%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Created by M. Rammaha, June 4, 2005
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[10pt,reqno]{amsart}
%\usepackage{amsfonts,amstext}
\usepackage{amsmath,amstext,amssymb,amscd,mathrsfs}
\usepackage{amsmath,amssymb,amsthm,mathrsfs}
\usepackage{amsmath,amstext,amssymb,amscd}

\usepackage{verbatim}
\usepackage{graphicx}
% This package allows the use of \begin{comment}
    % and \end{comment}.
\setlength{\unitlength}{.5cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\usepackage{amsmath}
\usepackage{amsthm}
%\usepackage{amssymb}
%\usepackage[notcite,notref]{showkeys}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{assumption}[theorem]{Assumption}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{xca}[theorem]{Exercise}


\newtheorem{prop}[theorem]{Proposition}
\newtheorem{remark}{Remark}[section]
\newtheorem{ass}{Assumption}[section]


\theoremstyle{remark}
%\newtheorem{remark}[theorem]{Remark}

\numberwithin{equation}{section}
%%%%%%%%%%%%%

%\numberwithin{equation}{section}
%\theoremstyle{plain}
%\newtheorem{lemma}{Lemma}[section]
%\newtheorem{theorem}[lemma]{Theorem}
%\newtheorem{corollary}[lemma]{Corollary}
%\newtheorem{assumption}[lemma]{Assumption}
%\newtheorem{proposition}[lemma]{Proposition}
%\newtheorem{definition}[lemma]{Definition}
%\newtheorem{remark}[lemma]{Remark}
%\newtheorem{example}[lemma]{Example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%

\setlength{\oddsidemargin}{.125in} \setlength{\evensidemargin}{.125in} 
\setlength{\textwidth}{6.5in}

\begin{document}

\newcommand{\sgn}{\operatorname{sgn}}

\def\a{\alpha}
\def\b{\beta}
\def\d{\delta}
\def\g{\gamma}
\def\l{\lambda}
\def\o{\omega}
\def\s{\sigma}
\def\t{\tau}
\def\th{\theta}
\def\r{\rho}
\def\D{\Delta}
\def\G{\Gamma}
\def\O{\Omega}
\def\e{\epsilon}
\def\p{\phi}
\def\P{\Phi}
\def\S{\Psi}
\def\E{\eta}
\def\m{\mu}


\newcommand{\into}{\hookrightarrow}

\def\grad{\nabla}
\def\bar{\overline}
\newcommand{\reals}{\mathbb{R}}
\newcommand{\naturals}{\mathbb{N}}
\newcommand{\ints}{\mathbb{Z}}
\newcommand{\complex}{\mathbb{C}}
\newcommand{\rationals}{\mathbb{Q}}
\newcommand{\dualprod}[1]{\left\langle #1 \right\rangle}
\newcommand{\innerprod}[1]{\big( #1\big)}
\newcommand{\norm}[1]{\left\|#1\right\|}
\newcommand{\abs}[1]{\left|#1\right|}


%%%%%%%%% Kevin

\section{Introduction}

In this paper we continue the study of the nabla discrete fractional calculus. We define several extensions of the Laplace tranform in the nabla case based on the work done by \cite{REU2010}. We also develop composition rules motivated by their homologues in the delta case defined by M. Holm \cite{HolmCompositions}. Finally, we make use of these developed tools to solve various fractional initial value problems.

\section{Preliminary Definitions of the Nabla Discrete Calculus}
Here we will give the basic definitions and notation from the nabla calculus that are used in this paper. The delta calculus analog has been studied in detail, and a general overview is given in \cite{KelleyPetersonDifferenceEquations}. The domains used in this paper are denoted by $\naturals_a$ where $a \in \reals$. This is a discrete time scale with a graininess of $1$, so it is defined as $\naturals_a = \{a,a+1,a+2,\cdots\}$. For $\naturals_a$, we use the terminology that a function's domain, in the case studied here, is based at $a$. We use the $\rho$-function, or backwards jump function, from the time scale calculus where it is defined for this time scale as $\rho : \naturals_a \to \naturals_a$, given by $\rho(t) = \max\{a,t-1\}$. We define the backwards difference operator, or nabla operator ($\nabla$), for a function $f : \naturals_a \to \reals$ by $(\nabla f)(t):= f(t) - f(\rho(t)) = f(t)-f(t-1)$. In this paper, we use the convention that $\nabla f(t) := (\nabla f)(t)$. We then define higher order differences recursively by $\nabla^n f(t):= \nabla(\nabla^{n-1} f(t))$ for $t \in \naturals_{a+n}$ where $n \in \naturals$. We take as convention that $\nabla^0$ is the identity operator.

Based on these preliminary definitions, we say $F$ is an anti-nabla difference of $f$ on $\naturals_a$ if and only if $\nabla F(t) = f(t)$ for $t \in \naturals_{a+1}$. We then define the definite nabla integral of $f:\naturals_a \to \reals$ by
\begin{align*}
\int_c^d f(t) \nabla t := \left\{
	\begin{array}{lr}
		\sum_{t=c+1}^d f(t), & \text{ if } c < d  \\
		0, & \text{ if } c=d \\
		-\sum_{t=d+1}^c f(t), & \text{ if } d< c
	\end{array}
, 
\right.
\text{where } c, d \in \naturals_a.
\end{align*}

We now give the fundamental theorem of nabla calculus.

\begin{theorem}[Fundamental Theorem of Nabla Calculus]
Let $f: \naturals_a \to \reals$ and let $F$ be a anti-nabla difference of $f$ on $\naturals_a$, then for any $c, d \in \naturals_a$ we have
\begin{equation*}
\int_c^d f(t) \nabla t = F(d) - F(c).
\end{equation*}
\end{theorem}



The nabla product rule for two functions $u,v : \naturals_a \to \reals$ and $t  \in \naturals_{a+1}$ is given by
\begin{equation*}
\nabla(u(t)v(t)) = u(t) \nabla v(t) + v(\rho(t))\nabla u(t).
\end{equation*}

This immediately leads to the summation by parts formula for nabla calculus 

\begin{equation*}
\sum_{s=b+1}^{c} u(t) \nabla v(t) = u(t)v(t)|_b^c - \sum_{s=b+1}^c v(\rho(t)) \nabla u(t).
\end{equation*}

It is useful for the work in this paper to use the summation by parts formula written in the following alternative form 

\begin{equation*}
\sum_{s=b+1}^c v(\rho(t)) \nabla u(t) = u(t)v(t)|_b^c -  \sum_{s=b+1}^{c} u(t) \nabla v(t).
\end{equation*}













%%%%% Kevin


\section{Fractional Sums and Differences}

Now that we have established the basic definitions of the nabla calculus, we will move on to extending these definitions to the fractional case and establish definitions for the fractional sum and fractional difference which are analogous to the continuous fractional integral and continuous fractional derivative. In order to do this, we remind the reader of the rising factorial function. For $n,k \in \naturals$, the rising factorial function is defined by
\begin{equation}
k^{\overline{n}}:= k(k+1)\cdots(k+n-1)= \frac{(k+n-1)!}{(k-1)!}.
\end{equation}


This definition can be extended for fractional values using the gamma function as follows.

\begin{definition}[Rising Function]
For $k, \nu \in \reals$, the rising function is defined by
\begin{equation}
k^{\overline{\nu}}:=  \frac{\Gamma(k+\nu)}{\Gamma(k)}.
\end{equation}
\end{definition}




To motivate the definition of a fractional sum, we look at the definition of the integral sum derived from the repeated summation rule. 

\begin{definition}[Integral Sum for Integers \cite{REU2010}] 
\begin{equation*}
\nabla_a^{-n} f(t) := \frac{1}{(n-1)!} \int_a^t (t-\rho(s))^{\overline{n-1}} f(s) \nabla s
\end{equation*}
\end{definition}

We can now extend this definition for fractional values by again using the gamma function and the generalized rising function.


\begin{definition}[Nabla Fractional Sum \cite{REU2010}]
Let $f : \naturals_a \to \reals$ and $\nu > 0$, then the $v^{th}$-order fractional sum is given by

\begin{equation}
\label{nablaSum}
\nabla_a^{-\nu} f(t) := \frac{1}{\Gamma(\nu)} \int_a^t (t-\rho(s))^{\overline{\nu-1}} f(s) \nabla s.
\end{equation}

\end{definition}

One should note that the fractional sum $\nabla_a^{-\nu}$ does not depend on its lower limit of integration, but rather its upper limit.

The definition for a fractional difference is achieved utilizing the definition for a fractional sum.


\begin{definition}[Nabla Fractional Difference \cite{REU2010}]
Let $f : \naturals_a \to \reals$, $\nu > 0$, and choose $N \in \naturals$ such that $N-1 < \nu \leq N$. Then, for $t \in \naturals_{a+N}$, the $\nu^{th}$-order fractional difference is given by

\begin{equation}
\label{nablaDifference}
\nabla_a^\nu f(t) := \nabla^N\nabla_a^{-(N-\nu)}f(t).
\end{equation}

\end{definition}

The fractional difference depends on where it is based. It will be shown later that this dependence goes away as $\nu$ approaches a whole order value. Therefore when we want to denote a fractional difference $\nabla_a^\nu f$, we employ the subscript $a$ to show where the difference is based. However, when dealing with a whole order difference, such as $\nabla^3 f$, the subscript is omitted as it no longer depends on the base.







%%%%%%%% Katy


\section{Fractional Taylor Monomials}

Here we define Taylor Monomials which will prove useful later in solving initial value problems.  We begin with the recursive definition.

\begin{definition}[Taylor Monomials \cite{REU2010}]
The $n$-th Taylor monomial is given by
\[
	\begin{cases}
		h_{0}(t,a) := 1 &\\
		h_{n}(t,a) := \int_{a}^{t}h_{n-1}(\tau,a)\nabla\tau, &n \in \naturals.
	\end{cases}
\]
\end{definition}


This definition leads to the following theorem which can be used to find specific positive integer Taylor monomials.

\begin{theorem}[\cite{REU2010}]
For $n \in \naturals_{0}$, 
\begin{equation*}h_{n}(t,a) = \frac{(t-a)^{\overline{n}}}{n!}.\end{equation*}
\end{theorem}


We now use the previous theorem to generalize Taylor monomials to include fractional orders.

\begin{definition}[Fractional Order Taylor Monomials \cite{REU2010}]
For $\nu \in \reals\backslash\{-1,-2,...\}$, the $\nu$-th Taylor Monomial is \begin{equation*}h_{\nu}(t,a) = \frac{(t-a)^{\overline{\nu}}}{\Gamma(\nu + 1)}.\end{equation*}
\end{definition}


Applying the generalized power rule to the definiton of fractional Taylor monomials results in the following theorem.

\begin{theorem} 
For $\mu$, $\nu \in \reals$ such that $\nu$ and $\nu + \mu$ are not negative integers, 
\begin{equation*}\nabla^{\mu}h_{\nu}(t,a) = h_{\nu-\mu}(t,a), \end{equation*}
where
\[
	\begin{cases}
		t \in \naturals_{a} &\mu < 0\\
		t \in \naturals_{a+N} &\mu > 0.
	\end{cases}
\] 
\end{theorem}

\begin{proof}
\begin{align*}
\nabla^{\mu}h_{\nu}(t,a) &= \frac{1}{\Gamma(\nu+1)}\nabla^{\mu} (t-a)^{\overline{\nu}} \\
&= \frac{1}{\Gamma(\nu+1)}\frac{\Gamma(\nu+1)}{\Gamma(\nu-\mu+1)} (t-a)^{\overline{\nu-\mu}} \\
&= h_{\nu-\mu}(t,a)
\end{align*}
\end{proof}


The next theorem relates Taylor monomials centered at values that differ by one.

