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\begin{document}

	\begin{center}

	\textbf{Nabla Fractional Calculus Formula Sheet (Revision 6)}
	
		\begin{multicols}{2}
		
		Nabla Difference
		
		For any function $f:\mathbb{N}\rightarrow\mathbb{R}$, then $\nabla f(t) = f(t)-f(t-1)$, $t\in\mathbb{N}_{a+1}$
		
		\bigskip
		
		Nabla Integration
		\begin{displaymath}   
		\int^{d}_{c}f(t)\nabla t = \left\{     
		\begin{array}{lr}       
		\sum\limits_{c+1}^{d}f(t) & c<d\\       
		0 & c=d \\
		$-$\sum\limits_{d+1}^{c}f(t) & c>d
		\end{array}   
		\right.
		\end{displaymath} 
		
		Fundamental Theorem of Nabla Calculus
		$$\int^{d}_{c}f(t)\nabla t = F(d) - F(c)$$
		
		Product Rule
		\begin{eqnarray*}
 		\nabla(f(t)g(t)) &=& f(t)\nabla g(t) + g(\rho(t))\nabla f(t) \\
   	&=& f(\rho(t))\nabla g(t) + g(t)\nabla f(t)\\
		\end{eqnarray*}
		
		Quotient Rule
		$\nabla(\frac{f(t)}{g(t)}) = \frac{g(t)\nabla f(t) + f(t)\nabla g(t)}{g(t)g(\rho(t))}$
		
		\bigskip
   	
   	Leibniz Rule
   	$$\nabla_{t}\int^{t}_{b}g(s,t)\nabla s = \int^{t}_{b}\nabla_{t}g(s,t)\nabla s + g(t,\rho(t))$$
   	$$\nabla_{t}\int^{t}_{b}g(s,t)\nabla s = \int^{t-1}_{b}\nabla_{t}g(s,t)\nabla s + g(t,t)$$
    for $t\in\mathbb{N}_{b+1}, b\in\mathbb{N}_{a}, g:\mathbb{N}_{a}\times\mathbb{N}_{a}\rightarrow\mathbb{R}$
    
    \bigskip
   	
		Integration By Parts
		$$\int^{d}_{c}u(t)\nabla v(t)\nabla t = u(t)v(t)|^{d}_{c} -\int^{d}_{c}v(\rho(t))\nabla u(t) \nabla t$$ 
  	$$\int^{d}_{c}\nabla u(t) v(\rho(t))\nabla t = u(t)v(t)|^{d}_{c} -\int^{d}_{c}u(t)\nabla v(t) \nabla t$$
	
		\bigskip
		
		Rising Function
		
		\medskip
		$k^{\overline{\nu}} = \frac{\Gamma(k+\nu)}{\Gamma(k)}$
		
		$(1+\nu)^{\overline{n}} = \frac{(-1)^{n}\Gamma(-\nu)}{\Gamma(-\nu-n)}$
		
		$\Gamma(x+1)=x\Gamma(x)$
		
		\bigskip
		
		Fractional Nabla Sum and Difference
		$$\nabla^{-\nu}_{a}f(t) = \frac{1}{\Gamma(\nu)}\int^{t}_{a}(t-\rho(s))^{\overline{\nu-1}}f(s)\nabla s$$
		$$\nabla^{\nu}_{a}f(t) = \frac{1}{\Gamma(-\nu)}\int^{t}_{a}(t-\rho(s))^{\overline{-\nu-1}}f(s)\nabla s$$
		$$\nabla^{\nu}_{a}f(t) = \nabla^{N}\nabla^{-(N-\nu)}_{a}f(t)$$ for $t\in\mathbb{N}_{a+N}$ and $N-1<\nu\leq N$
		
		$\nabla^{-\nu}_{a}f(a+1) = f(a+1)$
		
		\bigskip
		
		Power Rule
		
		$\nabla (t-a)^{\overline{\mu}} = \mu(t-a)^{\overline{\mu-1}}$
		$\nabla^{n} (t-a)^{\overline{\mu}} = (\mu-n+1)^{\overline{n}}(t-a)^{\overline{\mu-n}}$
		
		\bigskip
		
		Generalized Power Rule
		
		$\nabla^{-\nu}_{a} (t-a)^{\overline{\mu}} = (\frac{\Gamma(\mu+1)}{\Gamma(\mu+\nu+1)})(t-a)^{\overline{\mu+\nu}}$
		$\nabla^{\nu}_{a} (t-a)^{\overline{\mu}} = (\frac{\Gamma(\mu+1)}{\Gamma(\mu-\nu+1)})(t-a)^{\overline{\mu-\nu}}$
		
		\bigskip
		
		Exponential Function
		
		$e_{p}(t,a) = (1-p)^{a-t}, t\in\mathbb{N}_{a}, p\in\mathbb{C}/\{1\}$
		$e_{p}(t,a)e_{r}(t,a) = e_{p\oplus r}(t,a)$ where $p\oplus r = p+q-pq$
		$\ominus s = -\frac{s}{1-s}$
		
		\bigskip
		
		Laplace Transform
		
		\begin{eqnarray*}
 		\mathcal{L}_{a}\{f\}(s) &=& \int^{\infty}_{a}e^{\rho}_{\ominus s}(t,a)f(t)\nabla t\\
   	&=& \sum\limits^{\infty}_{k=1}(1-s)^{k-1}f(a+k)
		\end{eqnarray*}
		
