.

[The end-profile] Any $h$ that minimises $S(h,{\mathbb{T}}^{(N)})$ is an
element of ${\mathcal{H}}_{r,d,N}$ with $r=2^{K}$, where $K$ is the integer
closest to

$\tfrac{N-4}{3}+\tfrac{1}{3}\ln_{2}(3\epsilon^{2}/{\beta})+O(2^{-N/6})=\tfrac{N}{3}+O(1),$
(1.12)

and with $d=(2^{N}/r)-(r+1)$. See Figure 1.

######  Theorem 1.2.

[The post-profile] Any $h$ that minimises $S(h,{\mathbb{T}}^{(N)})$ is such
that $h({\boldsymbol{z}}^{(n)})\in{\mathcal{H}}_{r_{n},d_{n},n}$ for all
$N-K\leq n\leq N$, where

$r_{n}=2^{n-(N-K)},$ (1.13)

and $d_{n}=(2^{n}/r_{n})-(r_{n}+1)$. See Figure 2.

######  Theorem 1.3.

[The pre-profile] Any $h$ that minimises $S(h,{\mathbb{T}}^{(N)})$ is such
that

$\\{h({\boldsymbol{z}}^{(n)})\colon\,1\leq n\leq N-K\\}$ (1.14)

is the solution to the Dirichlet problem

$\displaystyle
h({\boldsymbol{z}}^{(n)}0)+h({\boldsymbol{z}}^{(n)}1)+h({\boldsymbol{z}}^{(n-1)})$
$\displaystyle=$ $\displaystyle 3h({\boldsymbol{z}}^{(n)}),\quad 1\leq n<N-K,$
$\displaystyle h(\emptyset)$ $\displaystyle=$ $\displaystyle 0,$
$\displaystyle h({\boldsymbol{z}}^{(N-K)})$ $\displaystyle=$ $\displaystyle
u({\boldsymbol{z}}^{(N-K)}),$ (1.15)

with boundary condition

$u({\boldsymbol{z}}^{(N-K)})=\begin{cases}\epsilon\sum_{\ell=2}^{N-K}2^{(N-K)-\ell}z_{\ell}+\tfrac{1}{2}\epsilon,\quad&z_{1}=1,\\\
-\epsilon\sum_{\ell=2}^{N-K}2^{(N-K)-\ell}z_{\ell}-\tfrac{1}{2}\epsilon,\quad&z_{1}=0.\end{cases}$
(1.16)

In particular, the position $h_{n}$ of the right-most particle at time $n$ is

$h_{n}=\begin{cases}{\epsilon}\,2^{N-K-1}(1-2^{-n-1}n),&2\leq n\ll N-K,\\\
{\epsilon}\,2^{N-K-3},&n=1.\end{cases}$ (1.17)

See Figure 3.

Figure 1. The end-profile of a configuration in ${\mathcal{H}}_{r,d,N}$.
Figure 2. The post-profile of a configuration in ${\mathcal{H}}_{r,d,N}$.
Figure 3. The pre-profile of a configuration in ${\mathcal{H}}_{r,d,N}$.

######  Theorem 1.4.

[Cost of the profile] For $N\to\infty$,

$\ln\min_{h\colon{\mathbb{T}}^{(N)}\to{\mathbb{R}}}S(h,{\mathbb{T}}^{(N)})=-2^{5/3}3^{-2/3}({\beta}\epsilon)^{2/3}2^{4N/3}+O(2^{7N/6}).$
(1.18)

The following rough picture emerges for the optimal profile:

  * (i)

Until time $2N/3$ particles spread out quickly according to a profile that is
captured by a Dirichlet problem.

  * (ii)

At time $2N/3$, there is a single particle on every site in

$\mathcal{L}_{1}=\epsilon\,\left\\{-2^{N-K-1}-\tfrac{1}{2},\dots,2^{N-K-1}+\tfrac{1}{2}\right\\}.$
(1.19)

  * (iii)

The particles in

$\mathcal{L}_{2}=\epsilon\,\left\\{-2^{N-K-1}-\tfrac{1}{2}+2^{K-1},\dots,2^{N-K-1}+\tfrac{1}{2}-2^{K-1}\right\\}$
(1.20)

and their descendants do not move between time $2N/3$ and time $N$.

  * (iv)

While the particles in $\mathcal{L}_{1}\setminus\mathcal{L}_{2}$ reproduce,
their offspring move minimally to remove increments of height larger than $1$.

###  1.6. Heuristics

Before closing this introduction we give a _rough heuristic_ for the optimal
strategy describing how the particles spread out over space in the course of
time. If at time $n$ the particles are evenly distributed over the interval
$[-r(n),r(n)]$, then each particle has roughly $\epsilon\,2^{n}/r(n)$
particles within distance $\epsilon$ (which is possible as long as
$\epsilon\,2^{n}/r(n)>1$), and so $J_{N,\epsilon}$ receives a contribution of
order $\epsilon\,2^{2n}/r(n)$. On the other hand, to spread out the offspring,
for all $x\in[-1,1]$, particles that were at a position $xr(n-1)$ at time
$n-1$ must move to position $xr(n)$ at time $n$, which entails a displacement
cost of order $\exp(\tfrac{1}{2}x^{2}[r(n)-r(n-1)]^{2})$ per particle, and
hence a total cost of order

$\exp\left(2^{n}\int_{-1}^{1}{\mathrm{d}}x\,\tfrac{1}{2}x^{2}(r(n)-r(n-1))^{2}\right)=\exp\left(2^{n}\tfrac{1}{3}(r(n)-r(n-1))^{2}\right).$
(1.21)

We would be tempted to think that, for the optimal choice of $r(n)$, both
costs must be equal, i.e.,

${\beta}\,\frac{\epsilon\,2^{2n}}{r(n)}=2^{n}\tfrac{1}{3}(r(n)-r(n-1))^{2}.$
(1.22)

Hence

$r(n)^{3}=2^{n}\,3{\beta}\epsilon\,\left(1-\frac{r(n-1)}{r(n)}\right)^{-2},$
(1.23)

which would give

$r(n)=C\,2^{n/3},\qquad C^{3}=\frac{3{\beta}\epsilon}{(1-2^{-1/3})^{2}}.$
(1.24)

However, the above reasoning is _flawed_ : it is _not at all_ optimal to
optimise the cost in each summand separately. Rather, in anticipation of the
fact that the process terminates at time $N$, it pays off to _prepare for
future expenses at an early stage_ , when there are not so many particles yet.

Figure 4. Sketch of an optimal configuration.

To get a better answer, we must minimise the functional

$S(r)=\sum_{n=1}^{N}\left({\beta}\,\frac{\epsilon\,2^{2n}}{r(n)}+2^{n}\tfrac{1}{3}(r(n)-r(n-1))^{2}\right).$
(1.25)

Differentiating w.r.t. $r(n)$, we get, for $1<n<N$,

$-3{\beta}\epsilon\,2^{2n}r(n)^{-2}+2^{n+1}(r(n)-r(n-1))-2^{n+2}(r(n+1)-r(n))=0$
(1.26)

or

$r(n)-r(n-1)=3{\beta}\epsilon\,2^{n-1}r(n)^{-2}+2(r(n+1)-r(n)),$ (1.27)

while, for $n=N$,

$r(N)-r(N-1)=3{\beta}\epsilon\,2^{N}r(N)^{-2},$ (1.28)

and, for $n=1$,

$r(1)=3{\beta}\epsilon\,r(1)^{-2}+2(r(2)-r(1)),$ (1.29)

with $r(0)=0$. Assume that $2^{n}r(n)^{-2}\ll 1$ for all $1\leq n\leq N$. Then
we can neglect the nonlinear term, to get the recursion

$r(n+1)-r(n)=\tfrac{1}{2}(r(n)-r(n-1)),$ (1.30)

which is the Laplace equation on a binary tree. The solution is

$r(n)=2r(1)(1-2^{-n}),$ (1.31)

with $r(1)$ a free parameter. Inserting this relation into (1.25), we get

$\displaystyle S_{N}(r)$
$\displaystyle=\sum_{n=1}^{N}\left({\beta}\,\frac{\epsilon\,2^{2n}}{2r(1)(1-2^{-n})}+2^{n}\tfrac{4}{3}r(1)^{2}2^{-2n}\right)$
(1.32)
$\displaystyle\sim\tfrac{1}{3}{\beta}\epsilon\,2^{2N+1}\,r(1)^{-1}+\tfrac{4}{3}r(1)^{2},$

which takes its minimum for

$r(1)^{3}={\beta}\epsilon\,2^{2N-2}.$ (1.33)

Hence, $2^{n}r(n)^{-2}\asymp 2^{n}2^{-4N/3}\leq 2^{-N/3}\ll 1$, so that our
initial assumption is indeed verified. Consequently,

$S_{N}(r)\sim\tfrac{2}{3}2^{2/3}({\beta}\epsilon)^{2/3}2^{4N/3}.$

This is the same scaling as in Theorem 1.4, but with a different constant (the
precise constant is lost in the heuristic argument).

Theorems 1.1–1.4 provide a detailed picture of the underlying particle system
(summarised in Figure 4).

###  1.7. Outline

The remainder of this paper is organised as follows. In Section 2 we identify
the minimisers of the functional $S_{\mathrm{spr}}(h,{\mathbb{T}}^{(N)})$ in
(1.10), which _ignores_ the repulsive interaction between the particles. This
identification involves a Dirichlet problem and leads to sharp estimates on
$S_{\mathrm{spr}}(h,{\mathbb{T}}^{(N)})$. In Section 3 we identify the
minimisers of the functional $S(h,{\mathbb{T}}^{(N)})$ in (1.10), which
_includes_ the repulsive interaction between the particles. In Section 4 we
collect our estimates and prove Theorems 1.1–1.4. The reason why the analysis
of $S_{\mathrm{spr}}(h,{\mathbb{T}}^{(N)})$ is crucial is that in the optimal
strategy the effect of the repulsive interaction between the particles is
simple, as argued in the heuristics in Section 1.6.

##  2\. Optimal strategy without interaction

In this section we identify the optimal strategy for the branching random walk
to reach a configuration at time $M\in{\mathbb{N}}$ where the $2^{M}$
particles are distributed at positions such that their distance is just large
enough to ensure that there is no penalty to pay yet. In Section 2.1 we
represent the optimal strategy as a Dirichlet problem. In Section 2.2 we find
the solution to this Dirichlet problem. In Section 2.3 we compute the cost of
spreading out the particles. The main results are collected in Lemmas 2.1–2.6
below.

###  2.1. Representation of the optimal strategy as a Dirichlet problem

When we optimise $S_{\mathrm{spr}}(h,{\mathbb{T}}^{(M)})$ under the _boundary
conditions_

$h(\emptyset)=0,\qquad
h({\boldsymbol{z}}^{(M)})=u({\boldsymbol{z}}^{(M)}),\qquad
u\colon\,{\mathbb{T}}^{(M)}\to{\mathbb{R}},$ (2.1)

the solution is the harmonic function with these boundary conditions, i.e., a
solution of the _Dirichlet problem_

$\displaystyle
h({\boldsymbol{z}}^{(n)}0)+h({\boldsymbol{z}}^{(n)}1)+h({\boldsymbol{z}}^{(n-1)})$
$\displaystyle=$ $\displaystyle 3h({\boldsymbol{z}}^{(n)}),\quad 1\leq n<M,$
$\displaystyle h(\emptyset)$ $\displaystyle=$ $\displaystyle 0,$
$\displaystyle h({\boldsymbol{z}}^{(M)})$ $\displaystyle=$ $\displaystyle
u({\boldsymbol{z}}^{(M)}).$ (2.2)

It will be convenient to express the latter in terms of the increments of $h$.
For ${\boldsymbol{z}}\in{\mathbb{T}}$ and $1\leq n\leq M$, set

$a({\boldsymbol{z}}^{(n)})=h({\boldsymbol{z}}^{(n)})-h({\boldsymbol{z}}^{(n-1)}).$
(2.3)

Then

$a(\emptyset)+\sum_{j=1}^{M}a({\boldsymbol{z}}^{(j)})=u({\boldsymbol{z}}^{(M)}),$
(2.4)

and, for $1\leq n\leq M$,

$a({\boldsymbol{z}}^{(n-1)})=a({\boldsymbol{z}}^{(n-1)}0)+a({\boldsymbol{z}}^{(n-1)}1).$
(2.5)

Define, for $0\leq n\leq M$,

$\Sigma({\boldsymbol{z}}^{(n)})=\sum_{{\boldsymbol{y}}^{(M-n)}\in{\mathbb{T}}^{(M-n)}}u({\boldsymbol{z}}^{(n)}\,{\boldsymbol{y}}^{(M-n)}).$
(2.6)

