The main aim of this paper to establish the relations between forward, backward and central finite (divided) differences (that is discrete analog of the derivative) and partial & ordinary high-order derivatives of the polynomials.
\begin
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pdftitle={On the link between finite difference and derivative of polynomials},
pdfsubject={Mathematics, Differential Calculus, Derivatives, Differential equations},
pdfauthor={Kolosov Petro},
pdfkeywords={Finite difference, Derivative, Divided difference, Ordinary differential equation, Partial differential equation, Partial derivative, Differential calculus, Difference Equations, Numerical Differentiation, Finite difference coefficient, Polynomial, Power function, Monomial, Exponential function, Exponentiation, arXiv, Preprint, Calculus, Mathematics, Mathematical analysis, Numerical methods, Applied Mathematics}
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\title{On the link between finite difference and derivative of polynomials}
\author{Kolosov Petro}
\begin{document}
\begin{abstract}
The main aim of this paper to establish the relations between forward, backward and central finite (divided) differences (that is discrete analog of the derivative) and partial \& ordinary high-order derivatives of the polynomials.
\\\\
\noindent \textbf{Keywords.} finite difference, divided difference, high order finite difference, derivative, ode, pde, partial derivative, partial difference, power, power function, polynomial, monomial, power series, high order derivative, mathematics, differential calculus, differential equations\\\\
\noindent \textbf{2010 Math. Subject Class.} 46G05, 30G25, 39-XX\\
\noindent \textbf{Personal website:} \ \href{https://goo.gl/P9Ttmu}{kolosovpetro.github.io}\\
\noindent \textbf{ORCID:} \ \href{https://goo.gl/qqxSYM}{0000-0002-6544-8880}\\
\noindent \textbf{e-mail:} \ \ \ \href{mailto:kolosovp94@gmail.com}{kolosovp94@gmail.com}\\
\end{abstract}
\maketitle
\tableofcontents
\section{Introduction}\label{d1}
Let introduce the basic definition of finite difference. Finite difference is difference between function values with constant increment. There are three types of finite differences: forward, backward and central. Generally, the first order forward difference could be noted as: $\Delta_hf(x)=f(x+h)-f(x)$ backward, respectively, is $\nabla_hf(x)=f(x)-f(x-h)$ and central $\delta_hf(x) = f(x+\tfrac12h)-f(x-\tfrac12h)$, where $h=const$, (see \cite{1}, \cite{2}, \cite{3}). When the increment is enough small, but constant, we can say that finite difference divided by increment tends to derivative, but not equals. The error of this approximation could be counted next: $\frac{\Delta_hf(x)}{h}-f^{'}(x)=\mathcal{O}(h)\rightarrow0$, where $h$ - increment, such that, $h\rightarrow0$. By means of induction as well right for backward difference. More exact approximation we have using central difference, that is: $\frac{\delta_hf(x)}{h}-f^{'}(x)=\mathcal{O}(h^2)$, note that function should be twice differentiable.
The finite difference is the discrete analog of the derivative (see \cite{4}), the main distinction is constant increment of the function's argument, while difference to be taken. Backward and forward differences are opposite each other. More generally, high order finite differences (forward, backward and central, respectively) could be denoted as (see \cite{7}):
\begin{equation}\label{f1}
\Delta_h^k f(x)=\Delta^{k-1} f(x+h)-\Delta^{k-1} f(x)=\sum_{k=0}^{n}{n \choose k}(-1)^{k}\cdot f(x+(n-k)h)
\end{equation}
\begin{equation}\label{f2}
\delta^n_h f(x)=\sum_{k = 0}^{n} (-1)^k \binom{n}{k}\cdot f\left(x + \left(\frac{n}{2} - k\right) h\right)
\end{equation}
\begin{equation}\label{f3}
\nabla_h^k f(x)=\nabla^{k-1} f(x)-\nabla^{k-1} f(x-h)=\sum_{k=0}^{n}{n \choose k}(-1)^{k}\cdot f(x-kh)
\end{equation}
Let describe the main properties of finite difference operator, they are next (see \cite{5})
\begin{enumerate}\label{f4}
\item Linearity rules $\Delta(f(x)+g(x))=\Delta f(x)+\Delta g(x)$
\newline $\delta(f(x)+g(x))=\delta f(x)+\delta g(x)$
\newline $\nabla(f(x)+g(x))=\nabla f(x)+\nabla g(x)$
\item $\Delta(C\cdot f(x))=C\cdot\Delta f(x), \ \nabla(C\cdot f(x))=C\cdot\nabla f(x), \ \newline \delta(C\cdot f(x))=C\cdot\delta f(x)$
\item Constant rule $\Delta C=\nabla C=\delta C=0$
\end{enumerate}
Strictly speaking, divided difference (see \cite{6}) with constant increment is discrete analog of derivative, when finite difference is discrete analog of function's differential. They are close related to each other. To show this, let define the divided difference.
