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\title[]{Discrete Sine Integral} % The short title appears at the bottom of every slide, the full title is only on the title page

\author[Treston Brown]{Treston Brown \\% Your name
Fairmont State University}
\date{\today} % Date, can be changed to a custom date

\renewcommand{\qedsymbol}{}
\newcommand{\si}{\mathrm{Si}}
\begin{document}

\begin{frame}
\titlepage % Print the title page as the first slide
\end{frame}

\begin{frame}{Classic Sine Integral}

\begin{equation}
\mathcal{SI}(x) = \int_0^x\frac{\sin t}{t}\,dt = \begin{array}{ll}
\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}{x^{2k-1}}}{(2k-1)(2k-1)!}
\end{array}
\end{equation}
\begin{center}

\includegraphics[scale=0.4]{classicsi.png}
\end{center}
\end{frame}

\begin{frame}{Differential Equation of Classic Sine Integral}
$\mathcal{SI}$ Satisfies the following differential equation: 
\begin{equation}ty'''(t) + 2y''(t) + ty'(t) = 0
\end{equation}
Note: this function has applications within signal processing  (e.g. step response of the sinc filter) and electronics.\\

\end{frame}


\begin{frame}{Discrete Calculus -- Discrete Derivative}
The forward difference operator $\Delta$ is defined by:
\begin{equation}\Delta f(x) = f(x+1) - f(x)
\end{equation}
The rising factorial $(a)_k$ is defined by: \begin{equation}(a)_k = a(a+1)(a+2)\ldots (a+k-1)
\end{equation} 
The discrete monomials are defined by: \begin{equation}(-1)^k(-t)_k
\end{equation}

\end{frame}

\begin{frame}{Discrete Calculus -- Discrete Integral}
We define the discrete integral ``from $a$ to $b$" of a function $f$ by
\begin{equation} \displaystyle\sum_{k=a}^{b-1} f(k)
\end{equation}
because a fundamental theorem of discrete calculus holds:
\begin{equation}\Delta\Bigg(\displaystyle\sum_{k=a}^{x-1} f(k)\Bigg) = f(x)
\end{equation}\\
\begin{equation}\displaystyle\sum_{k=a}^{b-1} \Delta f(k) = f(b)-f(a)
\end{equation}
\end{frame}

\begin{frame}{Discrete Calculus -- Discrete Power Rule}
\textbf{Theorem}: (Discrete Power Rule -- Discrete Difference)
\begin{equation}\Delta[{{(-1)^k}{(-t)_k}}] = k[{(-1)^{k-1}}{(-t)_{k-1}}]
\end{equation}
\textit{Proof}:
$$\begin{array}{ll}
\Delta[{{(-1)^k}{(-t)_k}}]= {{(-1)^k}{(-(t+1))_k}-{(-1)^k}{(-t)_k}} 
\\ 
= [{{(-1)^{k-1}}{(-t)_{k-1}}}][{(-1)(-t-1)-(-1)(-t+k-1)}]
\\
= [{(-1)^{k-1}}{(-t)_{k-1}}][{t+1-t-1+k}] 
\\
= k[{(-1)^{k-1}}{(-t)_{k-1}}]
\end{array}$$
\hrule 
\textbf{Theorem}: (Discrete Power Rule -- Discrete Integral)
\begin{equation}\displaystyle\sum_{k=0}^{t-1} (-1)^n(-k)_n =\dfrac{(-1)^{n+1}{(-t)_{n+1}}}{n+1}
\end{equation}
\end{frame}


\begin{frame}{Discrete Calculus -- Discrete Sine}
The discrete exponential $e_p(t)$ is defined by difference equation 
\begin{equation}\Delta y(t) = p(t)y(t), \quad   y(0)=1 
\end{equation}
The discrete sine is defined by
\begin{equation}\sin_1(t) = \frac{e_i(t)-e_{-i}(t)}{2i}
\end{equation}\\
and obeys the difference equation
\begin{equation}\Delta^2 y(t) = -y(t),\quad y(0)=0
\end{equation}

\end{frame}

\begin{frame}{Discrete Calculus -- Discrete Sine}

\begin{center}
\includegraphics[scale=.35]{integersine.png}
\end{center}
\end{frame}

\begin{frame}{Discrete Calculus -- Shifting}
It can be shown that if \begin{equation}f(t) = (-1)^n(-t)_n
\end{equation} then
\begin{equation} tf(t-1)=(-1)^{n+1}(-t)_{n+1}
\end{equation}
and
\begin{equation}\frac{f(t+1)}{t+1}=(-1)^{n-1}(-t)_{n-1}
\end{equation}


\end{frame}

\begin{frame}{Discrete Sine Integral Difference Equation}
The classical sine integral obeys the following differential equation:

$$ty'''(t) + 2y''(t) + ty'(t) = 0.$$

We propose the following analagous difference equation to define a discrete sine integral function:

\begin{equation}\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0.
\end{equation}