######  Lemma 2.1.

The unique solution of (2.4)–(2.5) satisfies the recursive equation

$\Sigma({\boldsymbol{z}}^{(n)})=2^{M-n}\left(\sum_{j=1}^{n-1}a({\boldsymbol{z}}^{(j)})\right)+\left(2^{M-n+1}-1\right)a({\boldsymbol{z}}^{(n)}).$
(2.7)

###### Proof.

Fix ${\boldsymbol{z}}\in{\mathbb{T}}$ and $0\leq n\leq M$. Sum (2.4) over the
descendants of ${\boldsymbol{z}}^{(n)}$. This gives

$\displaystyle\Sigma({\boldsymbol{z}}^{(n)})$ $\displaystyle=$
$\displaystyle\sum_{{\boldsymbol{y}}^{(M-n)}\in{\mathbb{T}}^{(M-n)}}\left(a(\emptyset)+\sum_{j=1}^{M}a\Big{(}({\boldsymbol{z}}^{(n)}\,{\boldsymbol{y}}^{(M-n)})^{(j)}\Big{)}\right)$
(2.8) $\displaystyle=$ $\displaystyle
2^{M-n}a(\emptyset)+2^{M-n}\sum_{j=1}^{n}a({\boldsymbol{z}}^{(j)})+\sum_{j=n+1}^{M}\sum_{{\boldsymbol{y}}^{(M-n)}\in{\mathbb{T}}^{(M-n)}}a({\boldsymbol{z}}^{(n)}\,{\boldsymbol{y}}^{(j-n)})$
$\displaystyle=$ $\displaystyle
2^{M-n}a(\emptyset)+2^{M-n}\sum_{j=1}^{n}a({\boldsymbol{z}}^{(j)})+\sum_{j=n+1}^{M}2^{M-j}\sum_{{\boldsymbol{y}}^{(j-n)}\in{\mathbb{T}}^{(j-n)}}a({\boldsymbol{z}}^{(n)}\,{\boldsymbol{y}}^{(j-n)})$
$\displaystyle=$ $\displaystyle
2^{M-n}\,a(\emptyset)+2^{M-n}\sum_{j=1}^{n}a({\boldsymbol{z}}^{(j)})+\sum_{j=n+1}^{M}2^{M-j}\,a({\boldsymbol{z}}^{(n)})$
$\displaystyle=$ $\displaystyle
2^{M-n}\,a(\emptyset)+2^{M-n}\sum_{j=1}^{n-1}a({\boldsymbol{z}}^{(j)})+\left(2^{M-n+1}-1\right)a({\boldsymbol{z}}^{(n)}),$

where the fourth equality uses (2.5). Since $a(\emptyset)=0$, this gives
(2.7). ∎

Note that

$a({\boldsymbol{z}}^{(M)})=h({\boldsymbol{z}}^{(M)})-\sum_{j=1}^{M}a({\boldsymbol{z}}^{(j)}).$
(2.9)

The next lemma gives an explicit form of the solution of the recursion in
(2.7).

######  Lemma 2.2.

Set $b_{n}=(2^{M-n+1}-1)^{-1}$. Then

$h({\boldsymbol{z}}^{(n)})=b_{n}\Sigma({\boldsymbol{z}}^{(n)})+\sum_{\ell=1}^{n-1}\frac{b_{n-\ell+1}b_{n-\ell}}{b_{n+1}}\,\Sigma({\boldsymbol{z}}^{(n-\ell)})$
(2.10)

and

$\displaystyle a({\boldsymbol{z}}^{(n)})$ $\displaystyle=$ $\displaystyle
b_{n}\Sigma({\boldsymbol{z}}^{(n)})+\sum_{\ell=1}^{n-1}b_{n-\ell}\Sigma({\boldsymbol{z}}^{(n-\ell)})\left(\frac{b_{n-\ell+1}}{b_{n+1}}-\frac{b_{n-\ell+1}}{b_{n}}\right).$
(2.11)

###### Proof.

The recursion (2.7) reads

$a({\boldsymbol{z}}^{(n)})=b_{n}\Sigma({\boldsymbol{z}}^{(n)})-b_{n}2^{M-n}\sum_{\ell=1}^{n-1}a({\boldsymbol{z}}^{(\ell)})=b_{n}\Sigma({\boldsymbol{z}}^{(n)})-b_{n}2^{M-n}h({\boldsymbol{z}}^{(n-1)}),$
(2.12)

or

$h({\boldsymbol{z}}^{(n)})=b_{n}\Sigma({\boldsymbol{z}}^{(n)})-\left(b_{n}2^{M-n}-1\right)h({\boldsymbol{z}}^{(n-1)})=b_{n}\Sigma({\boldsymbol{z}}^{(n)})+\frac{b_{n}}{b_{n+1}}h({\boldsymbol{z}}^{(n-1)}).$
(2.13)

A simple iteration gives that (note that $\Sigma(\emptyset)=0$)

$h({\boldsymbol{z}}^{(n)})=b_{n}\Sigma({\boldsymbol{z}}^{(n)})+\sum_{\ell=1}^{n-1}\frac{b_{n-\ell+1}b_{n-\ell}}{b_{n+1}}\Sigma({\boldsymbol{z}}^{(n-\ell)}).$
(2.14)

Hence

$\displaystyle a({\boldsymbol{z}}^{(n)})$ $\displaystyle=$ $\displaystyle
b_{n}\Sigma({\boldsymbol{z}}^{(n)})+\sum_{\ell=1}^{n-1}\frac{b_{n-\ell+1}b_{n-\ell}}{b_{n+1}}\Sigma({\boldsymbol{z}}^{(n-\ell)})$
(2.15) $\displaystyle-
b_{n-1}\Sigma({\boldsymbol{z}}^{(n-1)})-\sum_{\ell=1}^{n-2}\frac{b_{n-\ell}b_{n-1-\ell}}{b_{n}}\Sigma({\boldsymbol{z}}^{(n-1-\ell)})$
$\displaystyle=$ $\displaystyle
b_{n}\Sigma({\boldsymbol{z}}^{(n)})+\sum_{\ell=1}^{n-1}\Sigma({\boldsymbol{z}}^{(n-\ell)})\left(\frac{b_{n-\ell+1}b_{n-\ell}}{b_{n+1}}-\frac{b_{n-\ell+1}b_{n-\ell}}{b_{n}}\right),$

which gives (2.11). ∎

For future use, note that

$\frac{b_{n-\ell+1}}{b_{n+1}}-\frac{b_{n-\ell+1}}{b_{n}}=2^{-\ell}\frac{1}{1-2^{-M+n-\ell}}.$
(2.16)

###  2.2. Solution of the Dirichlet problem

In the following we are interested in the Dirichlet problem with boundary
conditions that are symmetric around zero and correspond to having single
particles at distance $\epsilon$. Hence, we look at

$u({\boldsymbol{z}}^{(M)})=\begin{cases}\epsilon\sum_{\ell=2}^{M}2^{M-\ell}z_{\ell}+\tfrac{1}{2}\epsilon,\quad&z_{1}=1,\\\
-\epsilon\sum_{\ell=2}^{M}2^{M-\ell}z_{\ell}-\tfrac{1}{2}\epsilon,\quad&z_{1}=0.\end{cases}$
(2.17)

For these boundary conditions we compute $a({\boldsymbol{z}}^{(n)})$ and
$h({\boldsymbol{z}}^{(n)})$. We write $h^{\mathrm{Dir}}$ for the solution of
the Dirichlet problem with boundary condition (2.17).

######  Lemma 2.3.

The solution of the Dirichlet problem with boundary condition (2.17) is given
by

$a({\boldsymbol{z}}^{(n)})=\epsilon\,2^{M-n-2}\left(2-n+2\sum_{k=2}^{n}z_{k}\right)+O(1),$
(2.18)

and, for $M\gg n$,

$h({\boldsymbol{z}}^{(n)})={\epsilon}\,2^{M-1}\left(n2^{-n-1}+2\sum_{k=2}^{n}(2^{-k}-2^{-n-1})z_{k}\right)(1+O(2^{-M+n}).$
(2.19)

In particular,

$h(1^{(n)})\sim\begin{cases}{\epsilon}\,2^{M-1}\left(1-2^{-n-1}n\right),&2\leq
n\leq M,\\\ {\epsilon}\,2^{M-3}&n=1.\end{cases}$ (2.20)

######  Remark.

In the course of the proof we obtain precise formulas for $a$ and $h$, but we
state only the approximate forms in the lemma.

###### Proof.

For $z_{1}=1$,

$\displaystyle\Sigma({\boldsymbol{z}}^{(n)})$ $\displaystyle=$
$\displaystyle\epsilon\sum_{{\boldsymbol{y}}^{(M-n)}\in{\mathbb{T}}^{(M-n)}}\left(\sum_{\ell=2}^{M}2^{M-\ell}z_{\ell}+2^{-1}\right)$
(2.21) $\displaystyle=$
$\displaystyle\epsilon\sum_{{\boldsymbol{y}}^{(M-n)}\in{\mathbb{T}}^{(M-n)}}\left(\sum_{\ell=2}^{n}2^{M-\ell}z_{\ell}+\sum_{\ell=n+1}^{M}2^{M-\ell}y_{\ell}+2^{-1}\right)$
$\displaystyle=$
$\displaystyle\epsilon\,2^{M}\sum_{\ell=2}^{n}2^{2M-n-\ell}z_{\ell}+\epsilon\,2^{M-n-1}\sum_{\ell=n+1}^{M}2^{-\ell}+\epsilon\,2^{M-n-1}$
$\displaystyle=$
$\displaystyle\epsilon\,2^{2M-n}\left(\sum_{\ell=2}^{n}2^{-\ell}z_{\ell}+2^{-n-1}\right).$

In particular,

$\Sigma(1)=-\Sigma(0)=\epsilon\,2^{2M-3}.$ (2.22)

Consequently

$b_{n-\ell}\Sigma({\boldsymbol{z}}^{(n-\ell)})=\frac{\epsilon}{2}2^{M}\frac{1}{1-2^{-M+n-\ell-1}}\left(\sum_{k=2}^{n-\ell}2^{-k}z_{k}+2^{-n-1}\right)$
(2.23)

and

$\displaystyle a({\boldsymbol{z}}^{(n)})$ $\displaystyle=$
$\displaystyle\frac{\epsilon}{2}\frac{1}{1-2^{-M+n-1}}\left(\sum_{k=2}^{n}2^{M-k}z_{k}+2^{M-n-1}\right)$
(2.24)
$\displaystyle-\frac{\epsilon}{2}\sum_{\ell=1}^{n-1}\frac{1}{1-2^{-M+n-\ell-1}}\left(\sum_{k=2}^{n-\ell}2^{M-k-\ell}z_{k}+2^{M-n-1}\right)\frac{1}{1-2^{-M+n-\ell}}.$

A re-ordering of the summations gives

$\displaystyle a({\boldsymbol{z}}^{(n)})$ $\displaystyle=$
$\displaystyle\frac{\epsilon}{2}2^{M-n-1}\left(\frac{1}{1-2^{-M+n-1}}-\sum_{\ell=1}^{n-1}c_{\ell}\right)$
(2.25)
$\displaystyle+\frac{\epsilon}{2}2^{M}\sum_{k=2}^{n}2^{-k}z_{k}\left(\frac{1}{1-2^{-M+n-1}}-\sum_{\ell=1}^{n-k}2^{-\ell}c_{\ell}\right)$

with

$c_{\ell}=\frac{1}{(1-2^{-M+n-\ell})(1-2^{-M+n-\ell-1})}.$ (2.26)

Using Wolfram Alpha, we get

$\frac{1}{1-2^{-M+n-1}}-\sum_{\ell=1}^{n-k}2^{-\ell}c_{\ell}=2^{k-n}\frac{1}{1-2^{-M+k-1}}.$
(2.27)

Hence

$\displaystyle a({\boldsymbol{z}}^{(n)})$ $\displaystyle=$
$\displaystyle\frac{\epsilon}{2}2^{M-n-1}\left(\frac{1}{1-2^{-M+n-1}}-\sum_{\ell=1}^{n-1}c_{\ell}\right)$
(2.28)
$\displaystyle+\frac{\epsilon}{2}2^{M-n}\sum_{k=2}^{n}z_{k}\left(\frac{1}{1-2^{-M+k-1}}\right).$