\begin{defn}
Divided difference of fixed increment definition (forward, centaral, backward respectively)
$$f^+[x_i, \ x_j]:=\frac{f(x_j)-f(x_i)}{x_j-x_i}, \ j>i, \ \Delta x\geq1$$
$$f^-[x_i, \ x_j]:=\frac{f(x_i)-f(x_j)}{x_i-x_j}, \ j<i, \ \nabla x\geq1$$
$$f^c[x_i]:=\frac{f(x_{i+m})-f(x_{i-m})}{2m}$$
\end{defn}
Hereby, divided diffrence could be represented from the finite difference, let be $j=i\pm \textrm{const}$, backward as $-$, respectively $+$ as forward and $c$ as centered
$$f^\pm[x_i, \ x_j]\equiv\frac{\Delta f(x_j)}{\Delta x}\equiv\frac{\nabla f(x_i)}{\nabla x}$$
$$f_c[x_i]\equiv\frac{\delta f(x_{i\pm m})}{2m}\equiv\frac{\delta f(x_{i\pm m})}{\delta x}$$
The $n$-order
$$f^\pm[x_i, \ x_j]^n\equiv\frac{\Delta^n f(x_j)}{\Delta x^n}\equiv\frac{\nabla^n f(x_i)}{\nabla x^n}$$
$$f_c[x_i]^n\equiv\frac{\delta^n f(x_{i\pm m})}{(2m)^n}\equiv\frac{\delta^n f(x_{i\pm m})}{\delta x^n}$$
Each properties, which holds for finite differences holds for divided differences as well.
\section{Definitions for discrete distribution}
Let be variable $x_g:x_g=g\cdot C, \ C=x_{g+1}-x_g=\textrm{const}, \ C\in \mathbb{R}_{>0}\rightarrow x_g\in\mathbb{R}_{>0}, \ g \in \mathbb{Z}$. To define the finite difference of function of such argument, we take $C=h$ and rewrite forward, backward and central differences of some analytically defined function $f(x_i)$ next way: $\Delta f(x_{i+1})=f(x_{i+1})-f(x_i), \ \nabla f(x_{i-1})=f(x_{i})-f(x_{i-1}), \ \delta f(x_i) = f\left(x_{i+\tfrac12}\right)-f\left(x_{i-\tfrac12}\right)$. The $n$-th differences of such a function could be written as
\begin{equation}\label{f5}
\Delta^nf(x_{i+1})=\Delta^{n-1}f(x_{i+1})-\Delta^{n-1}f(x_{i})=\sum_{k=0}^{n}{n \choose k}(-1)^{k}\cdot f(x_{i+n-k})
\end{equation}
\begin{equation}\label{f6}
\delta^nf(x_i)=\sum_{k = 0}^{n}\binom{n}{k}(-1)^k\cdot f\left(x_{i+\frac{n}{2}-k}\right)
\end{equation}
\begin{equation}\label{f7}
\nabla^nf(x_{i-1})=\nabla^{n-1}f(x_{i})-\nabla^{n-1}f(x_{i-1})=\sum_{k=0}^{n}{n \choose k}(-1)^{k}\cdot f(x_{i-n+k})
\end{equation}
Let be differences $\Delta f(x_{i+1}), \ \delta f(x_i), \ \nabla f(x_{i-1})$, such that $i\in\mathbb{Z}$ and differences is taken starting from point $i$, which divides the space $\mathbb{Z}$ into $\mathbb{Z}=\mathbb{Z}^-\cup\mathbb{Z}^+$ symmetrically (note that $+/-$ symbols mean the left and right sides of start point $i=0$, i.e backward and forward direction), this way we have $(i+1)\in\mathbb{Z}^+, \ (i-1)\in\mathbb{Z}^-, \ i\neq0\in(\mathbb{Z}^+,\mathbb{Z}^-)$. Let derive some properties of that distribution:
\begin{enumerate}\label{f8}
\item $\max(\mathbb{Z}^-)=\min(\mathbb{Z}^+)=i$
\item Forward difference is taken starting from $\min(\mathbb{Z}^+)$, while backward from $\max(\mathbb{Z}^-)$
\item $\mathrm{card}(\mathbb{Z}^+)=\mathrm{card}(\mathbb{Z}^-)$, i.e $\sum_{k\in\mathbb{Z}^+}1=\sum_{k\in\mathbb{Z}^-}1$
\item Maximal order of forward difference in which it is not equal to zero is $\max(\mathbb{Z}^+)$
\item Maximal order of backward difference in which it is not equal to zero is $\min(\mathbb{Z}^-)$
\item Maximal order of central difference in which it is not equal to zero is $\max(\mathbb{Z}^+)$
\item Forward and backward difference equal each other by absolute value, while to be taken from $i=0$
\end{enumerate}
\begin{limit}\label{lim1}
Note that most expression generated as case of $i=0$, so the initial start point of each difference and inducted expressions are $0$.