		\bigskip
		
		Fractional Taylor Monomials
		
		$h_{\nu}(t,a) = \frac{(t-a)^{\overline{\nu}}}{\Gamma(\nu+1)}$
		
		$\mathcal{L}\{h_{\nu}(\cdot,a)\}(s) = \frac{1}{s^{\nu+1}}$ for $|1-s| < 1$
		
		$h_{\nu-1}(t,a+1) + h_{\nu-2}(t,a) = h_{\nu-1}(t,a)$
		
		$h_{\nu}(a+1,a) = 1$
		
		\bigskip
		
		Convolution
		$$(f*g)(t) = \int^{t}_{a}f(t-\rho(s)+a)g(s)\nabla s$$
		$\mathcal{L}_{a}\{f*g\}(s) = \mathcal{L}_{a}\{f\}(s)\mathcal{L}_{a}\{g\}(s)$
		
		$\nabla^{-\nu}_{a}f(t)=h_{\nu-1}(\cdot,a)*f(\cdot)$
		
		\end{multicols}
		
		\newpage
		
		
		
		More Laplace Transforms
		
		$\mathcal{L}_{a}\{\nabla^{-\nu}_{a}f\}(s) = \frac{1}{s^{\nu}}\mathcal{L}_{a}\{f\}$
		
		$\mathcal{L}_{a+1}\{\nabla f\}(s) = s\mathcal{L}_{a+1}\{f\} - f(a+1)$
		
		$\mathcal{L}_{a+n}\{\nabla^{n} f\}(s) = s^{n}\mathcal{L}_{a+n}\{f\} - \sum\limits^{n-1}_{j=0}s^{j}\nabla^{n-j-1}f(a+n)$
		
		\bigskip
		
		Shifting Theorem
		
		\medskip
		
		$\mathcal{L}_{a+1}\{f\}(s) = \frac{1}{1-s}\mathcal{L}_{a}\{f\}(s) - \frac{1}{1-s}f(a+1)$
		
		$\mathcal{L}_{a+n}\{f\}(s) = (\frac{1}{1-s})^{n}\mathcal{L}_{a}\{f\}(s) - \sum\limits^{n-1}_{k=0}[(\frac{1}{1-s})^{n-k}f(a+k+1)]$
		
		$\mathcal{L}_{a}\{f\}(s) = (1-s)^{n}\mathcal{L}_{a+n}\{f\}(s) + (1-s)^{n}\sum\limits^{n-1}_{k=0}[(\frac{1}{1-s})^{n-k}f(a+k+1)]$
		
		\bigskip
		
		Taylor Monomial Shifting Theorem
		
		\begin{equation*}
		h_{\nu-N}(t,a+m) = \sum\limits_{k=0}^{m} \binom{m}{k}(-1)^k h_{\nu-N-k}(t,a)
		\end{equation*}
		
		\bigskip
		
		Laplace Transform of a $\nu$-th Order Nabla Sum
		
		\begin{align*}
		\mathcal{L}_{a+N} \{ \nabla^{\nu}_{a}f\}(s) &= s^{\nu}\mathcal{L}_{a+N}\{f\}(s)+\sum\limits_{k=0}^{N-1}[s^{\nu}(\frac{1}{1-s})^{N-k}f(a+k+1)\\
		&\qquad-s^{N}(\frac{1}{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}]
		\end{align*}

		$\mathcal{L}_{a+N} \{ \nabla^{\nu}_{a}f\}(s) =\frac{s^{\nu}}{(1-s)^{N}}\mathcal{L}_{a}\{f\}(s) - \sum\limits^{N-1}_{k=0}[\frac{s^{N}}{(1-s)^{N-k}}\nabla^{-(N-\nu)}_{a}f(a+k+1)+\nabla^{N-k-1}\nabla^{-(N-\nu)}_{a}f(a+N)s^{k}]$

		\bigskip
		
		Mittag-Leffler Function
		
		\medskip
		
		Note: For $|p| < 1,$ $e_{p}(t,a) = \sum\limits^{\infty}_{k=0}p^{k}h_{k}(t,a), t\in\mathbb{N}_{a}$
		
		$$E_{p,\alpha,\beta}(t,a) = \sum\limits^{\infty}_{k=0}p^{k}h_{\alpha k + \beta}(t,a),$$ $\beta\in\mathbb{R},\alpha>0$
		
		\medskip
		
		Composition Rules
		
		\medskip
		
		$\nabla^{-\nu}_{a}\nabla^{-\mu}_{a}f(t) = \nabla^{-\nu-\mu}_{a}f(t),\qquad t\in\mathbb{N}_{a+N}$
		
		$\nabla^{\nu}_{a}\nabla^{-\mu}_{a}f(t) = \nabla^{\nu-\mu}_{a}f(t),\qquad t\in\mathbb{N}_{a}$
		
		$\nabla^{-\nu}_{a+k}\nabla^{k}f(t) = \nabla^{k-\nu}_{a+k}f(t) - \sum\limits^{k-1}_{j=0}\frac{\nabla^{j}_{a}f(a+k)}{\Gamma(\nu-k+j+1)}(t-a-k)^{\overline{\nu-k+j}},\qquad t\in\mathbb{N}_{a+k}$
		
		\bigskip
		
		A Useful Lemma
		$$\nabla^{-\nu}_{a}f(a+N) = \sum\limits^{N-1}_{k=0}\frac{\nu^{\overline{k}}}{\Gamma(k+1)}f(a+N-k), \qquad t\in\mathbb{N}_{a}.$$
		
	\end{center}

\end{document}