We readily verify that, uniformly in $0\leq n\leq M$,

$a({\boldsymbol{z}}^{(n)})=\bar{a}({\boldsymbol{z}}^{(n)})+O(1),$ (2.29)

where we set

$\bar{a}({\boldsymbol{z}}^{(n)})=\epsilon
2^{M-n-2}\left((2-n)+2\sum_{k=2}^{n}z_{k}\right).$ (2.30)

Next, we turn to $h$. Combining (2.14) with (2.23), we get

$\displaystyle h({\boldsymbol{z}}^{(n)})$ $\displaystyle=$
$\displaystyle\frac{\epsilon}{2}2^{M}\frac{1}{1-2^{-M+n-1}}\left(\sum_{k=2}^{n}2^{-k}z_{k}+2^{-n-1}\right)$
(2.31)
$\displaystyle+\frac{\epsilon}{2}2^{M}\sum_{\ell=1}^{n-1}\frac{b_{n-\ell+1}}{b_{n+1}}\frac{1}{1-2^{-M+n-\ell-1}}\left(\sum_{k=2}^{n-\ell}2^{-k}z_{k}+2^{-n+\ell-1}\right)$
$\displaystyle=$
$\displaystyle\frac{\epsilon}{2}2^{M}\frac{1}{1-2^{-M+n-1}}\left(\sum_{k=2}^{n}2^{-k}z_{k}+2^{-n-1}\right)$
$\displaystyle+\frac{\epsilon}{2}2^{M}(1-2^{-M+n})\sum_{\ell=1}^{n-1}2^{-\ell}c_{\ell}\left(\sum_{k=2}^{n-\ell}2^{-k}z_{k}+2^{-n-1+\ell}\right),$

where we use that (recall the definition of $c_{\ell}$ in (2.26)

$\frac{b_{n-\ell+1}}{b_{n+1}}\frac{1}{1-2^{-M+n-\ell-1}}=2^{-\ell}(1-2^{-M+n})c_{\ell}.$
(2.32)

Re-organising terms, we arrive at

$\displaystyle
h({\boldsymbol{z}}^{(n)})=\frac{\epsilon}{2}2^{M-n-1}\left(\frac{1}{1-2^{-M+n-1}}+(1-2^{-M+n})\sum_{\ell=1}^{n-1}c_{\ell}\right)$
$\displaystyle\quad+\frac{\epsilon}{2}2^{M}\left(\frac{1}{1-2^{-M+n-1}}\sum_{k=2}^{n}2^{-k}z_{k}+(1-2^{-M+n})\sum_{k=2}^{n-1}2^{-k}z_{k}\sum_{\ell=1}^{n-k}2^{-\ell}c_{\ell}\right).$
(2.33)

By (2.27), this is equal to

$\displaystyle\frac{\epsilon}{2}2^{M-n-1}\left(\frac{1}{1-2^{-M+n-1}}+(1-2^{-M+n})\sum_{\ell=1}^{n-1}c_{\ell}\right)$
$\displaystyle\frac{\epsilon}{2}2^{M}\Biggl{(}\frac{1}{1-2^{-M+n-1}}\sum_{k=2}^{n}2^{-k}z_{k}+\frac{1-2^{-M+n}}{1-2^{-M+n-1}}\sum_{k=2}^{n-1}2^{-k}z_{k}$
$\displaystyle-2^{-n}\sum_{k=2}^{n-1}z_{k}\frac{1-2^{-M+n}}{1-2^{-M+k-1}}\Biggr{)}.$
(2.34)

Even though this looks rather complicated, it is true that
$h({\boldsymbol{z}}^{(M)})=u({\boldsymbol{z}}^{(M)})$, as it should be. For
$n\ll M$, this simplifies to

$h({\boldsymbol{z}}^{(n)})=\frac{\epsilon}{2}2^{M}\left(n2^{-n-1}+2\sum_{k=2}^{n}2^{-k}z_{k}-2^{-n}\sum_{k=2}^{n}z_{k}\right)(1+O(2^{-M+n})),$
(2.35)

and hence, for $n\geq 2$,

$\displaystyle h(1^{(n)})$ $\displaystyle=$
$\displaystyle\frac{\epsilon}{2}2^{M}\left(n2^{-n-1}+1-n2^{-n}\right)(1+O(2^{-M+n}))$
(2.36) $\displaystyle=$
$\displaystyle\frac{\epsilon}{2}2^{M}\left(1-n2^{-n-1}\right)(1+O(2^{-M+n})),$

and

$h(1^{(1)})=\epsilon 2^{M-3}.$ (2.37)

∎

Next, we want to ensure that the spacing between particles decreases as $n$
increases, for which we need the following lemma.

######  Lemma 2.4.

Suppose that $z_{1}=1$ and
$u({\boldsymbol{z}}^{(M)})=\epsilon\sum_{\ell=2}^{M}2^{M-\ell}z_{\ell}+\tfrac{1}{2}\epsilon$.
If $h$ is a solution of (2.1), then, for all $0\leq n\leq M$,

$\min_{{{\boldsymbol{z}},{\boldsymbol{y}}\in{\mathbb{T}}^{(n)}}\atop{{\boldsymbol{z}}\neq{\boldsymbol{y}}}}|h({\boldsymbol{z}}^{(n)})-h({\boldsymbol{y}}^{(n)})|\geq\epsilon.$
(2.38)

###### Proof.

Let ${\boldsymbol{z}}^{(n)}>{\boldsymbol{y}}^{(n)}$. From (2.2) we get, for
$n<M$,

$\displaystyle
h({\boldsymbol{z}}^{(n)})-h({\boldsymbol{y}}^{(n)})=\frac{\epsilon}{2}2^{M}\Biggl{(}\frac{1}{1-2^{-M+n-1}}\sum_{k=2}^{n}2^{-k}(z_{k}-y_{k})+\frac{1-2^{-M+n}}{1-2^{-M+n-1}}\sum_{k=2}^{n-1}2^{-k}(z_{k}-y_{k})$
$\displaystyle\quad\quad-2^{-n}\sum_{k=2}^{n-1}(z_{k}-y_{k})\frac{1-2^{-M+n}}{1-2^{-M+k-1}}\Biggr{)}$
$\displaystyle\quad\geq\frac{\epsilon}{2}2^{M}\frac{1}{1-2^{-M+n-1}}2^{-n}\geq
4\epsilon/3.$ (2.39)

For $n=M$, the result holds since
$h({\boldsymbol{z}}^{(M)})=u({\boldsymbol{z}}^{(M)})$. ∎

###  2.3. Cost of spreading out the particles

We close this section by deriving bounds on the asymptotics of
$S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(M)})$ for $h^{\mathrm{Dir}}$
satisfying _linear_ boundary conditions.

######  Lemma 2.5.

Let $h^{\mathrm{Dir}}$ be the solution to the Dirichlet problem (2.1) with
boundary condition $u$ as in Lemma 2.4. Then

$\epsilon^{2}\frac{2^{2M-6}}{(1-2^{-M})^{2}}\leq
S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(M)})\leq\epsilon^{2}\frac{(2^{M-1}+\tfrac{1}{2})^{2}}{1-2^{-M}}.$
(2.40)

###### Proof.

Obviously,

$\displaystyle
S_{\mathrm{spr}}(h,{\mathbb{T}}^{(M)})\geq\frac{1}{2}\sum_{x_{1}=0}^{1}a(x_{1})^{2}.$
(2.41)

Hence

$S_{\mathrm{spr}}(h,{\mathbb{T}}^{(M)})\geq\frac{\Sigma(0)^{2}+\Sigma(1)^{2}}{2\left(2^{M}-1\right)^{2}}.$
(2.42)

Since $\Sigma(0)=-\Sigma(1)=\epsilon\,2^{2M-3}$, this gives the claimed lower
bound.

Choose any function that satisfies the boundary condition (2.4). For example,
if $h({\boldsymbol{z}}^{(M)})$ is linear with $h(0\dots
0)=-2^{M-1}-\frac{1}{2}$ and $h(1\dots 1)=2^{M-1}+\frac{1}{2}$, then we can
choose

$a(0)=-\frac{2^{M-1}+\tfrac{1}{2}}{1-2^{-M}}2^{-1},\,\,a(00)=-\frac{2^{M-1}+\tfrac{1}{2}}{1-2^{-M}}2^{-2},\,\,\dots,\,\,a(0\dots
0)=-\frac{2^{M-1}+\tfrac{1}{2}}{1-2^{-M}}2^{-M}.$ (2.43)

Note that

$a(0)+\dots+a(0\dots 0)=-2^{M-1}-\tfrac{1}{2},$ (2.44)

and choose

$a(x_{1}\dots x_{n})=a(0\dots 0)\,(-1)^{\sum_{i=1}^{n}x_{i}}.$ (2.45)

This choice satisfies the boundary conditions and, since

$\sum_{j=1}^{M}2^{j}\left(\frac{2^{M-1}+\tfrac{1}{2}}{1-2^{-M}}\right)^{2}2^{-2j}=\frac{(2^{M-1}+\tfrac{1}{2})^{2}}{1-2^{-M}},$
(2.46)

gives the claimed upper bound. ∎

Lemma 2.5 shows that, to leading order in $M$, the lower bound is
$\epsilon^{2}\,2^{2M-4}$ and the upper bound is $\epsilon^{2}\,2^{2M-2}$,
which differ by a factor $4$. The next lemma closes this gap.

######  Lemma 2.6.

As $M\to\infty$,

$S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(M)})=\epsilon^{2}2^{2M-3}+O(M^{2}2^{M}).$
(2.47)

###### Proof.

We have that

$\displaystyle S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(M)})$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{M}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\tfrac{1}{2}\left(a({\boldsymbol{z}}^{(n)})\right)^{2}=\sum_{n=1}^{M}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\tfrac{1}{2}\left(\bar{a}({\boldsymbol{z}}^{(n)}+O(1)\right)^{2}$
(2.48) $\displaystyle=$
$\displaystyle\sum_{n=1}^{M}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\tfrac{1}{2}\left(\bar{a}({\boldsymbol{z}}^{(n)}\right)^{2}+O(1)\sum_{n=1}^{M}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\bar{a}({\boldsymbol{z}}^{(n)}).$

Since $|\bar{a}({\boldsymbol{z}}^{(n)})|\leq\epsilon 2^{M-n}n$, the last term
is smaller than

$O(1)\sum_{n=1}^{M}2^{M}n=O(1)2^{M}M^{2}.$ (2.49)

The first summand is equal to

$\displaystyle\sum_{n=1}^{M}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\tfrac{1}{2}\left(\bar{a}({\boldsymbol{z}}^{(n)})\right)^{2}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{M}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\left(\epsilon
2^{M-n-2}\left(2-n+2\sum_{k=2}^{n}z_{k}\right)\right)^{2}$ $\displaystyle=$
$\displaystyle\epsilon^{2}2^{2M-4}\sum_{n=1}^{M}2^{-2n}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\left(2-n+2\sum_{k=2}^{n}z_{k}\right)^{2}$
$\displaystyle=$
$\displaystyle\epsilon^{2}2^{2M-4}\sum_{n=1}^{M}2^{-2n}\sum_{{\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}}\left((2-n)^{2}+4(2-n)\sum_{k=2}^{n}z_{k}+4\sum_{k,k^{\prime}=2}^{n}z_{k}z_{k^{\prime}}\right)$

Computing the sum over ${\boldsymbol{z}}^{(n)}\in{\mathbb{T}}^{(n)}$, we see
that this is equal to

$\displaystyle\epsilon^{2}2^{2M-4}\sum_{n=1}^{M}2^{-2n}\left((2-n)^{2}2^{n-1}+(2-n)2^{n}(n-1)+(n-1)2^{n}+n(n-1)2^{n-1}\right)$
(2.51) $\displaystyle=$
$\displaystyle\epsilon^{2}2^{2M-4}\sum_{n=1}^{M}2^{-n}(\tfrac{3}{2}n-1)=\epsilon^{2}2^{2M-4}(2-2^{-M-1}(3M+4)).$

This, together with (2.49), gives the assertion of the lemma. ∎

##  3\. Optimal strategy with interaction

We next turn to the problem of minimising the full functional
$S(h,{\mathbb{T}}^{(N)})$ in (1.10). In Section 3.1 we derive properties of
the optimal configurations. In Section 3.2 we show that optimal configurations
are specific admissible configurations. In Section 3.3 we compute the cost of
the optimal configurations. The main results in this section are collected in
Lemmas 3.1–3.12 below.

###  3.1. Properties of optimal configurations

First, we show that optimal configurations are restricted to stay on
$\epsilon\,{\mathbb{Z}}$.