\end{limit}
\begin{defns} Generalized definitions complete this section
\begin{enumerate}
\item $\mathbb{Z}^+:=\mathbb{N}_1$ - positive integers
\item $\mathbb{Z}^-:=\{-1, \ -2, \ \ldots, \ \min(\mathbb{Z}^-)\}$ - negative integers
\item $\{f, \ f(x), \ f(x_i)\}:=x^n$ - power function, value of power function in point $i$ of difference table
\item $i=0$ - initial point of every differentiating process, $\delta f(x_0)$ exist only for operator of centered difference (as per limitation \hyperref[lim1]{2.4})
\item $x_i:=i\cdot\Delta x\equiv\nabla x\equiv(\delta x)/2=\sum\Delta x$ - value of function's argument in point $i$ of difference table
\item $\Delta x\equiv\nabla x\equiv(\delta x)/2$ - function's argument differentials, constant values $\in\mathbb{R}_{>0}$
\item $\Delta f(x_{i+1}), \ \nabla f(x_{i-1})$ - forward and backward finite differences in points $i+1$ and $i-1$ of difference table
\item $\delta f (x_i)$ - centered finite difference in point $i$ of difference table
\item $\Delta^0f\equiv\delta^0f\equiv\nabla^0f\equiv f$
\end{enumerate}
\end{defns}
\section{Difference and derivative of power function}
Since the $n$-order polynomial defined as summation of argument to power multiplied by coefficient, with higher power $n$, let describe a few properties of finite (divided) difference of power function.
\begin{lem}\label{lem1}
For each power function with natural number as exponent holds the equality between forward, backward and central divided differences, and derivative with order respectively to exponent and equals to exponent under factorial sign multiplied by argument differential to power.
\end{lem}
\begin{proof}
Let be function $f(x)=x^n, \ n\in \mathbb{N}$. The derivative of power function, $f^{'}(x)=nx^{n-1}$, so $k$-th derivative $f^{(k)}(x)=n\cdot(n-1)\cdots(n-k+1)\cdot x^{n-k}$, $n>k$. Using limit notation, we have:
$
\lim_{m \to n^{-}}f^{(m)}(x)=f^{(n)}(x)=n!.
$
Let rewrite expressions (\hyperref[f5]{2.1}, \hyperref[f6]{2.2}, \hyperref[f7]{2.3}) according to definition $x_i=i\cdot\Delta x$, note that $\Delta x\equiv\nabla x\equiv\delta x/2$. By means of power function multiplication property $(i\cdot \Delta x)^n=i^n\cdot \Delta x^n$, we can rewrite the $n$-th finite difference equations (\hyperref[f5]{2.1}, \hyperref[f6]{2.2}, \hyperref[f7]{2.3}) as follows
\begin{equation}\label{f-}
\Delta^m(x_{i+1}^n)=\sum_{k=0}^{m}{m \choose k}(-1)^k\cdot\big(i+m-k\big)^m\cdot\Delta x^m, \ m<n\in \mathbb{N}
\end{equation}
Using limit notation on divided by $\Delta x$ to power (\hyperref[f-]{3.2}), we obtain
\begin{equation}\label{s1}
\lim \limits_{m \to n^{-}}\frac{\Delta^m(x_{i+1}^n)}{\Delta x^m}=\lim \limits_{m \to n^{-}}\sum_{k=0}^{m}{m \choose k}(-1)^k\cdot\big(i+m-k\big)^m
\end{equation}
$$=\sum_{k=0}^{n}{n\choose k}(-1)^k\cdot\big(i+n-k\big)^{n-0}=n!$$\\
Similarly, going from (\hyperref[f7]{2.3}), backward $n$-th difference equals:
\begin{equation}\label{s2}
\lim \limits_{m \to n^{-}}\frac{\nabla^m(x_{i-1}^n)}{\nabla x^m}=\lim \limits_{m \to n^{-}}\sum_{k=0}^{m}{m \choose k}(-1)^k\cdot\big(i-m+k\big)^m
\end{equation}
$$=\sum_{k=0}^{n}{n\choose k}(-1)^k\cdot\big(i-n+k\big)^{n-0}=n!$$
And $n$-th central (\hyperref[f6]{2.2}), respectively
\begin{equation}\label{s3}
\lim \limits_{m \to n^{-}} \frac{\delta^mf(x_i)}{\delta x^m}=\lim \limits_{m \to n^{-}} \sum_{k = 0}^{m} \binom{m}{k}(-1)^k\cdot \left(i+\frac{m}{2}-k\right)^m
\end{equation}
$$
=\sum_{k = 0}^{n} \binom{n}{k}(-1)^k\cdot \left(i+\frac{n}{2}-k\right)^{n-0}=n!
$$
\\As we can see the next conformities hold
\begin{equation}\label{g1}
\lim_{\Delta x \to 0}\frac{\Delta^n f}{\Delta x^n}\equiv\lim_{\Delta x \to C}\frac{\Delta^n f}{\Delta x^n}\equiv n!