\begin{lemma}[One Step Taylor Monomial Shifting]
For $\nu \in \reals\backslash\{-1,-2,...\}$ and $N,m \in \naturals$,
\begin{equation*}h_{\nu-N}(t,a+1) = h_{\nu-N}(t,a) - h_{\nu-N-1}(t, a).\end{equation*}
\end{lemma}

\begin{proof}
\begin{align*}
h_{\nu-N}(t,a) - h_{\nu-N-1}(t,a)	&= \frac{(t-a)^{\overline{\nu-N}}}{\Gamma(\nu-N+1)} - \frac{(t-a)^{\overline{\nu-N-1}}}{\Gamma(\nu-N)} \\
	&= \frac{\Gamma(t-a+\nu-N-1)}{\Gamma(t-a)\Gamma(\nu-N+1)}(t-a+\nu-N-1-(\nu-N)) \\
	&= \frac{\Gamma(t-a+\nu-N-1)}{\Gamma(t-a-1)\Gamma(\nu-N+1)} \\
	&= h_{\nu-N}(t,a+1) 
\end{align*}
\end{proof}


The previous theorem can be extended to the following general formula.

\begin{theorem}[General Taylor Monomial Shifting]
For $\nu \in \reals\backslash\{-1,-2,...\}$ and $N,m \in \naturals$, 
\begin{equation}\label{taylorShifting}h_{\nu-N}(t,a+m) = \sum\limits_{k=0}^{m} {m \choose k}(-1)^k h_{\nu-N-k}(t,a).\end{equation}
\end{theorem}

\begin{proof}
The proof is by induction on $m$.  The base case, $m = 1$, is proven by the previous theorem.  Assume that $ h_{\nu-N}(t,a+m) = \sum\limits_{k=0}^{m} {m \choose k}(-1)^k h_{\nu-N-k}(t,a) $ for $m \geq 1$.  From the previous theorem, 
\begin{equation*}
	h_{\nu-N}(t,a+m+1) = h_{\nu-N}(t,a+m) - h_{\nu-N-1}(t, a+m).
\end{equation*}
Applying the induction hypothesis to both terms on the right side of the equation gives
\begin{align*}
	&= \sum\limits_{k=0}^{m}{m \choose k}(-1)^k h_{\nu-N-k}(t,a) - \sum\limits_{k=0}^{m} {m \choose k} (-1)^k h_{\nu-N-1-k}(t,a) \\
	&= \sum\limits_{k=0}^{m}{m \choose k}(-1)^k h_{\nu-N-k}(t,a) - \sum\limits_{k=1}^{m+1} {m \choose k-1} (-1)^{k-1} h_{\nu-N-k}(t,a) \\
	&= \sum\limits_{k=0}^{m+1}{m \choose k}(-1)^k h_{\nu-N-k}(t,a) + {m \choose m+1}(-1)^{m+1}h_{\nu-N-m-1}(t,a) \\
	&\qquad\qquad - \sum\limits_{k=0}^{m+1} {m \choose k-1} (-1)^{k-1} h_{\nu-N-k}(t,a) + {m \choose -1}(-1)^{-1} h_{\nu-N}(t,a) \\
	&=\sum\limits_{k=0}^{m+1}\left({m \choose k} + {m \choose k-1}\right) (-1)^k h_{\nu-N-k}(t,a) \\
	&=\sum\limits_{k=0}^{m+1}{m+1 \choose k} (-1)^k h_{\nu-N-k}(t,a).
\end{align*}
\end{proof}









%%%%%%% Katy


\section{Laplace Transforms}


One way to solve initial value problems is using Laplace transforms.  This method proves useful due to the linearity of the Laplace transform which is evident in the next few definitions and theorems.

\begin{definition}[Laplace Transform \cite{REU2010}]
For a function $f : \naturals_{a} \to \reals$ and $s \in \reals$, the Laplace transform of $f$ is 
\begin{equation*}\mathcal{L}_a \{f\}(s) := \sum\limits_{k=1}^{\infty}(1-s)^{k-1}f(a+k).\end{equation*}
\end{definition}




\begin{theorem}[Existence of Laplace Transform \cite{REU2010}]
Given a function of exponential order $\alpha$, its Laplace transform exists for $|\frac{1-s}{1-\alpha}| < 1$.
\end{theorem}


\begin{theorem}[Uniqueness of the Laplace Transform \cite{REU2010}]
\label{uniquenessOfLaplaceTransform}
$\mathcal{L}_{a}\{f\}(s) = 0$ if and only if $f(t) = 0$ for $t \in \naturals_{a+1}$.
\end{theorem}


Next we give the Laplace transform of fractional order Taylor monomials.

\begin{theorem}[Laplace Transform of a Taylor Monomial\cite{REU2010}]
For $\nu \in \reals\backslash\{-1,-2,...\}$, 
\begin{equation*}
\mathcal{L}\{h_{\nu}(\cdot,a)\}(s) = \frac{1}{s^{\nu+1}}.
\end{equation*}
\end{theorem}



Here we define the convolution of two functions.

\begin{definition}[Convolution \cite{REU2010}]
For $f, g : \naturals_a \to \reals$ and $t \in \naturals_{a+1}$, the convolution of $f$ and $g$ is 
\begin{equation*}(f * g)(t) := \int_{a}^{t}g(t-\rho(s)+a)g(s)\nabla s.\end{equation*}
\end{definition}


\begin{theorem}[Convolution Theorem \cite{REU2010}]
For $f, g : \naturals_{a} \to \reals$, 
\begin{equation*}\mathcal{L}_{a}\{f * g\}(s) = \mathcal{L}_{a}\{f\}(s)\mathcal{L}_{a}\{g\}(s).\end{equation*}
\end{theorem}


We next use the convolution to define fractional sums and their transforms.


\begin{theorem} \cite{REU2010}
For $\nu \in \reals\backslash\{0,-1,-2,...\}$ and $f : \naturals_{a} \to \reals$,
\begin{equation}
\label{convolution}
\nabla_{a}^{-\nu}f(t) = (h_{\nu-1}(\cdot,a) * f(\cdot))(t).
\end{equation}
\end{theorem}


\begin{theorem}[Transformation of Fractional Sums \cite{REU2010}]
\label{fractionalSumTransform}
For $f : \naturals_{a} \to \reals$ and $\nu \in \reals^{+}$, \begin{equation*}\mathcal{L}_{a}\{\nabla_{a}^{-\nu} f\}(s) = \frac{1}{s^{\nu}}\mathcal{L}_{a}\{f\}(s).\end{equation*}
\end{theorem}




The Mittag-Leffler function generalizes the exponential function.


\begin{definition}[Mittag-Leffler Function \cite{REU2010}]
For $\abs{p} < 1$, $\alpha > 0, \beta \in \reals$, and $t \in \naturals_{a}$, the Mittag-Leffler function is
\begin{equation*}
E_{p,\alpha, \beta}(t,a) := \sum\limits_{k=0}^{\infty} p^k h_{\alpha k + \beta}(t,a).
\end{equation*}
\end{definition}


The Laplace transform of the Mittag-Leffler fuction is useful when solving initial value problems using the Laplace transform method.


\begin{theorem}\cite{REU2010}
For $\abs{p} < 1, \alpha > 0, \beta \in \reals, \abs{1-s} < 1$, and $\abs{s^{\alpha}} > \abs{p}$, 
\begin{equation*}
\mathcal{L}_a \{E_{p,\alpha, \beta}(\cdot,a)\}(s) = \frac{s^{\alpha-\beta-1}}{s^{\alpha}-p}.
\end{equation*}
\end{theorem}



Shifting the bases of Laplace transforms is important in solving equations using Laplace transforms because different functions can be based at different values in the nabla case.


\begin{theorem}[Laplace Shifting Theorem]
Given $f : \naturals_a \to \reals$, we have
\begin{equation}\label{shiftingTheorem}
\mathcal{L}_{a+n} \{f\}(s) = \left( \frac{1}{1-s} \right)^n \mathcal{L}_a\{f\}(s) - \sum_{k=0}^{n-1} \left( \left( \frac{1}{1-s} \right)^{n-k} f(a+k+1) \right) 
\end{equation}

\begin{proof}
The proof follows from induction. For the base case $n=1$ we have the following,
\begin{eqnarray*}
\mathcal{L}_{a+1} \{f\}(s) &= &\left( \frac{1}{1-s} \right) \mathcal{L}_a\{f\}(s) - \sum_{k=0}^{0} \left( \left( \frac{1}{1-s} \right)^{1-k} f(a+k+1) \right) \\
& = & \left( \frac{1}{1-s} \right) \mathcal{L}_a\{f\}(s) - \left( \frac{1}{1-s} \right) f(a+1), \\
\end{eqnarray*}

which matches the Shifting Lemma proven in \cite[p.91]{REU2010}. \\

For the induction step, we want to show that the equation (\ref{shiftingTheorem}) implies,
\begin{equation*}
\mathcal{L}_{a+n+1} \{f\}(s) = \left( \frac{1}{1-s} \right)^{n+1} \mathcal{L}_a\{f\}(s) - \sum_{k=0}^{n} \left( \left( \frac{1}{1-s} \right)^{n+1-k} f(a+k+1) \right).
\end{equation*}

We look at $\mathcal{L}_{a+n+1} \{f\}(s)$ and apply the Shifting Lemma in \cite{REU2010},
\begin{align*}
\mathcal{L}_{a+n+1} \{f\}(s) & = \frac{1}{1-s} \mathcal{L}_{a+n} \{f\}(s) -  \frac{1}{1-s} f(a+n+1) \\
&=  \frac{1}{1-s} \left( \left( \frac{1}{1-s} \right)^n \mathcal{L}_a\{f\}(s) - \sum_{k=0}^{n-1} \left( \left( \frac{1}{1-s} \right)^{n-k} f(a+k+1) \right) \right)\\
&\qquad\qquad- \frac{1}{1-s} f(a+n+1) \\
&=  \left( \frac{1}{1-s} \right)^{n+1} \mathcal{L}_a\{f\}(s) - \sum_{k=0}^{n-1} \left( \left( \frac{1}{1-s} \right)^{n+1-k} f(a+k+1) \right)\\
&\qquad\qquad -  \frac{1}{1-s} f(a+n+1) \\
&= \left( \frac{1}{1-s} \right)^{n+1} \mathcal{L}_a\{f\}(s) - \sum_{k=0}^{n} \left( \left( \frac{1}{1-s} \right)^{n+1-k} f(a+k+1) \right).
\end{align*}

\end{proof}

\end{theorem}




We now provide the transformation of an $n$-th order nabla difference which is used in solving initial value problems in the fractional case as well as the whole order case.

\begin{theorem}[Transformation of $n^{th}$-Order Nabla Difference \cite{REU2010}]
For $f : \naturals_{a} \to \reals$, 
\begin{equation*}\mathcal{L}_{a+n}\{\nabla^{n}f\}(s) = s^n \mathcal{L}_{a+n}\{f\}(s) - \sum\limits_{k=0}^{n-1} s^{k}\nabla^{n-k}f(a+n).\end{equation*}
\end{theorem}



The next theorem provides the Laplace transform of any $\nu$-th order fractional difference.

\begin{theorem} Let $f:\mathbb{N}_{a} \in \mathbb{R}$. Pick $N\in\mathbb{Z}^{+}$ such that $N-1<\nu\leq N$. Then 

\begin{align}
\label{laplaceTransFracDifference}
\mathcal{L}_{a+N} \{ \nabla^{\nu}_{a}f\}(s) &=s^{\nu}\mathcal{L}_{a+N}\{f\}(s)+\sum\limits_{k=0}^{N-1}\left[s^{\nu}(\frac{1}{1-s})^{N-l}f(a+k+1)\right.\\
&\left.\qquad-s^{N}(\frac{1}{1-s})^{N-k}\nabla^{-(N-\nu)}_{a}f(a+k+1) \right.\\
&\left.\qquad- \nabla^{N-k-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{k}\right].
\end{align}