######  Lemma 3.1.

Fix a range
${\mathcal{R}}=\epsilon\,[-2^{K-1}-\tfrac{1}{2},2^{K-1}+\tfrac{1}{2}]$, and
let $n\geq K$. Then the minimiser of $I_{n}$ with range restricted to
${\mathcal{R}}$ is

$h^{*}({\boldsymbol{z}}^{(n)})=\epsilon\left(\sum_{\ell=0}^{K-1}2^{\ell}z_{\ell}-2^{K-1}+\tfrac{1}{2}\right).$
(3.1)

###### Proof.

Let $h$ be any configuration of the $2^{n}$ particles on ${\mathcal{R}}$. We
first show that $I_{n}(h)\geq I_{n}(h^{*})$, where

$h^{*}({\boldsymbol{z}}^{(n)})=\epsilon(\ell-\tfrac{1}{2})\quad\text{if}\quad|h({\boldsymbol{z}}^{(n)})-\epsilon(\ell-\tfrac{1}{2})|<\tfrac{1}{2}\epsilon.$
(3.2)

Indeed,

$\displaystyle I_{n}(h^{*})$ $\displaystyle=$
$\displaystyle\sum_{{{\boldsymbol{z}}^{(n)},{\boldsymbol{y}}^{(n)}\in{\mathbb{T}}^{(n)}}\atop{{\boldsymbol{z}}^{(n)}\neq{\boldsymbol{y}}^{(n)}}}\mathbbm{1}_{|h^{*}({\boldsymbol{z}}^{(n)})-h^{*}({\boldsymbol{y}}^{(n)})|<\epsilon}=\sum_{{{\boldsymbol{z}}^{(n)},{\boldsymbol{y}}^{(n)}\in{\mathbb{T}}^{(n)}}\atop{{\boldsymbol{z}}^{(n)}\neq{\boldsymbol{y}}^{(n)}}}\mathbbm{1}_{h^{*}({\boldsymbol{z}}^{(n)})=h^{*}({\boldsymbol{y}}^{(n)})}$
(3.3) $\displaystyle=$
$\displaystyle\sum_{{{\boldsymbol{z}}^{(n)},{\boldsymbol{y}}^{(n)}\in{\mathbb{T}}^{(n)}}\atop{{\boldsymbol{z}}^{(n)}\neq{\boldsymbol{y}}^{(n)}}}\mathbbm{1}_{h^{*}({\boldsymbol{z}}^{(n)})=h^{*}({\boldsymbol{y}}^{(n)})}\mathbbm{1}_{|h({\boldsymbol{z}}^{(n)})-h({\boldsymbol{y}}^{(n)})|<\epsilon}$
$\displaystyle\leq$
$\displaystyle\sum_{{{\boldsymbol{z}}^{(n)},{\boldsymbol{y}}^{(n)}\in{\mathbb{T}}^{(n)}}\atop{{\boldsymbol{z}}^{(n)}\neq{\boldsymbol{y}}^{(n)}}}\mathbbm{1}_{|h({\boldsymbol{z}}^{(n)})-h({\boldsymbol{y}}^{(n)})|<\epsilon}=I_{n}(h).$

The last inequality is strict if there is any point that does not lie in
$\epsilon\,{\mathbb{Z}}$ and there are at least $2^{K}$ points, which is the
case for $n\geq K$. This shows that any minimiser $h$ must be supported on
$\epsilon\,{\mathbb{Z}}$. Thus, for any minimiser there must be $a_{\ell}$
particles at site $\epsilon(\ell-\tfrac{1}{2})$, $0\leq\ell<2^{K}$ with
$\sum_{\ell=0}^{2^{K}-1}a_{\ell}=2^{n}$. For such a configuration the
interaction energy is equal to

$I_{n,\epsilon}(a)=\sum_{\ell=0}^{2^{K}-1}a_{\ell}(a_{\ell}-1).$ (3.4)

Clearly, the constrained minimiser is given by $a_{\ell}=2^{n-K}$, which
proves the claim. ∎

######  Lemma 3.2.

Define the set of configurations

${\mathcal{H}}_{K}=\left\\{h\colon\,{\mathbb{T}}^{(N)}\to\epsilon\,[-2^{K-1}+\tfrac{1}{2},2^{K-1}-\tfrac{1}{2}]\right\\}.$
(3.5)

Set

$S^{K}(h,{\mathbb{T}}^{(N)})=\sum_{n=N-K}^{N}[I_{n}(h)+W_{n}(h)].$ (3.6)

Let $\bar{h}\in{\mathcal{H}}_{K}$ be such that, for any $n\geq N-K$,

$\bar{h}({\boldsymbol{z}}^{(n)})=h^{*}({\boldsymbol{z}}^{(n)}).$ (3.7)

Then

$S^{K}(\bar{h},{\mathbb{T}}^{(N)})=\min_{h\in{\mathcal{H}}_{K}}S^{K}(h,{\mathbb{T}}^{(N)}).$
(3.8)

###### Proof.

By Lemma 3.1, $\bar{h}$ minimises $I_{n}$ for all $0\leq n\leq N-K$. Moreover,
$W_{n}(h)=0$ for all $n\geq N-K$ (recall (1.7)), which proves the claim. ∎

The next lemma controls the range of optimal configurations.

######  Lemma 3.3.

Suppose that $h$ minimises $S(h,{\mathbb{T}}^{(N)})$ and that the range of
$h({\boldsymbol{z}}^{(N)})$ is $2^{K}$. Then, for any $0\leq n\leq N$, the
range of $h({\boldsymbol{z}}^{(n)})$ is at most $2^{K}$.

###### Proof.

Assume that, for some $0\leq n<N$, there exist a ${\boldsymbol{z}}$ such that
$h({\boldsymbol{z}}^{(n)})>\epsilon\,(2^{M-1}+\tfrac{1}{2})$. Compare the cost
of _moving this particle and all its offspring_ back to positions
$\leq\epsilon\,(2^{M-1}+\tfrac{1}{2})$ to the cost of just leaving one
particle at $h({\boldsymbol{z}}^{(n)})$. Clearly, the former is smaller than
the latter, because it costs nothing not to move the particle, while the cost
of the interaction is lowered because the particle does not get with distance
$\epsilon$ of any other particle. ∎

The next lemma identifies the optimal configurations that also optimise the
interaction cost in the last generation.

######  Lemma 3.4.

Let

$u({\boldsymbol{z}}^{(M)})=\epsilon\left(\sum_{\ell=1}^{M}2^{M-\ell}z_{\ell}-2^{M-1}+\tfrac{1}{2}\right).$
(3.9)

Let $h^{\mathrm{Dir}}({\boldsymbol{z}})$ be the solution to the Dirichlet
problem in (2.1) with boundary condition $u$. Set

$h^{**}({\boldsymbol{z}}^{(n)})=\begin{cases}h^{\mathrm{Dir}}({\boldsymbol{z}}^{(n)}),&0\leq
n\leq M,\\\ u({\boldsymbol{z}}^{(M)}),&n\geq M.\end{cases}$ (3.10)

If $M=2N/3$, then

$S(h^{**},{\mathbb{T}}^{(N)})={\beta}\left({\textstyle{1\over
3}}2^{4N/3+2}-2^{N+1}+{\textstyle{2\over
3}}2^{2N/3}\right)+S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(2N/3)}),$
(3.11)

where

$S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(2N/3)})=\epsilon^{2}\,2^{4N/3-3}+O(N2^{2N/3}).$
(3.12)

###### Proof.

The claim is straightforward from Lemma 3.3. Use Lemma 2.6 with $M=2N/3$. ∎

Lemma 3.4 essentially states that there exists a configuration $h^{**}$ with
total cost

$S(h^{**},{\mathbb{T}}^{(N)})\sim\left(\tfrac{4}{3}\,{\beta}+2^{-3}\epsilon^{2}\right)2^{4N/3}.$
(3.13)

At this stage we may be tempted to believe that the configurations $h^{**}$ in
Lemma 3.4 are optimisers. However, this is _not_ the case.

######  Lemma 3.5.

Let $h$ be any configuration such that at time $N$ there exists a point
$\epsilon\,k$ for which $|a_{k}-a_{k+1}|\geq 2$. Then $h$ cannot minimise
$S(h,{\mathbb{T}}^{(N)})$.

###### Proof.

Assume w.l.o.g. that $a_{k}=a_{k+1}+m$ with $m\geq 2$. Then one of the $m$
extra particles can be moved at cost $\leq\epsilon^{2}$ from $k$ to $k+1$ from
time $N-1$ to time $N$. But then the contribution to $I_{N}$ from this site
changes from $(a_{k+1}+m)(a_{k+1}+m-1)+a_{k+1}(a_{k+1}-1)$ to
$(a_{k+1}+m-1)(a_{k+1}+m-2)+(a_{k+1}+1)a_{k+1}$. Consequently, the interaction
cost is reduced by $2(m-1){\beta}\geq 2{\beta}$, which offsets the cost of the
displacement because we assumed that ${\beta}>\epsilon^{2}/2$. Hence such a
configuration cannot be a minimiser. ∎

######  Remark.

Lemma (3.5) is the reason we impose the condition ${\beta}>\epsilon^{2}/2$. If
${\beta}$ is smaller than that, one cannot exclude steps size larger than one.
In fact, as ${\beta}$ gets smaller, increasingly larger steps become
admissible. This will lead to optimal configurations where the boundary
staircases become increasingly steep.

###  3.2. Admissible configurations

We have seen that without the restriction on the increments to be at most 1,
the optimal configurations are constants. The next lemma identifies the
optimal configurations under this restrictions.

######  Lemma 3.6.

Consider the functional $I_{n,\epsilon}(a)$ from (3.4). Then the minimiser of
$I_{n,\epsilon}(a)$ with range $\\{1,\dots,L\\}$ for $L\in{\mathbb{N}}_{0}$,
under the constraint $\sum_{\ell=0}^{L}a_{\ell}=2^{n}$ and with the
restriction $|a_{\ell+1}-a_{\ell}|\leq 1$, $1\leq\ell<L$, is given by

$a_{\ell}=\min\left(\tfrac{1}{2}({\lambda}^{*}+1),\min(\ell,L+1-\ell)\right),$
(3.14)

where

${\lambda}^{*}=L-\sqrt{(L+1)^{2}-2^{n+2}}.$ (3.15)

###### Proof.

The constrained optimisation problem without the restriction on the increments
amounts to finding the critical point of the functional

$F(a,{\lambda})=I_{n,\epsilon}(a)-{\lambda}\left(\sum_{\ell=1}^{L}a_{\ell}-2^{n}\right).$
(3.16)

The gradient of $F$ has the components

$\displaystyle\frac{{\partial}}{{\partial}a_{\ell}}F(a,{\lambda})$
$\displaystyle=2a_{\ell}-1-{\lambda},\quad 1\leq\ell\leq L,$ (3.17)
$\displaystyle\frac{{\partial}}{{\partial}{\lambda}}F(a,{\lambda})$
$\displaystyle=\sum_{\ell=1}^{L}a_{\ell}-2^{n}.$

We claim that in order to take the restriction on the increments into account,
it is enough to impose the weaker constraints

$0\leq a_{\ell}\leq\min(\ell,L+1-\ell).$ (3.18)

We search for $a$ and ${\lambda}$ such that the gradient vanishes except on
the boundary of the admissible region. For given ${\lambda}$ this means that
$a$ is given by

$a=a^{\lambda}_{\ell}=\min\left(\tfrac{1}{2}({\lambda}+1),\min(\ell,L+1-\ell)\right),$
(3.19)

and ${\lambda}$ by

$2\sum_{\ell=1}^{({\lambda}+1)/2}\ell+\tfrac{1}{2}(L-{\lambda}-1)(1+{\lambda})-2^{n}=0.$
(3.20)

The latter has the unique solution

${\lambda}^{*}=L-\sqrt{(L+1)^{2}-2^{n+2}},$ (3.21)

and $a=a^{{\lambda}^{*}}$. ∎

As a consequence, the configuration that minimises $I_{N}$ with range
${\mathcal{R}}_{N}=L$ is in ${\mathcal{H}}_{r_{N},d_{N},N}$ with

$r_{N}=\tfrac{1}{2}({\lambda}^{*}+1),$ (3.22)

and $d_{N}=L-2r_{N}$.

######  Lemma 3.7.

The configuration at time $n-1$ that optimises $I_{n-1,\epsilon}$ with a range
that allows to obtain the configuration in ${\mathcal{H}}_{r_{n},d_{n},n}$ for
some given $r_{n}$ at time $n$ is the configuration in
${\mathcal{H}}_{r_{n}/2,d_{n}+r_{n},n-1}$ with minimal spreading cost.