\end{frame}

\begin{frame}{Discrete Sine Integral}
\begin{equation}
\si(t) = \displaystyle\sum_{k=0}^{x-1} \dfrac{\sin_1(k+1)}{k+1} =
\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}[(-1)^{2k-1}(-t)_{2k-1}]}{(2k-1)(2k-1)!}
\end{equation}

\begin{center}
\includegraphics[scale=0.4]{discretesi-5to25.png}
\end{center}
\end{frame}

\begin{frame}{Discrete Sine Integral}

$$\begin{array}{ll}
\si(t) = \displaystyle\sum_{k=0}^{x-1} \dfrac{\sin_1(k+1)}{k+1} =
\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}[(-1)^{2k-1}(-t)_{2k-1}]}{(2k-1)(2k-1)!}
\end{array}$$

\begin{center}
\includegraphics[scale=0.4]{discretesi-5to50.png}
\end{center}

\end{frame}

\begin{frame}{Solution of Difference Equation}

$$\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0$$

$$\begin{array}{ll}
\Delta\si(t) &= \Delta\Bigg[\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}[(-1)^{2k-1}(-t)_{2k-1}]}{(2k-1)(2k-1)!}\Bigg] \\
&= \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-\-1}}{(2k-1)(2k-1)!} \Delta\Bigg[(-1)^{2k-1}(-t)_{2k-1}\Bigg] \\
&= \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{(2k-1)(2k-1)!} (2k-1)\Bigg[(-1)^{2k-2}(-t)_{2k-2}\Bigg] \\
&= \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-\-1}(-1)^{2k-\-2}(-t)_{2k-2}}{(2k-1)!}
\end{array}$$

\end{frame}

\begin{frame}{Solution of Difference Equation}

$$\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0$$

$$\begin{array}{ll}
t\Delta\si(t-1) &= t\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{(2k-1)!} \Bigg[(-1)^{2k-2}(-t+1)_{2k-2}\Bigg]
\\
&= \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{(2k-1)!} \Bigg[(-1)^{2k-1}(-t)_{2k-1}\Bigg]
\\
&= \displaystyle\sum_{k=2}^{\infty} \dfrac{(-1)^{k-2}}{(2k-3)!} \Bigg[(-1)^{2k-3}(-t)_{2k-3}\Bigg]
\end{array}$$
\end{frame}

\begin{frame}{Solution of Difference Equation}

$$\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0.$$

$$\begin{array}{ll}
\Delta^2\si(t) = \displaystyle\sum_{k=2}^{\infty} \dfrac{(-1)^{k-1}(2k-2)[(-1)^{2k-3}(-t)_{2k-3}]}{(2k-1)!}
\end{array}$$
\end{frame}

\begin{frame}{Solution of Difference Equation}

$$\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0.$$

$$\begin{array}{ll}
\Delta^3\si(t)&= \displaystyle\sum_{k=2}^{\infty} \dfrac{(-1)^{k-1}(2k-2)(2k-3)[(-1)^{2k-4}(-t)_{2k-4}]}{(2k-1)!}
\end{array}$$

\begin{eqnarray*}
\lefteqn{t\Delta^3\si(t-1)} \\
&& = t\displaystyle\sum_{k=2}^{\infty} \dfrac{(-1)^{k-1}(2k-2)(2k-3)}{(2k-1)!} \Bigg[(-1)^{2k-4}(-t+1)_{2k-4}\Bigg] \\
& &= \displaystyle\sum_{k=2}^{\infty} \dfrac{(-1)^{k-1}(2k-2)(2k-3)}{(2k-1)!} \Bigg[(-1)^{2k-3}(-t)_{2k-3}\Bigg]
\end{eqnarray*}
\end{frame}

\begin{frame}{Solution of Difference Equation}

$$\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0.$$

Combining these results:
\begin{eqnarray*}
\lefteqn{\displaystyle\sum_{k=2}^{\infty} \Bigg[\dfrac{(-1)^{k-\-1}(2k-2)(2k-3)}{(2k-1)!}} \\
& & + \dfrac{2(-1)^{k-\-1}(2k-2)}{(2k-1)!} + \dfrac{(-1)^{k-\-2}}{(2k-3)!}\Bigg]\Bigg[(-1)^{2k-3}(-t)_{2k-3} \Bigg]
\end{eqnarray*}
\end{frame}

\begin{frame}{Solution of Difference Equation}
$$\begin{array}{l}
\Bigg[\dfrac{(-1)^{k-\-1}(2k-2)(2k-3)}{(2k-1)!}
 + \dfrac{2(-1)^{k-\-1}(2k-2)}{(2k-1)!} + \dfrac{(-1)^{k-2}}{(2k-3)!}\Bigg] \\
=  \dfrac{(-1)^{k-\-1}}{(2k-1)!}\Bigg[(2k-2)(2k-3)
 + {2(2k-2)} - {(2k-2)(2k-1)}\Bigg] \\
 =   \dfrac{(-1)^{k-\-1}}{(2k-1)!}\Bigg[({4k^2}-6k-4k+6)
 + {(4k-4)} - {({4k^2}-2k-4k+2)}\Bigg] \\
 =0,
 \end{array}$$ 
 proving that $\si$ satisfies the difference equation 
 $$\mathrm{t}\Delta^3\si(\mathrm{t}-1) +  \mathrm{2}\Delta^2\si(\mathrm{t})+\mathrm{t}\Delta\si(\mathrm{t}-1)=0.$$
\end{frame}

\begin{frame}{Future Work}
More properties of discrete sine integral \\
Extend theory to discrete exponential integrals \\
Extend theory to discrete cosine integrals \\
\end{frame}
%------------------------------------------------

\begin{frame}{Sayonara}

Thank you for coming

\end{frame}
\end{document} \grid