\end{theorem}

\begin{proof}
We begin by applying the definition of the $\nu$-th order nabla difference.
\begin{equation*}
\mathcal{L}_{a+N}\{ \nabla^{\nu}_{a}f\}(s) = \mathcal{L}_{a+N}\{ \nabla^{N}\nabla^{-(N-\nu)}_{a}f\}(s).\\
\end{equation*}

Next, we use the Laplace transform of an N-th order difference.

\begin{equation*}
=s^{N}\mathcal{L}_{a+N}\{\nabla^{-(N-\nu)}_{a} f\}(s) - \sum\limits^{N-1}_{j=0}\nabla^{N-j-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{j}.\\
\end{equation*}

Now applying the shifting theorem, we get that

\begin{align*}
&=s^{N}[(\frac{1}{1-s})^{N}\mathcal{L}_{a}\{\nabla^{-(N-\nu)}_{a} f\}(s) - \sum\limits^{N-1}_{k=0}(\frac{1}{1-s})^{N-k}\nabla^{-(N-\nu)}_{a}f(a+k+1)]\\
&\qquad-\sum\limits^{N-1}_{j=0}\nabla^{N-j-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{j}.\\
\end{align*}

Applying the Laplace transform of a $\nu$-th order difference we have

\begin{align*}
&=s^{N}[(\frac{1}{1-s})^{N}(\frac{1}{s^{N-\nu}})\mathcal{L}_{a}\{f\}(s) - \sum\limits^{N-1}_{k=0}(\frac{1}{1-s})^{N-k}\nabla^{-(N-\nu)}_{a}f(a+k+1)]\\
&\qquad-\sum\limits^{N-1}_{j=0}\nabla^{N-j-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{j}\\
&=\frac{s^{\nu}}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s)-s^{N}\sum\limits^{N-1}_{k=0}(\frac{1}{1-s})^{N-k}\nabla^{-(N-\nu)}_{a}f(a+k+1)\\
&\qquad-\sum\limits^{N-1}_{j=0}\nabla^{N-j-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{j}\\
&=\frac{s^{\nu}}{(1-s)^{N}}[(1-s)^{N}\mathcal{L}_{a+N}\{f\}(s)+(1-s)^N\sum\limits^{N-1}_{l=0}(\frac{1}{1-s})^{N-l}f(a+l+1)]\\
&\qquad-s^{N}\sum\limits^{N-1}_{k=0}(\frac{1}{1-s})^{N-k}\nabla^{-(N-\nu)}_{a}f(a+k+1)-\sum\limits^{N-1}_{j=0}\nabla^{N-j-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{j}\\
&\qquad=s^{\nu}\mathcal{L}_{a+N}\{f\}(s)+\sum\limits_{k=0}^{N-1}[s^{\nu}(\frac{1}{1-s})^{N-l}f(a+k+1)-s^{N}(\frac{1}{1-s})^{N-k}\\
&\qquad\qquad-\nabla^{-(N-\nu)}_{a}f(a+k+1) - \nabla^{N-k-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{k}].\qquad
\end{align*}
\end{proof}























%%%%%% Kevin


\section{Unifying Nabla Sums and Differences}
We now have definitions for fractional sums and fractional differences, however they are not unified in a similar form. We will show here that the traditional definition of a fractional difference can be rewritten in a form similar to the definition for a fractional sum. In order to do so, we must employ Leibniz's rule in the discrete case, as shown in \cite{REU2010}.

\begin{theorem}[Leibniz's Rule \cite{REU2010}]
Let $g : \naturals_a \times \naturals_a \to \reals$, then for $t \in \naturals_{b+1}, b \in \naturals_a$,

\begin{equation}
\label{leibnizRule}
\nabla_t \sum_{s=b+1}^t g(s,t) = g(t, t-1)+ \sum_{s=b+1}^t \nabla_t g(s,t).
\end{equation}

\end{theorem}


With this theorem in hand, we can now go on and show that the fractional difference can be rewritten in a more useful form.





\begin{theorem}[Alternative Definition of a Fractional Difference]
Let $f: \naturals_a \to \reals, \nu > 0$, $\nu \not \in \naturals$ be given and choose $N \in \naturals$ such that $N-1<\nu < N.$ Then for $t \in \naturals_{a+N}$,
\begin{equation*}
\nabla_a^{\nu} f(t) = \nabla^N \nabla^{-(N-\nu)}_a f(t),
\end{equation*}
implies

\begin{equation}
\label{alternativeNablaDifference}
\nabla_a^{\nu} f(t) = \frac{1}{\Gamma(-\nu)} \sum_{s=a+1}^t (t-\rho(s))^{\overline{-\nu-1}} f(s),
\end{equation}
for $\nu \not \in \naturals_0$.

\end{theorem}

\begin{proof}
\begin{align*}
\nabla_a^{\nu} f(t)& =  \nabla^N \nabla^{-(N-\nu)}_a f(t) & \text{from }(\ref{nablaDifference}) \\
=& \nabla^N \frac{1}{\Gamma(N-\nu)} \int_a^t (t-\rho(s))^{\overline{N-\nu-1}}f(s) \nabla s  & \text{from }(\ref{nablaSum})\\
=& \nabla^N \frac{1}{\Gamma(N-\nu)} \sum_{s=a+1}^t (t-\rho(s))^{\overline{N-\nu-1}}f(s)  \\
=& \nabla^{N-1} \frac{1}{\Gamma(N-\nu)} \nabla_t \sum_{s=a+1}^t (t-\rho(s))^{\overline{N-\nu-1}}f(s) \\
=& \frac{\nabla^{N-1}}{\Gamma(N-\nu)} \left( ((t-1)-(t-1))^{\overline{N-\nu-1}} + \sum_{s=a+1}^t (N-\nu-1)(t-\rho(s))^{\overline{N-\nu-2}}f(s) \right) & \text{from } (\ref{leibnizRule})\\
=& \frac{\nabla^{N-1}}{\Gamma(N-\nu-1)} \sum_{s=a+1}^t (t-\rho(s))^{\overline{N-\nu-2}}f(s) \nabla s. \\
\end{align*}

By applying Leibniz's Rule $N-1$ more times we get
\begin{equation*}
\nabla_a^{\nu} f(t) = \frac{1}{\Gamma(-\nu)} \sum_{s=a+1}^t(t-\rho(s))^{\overline{-\nu-1}} f(s).
\end{equation*}

\end{proof}

One should note that $\nu \not \in \naturals$ is required due to the $\frac{1}{\Gamma(-\nu)}$ term being undefined for negative integers. Also, with this definition it appears that a fractional difference can be defined on all of $\naturals_a$, however it should be emphasized that $t$ must be in $\naturals_{a+N}$. 

With this theorem proven, we can now give a unified definition for fractional sums and differences.



\begin{definition}[Unified Definition for Nabla Sums and Differences]
\label{unifiedDefinition}


Let $f : \naturals_a \to \reals$, $\nu > 0$, and $N \in \naturals$ such that $N-1 < \nu \leq N$ be given. Then

\begin{enumerate}

\item the $\nu^{th}$-order fractional sum of $f$ is given by
\begin{equation}
\label{dualSum}
\nabla_a^{-\nu} f(t) := \frac{1}{\Gamma(\nu)} \sum_{s=a+1}^t (t-\rho(s))^{\overline{\nu-1}} f(s), \text{ for } t \in \naturals_a.
\end{equation}

\item the $\nu^{th}$-order fractional difference of $f$ is given by

\begin{equation}
\label{dualDifference}
\nabla_a^{\nu} f(t) := \left\{
	\begin{array}{lr}
		\frac{1}{\Gamma(-\nu)} \sum_{s=a+1}^t (t-\rho(s))^{\overline{-\nu-1}} f(s), & \nu \notin \naturals \\
		\nabla^N f(t), & \nu = N \in \naturals \\
	\end{array}
, \text{ for } t \in \naturals_{a+N}.
\right.
\end{equation}

\end{enumerate}


\end{definition}




One would want the $\nu^{th}$-order fractional difference to be continuous with respect to $\nu$, however using this definition, as $\nu$ approaches a whole number, the term $\frac{1}{\Gamma(-\nu)}$ approaches infinity. This following theorem shows that the fractional difference is indeed continuous with respect to $\nu$.






\begin{theorem}[Continuity of the Nabla Fractional Difference]
Let $f: \naturals_a \to \reals$ be given. Then the fractional difference $\nabla_a^\nu f$ is continuous with respect to $\nu \geq 0$.
\end{theorem}

\begin{proof}
It is sufficient for this proof to show that for $f:\naturals_a \to \reals$, $N - 1 < \nu \leq N$, and $m \in \naturals_0$, that the following hold:
\begin{equation}
\label{continuousInMiddle}
\nabla_a^\nu f (a+N+m) \text{ is continuous with respect to $\nu$ on $(N-1,N)$,}
\end{equation}

\begin{equation}
\label{continuousToRight}
\nabla_a^\nu f(a+N+m) \to \nabla^N f(a+m) \text{ as } \nu \to N^-,
\end{equation}

\begin{equation}
\label{continuousToLeft}
\text{ and } \nabla_a^\nu f(a+N+m) \to \nabla^{N-1} f(a+m) \text{ as } \nu \to (N-1)^+.
\end{equation}



Let $\nu$ be fixed such that $N-1 < \nu < N$. Then

\begin{align*}
\nabla_a^\nu f(a+N+m) =  &\frac{1}{\Gamma(-\nu)} \sum_{s=a+1}^t (t - \rho(s))^{\overline{-\nu-1}} |_{t=a+N+m} \\
=& \frac{1}{\Gamma(-\nu)} \sum_{s=a+1}^{a+N+m} (a+N+m - \rho(s))^{\overline{-\nu-1}} \\
=& \sum_{s=a+1}^{a+N+m} \frac{\Gamma(a+N+m-s+1-\nu-1)}{\Gamma(a+N+m-s+1)\Gamma(-\nu)} f(s) \\
=& \sum_{s=a+1}^{a+N+m} \frac{(a+N+m-s-\nu-1)(a+N+m-s-\nu-2)\cdots(-\nu)\Gamma(-\nu)}{(a+N+m-s)! \Gamma(-\nu)} f(s) \\
=& \sum_{s=a+1}^{a+N+m-1} \frac{(a+N+m-s-\nu-1)\cdots(-\nu)}{(a+N+m-s)!} f(s)+f(a+N+m). \\
&\text{Let $i := a+N+m-s$. Then,} \\
=& \sum_{i=1}^{N+m} \frac{(i-1-\nu)\cdots(1-\nu)(-\nu)}{i!}f(a+N+M-i) + f(a+N+M).
\end{align*}

This shows that the $\nu^{th}$ order fractional difference is continuous on $N-1<\nu<N$, showing (\ref{continuousInMiddle}). \\

Now we look at $\nu \to N^-$ in order to show (\ref{continuousToRight}).
\begin{align*}
\lim_{\nu \to N^{-}} \nabla_a^\nu f(a+N-m) =& \lim_{\nu \to N^{-}} \sum_{i=1}^{N+m} \frac{(i-1-\nu)\cdots(1-\nu)(-\nu)}{i!}f(a+N+M-i) + f(a+N+M) \\
=&\sum_{i=1}^{N+m} \left( \frac{(i-1-N)\cdots(-N)}{i!} f(a+N+m-i) \right) + f(a+N+m) \\
=&\sum_{i=0}^{N+m} \left( (-1)^i \frac{(N+1-i)\cdots(N)}{i!} f(a+N+m-i) \right)\\
=&\sum_{i=0}^{N+m} (-1)^i {N \choose i} f(a+N+m-i) \\
=&\sum_{i=0}^{N} (-1)^i {N \choose i} f(a+m+N-i) \\
=& \nabla^N f(a+m).
\end{align*}

Finally we want to show (\ref{continuousToLeft}).

\begin{align*}
\lim_{\nu \to (N-1)^+}  \nabla_a^\nu f(a+N-m) =& \lim_{\nu \to (N-1)^{+}} \sum_{i=1}^{N+m} \frac{(i-1-\nu)\cdots(1-\nu)(-\nu)}{i!}f(a+N+M-i) + f(a+N+M) \\
= & \sum_{i=1}^{N+m} \frac{(i-N)(i-N-1)\cdots(1-N)}{i!}f(a+N+m-i)+f(a+N+m) \\
=&  \sum_{i=0}^{N+m} (-1)^i \frac{(N-i)(N+1-i)\cdots(N-1)}{i!}f(a+N+m-i)\\
=& \sum_{i=0}^{N+m} (-1)^i {N-1 \choose i} f(a+m+1+N-1-i) \\
=& \nabla^{N-1} f(a+m+1).
\end{align*}


\end{proof}

These three proven statements show that the nabla fractional derivative is continuous for all $\nu > 0$, and furthermore, as $\nu \to M$ for some $M \in \naturals_0$, the fractional derivative depends less and less on the base and behaves more and more like a whole order difference.