###### Proof.

Consider a configuration in ${\mathcal{H}}_{r_{n}/2,d_{n}+r_{n},n-1}$. At time
$n$, each of the particles splits into two and without moving (i.e., without
any spreading cost), we obtain a configuration where the occupation number at
each site is doubled. This configuration is not smooth. Now, we can apply the
procedure described in Lemma 3.5 to move particles in steps each of which
lowers the interaction energy by more than the spreading cost involved in this
move. This involves moving particles only in the staircase and a strip of
width $\tfrac{1}{4}r_{n}$ from each side of the central part.

The total number of particles in the staircase on the $r_{n}$ sites is given
by $\tfrac{1}{2}r_{n}(r_{n}+1)$. As $r_{n-1}=\tfrac{1}{2}r_{n}$, a staircase
of width $r_{n-1}$ contains $\tfrac{1}{4}r_{n}(\tfrac{1}{2}r_{n}+1)$
particles, which produce $\tfrac{1}{2}r_{n}(\tfrac{1}{2}r_{n}+1)$ offsprings
in the next generation. The remaining $\tfrac{1}{2}r_{n}^{2}$ particles come
from the central part. In total, these are

$\tfrac{1}{2}r_{n}(\tfrac{1}{2}r_{n}+1)+\tfrac{1}{4}r_{n}^{2}=\tfrac{1}{2}r_{n}(r_{n}+1)$
(3.23)

particles for each of the two staircases of width $r_{n}$. The resulting
configuration is in ${\mathcal{H}}_{r_{n},d_{n},n}$, in particular it is the
optimal configuration for $I_{n,\epsilon}$ with range $2r_{n}+d_{n}$. Thus,
moving any more particles would increase the spreading cost and could not
reduce the interaction cost without increasing the range. Note that

$d_{n-1}=d_{n}+\tfrac{1}{2}r_{n},$ (3.24)

and hence

$L_{n-1}=d_{n-1}+2r_{n-1}=d_{n}+\tfrac{3}{2}r_{n}=L_{n}-\tfrac{1}{2}r_{n}.$
(3.25)

∎

######  Lemma 3.8.

Let $h$ be a minimiser of $S(h,{\mathbb{T}}^{(N)})$ such that its range is
$L_{N}$. Then for $M\leq n\leq N$, with $M$ such that $2^{-(N-M)}r_{N}=1$,
$h({\mathbb{T}}^{(n)})$ is as described above, i.e., it is the minimising
admissible configuration with range

$L_{n}=L_{N}-2r_{N}+(1-2^{-(N-n)})\tfrac{1}{2}r_{N}+2^{-(N-n)+1}r_{N}.$ (3.26)

###### Proof.

The structure of the proof is as follows.

  * •

Let $h$ be a minimiser of $S(h,{\mathbb{T}}^{(N)})$. Assume that the range of
$h$ at time $N$ is $L_{N}$. We will show that then the range of $h$ at times
$n<N$ is not larger than some $L_{n}$. This $L_{n}$ will be determined by the
fact that if the range at time $n$ were larger than $L_{n}$, then we could
construct a configuration with a smaller total cost and a range at time $N$
larger than $L_{N}$, in contradiction with the assumption that the optimal
configuration has range $L_{N}$. This procedure ends at time $M$ when
$L_{M}=\epsilon\,2^{M}$ and the interaction cost is zero for the configuration
with a single particle at each site.

  * •

At times before time $M$, the best configuration is given by the solution of
the Dirichlet problem discussed in Section 2. Given these constraints on the
ranges, the configuration that minimises the total interaction cost is
realised if, for any $M\leq n\leq N$, $h({\boldsymbol{z}}^{(n)})$ is such that
it minimises $I_{n,\epsilon}$ over all admissible configurations with range at
most $L_{n}$, which implies that the range is equal to $L_{n}$.

  * •

Finally, we show that from such a configuration at time $n$ after doubling all
particles, we can arrive at the configuration at time $n+1$ by moving
particles in such a way that total cost is reduced in each step.

We will first establish the existence of a sequence $L_{n}$ and a
configuration $h$ as described above. If $h$ minimises $I_{n}$ with range
$L_{n}$, then we know that $L_{n}=d_{n}+2r_{n}$ and
$h({\boldsymbol{z}}^{(n)})\in{\mathcal{H}}_{r_{n},d_{n},n}$. It follows from
Lemma 3.7 that $r_{n}=r_{N}2^{-N+n}$. It is also clear that $r_{N}$ is a power
of $2$, say $r_{N}=2^{K}$, and then $M=N-K$. Following the procedure in Lemma
3.7, we obtain $h$ such that, for each $M\leq n\leq N$,
$h({\boldsymbol{z}}^{(n)})\in{\mathcal{H}}_{r_{n},d_{n},n}$. Since there is no
way to further reduce the interaction energies without increasing the final
range $L_{N}$, and any further move would increase the spreading costs, this
$h$ optimises the total cost from time $M$ to time $N$. Finally, for times up
to $M$, choosing for $h$ the solution of the Dirichlet problem also minimises
the cost during this time. As a consequence, we have obtained the minimiser
under the assumption that its range is $L_{N}$. ∎

###  3.3. Cost of the optimal configurations

In Lemma 3.8, we have identified the candidate minimisers $h$ assuming that
the range of $h$ at time $N$ is $2r_{N}+d_{N}=L_{N}$. We denote these
configurations by $h^{*}_{r}$, is $r_{N}=r$. We still have to determine the
value of $r_{N}$. To do so, we must compute very precisely the total cost of
such a configuration.

######  Lemma 3.9.

The interaction cost of a configuration $h^{*}_{r}$ is given by

$J_{N,\epsilon}(h^{*}_{r},{\mathbb{T}}^{(N)})=\tfrac{1}{3}2^{N+2}+2r-2^{N+1}+O(r^{2}).$
(3.27)

###### Proof.

The interaction energy at generation $n$ of a configuration in
${\mathcal{H}}_{r,d,n}$ is

$2\sum_{j=1}^{r}(j^{2}-j)+dr(r-1)=\tfrac{1}{3}r(r+1)(2r+1)-r(r+1)+dr(r-1).$
(3.28)

Hence, by (1.11) and (3.24),

$\displaystyle I_{N-k,\epsilon}(h^{*}_{r})$ $\displaystyle=$
$\displaystyle\tfrac{1}{3}r_{k}(r_{k}+1)(2r_{k}+1)-r_{k}(r_{k}+1)+d_{k}r_{k}(r_{k}-1)$
(3.29) $\displaystyle=$
$\displaystyle\tfrac{2}{3}r_{k}^{3}-\tfrac{2}{3}r_{k}+d_{k}r_{k}(r_{k}-1)$
$\displaystyle=$
$\displaystyle\tfrac{2}{3}r^{3}2^{-3k}-\tfrac{2}{3}r2^{-k}+(d+r(1-2^{-k})(r^{2}2^{-2k}-r2^{-k}).$

Summing this over $k$ from $0$ to $N-M$, we get the total interaction energy

$\displaystyle
J_{N,\epsilon}(h^{*}_{r},{\mathbb{T}}^{(N)})=\sum_{k=0}^{{N-M}}I_{N-k}$
$\displaystyle=\tfrac{2}{3}r^{3}\tfrac{1}{7}(8-8^{-{(N-M)}})-\tfrac{2}{3}r(2-2^{-{(N-M)}})+(d+r)\left(r^{2}\tfrac{1}{3}(4-4^{-{(N-M)}})-r(2-2^{-{(N-M)}})\right)$
$\displaystyle\qquad-r^{3}\tfrac{1}{7}(8-8^{-{(N-M)}})+r^{2}\tfrac{1}{3}(4-4^{-{(N-M)}}).$
(3.30)

Since $2^{-{(N-M)}}=1/r$, we can neglect all terms with powers $-{(N-M)}$.
Using that $d=(2^{N}/r)-r-1$, we get

$J_{N,\epsilon}(h^{*}_{r},{\mathbb{T}}^{(N)})=\tfrac{4}{3}\,2^{N}r-\tfrac{8}{21}\,r^{3}-2^{N+1}+O(r^{2}).$
(3.31)

∎

To compute the total cost of a configuration $h^{*}_{r}$, it remains to
compute the spreading cost. This consists of two parts. First, there is the
spreading cost from the Dirichlet problem up to time $M$, which is estimated
in Lemma 2.6. The second part is the spreading cost from time $M$ to time $N$.
We do not have a precise formula for this, but we know that it is smaller than
the difference between the interaction cost of $h^{*}_{r}$ and the
configuration obtained from not moving particles any more after time $M$.

######  Lemma 3.10.

With the notation above,

$0\leq
S_{\mathrm{spr}}(h^{*}_{r},{\mathbb{T}}^{(N)})-S_{\mathrm{spr}}(h^{\mathrm{Dir}},{\mathbb{T}}^{(M)})\leq
O(r^{3}).$ (3.32)

###### Proof.

The cost of a configuration where particles only reproduce and do not move
from time $M$ to time $N$ is

$2^{M}\sum_{k=1}^{N-M}2^{k}(2^{k}-1)=\tfrac{1}{3}2^{2N-M+2}-2^{N+1}+\tfrac{2}{3}=\tfrac{1}{3}2^{N+2}r-2^{N+1}+\tfrac{2}{3},$
(3.33)

where $r=2^{{N-M}}$. This yields the assertion of the lemma. ∎

Collecting these result, we get the following lemma.

######  Lemma 3.11.

$S_{N}(h^{*}_{r},{\mathbb{T}}^{(N)})={\beta}\left(\tfrac{4}{3}2^{N}r-2^{N+1}\right)+\epsilon^{2}2^{2N-3}r^{-2}+O(r^{3}).$
(3.34)

Lemma 3.11 allows us to obtain sharper bounds on $r$ as stated in Theorem 1.1.

######  Lemma 3.12.

Let $h$ be any minimiser of $S(h,{\mathbb{T}}^{(N)})$. Then
$h\in{\mathcal{H}}_{r,d,N}$ with

$|r-r^{*}|\leq O(2^{N/6}),$ (3.35)

where

$r^{*}=\arg\min_{r}\left({\beta}\left(\tfrac{1}{3}2^{N+2}r-2^{N+1}\right)+\epsilon^{2}2^{2N-3}r^{-2}\right).$
(3.36)

Moreover,

$r^{*}=(3\epsilon^{2}/{\beta})^{1/3}2^{(N-4)/3}$ (3.37)

and

$S(h^{*}_{r^{*}},{\mathbb{T}}^{(N)})=2({\beta}\epsilon/3)^{2/3}2^{4N/3+2/3}+O(2^{N}).$
(3.38)

###### Proof.

The computing of $r^{*}$ is straightforward, while (3.35) follows from an
elementary perturbation argument. Note that, for the minimising $h$, the
interaction cost and the spreading cost are the same. ∎

##  4\. Proof of the main theorems

We are now ready to prove the main theorems.

###### Proof of Theorem 1.1.

Lemma 3.1 shows that any optimising configuration at time $N$ has particles
only on $\epsilon\,{\mathbb{Z}}$. Lemmas 3.5–3.6 shows that such
configurations must be admissible and belong to some ${\mathcal{H}}_{r,d,N}$.
Lemma 3.12 provides a control on the value of $r$. Taking the logarithm, we
get

$\ln_{2}(r)=\ln_{2}\left(3\epsilon^{2}/{\beta})^{1/3}2^{(N-4)/3}+O(2^{N/6})\right)=\tfrac{1}{3}(N-4)+\tfrac{1}{3}\ln_{2}(3\epsilon^{2}/{\beta})+\ln_{2}\left(1+O(2^{-N/6})\right),$
(4.1)

which implies Theorem 1.1 as $\ln_{2}\left(1+O(2^{-N/6})\right)=O(2^{-N/6})$.
∎

###### Proof of Theorem 1.2.

Theorem 1.2 is a direct consequence of Lemmas 3.7 and 3.12. ∎

###### Proof of Theorem 1.3.

Theorem 1.3 is a direct consequernce of Lemmas 3.7 and 2.3. ∎

###### Proof of Theorem 1.4.

Theorem 1.4 is a direct consequence of Lemma 3.12. ∎

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# Fractional moments

Óscar Ciaurri  Departamento de Matemáticas y Computación, Universidad de La
Rioja, Complejo Científico-Tecnológico, Calle Madre de Dios 53, 26006 Logroño,
Spain<EMAIL_ADDRESS>
###### Abstract.

