fractional differentiationFractionalDifferentiation2006-10-09 13:29:532008-03-30 18:02:30Definitionfractional derivativeleft-hand Grunwald-Letnikov derivativeleft hand Grundwald Letnikov derivativeright-hand Grundwald-Letnikov derivativeright hand Grundwald-Letnikov derivative% this is the default PlanetMath preamble. as your knowledge
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The idea of Grunwald-Letnikov differentiation comes from the following formulas of \PMlinkname{backward}{BackwardDifference} and forward difference \PMlinkescapetext{equations}. Within this entry, $[ \cdot ]$ will be used to denote the greatest integer function and $\Gamma$ will be used to denote the gamma function.
{\bf Backward difference}
\begin{equation}D_{-}(f)(x) = \lim_{h\to 0}\frac{f(x)-f(x-h)}{h}
\end{equation}
\begin{equation}D^n_{-}(f)(x)=\lim_{h\to 0}\frac{1}{h^n}\sum_{k=0}^n
\frac{(-1)^k n!}{k! (n-k)!}f(x-kh) \end{equation}
For derivatives of integer orders, we only requires to specifies one point $x\in {\mathbb R}$. Fractional derivatives, like fractional definite integrals, require an interval $[a,b]$ to be specified for the function $f:{\mathbb R}\to {\mathbb R}$ we are talking about.
{\bf Definition 1: Left-hand Grunwald-Letnikov derivative}
\begin{equation}D^p_{-}(f)(x)=
\lim_{h\to 0}\frac{1}{h^p}\sum_{k=0}^{\left[\frac{b-a}{h}\right]}
\frac{(-1)^k\Gamma (p+1)}{k! \Gamma (p-k+1)} f(x-kh) \end{equation}
{\bf Forward difference}
\begin{equation}D_{+}(f)(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
\end{equation}
\begin{equation}D^n_{+}(f)(x) = \lim_{h\to 0}\frac{1}{h^n}\sum^n_{k=0}
\frac{(-1)^k n!}{k! (n-k)!} f(x+(n-k-1)h) \end{equation}
{\bf Definition 2: Right-hand Grunwald-Letnikov derivative}
\begin{equation}D^p_{+}(f)(x)=
\lim_{h\to 0}\frac{1}{h^p}\sum_{k=0}^{\left[\frac{b-a}{h}\right]}
\frac{(-1)^k\Gamma (p+1)}{k! \Gamma (p-k+1)} f(x+(m-k-1)h) \end{equation}
{\bf Theorem 1: Properties of fractional derivatives}
\begin{itemize}
\item {\rm Linearity}: $D^p_{\pm}(a f+ b g)(x) = a D^p_{\pm}(f)(x) + b D^p_{\pm}(g)(x)$ where $a,b\in {\mathbb R}$ are any real constants
\item {\rm Iteration}: $D^p_{\pm}D^q_{\pm}(f)(x) = D^{p+q}_{\pm}(f)(x)$
\item {\rm Chain rule}: $\displaystyle{\frac{d^\beta f(g(x))}{dx^{\beta}}
=\sum_{k=0}^{\infty}\frac{\Gamma(1+\beta)}{\Gamma(1+k)\Gamma(1-k+\beta)}
\frac{d^{\beta-k}1}{dx^{\beta-k}}
\frac{d^k f(g(x))}{dx^k} }$
\item {\rm Leibniz Rule}: $\displaystyle{\frac{d^\beta (f(x)g(x))}{dx^\beta}
=\sum_{k=0}^{\infty}\frac{\Gamma(1+\beta)}{\Gamma(1+k)\Gamma(1-k+\beta)}
\frac{d^k f(x)}{dx^k} \frac{d^{\beta-k}g(x)}{dx^{\beta-k}} }$
\end{itemize}
{\bf Theorem 2: Table of fractional derivatives}
\begin{itemize}
\item $\displaystyle{ D^{\alpha}_{\pm}(x^p)
=\frac{\Gamma (p+1)x^{p-\alpha}}{\Gamma (p-\alpha+1)} }$ where $\alpha,p\in {\mathbb R}$ and $\Gamma(x)$
\item $\displaystyle{ D^{\alpha}_{\pm}( e^{\lambda x} )
=\lambda^{\alpha} e^{\lambda x} }$ for all $\lambda\in {\mathbb R}$
\item $\displaystyle{ D^{\alpha}_{\pm} (\sin x) = \sin \left(x+\frac{\alpha \pi}{2}\right)}$
\item $\displaystyle{ D^{\alpha}_{\pm} (\cos x) = \cos \left(x+\frac{\alpha \pi}{2}\right)}$
\item $\displaystyle{ D^{\alpha}_{\pm} (e^{i x}) =\cos \left(x+\frac{\pi\alpha}{2}\right)+i\sin \left(x+\frac{\pi\alpha}{2}\right) }$
\end{itemize}