%%%%%%%%%%%% Kevin




\section{Composition Rules}

We are now interested in the composition of fractional sums and fractional differences. The unified definition for fractional sums and differences allows us to calculate exactly such a composition, however it is not very useful to use in a more general case. The following composition rules give us more useful tools in evaluating compositions of sums and differences.

\begin{theorem}[Fractional Nabla Sums and Differences Composition Rules]
Let $\mu, \nu > 0$ and $k \in \naturals_0$ be given, and choose $N \in \naturals$ such that $N-1 < \nu \leq N$, then we have
\begin{equation}
\label{sumWithSum}
\nabla_a^{-\nu} \nabla_a^{-\mu} f(t) = \nabla_a^{-\nu-\mu} f(t), \text{ for } t \in \naturals_a,
\end{equation}

\begin{equation}
\label{differenceWithSum}
\nabla_{a}^{\nu} \nabla_{a}^{-\mu} f(t) = \nabla_{a}^{\nu-\mu} f(t), \text{ for } t \in \naturals_{a+N},
\end{equation}

\begin{equation}
\label{sumWithWholeOrderDifference}
\nabla_{a+k}^{-\nu} \nabla^{k} f(t)= \nabla_{a+k}^{k-\nu}f(t) - \sum_{j=0}^{k-1} \nabla^{j}f(a+k) \frac{(t-a-k)^{\overline{\nu -k + j}}}{\Gamma(\nu -k +j +1)}, \text{ for } t \in \naturals_{a+k},
\end{equation}

\begin{equation}
\label{differenceWithWholeOrderDifference}
\nabla_{a+k}^\nu \nabla^k = \nabla_{a+k}^{k+\nu} f(t) - \sum_{j=0}^{k-1}\nabla^j f(a+k) \frac{(t-a-k)^{\overline{-\nu-k+j}}}{\Gamma(-v-k+j+1)} ,\text{ for } t \in \naturals_{a+k+N}.
\end{equation}
\end{theorem}






\subsection{Nabla Sum Composed with a Nabla Sum}

We will utilize the definition of the Laplace transforms of a fractional sum and the uniqueness of the Laplace transforms for fractional sums given in the previous section in order to prove the following theorem for a fractional sum composed with another fractional sum.

\begin{theorem}
Let $\nu, \mu > 0$ be given. Then

\begin{equation*}
\nabla_a^{-\nu} \nabla_a^{-\mu} f(t) = \nabla_{a}^{-\nu - \mu} f(t).
\end{equation*}
\end{theorem}


\begin{proof}
We will use the definition of the Laplace transformation of a fractional sum from %(\ref{fractionalSumTransform}).

\begin{align*}
\mathcal{L}\{\nabla_a^{-\nu} \nabla_a^{-\mu} f \}(s) =& \frac{1}{s^\nu} \mathcal{L}\{ \nabla_a^{-\mu} \} (s) \\
=& \frac{1}{s^{\nu+\mu}} \mathcal{L} \{ f\}(s) \\
=& \mathcal{L} \{ \nabla_a^{-\nu-\mu} f\}(s). \\
\end{align*}

Due to the uniqueness of the Laplace transforms from %(\ref{uniquenessOfLaplaceTransform})}
, we have

\begin{equation*}
\nabla_a^{-\nu} \nabla_a^{-\mu} f(t) = \nabla_{a}^{-\nu - \mu} f(t),
\end{equation*}

proving (\ref{sumWithSum}).

\end{proof}








\subsection{Nabla Difference Composed with a Nabla Sum}

We will now provided a general form for a fractional difference composed with a fractional sum, however in doing so we will first consider the case of whole order differences composed with fractional sums and fractional differences.



\begin{lemma}[Whole Order Differences Composed with Fractional Sums and Differences]
Let $k \in \naturals_0$, $\mu > 0$, and choose $M \in \naturals$ such that $M-1 < \mu \leq M$ be given. Then

\begin{equation}
\label{wholeOrderDifferenceWithSum}
\nabla^k \nabla_a^{-\mu} f(t) = \nabla_a^{k-\mu} f(t),
\end{equation}

and

\begin{equation}
\label{wholeOrderDifferenceWithDifference}
\nabla^k \nabla_{a}^{\mu} f(t) = \nabla_{a}^{k+\mu} f(t).
\end{equation}
\end{lemma}

\begin{proof}
Let $k \in \naturals_0$, $\mu > 0$, and choose $M \in \naturals$ such that $M-1 < \mu \leq M$ be given. \\


Case 1: $\mu = M$.

First note the following: 
\begin{align*}
 \nabla \nabla^{-1} f(t) &= \nabla \frac{1}{\Gamma(1)} \sum_{s=a+1}^t (t - \rho(s))^{\overline{1-1}} f(s) & \text{from } (\ref{nablaSum} )\\
 = & \nabla \sum_{s=a+1}^t f(s) \\
 = & \sum_{s=a+1}^t f(s) - \sum_{s=a+1}^{t-1} f(s) \\
 = & f(t).
\end{align*}

So then for the case of $\mu = M$ we have 
\begin{align*}
\nabla^k \nabla^{-M} f(t) = \nabla^{k-1} (\nabla \nabla^{-1} (\nabla^{-(M-1)} f(t))) &=   \nabla^{k-1} \nabla^{-(M-1)} f(t) \\
&=  \nabla^{k-2} \nabla^{-(M-2)} f(t) \\
& \dots \\
&=  \nabla^{-(M-k)} f(t) \\
&=  \nabla^{k-M} f(t). \\
\end{align*}
This shows for this particular case that (\ref{wholeOrderDifferenceWithSum}) holds. \\

And it follows trivially that for a whole order difference composed with another whole order difference that (\ref{wholeOrderDifferenceWithDifference}) holds. \\

Case 2: $M-1 < \mu < M$ \\
First we will show $\nabla \nabla_a^{\mu} f(t) = \nabla_a^{1+\mu}$.

\begin{align*}
\nabla \nabla_a^{\mu} f(t) &= \nabla \left( \frac{1}{\Gamma(-\mu)} \sum_{s=a+1}^t (t - \rho(s))^{\overline{-\mu-1}} \right) & \text{from } (\ref{alternativeNablaDifference}) \\
&= \frac{1}{\Gamma(-\mu)} \sum_{s=a+1}^t (-\mu -1) (t-\rho(s))^{\overline{-\mu-2}} f(s) & \text{from } (\ref{leibnizRule}) \\
& = \frac{1}{\Gamma(-\mu-1)} \sum_{s=a+1}^t (t-\rho(s))^{\overline{-\mu-2}} \\
& = \nabla_a^{-(-\mu-1)} f(t) & \text{from } (\ref{alternativeNablaDifference}) \\
& = \nabla_a^{1+\mu} f(t). \\
\end{align*}

So for any $k \in \naturals_0$,

\begin{align*}
\nabla^k \nabla_a^\mu f(t) &= \nabla^{k-1} (\nabla \nabla_a^\mu f(t)) \\
& = \nabla^{k-1} \nabla_a^{1 + \mu} f(t) \\
& = \nabla^{k-2} \nabla_a^{2 + \mu} f(t) \\
& \dots \\
& = \nabla_a^{k+\mu} f(t), \\
\end{align*}

which shows (\ref{wholeOrderDifferenceWithSum}) holds for this case.

Due to the duality of definition \ref{unifiedDefinition}, a similar argument where one replaces $-\mu$ in place of $\mu$ gives the wanted result of $\nabla^k \nabla_a^{-\mu} f(t) = \nabla_a^{k - \mu}$, proving (\ref{wholeOrderDifferenceWithDifference}).


\end{proof}


With this lemma in hand, we can now prove the more general case.


\begin{theorem}
For $\nu, \mu > 0$,
\begin{equation*}
\nabla_{a}^\nu \nabla_a^{-\mu} f(t) = \nabla_{a}^{\nu-\mu} f(t).
\end{equation*}
\end{theorem}

\begin{proof}
Let $\nu, \mu > 0$ be given, and $N \in \naturals$ such that $N-1 < \nu \leq N$. Then we have

\begin{align*}
\nabla_{a}^\nu \nabla_a^{-\mu} &= \nabla^N \nabla_{a}^{-(N-\nu)} \nabla_a^{-\mu} & \text{from } (\ref{nablaDifference}) \\
& = \nabla^N \nabla_a^{-(N-\nu)-\mu} f(t) & \text{from } (\ref{sumWithSum}) \\
& = \nabla_a^{N-N+\nu-\mu} f(t) & \text{from previous lemma} \\
& = \nabla_a^{\nu-\mu} f(t).
\end{align*}

This shows (\ref{differenceWithSum}).
\end{proof}









\subsection{Nabla Sum Composed with a Whole Order Nabla Difference}
For the next two composition rules, where the inner operation is differentiation, the orders do not simply add up. The restriction on the inner differentiation being whole order is due to unresolved issues dealing with compositions that have mismatching bases. With whole order operations, there is no dependence on a base, allowing us to prove the following results.




\begin{theorem}
Let $\nu >0$ and $k \in \naturals_0$ be given. Then

\begin{equation*}
\nabla_{a+k}^{-\nu} \nabla^{k} f(t)= \nabla_{a+k}^{k-\nu}f(t) - \sum_{j=0}^{k-1} \nabla^{j}f(a+k) \frac{(t-a-k)^{\overline{\nu -k + j}}}{\Gamma(\nu -k +j +1)}.
\end{equation*}

\end{theorem}

\begin{proof}

\begin{align*}
\nabla_{a+k}^{-\nu} \nabla^{k} f(t)&= \frac{1}{\Gamma(\nu)} \sum_{s=a+k+1}^t (t - \rho(s))^{\overline{\nu-1}} \nabla^k f(s) & \text{from } (\ref{nablaSum})  \\
& =  \frac{1}{\Gamma(\nu)} \sum_{s=a+k+1}^t (t - \rho(s))^{\overline{\nu-1}} \nabla \nabla^{k-1} f(s)  \\
& = \frac{1}{\Gamma(\nu)} \left(\nabla^{k-1}f(s) (t-s)^{\overline{\nu-1}} |_{s=a}^t - \sum_{s=a+k+1}^t -(\nu-1)(t- \rho(s))^{\overline{\nu-2}} \nabla^{k-1} f(s)   \right) & \text{from summation by parts} \\
& = -\nabla^{k-1}f(a+k) \frac{(t-a-k)^{\overline{\nu-1}}}{\Gamma(\nu)} + \frac{1}{\Gamma(\nu-1)} \sum_{s=a+k+1}^t (t - \rho(s))^{\overline{\nu-2}} \nabla^{k-1} f(s) \\
& = \nabla_{a+k}^{-(\nu-1)} \nabla^{k-1} f(t) - \nabla^{k-1}f(a+k) \frac{(t-a-k)^{\overline{\nu-1}}}{\Gamma(\nu)} & \text{from } (\ref{nablaSum}) \\
& = \nabla( \nabla_{a+k}^{-\nu} \nabla^{k-1} f(t)) - \nabla^{k-1}f(a+k) \frac{(t-a-k)^{\overline{\nu-1}}}{\Gamma(\nu)} \\
& = \nabla_{a+k}^{-(\nu-2)} \nabla^{k-2} f(t) -\nabla^{k-2}f(a+k) \frac{(t-a-k)^{\overline{\nu-2}}}{\Gamma(\nu-1)} - \nabla^{k-1}f(a+k) \frac{(t-a-k)^{\overline{\nu-1}}}{\Gamma(\nu)}. \\
& \text{Doing summation by parts $k-2$ more times gives} \\
& = \nabla_{a+k}^{k-\nu}f(t) - \sum_{j=0}^{k-1} \nabla^{j}f(a+k) \frac{(t-a-k)^{\overline{\nu -k + j}}}{\Gamma(\nu -k +j +1)}.
\end{align*}

This proves (\ref{sumWithWholeOrderDifference}).