We evaluate the moments of some functions composed with the fractional part of
$1/x$. We name them fractional moments. In particular, we obtain expressions
for the fractional moments of some trigonometric functions, the Bernoulli
polynomials and the functions $x^{m}$ and $x^{m}(1-x)^{m}$.

###### Key words and phrases:

Moments, fractional part, polygamma, Bernoulli polynomials.

###### 2010 Mathematics Subject Classification:

Primary: 33B10, 33B15, 11B68.

The author was supported by grant PGC2018-096504-B-C32 AEI, from Spanish
Government.

##  1\. Introduction

The main purpose of this paper is the evaluation of moments of some functions
composed with the fractional part of $1/x$. In fact, we will analyze the
integrals

$I_{k}f=\int_{0}^{1}x^{k}f\left(\bigg{\\{}\frac{1}{x}\bigg{\\}}\right)\,dx,\qquad
k=0,1,2,\dots.$

We name this integrals the fractional moments of the function $f$.

Up to our knowledge, the particular case $f(x)=x^{m}$ appears in the
literature in different references (see [2, 3, 4, 7, 9, 10]). For example, in
[4, Theorem 2.1] or [3, Problem 22.2] we can see the identity

(1)
$\int_{0}^{1}x^{k}\left\\{\frac{1}{x}\right\\}^{m}\,dx=\frac{m!}{(k+1)!}\sum_{j=1}^{\infty}\frac{(k+j)!}{(m+j)!}(\zeta(k+j+1)-1),\qquad
m,k\in\mathbb{Z}^{+},$

where $\zeta$ denotes the zeta Riemann function. Moreover, for the particular
case $m=k+1$, by using the identity

$\sum_{j=1}^{\infty}\frac{\zeta(j+1)-1}{j+1}=1-\gamma,$

with $\gamma$ being the Euler-Mascheroni constant, it is proved that

$\int_{0}^{1}x^{k}\left\\{\frac{1}{x}\right\\}^{k+1}\,dx=H_{k+1}-\gamma-\sum_{j=2}^{k+1}\frac{\zeta(j)}{j},$

where $H_{n}$ is the $n$-th harmonic number. In this paper we will make a
systematic analysis of this kind of integrals and we will obtain appropriated
closed forms for them.

The main tool to evaluate fractional moments will be an identity relating them
with an integral involving the function $\log\Gamma(x+1)$ and the derivatives
of the function $f$ (see Lemma 1 in the next section). From this result, we
deduce the fractional moments for some functions. In particular, we analyze
the fractional moments for some trigonometric functions, the Bernoulli
polynomials and the functions $x^{m}$ and $x^{m}(1-x)^{m}$. The integrals
containing $\log\Gamma(x+1)$ will be evaluated by using some results in [1].
In that paper, we see that the integrals

$\int_{0}^{1}B_{n}(x)\log\Gamma(x)\,dx,$

where $B_{n}$ are the Bernoulli polynomials, are simpler to evaluate than the
integrals

$\int_{0}^{1}x^{n}\log\Gamma(x)\,dx.$

In fact, the latter integrals are evaluated in terms of the former ones. By
this reason, to analyze the fractional moments of $x^{m}(1-x)^{m}$ we start
giving an expansion for these functions and their derivatives in terms of the
Bernoulli polynomials (see Lemma 7 and the remark following it). We believe
that this result has its own interest.

The paper is organized as follows. In Section 2 we present the lemma allowing
us to obtain the fractional moments and in the rest of the paper we show some
examples and applications related to trigonometric functions, Bernoulli
polynomials and the functions $x^{m}$ and $x^{m}(1-x)^{m}$.

All along the paper when an empty sum appears it must be taken as zero.

##  2\. The main lemma

The following lemma will be the main tool to evaluate fractional moments.
Before, we define the sequence $\alpha_{n}=\zeta(n+1)$, for $n>0$, and
$\alpha_{0}=\gamma$.

######  Lemma 1.

Let $f$ be a function having $k+2$ derivatives in the interval $[0,1]$. Then

$I_{k}f=\frac{1}{(k+1)!}\Bigg{(}\sum_{j=0}^{k}(k-j)!\left(f^{(j)}(0)\alpha_{k-j}-f^{(j)}(1)(\alpha_{k-j}-1))\right)\\\
+\int_{0}^{1}f^{(k+2)}(x)\log\Gamma(x+1)\,dx\Bigg{)},\qquad k\geq 0.$

The polygamma function will play a crucial role in the proof of the following
lemma. It is defined as the $(m+1)$-th derivative of the logarithm of the
gamma function

$\psi^{(m)}(x)=\frac{d^{m+1}}{dx^{m+1}}\log\Gamma(x),\qquad m=0,1,2,\dots.$

Two facts about the polygamma function will be fundamental. Its representation
as a series

(2) $\psi^{(m)}(x)=(-1)^{m+1}m!\sum_{k=0}^{\infty}\frac{1}{(x+k)^{m+1}},$

and its values in the positive integers

$\displaystyle\psi^{(m)}(n)$
$\displaystyle=(-1)^{m+1}m!\left(\zeta(m+1)-\sum_{k=1}^{n-1}\frac{1}{k^{m+1}}\right)$
(3) $\displaystyle=(-1)^{m+1}m!\sum_{k=n}^{\infty}\frac{1}{k^{m+1}},\qquad
m,n=1,2,\dots,$

and

(4) $\phi^{(0)}(n)=-\gamma+\sum_{k=1}^{n-1}\frac{1}{k},\qquad n=1,2,\dots.$

###### Proof of Lemma 1..

With the change of variable $w=1/x$, we have

$\displaystyle I_{k}f$
$\displaystyle=\int_{1}^{\infty}f(\\{w\\})\,\frac{dw}{w^{k+2}}=\sum_{j=1}^{\infty}\int_{j}^{j+1}f(w-j)\,\frac{dw}{w^{k+2}}$
$\displaystyle=\sum_{j=1}^{\infty}\int_{0}^{1}f(s)\,\frac{ds}{(j+s)^{k+2}}=\frac{(-1)^{k}}{(k+1)!}\int_{0}^{1}f(s)\psi^{(k+1)}(s+1)\,ds,$

where in the last step we have used (2). Now, applying integration by parts
$k+2$ times and taking into account that $\log\Gamma(2)=\log\Gamma(1)=0$, we
arrive at

$I_{k}f=\frac{(-1)^{k}}{(k+1)!}\Bigg{(}\sum_{j=0}^{k}(-1)^{j}(f^{(j)}(1)\psi^{(k-j)}(2)-f^{(j)}(0)\psi^{(k-j)}(1))\\\
+(-1)^{k+2}\int_{0}^{1}f^{(k+2)}(s)\log\Gamma(s+1)\,ds\Bigg{)}.$

Finally, we conclude the proof applying (3) and (4). ∎

##  3\. Fractional moments for trigonometric functions

We start our examples given the fractional moments for the sine and cosine
functions. The expression that we obtain for them involve the classical
functions sine integral

$\operatorname{Si}(x)=\int_{0}^{x}\frac{\sin t}{t}\,dt$

and cosine integral

$\operatorname{Ci}(x)=-\int_{x}^{\infty}\frac{\cos t}{t}\,dt.$

The following lemma contains the evaluation of two integrals for
$\log\Gamma(x+1)$ with trigonometric functions.

######  Lemma 2.

It is verified that

$\int_{0}^{1}\sin(2\pi
x)\log\Gamma(x+1)\,dx=\frac{\operatorname{Ci}(2\pi)}{2\pi}$

and

$\int_{0}^{1}\cos(2\pi
x)\log\Gamma(x+1)\,dx=\frac{1}{4}-\frac{\operatorname{Si}(2\pi)}{2\pi}.$

###### Proof.

Taking $n=1$ in [5, 6.443.1 and 6.443.3], we have

$\int_{0}^{1}\sin(2\pi x)\log\Gamma(x)\,dx=\frac{\log 2\pi+\gamma}{2\pi}$

and

$\int_{0}^{1}\cos(2\pi x)\log\Gamma(x)\,dx=\frac{1}{4}.$

Then

$\int_{0}^{1}\sin(2\pi x)\log\Gamma(x+1)\,dx=\frac{\log
2\pi+\gamma}{2\pi}+\int_{0}^{1}\sin(2\pi x)\log x\,dx$

and

$\int_{0}^{1}\cos(2\pi x)\log\Gamma(x+1)\,dx=\frac{1}{4}+\int_{0}^{1}\cos(2\pi
x)\log x\,dx.$

Now, applying integration by parts and the identity [5, 8.230.2]

$\operatorname{Ci}(x)=\gamma+\log x+\int_{0}^{x}\frac{\cos t-1}{t}\,dt,$

we deduce

$\displaystyle\int_{0}^{1}\sin(2\pi x)\log x\,dx$
$\displaystyle=\frac{1}{2\pi}\int_{0}^{1}\frac{\cos(2\pi x)-1}{x}\,dx$
$\displaystyle=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\cos t-1}{t}\,dt=-\frac{\log
2\pi+\gamma}{2\pi}+\frac{\operatorname{Ci}(2\pi)}{2\pi}$

and the result for the integral with the sine follows. The integral with the
cosine can be obtained by using integration by parts only. Indeed,

$\int_{0}^{1}\cos(2\pi x)\log x\,dx=-\frac{1}{2\pi}\int_{0}^{1}\frac{\sin(2\pi
x)}{x}\,dx=-\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\sin
t}{t}\,dt=-\frac{\operatorname{Si}(2\pi)}{2\pi}.\qed$

With the notation

$f_{s}(x)=\sin(2\pi x)\qquad\text{ and }\qquad f_{c}(x)=\cos(2\pi x),$

we have

$f_{s}^{(2j)}(x)=(-1)^{j}(2\pi)^{2j}\sin(2\pi x),\qquad
f_{s}^{(2j+1)}(x)=(-1)^{j}(2\pi)^{2j+1}\cos(2\pi x),$
$f_{c}^{(2j)}(x)=(-1)^{j}(2\pi)^{2j}\cos(2\pi x),\quad\text{and}\quad
f_{c}^{(2j+1)}(x)=(-1)^{j+1}(2\pi)^{2j+1}\sin(2\pi x).$

Moreover, $f_{s}^{(2j)}(0)=f_{s}^{(2j)}(1)=0$,
$f_{s}^{(2j+1)}(0)=f_{s}^{(2j+1)}(1)=(-1)^{j}(2\pi)^{2j+1}$,
$f_{c}^{(2j+1)}(0)=f_{c}^{(2j+1)}(1)=0$, and
$f_{c}^{(2j)}(0)=f_{c}^{(2j)}(1)=(-1)^{j}(2\pi)^{2j}$.

Taking

$\mathcal{S}_{k}=\int_{0}^{1}x^{k}f_{s}\left(\left\\{\frac{1}{x}\right\\}\right)\,dx\qquad\text{
and
}\qquad\mathcal{C}_{k}=\int_{0}^{1}x^{k}f_{c}\left(\left\\{\frac{1}{x}\right\\}\right)\,dx$

we have the following result.

