The Relation Between Nabla Fractional Differences and Nabla Integer Differences

Jia Baoguo, Lynn Erbe, Christopher Goodrich, Allan Peterson

Abstract


In this paper we obtain two interrelated results.  The first result is the following inequality:\\

{\bf{Theorem.}} Assume that $f:\N_a\rightarrow\R$ satisfies $\nabla_a^\nu
f(t)\geq 0$, for each $t\in\N_{a+1}$, $\nu>0,
 \nu\notin\N_1$, and choose $N\in \N_1$, such that $N-1<\nu<N$. Then for each $k\in\N_{a+N}$, we have
\begin{equation}
\begin{split}
\nabla^{N-1}f(a+k)&\geq -\sum^{N-2}_{i=0}H_{-\nu+i}(a+k,a+i)\nabla^{i}f(a+i+1)\\
&-\sum^{k-1}_{i=N}H_{-\nu+N-2}(a+k,a+i-1)\nabla^{N-1}f(a+i),\notag
\end{split}
\end{equation}
where
$$H_{-\nu+N-2}(a+k,a+i-1)=\frac{(k-i+1)^{\overline{-\nu+N-2}}}{\Gamma(-\nu+N-1)}<0.$$
As an application of the above inequality we prove the following result:\\

{\bf{Theorem.}} Assume that $f:\N_a\rightarrow\R$ satisfies $\nabla^\nu_af(t)\geq 0$, for each $t\in\N_{a+1}$, where $5<\nu<6$.
Then $\nabla^5f(t)\geq 0$, for $t\in \N_{a+6}$.\\

\noindent This demonstrates that, in some sense, the positivity of the $\nu$-th
order fractional difference has a strong connection to the positivity of
an integer-order difference of the function $f$.


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