\end{proof}




\subsection{Nabla Difference Composed with a Whole Order Nabla Difference}

\begin{theorem}
Let $\nu > 0$ and $k \in \naturals_0$ be given and choose $N \in \naturals$ such that $N-1 < \nu \leq N$. Then
\begin{equation*}
\nabla_{a+k}^\nu \nabla^k = \nabla_{a+k}^{k+\nu} f(t) - \sum_{j=0}^{k-1}\nabla^j f(a+k) \frac{(t-a-k)^{\overline{-\nu-k+j}}}{\Gamma(-v-k+j+1)}.
\end{equation*}

\end{theorem}


\begin{proof}


\begin{align*}
\nabla_{a+k}^\nu \nabla^k &= \nabla^N(\nabla_{a+k}^{-(N-\nu)} \nabla^k f(t))\\
\\
&= \nabla^N \left(\nabla_{a+k}^{k-(N-\nu)} f(t) - \sum_{j=0}^{k-1} \nabla^j f(a+k) \frac{(t-a-k)^{\overline{N-\nu-k+j}}}{\Gamma(N-\nu-k+j+1)}  \right) \\
&= \nabla_{a+k}^{k+\nu} f(t) - \sum_{j=0}^{k-1} \nabla^j f(a+k) \nabla^{N}\frac{(t-a-k)^{\overline{N-\nu-k+j}}}{\Gamma(N-\nu-k+j+1)} \\
&= \nabla_{a+k}^{k+\nu} f(t) - \sum_{j=0}^{k-1} \nabla^j f(a+k) \nabla^{N-1} \frac{(N-\nu-k+j)(t-a-k)^{\overline{N-\nu-k+j-1}}}{\Gamma(N-\nu-k+j+1)}\\
&= \nabla_{a+k}^{k+\nu} f(t) - \sum_{j=0}^{k-1} \nabla^j f(a+k) \nabla^{N-1} \frac{(t-a-k)^{\overline{N-\nu-k+j-1}}}{\Gamma(N-\nu-k+j)}.\\
&\text{Then by taking the difference inside the summation $N-1$ more times, we get} \\
&=\nabla_{a+k}^{k+\nu} f(t) - \sum_{j=0}^{k-1}\nabla^j f(a+k) \frac{(t-a-k)^{\overline{-\nu-k+j}}}{\Gamma(-v-k+j+1))}.
\end{align*}

This proves (\ref{differenceWithWholeOrderDifference}).

\end{proof}






%%%%%%%%%%    Luke



\section{Solutions to Initial Value Problems}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We now will consider a general $\nu^{th}$ order fractional nabla initial-value problem and give a formula for its solution, $1<\nu\leq2$. To accomplish this, we will first look at the homogenous case.

\begin{theorem} Let $f:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $1<\nu\leq2$. Then, for $t\in\mathbb{N}_{a+2}$, the fractional initial value problem
	
\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t)=0, &t\in\naturals_{a+2}\\
		f(a+2)=A_{0}, &A_{0}\in\mathbb{R}\\
		\nabla f(a+2)=A_{1}, &A_{0}\in\mathbb{R}\\
	\end{cases}
	\]	
		
has the solution
	
\begin{align}	
\label{homogenousEquation1}	 
f(t) = [(2-\nu)A_{0}+(\nu-1)A_{1}]h_{\nu-1}(t,a) + [(\nu-1)A_{0}-\nu A_{1}]h_{\nu-2}(t,a).
\end{align}

\begin{proof}	 We begin by taking the Laplace transform of both sides, starting at a+2. Applying Theorems (\ref{laplaceTransFracDifference}) and (\ref{shiftingTheorem}) , we obtain


\begin{align*}
\mathcal{L}_{a+2} \{ \nabla^{\nu}_{a}f\}(s) &=\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - \sum\limits^{1}_{k=0}[\frac{s^{2}}{(1-s)^{2-k}}\nabla^{-(2-\nu)}_{a}f(a+k+1)\\ 
&\qquad+\nabla^{2-k-1}\nabla^{-(2-\nu)}_{a}f(a+2)s^{k}] = 0\\
&=\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - (\frac{s}{1-s})^{2}\nabla^{-(2-\nu)}_{a}f(a+1)-\nabla\nabla^{-(2-\nu)}_{a}f(a+2)\\
&\qquad-(\frac{s^{2}}{1-s})\nabla^{-(2-\nu)}_{a}f(a+2)-s\nabla^{-(2-\nu)}_{a}f(a+2) = 0.\\
\end{align*}

Now we substitute $f(a+2) = A_{0}$ and $\nabla f(a+2) = A_{1}$, where $f(a+1) = \nabla f(a+2) 
\\*-\nabla f(a+2)  = A_{0} - A_{1}$.

Also, note that $\nabla^{-(2-\nu)}_{a}f(a+2) = [(2-\nu)[A_{0}-A_{1}]+A_{0}]$ and $\nabla\nabla^{-(2-\nu)}_{a}f(a+2) = \nabla^{-(1-\nu)}_{a}f(a+2) = [(1-\nu)[A_{0}-A_{1}]+A_{0}]$.

\begin{align*}
\mathcal{L}_{a+2} \{ \nabla^{\nu}_{a}f\}(s) &=\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s)-(\frac{s}{1-s})^{2}[A_{0}-A_{1}]-[(1-\nu)[A_{0}-A_{1}]+A_{0}]\\
&\qquad-(\frac{s}{1-s})[(2-\nu)[A_{0}-A_{1}]+A_{0}] = 0.
\end{align*}

Next, we combine all terms with respect to $A_{0}$ and $A_{1}$ with the common denominator of $(1-s)^{2}$.

\begin{align*}
&\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - \frac{\nu s - s - \nu + 2}{(1-s)^2}A_{0} + \frac{\nu-\nu s - 1}{(1-s)^{2}}A_{1} = 0\\
&\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) = \frac{\nu s - s - \nu + 2}{(1-s)^2}A_{0} - \frac{\nu-\nu s - 1}{(1-s)^{2}}A_{1}.\\
\end{align*}

We now rearrange the terms and solve for the Laplace transform of $f$.

\begin{align*}
\mathcal{L}_{a}\{f\}(s) &= [(2-\nu)A_{0} + (\nu-1)A_{1}]\frac{1}{s^{\nu}}+[(\nu-1)A_{0} + \nu A_{1}]\frac{1}{s^{\nu-1}}.
\end{align*}

Finally, we take the inverse Laplace transform to obtain the desired result.

\begin{align*}
f(t) = [(2-\nu)A_{0}+(\nu-1)A_{1}]h_{\nu-1}(t,a) + [(\nu-1)A_{0}-\nu A_{1}]h_{\nu-2}(t,a)\\ 
\end{align*}

for $t\in\naturals_{a+2}.$ 

	\end{proof}	
	\end{theorem} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Next, we look at the non-homogenous equation with zero initial conditions.

\begin{theorem} Let $f,g:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $1<\nu\leq2$. Then, for $t\in\mathbb{N}_{a+2}$, the fractional initial value problem
	
\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t) = g(t), &t\in\naturals_{a+2}\\
		f(a+2) = 0\\
		\nabla f(a+2)=0\\
	\end{cases}
	\]	
		
has the solution

\begin{align}
\label{nonhomogenousEquation1}	
f(t) = \nabla^{-\nu}_{a}g(t)-[g(a+1)+g(a+2)]h_{\nu-1}(t,a)+g(a+2)h_{\nu-2}(t,a)
\end{align}

\bigskip

\begin{proof}	 We take the Laplace transform based at a+2 of both sides of the equation.

\begin{align*}
\mathcal{L}_{a+2}\{f\}(s) &= \mathcal{L}_{a+2}\{g\}(s).\\
\end{align*}

Next, we use Theorems (\ref{laplaceTransFracDifference}) and (\ref{shiftingTheorem}) on the left hand side, and Theorem (\ref{shiftingTheorem}) on the right hand side of the equation.

\begin{align*}
&\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - (\frac{s}{1-s})^{2}\nabla^{-(2-\nu)}_{a}f(a+1)-\nabla\nabla^{-(2-\nu)}_{a}f(a+2) - (\frac{s^{2}}{1-s})\nabla^{-(2-\nu)}_{a}f(a+2)-s\nabla^{-(2-\nu)}_{a}f(a+2)\\
&\qquad = \frac{1}{(1-s)^2}\mathcal{L}_{a}\{g\}(s) - \frac{1}{(1-s)^2}g(a+1) - \frac{1}{1-s}g(a+1).
\end{align*}

Now, we plug in $f(a+2) = 0$. Also, all of the fractional sums and f(a+1) terms can be written in terms of sums of $f(a+2)$ and $\nabla f(a+2)$. Thus, we can plug in zero for them as well.

\begin{align*}
\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) &= \frac{1}{(1-s)^2}\mathcal{L}_{a}\{g\}(s) - \frac{1}{(1-s)^2}g(a+1) - \frac{1}{1-s}g(a+1).
\end{align*}

Next, we solve for the Laplace transform of f(t).

\begin{align*}
\mathcal{L}_{a}\{f\}(s) &= \frac{1}{s^{\nu}}\mathcal{L}_{a}\{g\}(s) - \frac{1}{s^{\nu}}g(a+1)-\frac{(1-s)}{s^{\nu}}g(a+2)\\
&= [\mathcal{L}_{a}\{h_{\nu-1}(t,a)\}(s)\mathcal{L}_{a}\{g\}(s)] - \frac{1}{s^{\nu}}g(a+1)-\frac{1}{s^{\nu}}g(a+2) + \frac{1}{s^{\nu-1}}g(a+2).\\
\end{align*}

Finally, we take the inverse Laplace transform and note that $\nabla^{-\nu}_{a}g(t) = h_{\nu-1}(t,a)*g(t)$ (\ref{convolution}).

\begin{align*}
f(t) &= [h_{\nu-1}(\cdot,a)*g(\cdot)]-[g(a+1)+g(a+2)]h_{\nu-1}(t,a)+g(a+2)h_{\nu-2}(t,a)\\
&= \nabla^{-\nu}_{a}g(t)-[g(a+1)+g(a+2)]h_{\nu-1}(t,a)+g(a+2)h_{\nu-2}(t,a).
\end{align*}

for $t\in\naturals_{a+2}$

	\end{proof}
	\end{theorem}
	
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
	
	Given (\ref{homogenousEquation1}) and (\ref{nonhomogenousEquation1}), we can now solve the non-homogenous equation with non-homogenous initial conditions.
	
	\begin{theorem} Let $f,g:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $1<\nu\leq2$. Then, for $t\in\mathbb{N}_{a+2}$, the fractional initial value problem

\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t)=g(t), &t\in\naturals_{a+2}\\
		f(a+2) = A_{0},&A_{0}\in\mathbb{R}\\
		\nabla f(a+2) = A_{1},&A_{1}\in\mathbb{R}
	\end{cases}
	\]	
	
has the solution	
	
\medskip

\begin{align*}	
f(t) &= \nabla^{-\nu}_{a}g(t)+\left[(2-\nu)A_{0}+(\nu-1)A_{1}-g(a+1)-g(a+2)\right]h^{a}_{\nu-1}(t)\nonumber\\ 
&\qquad+\left[(\nu-1)A_{0}-\nu A_{1}+g(a+2)\right]h^{a}_{\nu-2}(t).\\
\end{align*}

\medskip

\begin{proof} The solution can be shown to the sum of the solutions from Theorem (\ref{homogenousEquation1}) and Theorem (\ref{nonhomogenousEquation1}). This can be done trivially.