######  Theorem 1.

For $n\geq 0$, it is verified that

$\mathcal{S}_{2n}=\frac{(-1)^{n+1}(2\pi)^{2n+1}}{(2n+1)!}\left(\sum_{j=0}^{n-1}\frac{(-1)^{j}(2j+1)!}{(2\pi)^{2j+2}}+\operatorname{Ci}(2\pi)\right),$
$\mathcal{S}_{2n+1}=\frac{(-1)^{n}(2\pi)^{2n+2}}{(2n+2)!}\left(\sum_{j=0}^{n}\frac{(-1)^{j}(2j)!}{(2\pi)^{2j+1}}-\frac{\pi}{2}+\operatorname{Si}(2\pi)\right),$
$\mathcal{C}_{2n}=\frac{(-1)^{n}(2\pi)^{2n+1}}{(2n+1)!}\left(\sum_{j=0}^{n}\frac{(-1)^{j}(2j)!}{(2\pi)^{2j+1}}-\frac{\pi}{2}+\operatorname{Si}(2\pi)\right),$

and

$\mathcal{C}_{2n+1}=\frac{(-1)^{n}(2\pi)^{2n+2}}{(2n+2)!}\left(\sum_{j=0}^{n}\frac{(-1)^{j}(2j+1)!}{(2\pi)^{2j+2}}+\operatorname{Ci}(2\pi)\right).$

###### Proof.

The identities can be deduced immediately by using Lemma 1, Lemma 2, and the
given properties about the derivatives of the functions $f_{s}$ and $f_{c}$. ∎

##  4\. Fractional moments for Bernoulli polynomials

Now, we analyze the fractional moments for the Bernoulli polynomials
$B_{n}(x)$. They can be defined through its exponential generating function

$\frac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n}(x)\frac{t^{n}}{n!},$

converging for $|t|<\pi$. It is well known that
$B_{n}(0)=(-1)^{n}B_{n}(1)=B_{n}$, where $B_{n}$ are the Bernoulli numbers.
Remember that $B_{2k+1}=0$, for $k\geq 1$, and $B_{1}=-1/2$. A main tool in
our approach will be the identity $B_{n}^{\prime}(x)=nB_{n-1}(x)$ (Bernoulli
polynomials are, in fact, a particular case of Appell polynomials). More
generally, it is verified

(5) $B_{n}^{(k)}(x)=\frac{n!}{(n-k)!}B_{n-k}(x),\qquad n\geq k.$

A crucial point to obtain a proper expression for the fractional moments of
the Bernoulli polynomials is the following identity (see [1, (6.5) and (6.6)])

(6) $\int_{0}^{1}B_{n}(x)\log\Gamma(x)\,dx=a_{n},\qquad n\geq 0,$

where the sequence $a_{n}$ is defined by

(7) $a_{n}=\begin{cases}-\zeta^{\prime}(-n),&n=0,2,4,\dots,\\\
\frac{B_{n+1}}{n+1}\left(\frac{\zeta^{\prime}(n+1)}{\zeta(n+1)}-\log(2\pi)-\gamma\right),&n=1,3,5\dots,\end{cases}$

with $\zeta^{\prime}$ being the derivative of the Riemann function. It is
convenient to remember that $\zeta^{\prime}(0)=-\log\sqrt{2\pi}$ and

$\zeta^{\prime}(2n)=(-1)^{n}\frac{(2n)!\zeta(2n+1)}{2(2\pi)^{2m}}.$

Moreover, we consider the sequence

$b_{n}=a_{n}-\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{B_{n+1-k}}{k},\qquad
n\geq 0.$

######  Lemma 3.

For $n\geq 0$, it is verified that

$\int_{0}^{1}B_{n}(x)\log\Gamma(x+1)\,dx=b_{n}.$

###### Proof.

From (6), we have

$\int_{0}^{1}B_{n}(x)\log\Gamma(x+1)\,dx=\int_{0}^{1}B_{n}(x)\log
x\,dx+a_{n}.$

Now, applying integration by parts to the first integral and using the
identity

(8) $B_{m}(x)=\sum_{k=0}^{m}\binom{m}{k}B_{m-k}x^{k}$

(which can be deduced from (5) by using that $B_{n}(0)=B_{n}$), we obtain that

$\displaystyle\int_{0}^{1}B_{n}(x)\log x\,dx$
$\displaystyle=-\frac{1}{n+1}\int_{0}^{1}\frac{B_{n+1}(x)-B_{n+1}}{x}\,dx$
$\displaystyle=-\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}B_{n+1-k}\int_{0}^{1}x^{k-1}\,dx$
$\displaystyle=-\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{B_{n+1-k}}{k}$

and the proof is completed. ∎

Taking

$J_{k}^{n}=\int_{0}^{1}x^{k}B_{n}\left(\left\\{\frac{1}{x}\right\\}\right)\,dx,$

we have the following result.

######  Theorem 2.

For $n\geq 1$, it is verified that

$J_{k}^{n}=\frac{1}{(k+1)\binom{k}{n}}\Bigg{(}\sum_{j=0}^{[n/2]}\binom{k-n+2j}{2j}B_{2j}+(k-n+1)\bigg{(}\frac{1}{2}-\zeta(k-n+2)\bigg{)}\Bigg{)},$

for $k\geq n$,

$J_{n-1}^{n}=\sum_{j=1}^{[n/2]}\frac{B_{2j}}{2j}+\frac{1}{2}-\gamma,$

and,

$J_{k}^{n}=\frac{1}{k+1}\binom{n}{k}\Bigg{(}\sum_{j=[(n-k+1)/2]}^{[n/2]}\frac{B_{2j}}{\binom{2j}{k-n+2j}}+(n-k)(n-k-1)b_{n-k-2}\Bigg{)},$

for $0\leq k\leq n-2$.

###### Proof.

The result follows by applying Lemma 1, Lemma 3, and the properties of
Bernoulli polynomials and Bernoulli numbers. ∎

##  5\. Fractional moments for the functions $x^{m}$

Considering the functions $h_{m}(x)=x^{m}$, with $m\geq 1$, in this section we
evaluate their fractional moments. More precisely, we calculate the integrals

$\mathcal{C}_{k}^{m}=\int_{0}^{1}x^{k}h_{m}\left(\left\\{\frac{1}{x}\right\\}\right)\,dx.$

To this end, we need the two following lemmas.

######  Lemma 4.

For $n\geq 0$, it is verified that

$\int_{0}^{1}x^{n}\log\Gamma(x+1)\,dx=-\frac{1}{(n+1)^{2}}+\frac{1}{n+1}\sum_{k=0}^{n}\binom{n+1}{k}a_{k},$

where the sequence $a_{k}$ was defined in (7).

###### Proof.

By using the identity111This fact follows from (8) applying the identity
$\sum_{k=0}^{m}\binom{m+1}{k}=\begin{cases}1,&m=0\\\ 0,&m\not=0\end{cases},$
which can be proved by using the exponential generating function
$\frac{t}{e^{t}-1}=\sum_{k=0}^{\infty}\frac{B_{k}}{k!}t^{k}.$

(9) $x^{n}=\frac{1}{n+1}\sum_{j=0}^{n}\binom{n+1}{j}B_{j}(x),$

it is clear that

$\displaystyle\int_{0}^{1}x^{n}\log\Gamma(x+1)\,dx$
$\displaystyle=\int_{0}^{1}x^{n}\log x\,dx+\int_{0}^{1}x^{n}\log\Gamma(x)\,dx$
$\displaystyle=-\frac{1}{(n+1)^{2}}+\frac{1}{n+1}\sum_{j=0}^{n}\binom{n+1}{j}\int_{0}^{1}B_{j}(x)\log\Gamma(x)\,dx$

and the results is obtained applying (6). ∎

######  Lemma 5.

The identities

$\sum_{j=0}^{m}\frac{(k-j)!}{(m-j)!}=\frac{(k+1)!}{m!(k+1-m)},\qquad k\geq m$

and

$\sum_{j=0}^{k}\frac{(k-j)!}{(m-j)!}=\frac{(k+1)!}{m!(k+1-m)}\left(1-\binom{m}{k+1}\right),\qquad
0\leq k\leq m-2,$

hold.

###### Proof.

The first identity is equivalent to

$\sum_{j=0}^{m}\binom{k-m+j}{j}=\binom{k+1}{m}$

and this is a consequence of the relation

$\sum_{j=0}^{m}\binom{n+j}{j}=\binom{n+m+1}{m},\qquad n\geq 0,$

which can be proved in an elementary way by using induction on $m$.

The second identity is equivalent to

$\frac{m-k-1}{k+1}\sum_{j=0}^{k}\frac{\binom{m}{j}}{\binom{k}{j}}=\binom{m}{k+1}-1.$

In this case, it is enough to check that both sides of the identity satisfy
the recurrence relation

$(m-k)a_{m+1,k}-(m+1)a_{m,k}=k+1,\qquad 0\leq k\leq m-2,$

and they coincide for $m=2$ and $k=0$. ∎

In the proof of our result for $\mathcal{C}_{k}^{m}$, we will apply that
$h_{m}^{(k)}(x)=m!x^{m-k}/(m-k)!$, for $0\leq k\leq m$, and
$h_{m}^{(k)}(x)=0$, for $k>m$. Moreover, $h_{m}^{(k)}(0)=0$, for $k\not=m$,
$h_{m}^{(m)}(0)=m!$, and $h_{m}^{(k)}(1)=m!/(m-k)!$, for $0\leq k\leq m$.

Now, we have the following result.

######  Theorem 3.

For $m\geq 1$, it is verified that

$\mathcal{C}_{k}^{m}=\frac{1}{k+1-m}-\frac{1}{(k+1)\binom{k}{m}}\sum_{j=k-m+1}^{k}\binom{j}{k-m}\zeta(j+1),\qquad
k\geq m,$
$\mathcal{C}_{m-1}^{m}=H_{m}-\gamma-\sum_{j=1}^{m-1}\frac{\zeta(j+1)}{j+1},$

and

$\mathcal{C}_{k}^{m}=\frac{1}{k+1-m}-\frac{1}{k+1}\binom{m}{k}\left(\sum_{j=1}^{k}\frac{\zeta(j+1)}{\binom{m-k+p}{p}}+\gamma\right)\\\
+\binom{m}{k+1}\sum_{j=0}^{m-k-2}\binom{m-k-1}{j}a_{j},\qquad 0\leq k\leq
m-2.$

###### Proof.

The result can de deduced by applying Lemma 1, Lemma 4, Lemma 5, and the given
elementary properties for the derivatives of $h_{m}$. ∎

######  Remark 1.

We have to observe that a closed form for $\mathcal{C}_{k}^{m}$, when $k\geq
m$, appears in [10, Problem 1.47] and it is equivalent to the given one in our
previous theorem.

Note that, taking $k=m$, we obtain that

$\mathcal{C}_{m}^{m}=1-\frac{1}{m+1}\sum_{j=1}^{m}\zeta(j+1),$

which is the result in [4, Corollary 2.2]. Moreover, the identity (1) and the
previous theorem imply the following corollary.

######  Corollary 1.

For $m\geq 1$, it is verified that

$\frac{m!}{(k+1)!}\sum_{j=1}^{\infty}\frac{(k+j)!}{(m+j)!}(\zeta(k+j+1)-1)=\frac{1}{k+1-m}\\\
-\frac{1}{(k+1)\binom{k}{m}}\sum_{j=k-m+1}^{k}\binom{j}{k-m}\zeta(j+1),\qquad
k\geq m,$
$\sum_{j=1}^{\infty}\frac{\zeta(m+j)-1}{m+j}=H_{m}-\gamma-\sum_{j=1}^{m-1}\frac{\zeta(j+1)}{j+1},$

and

$\frac{m!}{(k+1)!}\sum_{j=1}^{\infty}\frac{(k+j)!}{(m+j)!}(\zeta(k+j+1)-1)=\frac{1}{k+1-m}-\frac{1}{k+1}\binom{m}{k}\\\
\times\left(\sum_{j=1}^{k}\frac{\zeta(j+1)}{\binom{m-k+p}{p}}+\gamma\right)+\binom{m}{k+1}\sum_{j=0}^{m-k-2}\binom{m-k-1}{j}a_{j},\qquad
0\leq k\leq m-2.$

By using the values

$a_{0}=\log\sqrt{2\pi}\qquad\text{ and }\qquad
a_{1}=-\frac{1}{4}+\frac{\zeta^{\prime}(2)}{2\pi^{2}}+\frac{1}{3}\log\sqrt{2\pi}-\frac{\gamma}{12},$

we can see some particular cases. Taking $k=m-2$ and $k=m-3$, we have,
respectively,

$\sum_{j=1}^{\infty}\frac{\zeta(m+j-1)-1}{(m+j)(m+j-1)}=-\frac{1}{m}+\log\sqrt{2\pi}-\frac{\gamma}{2}-\sum_{n=2}^{m-1}\frac{\zeta(n)}{n(n+1)},\qquad
m\geq 2,$

and

$\sum_{j=1}^{\infty}\frac{\zeta(m+j-2)-1}{(m+j)(m+j-1)(m+j-2)}\\\
=-\frac{1}{2m(m-1)}+\frac{\zeta^{\prime}(2)}{2\pi^{2}}+\frac{1}{3}\log\sqrt{2\pi}-\frac{\gamma}{4}-\sum_{n=2}^{m-2}\frac{\zeta(n)}{n(n+1)(n+2)},\qquad
m\geq 3.$

######  Remark 2.

From the well-known Hermite identity

$\sum_{k=0}^{n-1}\left[x+\frac{k}{n}\right]=[nx],\qquad x\in\mathbb{R},$

we obtain that

$\sum_{k=0}^{n-1}\left\\{x+\frac{k}{n}\right\\}=\\{nx\\}+\frac{n-1}{2},\qquad
x\in\mathbb{R}.$

Then, applying Theorem 3, we can deduce that

$\int_{0}^{n}x^{k}\sum_{k=0}^{n-1}\left\\{\frac{1}{x}+\frac{k}{n}\right\\}\,dx=n^{k+1}\left(\frac{1}{k}+\frac{n-1-2\zeta(k+1)}{2(k+1)}\right),\qquad
n,k\geq 1,$

and

$\int_{0}^{n}\sum_{k=0}^{n-1}\left\\{\frac{1}{x}+\frac{k}{n}\right\\}\,dx=n\left(\frac{n+1}{2}-\gamma\right),\qquad
n\geq 1.$

######  Remark 3.

In [4, Theorem 3.1], it is proved that

$\int_{0}^{1}\int_{0}^{1}\left\\{\frac{x}{y}\right\\}^{m}\left\\{\frac{y}{x}\right\\}^{k}\,dx\,dy=\frac{\mathcal{C}_{k}^{m}+\mathcal{C}_{m}^{k}}{2}.$

In such paper, the double integral is written as an infinite sum of values of
the Riemann zeta function by using (1). With Theorem 3 it is possible to
obtain an appropriate closed form for the integral.