\end{proof}
	
	\end{theorem} 
	
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	

The result to the following initial-value problem can be obtained similarly.

\begin{theorem} Let $f,g:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $1<\nu\leq2$. Then, for $t\in\mathbb{N}_{a+2}$, the fractional initial value problem
	
	\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t)+cf(t)=g(t), &t\in\naturals_{a+2}\\
		f(a+2) = A_{0}, &A_{0}\in\mathbb{R}\\
		\nabla f(a+2) = A_{1}, &A_{1}\in\mathbb{R}\\
	\end{cases}
	\]
has the solution

\begin{align*}
f(t) &= [E_{-c,\nu,\nu-1}(\cdot,a)*g(\cdot)] -[g(a+1) + g(a+2) + (\nu-2c-2)A_{0}-(\nu-c-1)A_{1}]E_{-c,\nu,\nu-1}(t,a)\\
&\qquad + [g(a+2)+(\nu-c-1)A_{0}-\nu A_{1}]E_{-c,\nu,\nu-2}(t,a).
\end{align*}

\begin{proof} We begin by taking the Laplace transform based at a+2 of both sides of the equation.

\begin{align*}
\mathcal{L}_{a+2}\{\nabla^{\nu}_{a}\}(s) + c\mathcal{L}_{a+2}\{f\}(s) &= \mathcal{L}_{a+2}\{g\}(s).
\end{align*}

Now we apply Theorems (\ref{laplaceTransFracDifference}) and (\ref{shiftingTheorem}) to the first term and Theorem (\ref{shiftingTheorem}) the remaining terms.

\begin{align*}
&\frac{s^{\nu}}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - (\frac{s}{1-s})^{2}\nabla^{-(2-\nu)}_{a}f(a+1)-\nabla\nabla^{-(2-\nu)}_{a}f(a+2) - (\frac{s^{2}}{1-s})\nabla^{-(2-\nu)}_{a}f(a+2)-s\nabla^{-(2-\nu)}_{a}f(a+2)\\ 
&\qquad+ \frac{c}{(1-s)^2}\mathcal{L}_{a}\{f\}(s) - \frac{c}{(1-s)^2}f(a+1) - \frac{c}{1-s}f(a+2) = \frac{1}{(1-s)^2}\mathcal{L}_{a}\{g\}(s) - \frac{1}{(1-s)^2}g(a+1) - \frac{1}{1-s}g(a+2).
\end{align*}

Adding up like terms and simplifying gives

\begin{align*}
&\frac{s^{\nu}+c}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - \frac{(s^{2}+c)}{(1-s)^{2}}f(a+1) - \frac{c}{1-s}f(a+2) - \nabla^{-(1-\nu)}_{a}f(a+2) - \frac{s}{1-s}\nabla^{-(2-\nu)}_{a}f(a+2)\\
&\qquad= \frac{1}{(1-s)^2}\mathcal{L}_{a}\{g\}(s)- \frac{1}{(1-s)^2}g(a+1) - \frac{1}{1-s}g(a+2).
\end{align*}

Now we substitute $f(a+2) = A_{0}$ and $\nabla f(a+2) = A_{1}$, where $f(a+1) = \nabla f(a+2) 
\\*-\nabla f(a+2)  = A_{0} - A_{1}$.

Also, note that $\nabla^{-(2-\nu)}_{a}f(a+2) = [(2-\nu)[A_{0}-A_{1}]+A_{0}]$ and $\nabla^{-(1-\nu)}_{a}f(a+2) = [(1-\nu)[A_{0}-A_{1}]+A_{0}]$.

\begin{align*}
&\frac{s^{\nu}+c}{(1-s)^{2}}\mathcal{L}_{a}\{f\}(s) - \frac{(s^{2}+c)}{(1-s)^{2}}[A_{0}-A_{1}] - \frac{c}{1-s}A_{0} - [(1-\nu)[A_{0}-A_{1}]+A_{0}]- \frac{s}{1-s}[(2-\nu)[A_{0}-A_{1}]+A_{0}])\\ 
&\qquad= \frac{1}{(1-s)^2}\mathcal{L}_{a}\{g\}(s) - \frac{1}{(1-s)^2}g(a+1) - \frac{1}{1-s}g(a+2)
\end{align*}

Now we combine like terms with respect to $A_{i}$'s (Note: the common denominator of $\frac{1}{(1-s)^2}$ has been cancelled).

\begin{align*}
(s^{\nu}+c)\mathcal{L}\{f\}(s) +[\nu(1-s)+(s-2)(c+1)]A_{0} + [c+1+\nu(s-1)]A_{1} &= \frac{1}{(1-s)^2}\mathcal{L}_{a}\{g\}(s)- \frac{1}{(1-s)^2}g(a+1)\\
&\qquad - \frac{1}{1-s}g(a+2)
\end{align*}

Now solve for the Laplace transform of f(t).

\begin{align*}
\mathcal{L}\{f\}(s) &= \frac{1}{s^{\nu}+c}\mathcal{L}_{a}\{g\}(s) + \frac{1}{s^{\nu}+c}[g(a+1) + g(a+2) + (\nu-2c-2)A_{0}+(c+1-\nu)A_{1}]\\
&\qquad+\frac{s}{s^{\nu}+c}[g(a+2)+(\nu-c-1)A_{0}-\nu A_{1}].
\end{align*}

Finally, we take the inverse Laplace transform to obtain the desired result.

\begin{align*}
f(t) &= [E_{-c,\nu,\nu-1}(\cdot,a)*g(\cdot)] -[g(a+1) + g(a+2) + (\nu-2c-2)A_{0}\\
&\qquad-(\nu-c-1)A_{1}]E_{-c,\nu,\nu-1}(t,a) + [g(a+2)+(\nu-c-1)A_{0}-\nu A_{1}]E_{-c,\nu,\nu-2}(t,a). 
\end{align*}

\end{proof}

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Next, we will consider the non-homogenous initial value problem with homogenous initial conditions, $\nu>0$.

\begin{theorem} Let $f,g:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $\nu>0$. Then, for $t\in\mathbb{N}_{a+N}$, the fractional initial value problem

	\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t)=g(t), &t\in\naturals_{a+N}\\
		\nabla^{i}f(a+N)=0
	\end{cases}
	\]
	
has the solution

\begin{align}	
\label{nonhomogenousEquation2}
f(t) = \nabla^{-\nu}_{a}g(t)-\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}h_{\nu-1-i}(t,a).
\end{align}


\begin{proof} Pick N such that $N-1<\nu\leq N$ and take the Laplace transform based at N of both sides of the equation.

\begin{align*}
\mathcal{L}_{a+N}\{\nabla^{\nu}_{a}f\}(s) &= \mathcal{L}_{a+N}\{g\}(s).
\end{align*}

Next, apply Theorems (\ref{laplaceTransFracDifference}) and (\ref{shiftingTheorem})to the left hand side and theorem (\ref{shiftingTheorem}) to the right hand side.

\begin{align*}
&\frac{s^{\nu}}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s)-\sum\limits^{N-1}_{k=0}[\frac{s^{N}}{(1-s)^{N-k}}\nabla^{-(N-\nu)}_{a}f(a+k+1) + \nabla^{N-k-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{k}]\\
&\qquad= \frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}g(a+k+1).
\end{align*}

Note the sum on the left hand side goes to zero as a result of the initial conditions (the nabla sums and differences can be written in terms of sums of initial conditions which all are zero).

\begin{align*}
\frac{s^{\nu}}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s) &= \frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}g(a+k+1).
\end{align*}

Sovling for the Laplace transform of f gives

\begin{align*}
\mathcal{L}_{a}\{f\}(s) &= \frac{1}{s^{\nu}}\mathcal{L}_{a}\{g\}(s)-\frac{1}{s^{\nu}}\sum\limits^{N-1}_{k=0}(1-s)^{k}g(a+k+1).\\
\end{align*}

Next we use the fact that $(1-s)^{k} = \sum\limits^{k}_{i=0}\binom{k}{i}(-1)^{i}s^{i}$ to give

\begin{align*}
\mathcal{L}_{a}\{f\}(s) &= \frac{1}{s^{\nu}}\mathcal{L}_{a}\{g\}(s)-\frac{1}{s^{\nu}}\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}\binom{k}{i}(-1)^{i}s^{i}g(a+k+1)\\
&=\frac{1}{s^{\nu}}\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}\frac{1}{s^{\nu-i}}\\
&=\mathcal{L}_{a}\{h_{\nu-1}(t,a)\}(s)\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}\mathcal{L}_{a}\{h_{\nu-1-i}(t,a)\}(s).\\
\end{align*}

Finally, take the inverse Laplace transform and note that $\nabla^{-\nu}_{a}g(t) = h_{\nu-1}(t,a)*g(t)$ \ref{convolution}.

\begin{align*}
\mathcal{L}_{a}\{f\}(s) &= [h_{\nu-1}(t,a)*g(t)]-\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}h_{\nu-1-i}(t,a)\\
&= \nabla^{-\nu}_{a}g(t)-\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}h_{\nu-1-i}(t,a).
\end{align*}

\end{proof}

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{remark} We can also obtain this solution by applying the composition rules.

\begin{proof}

We apply the definition of a nabla fractional difference \cite{REU2010}

\begin{align*}
\nabla^{N}\nabla^{-(N-\nu)}_{a}f(t) &= g(t)
\end{align*}

Next, we apply a $\nu^{th}$ order nabla sum based at a+N to both sides of the equation. 

\begin{align*}
\nabla^{-\nu}_{a+N}\left(\nabla^{N}\nabla^{-(N-\nu)}_{a}f(t)\right) &= \nabla^{-\nu}_{a+N}g(t)
\end{align*}

Next, we apply the composition rule for a sum composed with a whole order difference based at a+N to the left hand side \ref{sumWithWholeOrderDifference}. The right hand side will simply be evaluated by the definition of a nabla sum \cite{REU2010} 

\begin{align*}
\nabla^{N-\nu}_{a+N}\nabla^{-(N-\nu)}_{a}f(t) - \sum\limits^{N-1}_{j=0}\frac{\nabla^{j}_{a}f(a+N)}{\G(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} = \nabla^{-\nu}_{a+N}g(t)
\end{align*}

Next, we want to manipulate the first term such that we can apply a composition rule to obtain f(t) (i.e. have the nabla sum and difference based at the same point). We do this by using the defintion of a fractional nabla sum \cite{REU2010}.

\begin{align*}
\nabla^{N-\nu}_{a+N}[\frac{1}{\Gamma(N-\nu)}\int^{t}_{a}(t-\rho(s))^{\overline{N-\nu-1}}f(s)\nabla s]=\nabla^{-\nu}_{a+N}g(t).
\end{align*}

Then we split the integral into two parts.