##  6\. Fractional moments for the functions $x^{m}(1-x)^{m}$

To finish with our examples, we study the fractional moments for the functions
$f_{m}(x)=x^{m}(1-x)^{m}$. To do this, we start obtaining an expression for
$f_{m}(x)$ in terms of Bernoulli polynomials because, as we said in the
Introduction, the integrals of $\log\Gamma(x+1)$ with polynomials behave
better with the Bernoulli polynomials $B_{k}(x)$ than with the usual powers
$x^{k}$. In fact, in [1, Example 6.4] to evaluate the integrals

$\int_{0}^{1}x^{k}\log\Gamma(x)\,dx$

the authors write them in terms of Bernoulli polynomials by using (9), as we
did in Lemma 4. To obtain the expansion of $f_{m}(x)$ in terms of Bernoulli
polynomials we need some sums of combinatorial numbers that are contained in
the following lemma.

######  Lemma 6.

The identities

(10)
$\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}=\frac{1}{(2m+1)\binom{2m}{m}},$
(11)
$\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}=0,\qquad
j=1,\dots,m,$

and

(12)
$\sum_{k=j-m}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}=\frac{(-1)^{m}(1+(-1)^{j})}{j}\binom{m}{j-m-1},$

for $j=m+1,\dots,2m$, hold

###### Proof.

To prove (10) we apply integration. Indeed, it is easy to check that

$\displaystyle\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}$
$\displaystyle=\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}\int_{0}^{1}x^{m+k}\,dx$
$\displaystyle=\int_{0}^{1}x^{m}(1-x)^{m}\,dx$
$\displaystyle=\frac{(\Gamma(m+1))^{2}}{\Gamma(2m+2)}=\frac{1}{(2m+1)\binom{2m}{m}}.$

In the proof of (11) we use the falling factorial, and it is given by
$(a)_{n}=a(a-1)\cdots(a-n+1)$, for $n>1$, and $(a)_{0}=1$. It is clear that
$\binom{m+k+1}{j}$ is a polynomial in the variable $k$ of degree $j$ and, for
$j\geq 1$,

$\displaystyle\binom{m+k+1}{j}\frac{1}{m+k+1}$
$\displaystyle=\frac{(m+k)(m+k-1)\cdots(m+k+2-j)}{j!}$
$\displaystyle=\sum_{\ell=0}^{j-1}a_{j,m,\ell}(k)_{\ell}$

for some coefficients $a_{j,m,\ell}$. From the identity

$\binom{m}{k}(k)_{\ell}=\binom{m-\ell}{k-\ell}(m)_{\ell},$

we have

$\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}=\sum_{\ell=0}^{j-1}a_{j,m,\ell}\sum_{k=\ell}^{m}(-1)^{k}\binom{m}{k}(k)_{\ell}\\\
\begin{aligned}
&=\sum_{\ell=0}^{j-1}a_{j,m,\ell}\sum_{k=\ell}^{m}(-1)^{k}\binom{m-\ell}{k-\ell}(m)_{\ell}\\\
&=\sum_{\ell=0}^{j-1}(-1)^{\ell}a_{j,m,\ell}(m)_{\ell}\sum_{k=0}^{m-\ell}(-1)^{k}\binom{m-\ell}{k}=0,\end{aligned}$

where in the last step we have applied the identity
$\sum_{k=0}^{p}(-1)^{k}\binom{p}{k}=0$ with $p=m-j+1,\dots,m$.

The identity (12) is equivalent to

$\sum_{k=j-m}^{m}(-1)^{k}\binom{m}{k}\binom{m+k}{j-1}=(-1)^{m}(1+(-1)^{j})\binom{m}{j-m-1}.$

Obviously,

$\sum_{k=j-m}^{m}(-1)^{k}\binom{m}{k}\binom{m+k}{j-1}\\\
=\sum_{k=j-m-1}^{m}(-1)^{k}\binom{m}{k}\binom{m+k}{j-1}-(-1)^{j-m-1}\binom{m}{j-m-1}.$

Now, applying [6, (5.24)], we have

$\sum_{k=j-m-1}^{m}(-1)^{k}\binom{m}{k}\binom{m+k}{j-1}=(-1)^{m}\binom{m}{j-m-1}$

and

$\sum_{k=j-m}^{m}(-1)^{k}\binom{m}{k}\binom{m+k}{j-1}=(-1)^{m}(1+(-1)^{j})\binom{m}{j-m-1}$

finishing the proof. ∎

######  Lemma 7.

For each $m\geq 1$, it is verified

$f_{m}(x)=\frac{1}{(2m+1)\binom{2m}{m}}+(-1)^{m}\sum_{j=m+1}^{2m}\frac{1+(-1)^{j}}{j}\binom{m}{j-m-1}B_{j}(x).$

###### Proof.

To obtain the identity, we use Newton’s binomial identity and (9). Indeed,

$\displaystyle f_{m}(x)$
$\displaystyle=x^{m}\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}x^{k}=\sum_{k=0}^{m}(-1)^{k}\binom{m}{k}x^{m+k}$
$\displaystyle=\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\sum_{j=0}^{m+k}\binom{m+k+1}{j}B_{j}(x)$
$\displaystyle=\sum_{j=0}^{m}B_{j}(x)\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}$
$\displaystyle\kern
25.0pt+\sum_{j=m+1}^{2m}B_{j}(x)\sum_{k=j-m}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}$
$\displaystyle=\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}+\sum_{j=1}^{m}B_{j}(x)\sum_{k=0}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}$
$\displaystyle\kern
25.0pt+\sum_{j=m+1}^{2m}B_{j}(x)\sum_{k=j-m}^{m}\frac{(-1)^{k}}{m+k+1}\binom{m}{k}\binom{m+k+1}{j}$

We finish applying the identities (10), (11), and (12) in the previous lemma.
∎

######  Remark 4.

It is interesting to observe that, by using (5), we have

$\displaystyle f_{m}^{(k)}(x)$
$\displaystyle=(-1)^{m}\sum_{j=\max\\{k,m+1\\}}^{2m}\frac{1+(-1)^{j}}{j}\binom{m}{j-m-1}\frac{j!}{(j-k)!}B_{j-k}(x),$
(13)
$\displaystyle=(-1)^{m}k!\sum_{j=\max\\{k,m+1\\}}^{2m}\frac{1+(-1)^{j}}{j}\binom{m}{j-m-1}\binom{j}{k}B_{j-k}(x),$

for $1\leq k\leq 2m$.

######  Remark 5.

Let $\mathcal{P}_{m}$ be the shifted Legendre polynomial of degree $m$; i. e,
$\mathcal{P}_{m}(x)=P_{m}(2x-1)$, where $P_{m}$ is the standard Legendre
polynomial. By Rodrigues’ formula

$\mathcal{P}_{m}(x)=\frac{(-1)^{m}}{m!}f_{m}^{(m)}(x)$

and applying (4) we deduce that, for $m\geq 1$,

$\displaystyle\mathcal{P}_{m}(x)$
$\displaystyle=\frac{1}{m!}\sum_{j=m+1}^{2m}\frac{1+(-1)^{j}}{j}\binom{m}{j-m-1}\frac{j!}{(j-m)!}B_{j-m}(x)$
$\displaystyle=\sum_{j=1}^{m}(1+(-1)^{m+j})\frac{(m+j-1)!}{j!\,(j-1)!\,(m-j+1)!}B_{j}(x).$

The previous identity for shifted Legendre polynomials matches with the given
ones in [8, Theorem 2.2 and Theorem 2.4].

To evaluate the fractional moments of the functions $f_{m}$ we need the
combinatorial identities contained in the next lemma.

######  Lemma 8.

For $m\geq 1$, the identities

(14)
$\sum_{j=m}^{2m}\frac{\binom{m}{j-m}}{\binom{k}{j}}=\frac{m!(k-2m)!(k+1)}{(k+1-m)!},\qquad
k\geq 2m,$

and

(15)
$\sum_{j=m}^{2m-1}\frac{\binom{m}{j-m}}{\binom{2m-1}{j}}=2m(H_{2m}-H_{m})$

hold.

###### Proof.

The first identity is equivalent to

$\sum_{n=0}^{m}\binom{m+n}{m}\binom{k-m-n}{k-2m}=\binom{k+1}{m}$

and it follows from [6, (5.26)]. To prove the second one we check that both
sides satisfy the recurrence relation

$ma_{m+1}=(m+1)a_{m}+\frac{m}{2m+1}$

and they coincide for $m=1$. ∎

To complete the evaluation of the fractional moments for the functions
$f_{m}$, we will need some elementary facts about them. More exactly,
$f_{m}^{(j)}(x)=0$, for $j>2m$, $f_{m}^{(j)}(0)=f_{m}^{(j)}(1)=0$, for $0\leq
j<m$, and, by using Leibniz rule for the derivative of a product,

(16) $(-1)^{j}f^{(j)}_{m}(0)=f_{m}^{(j)}(1)=(-1)^{m}j!\binom{m}{j-m},\qquad
m\leq j\leq 2m,$

Then, denoting

$\mathcal{I}_{k}^{m}=\int_{0}^{1}x^{k}f_{m}\left(\left\\{\frac{1}{x}\right\\}\right)\,dx,$
$\delta(x,y)=\begin{cases}1,&x=y,\\\ 0,&x\not=y,\end{cases}\qquad\text{ and
}\qquad p_{m,k}=\sum_{j=m}^{k}\frac{\binom{m}{j-m}}{\binom{k}{j}},$

we have the following result.

######  Theorem 4.

For $m\geq 1$, it is verified that

$\mathcal{I}_{k}^{m}=(-1)^{m}\Bigg{(}\frac{1}{(k+1-m)\binom{k-m}{m}}-\frac{2}{k+1}\sum_{j=[m/2]}^{m-1}\frac{\binom{m}{2j+1-m}}{\binom{k}{2j+1}}\zeta(k-2j)\Bigg{)},$

for $k\geq 2m$,

$\mathcal{I}_{2m-1}^{m}=(-1)^{m}\Bigg{(}H_{2m}-H_{m}-\gamma-\frac{1}{m}\sum_{j=[m/2]}^{m-2}\frac{\binom{m}{2j+1-m}}{\binom{2m-1}{2j+1}}\zeta(2m-1-2j)\Bigg{)},$
$\mathcal{I}_{k}^{m}=\frac{(-1)^{m}}{k+1}\Bigg{(}p_{m,k}-2\sum_{j=[m/2]}^{[k/2]-1}\frac{\binom{m}{2j+1-m}}{\binom{k}{2j+1}}\zeta(k-2j)\\\
-2\gamma\binom{m}{2m-k}\delta((k+1)/2,[(k+1)/2])\Bigg{)}\\\
+(-1)^{m}(k+2)\sum_{[m/2]+1}^{m}\binom{m}{2j-m-1}\binom{2j}{k+2}\frac{b_{2j-k-2}}{j},$

for $m\leq k\leq 2m-2$, and

$\mathcal{I}_{k}^{m}=(-1)^{m}(k+2)\sum_{j=[m/2]+1}^{m}\binom{m}{2j-m-1}\binom{2j}{k+2}\frac{b_{2j-k-2}}{j},$

for $0\leq k\leq m-1$.

###### Proof.

The result is a consequence of Lemma 1, (4), Lemma 3, and Lemma 8. In
particular, we have to use (14) for the case $k\geq 2m$ and (15) for $k=2m-1$.
Moreover, the values of the derivatives of $f_{m}$ in $x=0$ and $x=1$ given in
(16) are also used. ∎

######  Remark 6.

Unfortunately, we could not find a closed form for the sum

$p_{m,k}=\sum_{j=m}^{k}\frac{\binom{m}{j-m}}{\binom{k}{j}}$

appearing in the case $m\leq k\leq 2m-2$. This remains as an open question.

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