\begin{align*}
\nabla^{N-\nu}_{a+N}[\frac{1}{\Gamma(N-\nu)}(\int^{t}_{a+N}(t-\rho(s))^{\overline{N-\nu-1}}f(s)\nabla s+\int^{a+N}_{a}(t-\rho(s))^{\overline{N-\nu-1}}f(s)\nabla s)]= \nabla^{-\nu}_{a+N}g(t).
\end{align*}

We now apply the definition of a nabla sum and covert the second integral to a sum.

\begin{align*}
\nabla^{N-\nu}_{a+N}[\nabla^{-(N-\nu)}_{a+N}f(t)+\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)] &= \nabla^{-\nu}_{a+N}g(t)\\
\nabla^{N-\nu}_{a+N}[\nabla^{-(N-\nu)}_{a+N}f(t)]+\nabla^{N-\nu}_{a+N}[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)]&= \nabla^{-\nu}_{a+N}g(t).
\end{align*}

Applying the composition rule for composing fractional differences with fractional sums based at the same point \ref{differenceWithSum} gives

\begin{align*}
f(t)+\nabla^{N-\nu}_{a+N}[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)]&= \nabla^{-\nu}_{a+N}g(t).
\end{align*}

The second summation goes to zero with the application of the initial conditions. Next we show that this solution is equivalent to the solution obtained in Theorem (\ref{nonhomogenousEquation2})

\begin{align*}
f(t)&= \nabla^{-\nu}_{a+N}g(t)\\
&=\frac{1}{\G(\nu)}\int^{t}_{a+N}(t-\rho(s))^{\overline{\nu-1}}g(s)\nabla s\\
&=\frac{1}{\G(\nu)}[\int^{t}_{a}(t-\rho(s))^{\overline{\nu-1}}g(s)\nabla s-\int^{a}_{a+N}(t-\rho(s))^{\overline{\nu-1}}g(s)\nabla s] \\
&=\nabla^{-\nu}_{a}g(t) - \int^{a}_{a+N}\frac{(t-s+1)^{\overline{\nu-1}}}{\G(\nu)}g(s)\nabla s\\
&=\nabla^{-\nu}_{a}g(t) - \sum\limits^{a+1}_{a+N}h_{\nu-1}(t,s-1)g(s).
\end{align*}

We next change the index of summation by replacing $s\rightarrow s+a+1$.

\begin{align*}
&=\nabla^{-\nu}_{a}g(t) - \sum\limits^{N-1}_{s=0}h_{\nu-1}(t,s+a)g(s+a+1)\\
\end{align*}

Now we apply the Taylor Monomial shifting theorem (\ref{taylorShifting}) to obtain the desired result.

\begin{align*}
&=\nabla^{-\nu}_{a}g(t) - \sum\limits^{N-1}_{s=0}\sum\limits^{s}_{i=0}\binom{s}{i}(-1)^{i}h_{\nu-i-1}(t,a)g(a+s+1).\\
\end{align*}

\end{proof}

\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{theorem} Let $f,g:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $\nu>0$. Then, for $t\in\mathbb{N}_{a+N}$ and $|c|<1$, the fractional initial value problem
	
\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t)+cf(t)=g(t), &t\in\naturals_{a+N}\\
		\nabla^{i}f(a+N)=0
	\end{cases}
	\]	

has the solution

\begin{align*}
f(t) &= [E_{-c,\nu,\nu-1}(\cdot,a)*g(\cdot)] -\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}E_{-c,\nu,\nu-1-i}(t,a).
\end{align*}

\begin{proof} We begin by taking the Laplace transform based at a+N of both sides of the equation.

\begin{align*}
\mathcal{L}_{a+N}\{\nabla^{\nu}_{a}f\}(s)+c\mathcal{L}_{a+N}\{f\}(s) = \mathcal{L}_{a+N}\{g\}(s)
\end{align*}

We apply Theorems (\ref{laplaceTransFracDifference}) and (\ref{shiftingTheorem}) to the first term.

\begin{align*}
&\frac{s^{\nu}}{(1-s)^N}\mathcal{L}_{a}\{f\}(s) - \sum\limits^{N-1}_{k=0}\left[\frac{s^{N}}{(1-s)^{N-k}}\nabla^{-(N-\nu)}_{a}f(a+k+1)+\nabla^{N-k-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{k}\right]+c\mathcal{L}_{a+N}\{f\}(s) \\
&\qquad= \mathcal{L}_{a+N}\{g\}(s).
\end{align*}

Applying initial conditions causes the summation above to go to zero. Thus we have the following.

\begin{align*}
&\frac{s^{\nu}}{(1-s)^N}\mathcal{L}_{a}\{f\}(s)+c\mathcal{L}_{a+N}\{f\}(s) = \mathcal{L}_{a+N}\{g\}(s).
\end{align*}

Now we apply the shifting theorem (\ref{shiftingTheorem}) to the terms based at a+N

\begin{align*}
\frac{s^{\nu}}{(1-s)^N}\mathcal{L}_{a}\{f\}(s)+c[\frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s)-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}f(a+k+1)]&= \frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{g\}(s)\\
&\qquad-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}g(a+k+1).
\end{align*}

Applying the initial conditions again causes the first summation to go to zero as well.

\begin{align*}
\frac{s^{\nu}}{(1-s)^N}\mathcal{L}_{a}\{f\}(s)+\frac{c}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s)&= \frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}g(a+k+1)\\
\frac{s^{\nu}+c}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s)&= \frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}g(a+k+1).
\end{align*}

Next, we solve for the Laplace transform of f.

\begin{align*}
\mathcal{L}_{a}\{f\}(s)&= \frac{(1-s)^{N}}{s^{\nu}+c}[\frac{1}{(1-s)^{N}}\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\frac{1}{(1-s)^{N-k}}g(a+k+1)]\\
&=\mathcal{L}_{a}\{E_{-c,\nu,\nu-1}(\cdot,a)\}(s)\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\frac{(1-s)^{k}}{{s^{\nu}+c}}g(a+k+1).
\end{align*}

Now we apply the binomial theorem to $(1-s)^k$.

\begin{align*}
\mathcal{L}_{a}\{f\}(s)&=\mathcal{L}_{a}\{E_{-c,\nu,\nu-1}(\cdot,a)\}(s)\mathcal{L}_{a}\{g\}(s)-\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}\frac{s^{i}}{{s^{\nu}+c}}.
\end{align*}

Finally, take the inverse Laplace transform to obtain the desired result.

\begin{align*}
\mathcal{L}_{a}\{f\}(s)&= [E_{-c,\nu,\nu-1}(\cdot,a)*g(\cdot)] -\sum\limits^{N-1}_{k=0}\sum\limits^{k}_{i=0}g(a+k+1)\binom{k}{i}(-1)^{i}E_{-c,\nu,\nu-1-i}(t,a).
\end{align*}


\end{proof}

\end{theorem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Finally, we look at the homogenous initial value problem corresponding to Theorem (\ref{nonhomogenousEquation2}) with nonhomogenous initial conditions.

\begin{conjecture}Let $f:\mathbb{N}_{a}\rightarrow\mathbb{R}$ and $\nu>0$. Then, for $t\in\mathbb{N}_{a+N}$, the fractional initial value problem

\[
	\begin{cases}
		\nabla^{\nu}_{a}f(t)=0, &t\in\naturals_{a+N}\\
		\nabla^{i}f(a+N)=A_{i}, &A_{i}\in\mathbb{R},\qquad i=0,1,2...N-1
	\end{cases}
	\]	
	
has the solution

\begin{align}
\label{homogenousEquation2}	
f(t) = \sum\limits^{N-1}_{i=0}h_{\nu-1-i}(t,a)\sum\limits^{N-1}_{j=0}\sum\limits^{N-j-1}_{l=0}(-1)^{l}\binom{N-j-1}{l}A_{l}\sum\limits^{i}_{k}\frac{(i-\nu+1-k)^{\overline{N-1-j}}}{\Gamma(N-j)}\binom{N}{k}(-1)^{k}.
\end{align}

\begin{remark} Pick N such that $N-1<\nu\leq N$. Then

\begin{align*}
\nabla^{-\nu}_{a+N}\nabla^{\nu}_{a}f(t)&=0.
\end{align*}

We next apply the definition of a $\nu$-th order difference (\ref{nablaDifference}).

\begin{align*}
\nabla^{-\nu}_{a+N}\nabla^{N}(\nabla^{-(N-\nu)}_{a}f(t))&=0.
\end{align*}

Next, we apply the composition rule for a $\nu$-th order nabla sum based at $a+N$ of a whole order difference (\ref{sumWithWholeOrderDifference}).

\begin{align*}
\nabla^{N-\nu}_{a+N}\nabla^{-(N-\nu)}_{a}f(t) - \sum\limits^{N-1}_{j=0}\frac{\nabla^{j}\nabla^{-(N-\nu)}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} &= 0.
\end{align*}

Next, we use the definition of a fractional sum (\ref{nablaSum}).

\begin{align*}
\nabla^{N-\nu}_{a+N}[\frac{1}{\Gamma(N-\nu)}\int^{t}_{a}(t-\rho(s))^{\overline{N-\nu-1}}f(s)\nabla s] - \sum\limits^{N-1}_{j=0}\frac{\nabla^{j-(N-\nu)}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} &= 0.\\
\end{align*}

Then we split the integral into two parts.

\begin{align*}
\nabla^{N-\nu}_{a+N}\left[\frac{1}{\Gamma(N-\nu)}\left(\int^{t}_{a+N}(t-\rho(s))^{\overline{N-\nu-1}}f(s)\nabla s+\int^{a+N}_{a}(t-\rho(s))^{\overline{N-\nu-1}}f(s)\nabla s\right)\right]\\
- \sum\limits^{N-1}_{j=0}\frac{\nabla^{j-N+\nu}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} = 0.
\end{align*}

We now apply the definition of a nabla sum and covert the second integral to a sum (\ref{nablaSum}).

\begin{align*}
\nabla^{N-\nu}_{a+N}[\nabla^{-(N-\nu)}_{a+N}f(t)+\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)]- \sum\limits^{N-1}_{j=0}\frac{\nabla^{j-N+\nu}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} &= 0\\
\nabla^{N-\nu}_{a+N}[\nabla^{-(N-\nu)}_{a+N}f(t)]+\nabla^{N-\nu}_{a+N}[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)]- \sum\limits^{N-1}_{j=0}\frac{\nabla^{j-N+\nu}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} &= 0.
\end{align*}

Applying the composition rule for composing fractional differences with fractional sums based at the same point (\ref{differenceWithSum}) gives

\begin{align*}
f(t)+\nabla^{N-\nu}_{a+N}[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)]- \sum\limits^{N-1}_{j=0}\frac{\nabla^{j-N+\nu}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}} &= 0.
\end{align*}

Now we solve for f(t).

\begin{align*}
f(t) &= -\nabla^{N-\nu}_{a+N}[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)]+  \sum\limits^{N-1}_{j=0}\frac{\nabla^{j-N+\nu}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}}.
\end{align*}

Next, we focus our attention on the first summation.

\begin{align*}
\nabla^{N-\nu}_{a+N}\left[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)\right] &= \nabla^{N-\nu}_{a+N}\left[\frac{(t-a-1+1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+1)+\frac{(t-a-2+1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+2)\right.\\
&\left.\qquad+\cdots+\frac{(t-a-N+1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+N)\right]\\
&=\nabla^{N-\nu}_{a+N}\left[\frac{(t-a)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+1)+\frac{(t-a-1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+2)+\cdots\right.\\
&\qquad\left.+\frac{(t-a-N+1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+N)\right]\\
&=\nabla^{N-\nu}_{a+N}\frac{(t-a)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+1)+\nabla^{N-\nu}_{a+N}\frac{(t-a-1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+2)+\cdots\\
&\qquad+\nabla^{N-\nu}_{a+N}\frac{(t-a-N+1)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(a+N).
\end{align*}

Apply the definition of a fractional order nabla difference (\ref{nablaDifference}) to all terms. We then will split up the integrals similarly to earlier steps.


\begin{align*}
\nabla^{N-\nu}_{a+N}\left[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)\right] &=\frac{f(a+1)}{\Gamma(N-\nu)}\frac{1}{\Gamma(\nu-N)}\int^{t}_{a+N}(t-a)^{\overline{N-\nu-1}}(s-a)^{\overline{N-\nu-1}}\nabla s\\
&\qquad+\frac{f(a+2)}{\Gamma(N-\nu)}\frac{1}{\Gamma(\nu-N)}\int^{t}_{a+N}(t-a)^{\overline{N-\nu-1}}(s-a-1)^{\overline{N-\nu-1}}\nabla s +\cdots\\
&\qquad+\frac{f(a+N)}{\Gamma(N-\nu)}\frac{1}{\Gamma(\nu-N)}\int^{t}_{a+N}(t-a)^{\overline{N-\nu-1}}(s-a-N+1)^{\overline{N-\nu-1}}\nabla s\\
&=\frac{f(a+1)}{\Gamma(N-\nu)}\frac{1}{\Gamma(\nu-N)}\Bigg[\int^{t}_{a}(t-a)^{\overline{N-\nu-1}}(s-a)^{\overline{N-\nu-1}}\nabla s\\
&\qquad-\int^{a+N}_{a}(t-a)^{\overline{N-\nu-1}}(s-a)^{\overline{N-\nu-1}}\nabla s\Bigg]\\
&\qquad+\frac{f(a+2)}{\Gamma(N-\nu)}\frac{1}{\Gamma(\nu-N)}\Bigg[\int^{t}_{a+1}(t-a)^{\overline{N-\nu-1}}(s-a-1)^{\overline{N-\nu-1}}\nabla s\\
&\qquad-\int^{a+N}_{a+1}(t-a)^{\overline{N-\nu-1}}(s-a-1)^{\overline{N-\nu-1}}\nabla s\Bigg] +\cdots\\
&\qquad+\frac{f(a+N)}{\Gamma(N-\nu)}\frac{1}{\Gamma(\nu-N)}\Bigg[\int^{t}_{a+N-1}(t-a)^{\overline{N-\nu-1}}(s-a-N+1)^{\overline{N-\nu-1}}\nabla s\\
&\qquad-\int^{a+N}_{a+N-1}(t-a)^{\overline{N-\nu-1}}(s-a-N+1)^{\overline{N-\nu-1}}\nabla s\Bigg].
\end{align*}

Now we apply the definition of the fractional order nabla sum (\ref{nablaSum}) and convert the integrals to sums.

\begin{align*}
\nabla^{N-\nu}_{a+N}\left[\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s)\right] &=\frac{f(a+1)}{\Gamma(N-\nu)}\Bigg[\nabla^{N-\nu}_{a}(t-a)^{N-\nu-1}-\frac{1}{\Gamma(\nu-N)}\sum\limits^{a+N}_{s=a+1}(t-\rho(s))^{\overline{N-\nu-1}}(s-a)^{\overline{N-\nu-1}}\Bigg]\\
&\qquad+\frac{f(a+2)}{\Gamma(N-\nu)}\Bigg[\nabla^{N-\nu}_{a+1}(t-a-1)^{N-\nu-1}\\
&\qquad-\frac{1}{\Gamma(\nu-N)}\sum\limits^{a+N}_{s=a+2}(t-\rho(s))^{\overline{N-\nu-1}}(s-a)^{\overline{N-\nu-1}}\Bigg]+\cdots\\
&\qquad+\frac{f(a+N)}{\Gamma(N-\nu)}\Bigg[\nabla^{N-\nu}_{a+N-1}(t-a-N+1)^{N-\nu-1}\\
&\qquad-\frac{1}{\Gamma(\nu-N)}\sum\limits^{a+N}_{s=a+N}(t-\rho(s))^{\overline{N-\nu-1}}(s-a-N+1)^{\overline{N-\nu-1}}\Bigg].
\end{align*}

By the generalized power rule \cite{REU2010} that every nabla difference term can be taken. Also notice these terms go to zero. This leaves the following summation.

\begin{align*}
\nabla^{N-\nu}_{a+N}\sum\limits^{a+N}_{a+1}\frac{(t-\rho(s))^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}f(s) &= -\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{a+N}_{s=a+j+1}\frac{{(t-\rho(s))^{\overline{\nu-N-1}}(s-a-j)^{\overline{N-\nu-1}}}}{{\Gamma(\nu-N)}}\\
&=-\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{a+N}_{s=a+j+1}h_{\nu-N-1}(t,s-1)(s-a-j)^{\overline{N-\nu-1}}.\\
\end{align*}

Next we replace $s\rightarrow s+a+1$

\begin{align*}
&=-\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}h_{\nu-N-1}(t,s+a)(s+1-j)^{\overline{N-\nu-1}}.\\
\end{align*}

Substitute this summation into the original equation to give the following.

$$f(t)=\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}h_{\nu-N-1}(t,s+a)(s+1-j)^{\overline{N-\nu-1}}\\+\sum\limits^{N-1}_{j=0}\frac{\nabla^{j-N+\nu}_{a}f(a+N)}{\Gamma(\nu-N+j+1)}(t-a-N)^{\overline{\nu-N+j}}$$

Next, we rewrite $\nabla^{j-N+\nu}_{a}f(a+N)$.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}h_{\nu-N-1}(t,s+a)(s+1-j)^{\overline{N-\nu-1}}\\
&\qquad+\sum\limits^{N-1}_{j=0}\sum\limits^{N-1}_{l=0}\frac{(N-\nu-j)^{\overline{l}}}{\G(l+1)}f(a+N-l)h_{\nu-N+j}(t,a+N)
\end{align*}

Next, we will reverse the order of summation in the last summation.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}h_{\nu-N-1}(t,s+a)(s+1-j)^{\overline{N-\nu-1}}\\
&\qquad+\sum\limits^{N-1}_{j=0}\sum\limits^{N-1}_{l=0}\frac{(N-\nu-j)^{\overline{N-1-l}}}{\G(N-l)}f(a+l+1)h_{\nu-N+j}(t,a+N)
\end{align*}

Now we will swap the order of summation of the last two sums (note: we will also swap the j and l indices).

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}\frac{f(a+j+1)}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}h_{\nu-N-1}(t,s+a)(s+1-j)^{\overline{N-\nu-1}}\\
&\qquad+\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}h_{\nu-N+l}(t,a+N)\\
&=\sum\limits^{N-1}_{j=0}{f(a+j+1)}\left[\frac{(s+1-j)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}h_{\nu-N-1}(t,s+a)\right.\\
&\left.\qquad+\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}h_{\nu-N+l}(t,a+N)\right].
\end{align*}

Now we apply the shifting theorem for Taylor monomials based at a+m (\ref{taylorShifting}).

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}{f(a+j+1)}\left[\frac{(s+1-j)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}\sum\limits^{s}_{k=0}\binom{s}{k}(-1)^{k}h_{\nu-N-1-k}(t,a)\right.\\
&\left.\qquad+\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\sum\limits^{N}_{k=0}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a)\right].
\end{align*}

Next, we split the second summation into two parts.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}{f(a+j+1)}\left[\frac{(s+1-j)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}\sum\limits^{s}_{k=0}\binom{s}{k}(-1)^{k}h_{\nu-N-1-k}(t,a)\right.\\
&\left.\qquad+\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\left(\sum\limits^{l}_{k=0}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a)\right.\right.\\
&\left.\left.\qquad+\sum\limits^{N}_{k=l+1}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a)\right)\right]\\
&=\sum\limits^{N-1}_{j=0}{f(a+j+1)}\left[\frac{(s+1-j)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}\sum\limits^{s}_{k=0}\binom{s}{k}(-1)^{k}h_{\nu-N-1-k}(t,a)\right.\\
&\left.\qquad+\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\sum\limits^{l}_{k=0}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a)\right.\\
&\left.\qquad+\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\sum\limits^{N}_{k=l+1}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a)\right].
\end{align*}

By inspection, we notice that only the second summation contains the Taylor monomials we expect in the solution. Therefore, we obtain the following.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\sum\limits^{l}_{k=0}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a).
\end{align*}

Now we swap the order of the last two summations.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{N-1}_{k=0}\sum\limits^{N-1}_{l=k}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a).
\end{align*}

Next, we reverse the order of summation in the last sum by the change of variable $i=N-1-l+k$

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{N-1}_{k=0}\sum\limits^{N-1}_{i=k}\frac{(i-\nu+1-k)^{\overline{N-1-j}}}{\G(N-j)}\binom{N}{k}(-1)^{k}h_{\nu-1-i}(t,a).
\end{align*}

Now we want to swap the order of the last two sums, followed by the first two sums.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{N-1}_{i=0}\sum\limits^{i}_{k=0}\frac{(i-\nu+1-k)^{\overline{N-1-j}}}{\G(N-j)}\binom{N}{k}(-1)^{k}h_{\nu-1-i}(t,a)\\
&=\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{N-1}_{i=0}h_{\nu-1-i}(t,a)\sum\limits^{i}_{k=0}\frac{(i-\nu+1-k)^{\overline{N-1-j}}}{\G(N-j)}\binom{N}{k}(-1)^{k}\\
&=\sum\limits^{N-1}_{i=0}h_{\nu-1-i}(t,a)\sum\limits^{N-1}_{j=0}f(a+j+1)\sum\limits^{i}_{k=0}\frac{(i-\nu+1-k)^{\overline{N-1-j}}}{\G(N-j)}\binom{N}{k}(-1)^{k}.
\end{align*}

Finally, we rewrite f(a+j+1) in terms of initial conditions to obtain the desired result.

\begin{align*}
f(t)&=\sum\limits^{N-1}_{i=0}h_{\nu-1-i}(t,a)\sum\limits^{N-1}_{j=0}\sum\limits^{N-j-1}_{l=0}\binom{N-j-1}{l}(-1)^{l}A_{l}\sum\limits^{i}_{k=0}\frac{(i-\nu+1-k)^{\overline{N-1-j}}}{\G(N-j)}\binom{N}{k}(-1)^{k}.
\end{align*}

It is left to be shown that the remaining terms sum to zero.

\begin{align*}
&\sum\limits^{N-1}_{j=0}{f(a+j+1)}\left[\frac{(s+1-j)^{\overline{N-\nu-1}}}{\Gamma(N-\nu)}\sum\limits^{N-1}_{s=j}\sum\limits^{s}_{k=0}\binom{s}{k}(-1)^{k}h_{\nu-N-1-k}(t,a)\right.\\
&\left.\qquad+\sum\limits^{N-1}_{l=0}\frac{(N-\nu-l)^{\overline{N-1-j}}}{\G(N-j)}\sum\limits^{N}_{k=l+1}\binom{N}{k}(-1)^{k}h_{\nu-N+l-k}(t,a)\right]=0.
\end{align*}

\end{remark}

\end{conjecture}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%








\begin{thebibliography}{9}

\bibitem{REU2010}
J. Hein, S. McCarthy, N. Gaswick, B. McKain, and K. Spear,
\emph{Laplace Transforms for the Nabla-Difference Operator.}
PanAmerican Mathematical Journal
Volume 21(2011), Number 3, 79-96

\bibitem{HolmCompositions}
M. Holm, Sum and Difference Compositions in Discrete Fractional Calculus, \emph{CUBO Mathematical Journal},\textbf{13} (3),(2011)

\bibitem{KelleyPetersonDifferenceEquations}
W. Kelley and A. Peterson,
\emph{Difference Equations: An Introduction With Applications,} Second Edition.
Haircourt/Academic Press 2001.

\end{thebibliography}


\end{document}