\end{equation}
\begin{equation}\label{g2}
\lim_{\delta x \to 0}\frac{\delta^n f}{\delta x^n}\equiv\lim_{\delta x \to C}\frac{\delta^n f}{\delta x^n}\equiv n!
\end{equation}
\begin{equation}\label{g3}
\lim_{\nabla x \to 0}\frac{\nabla^n f}{\nabla x^n}\equiv\lim_{\nabla x \to C}\frac{\nabla^n f}{\nabla x^n}\equiv n!
\end{equation}
\begin{equation}\label{g4}
\lim_{\Delta x \to C}\frac{\Delta^n f}{\Delta x^n}\equiv\lim_{\delta x \to C}\frac{\delta^n f}{\delta x^n}\equiv\lim_{\nabla x \to C}\frac{\nabla^n f}{\nabla x^n} \ \ \ \ \ \ \ \ \ \forall(C\in\mathbb{R}^+)
\end{equation}\\
In partial case when $C=0$
\begin{equation}\label{g5}\\
\frac{d^n f}{dx^n}\equiv\lim_{\delta x \to 0}\frac{\delta^n f}{\delta x^n}\equiv\lim_{\nabla x \to 0}\frac{\nabla^n f}{\nabla x^n}
\end{equation}
As well holds
\begin{equation}\label{g6}
\frac{df}{dx}(x_0)=\left|\lim_{\nabla x \to 0}\frac{\nabla f}{\nabla x}(x_0)\right|
\end{equation}
\begin{equation}\label{g7}
\frac{d^n f}{dx^n}\equiv\lim_{\Delta x \to C}\frac{\Delta^n f}{\Delta x^n}\equiv\lim_{\delta x \to C}\frac{\delta^n f}{\delta x^n}\equiv\lim_{\nabla x \to C}\frac{\nabla^n f}{\nabla x^n}, \ \ \forall(C \in \mathbb{R}^+)
\end{equation}
\\where $f=x^n$. And there is exist the continuous derivative and difference of order $k\leq n$ since $f\in C^n$ class of smoothness.
Thus, from (\hyperref[g1]{3.6}, \hyperref[g2]{3.7}, \hyperref[g3]{3.8}), we can conclude
\begin{equation}\label{f9}
\frac{d^nx^n}{dx^n}=\frac{\Delta^n(x_{i+1}^n)}{\Delta x^n}=\frac{\delta^n(x_i^n)}{\delta x^n}=\frac{\nabla^n(x_{i-1}^n)}{\nabla x^n}=n!, \ (\Delta x, \ \delta x, \ \nabla x)\not\rightarrow dx
\end{equation}
This completes the proof.
\end{proof}
\begin{defn}\label{def2}
We introduce the difference equality operator $E(f)$, such that
\begin{equation}\label{f10}
E(f)\stackrel{\rm def}{=}\left(\frac{\Delta^nf}{\Delta x^n}=\frac{\delta^nf}{\delta x^n}=\frac{\nabla^nf}{\nabla x^n}\right)
\end{equation}
\end{defn}
\begin{ppty}
Let be central difference written as $\delta_m f(x_i)=f(x_{i+m} )-f(x_{i-m})$ the $n$-th central difference of $n$-th power is $\delta^n_{ \ m} (x_i^n)=n!\cdot2m\cdot\delta x^n$, where $\delta x= x_{i+1}-x_i=\mathrm{const}$.
\end{ppty}
Going from lemma (\hyperref[lem1]{3.1}), we have next properties
\begin{enumerate}\label{f11}
\item $\Delta^k(x_{i+1}^k)=\mathrm{const}, \ (i+1)\in \mathbb{Z+}:\max(\mathbb{Z+})>k\longrightarrow\Delta^k(x_{i+1}^k)\equiv\Delta^k(x_{i}^k)$
\item $\nabla^k(x_{i-1}^k)=\mathrm{const}, \ (i-1)\in \mathbb{Z-}:-\min(\mathbb{Z-})\ll k\longrightarrow\nabla^k(x_{i-1}^k)\equiv\nabla^k(x_{i}^k)$
\item $\delta^k(x_{i}^k)=\mathrm{const}, \ i\in \mathbb{Z+}:\max(\mathbb{Z+})>k\longrightarrow\delta^k(x_{i}^k)\equiv\delta^k(x_{i+j}^k)$
\item $\forall ([i+1]\in\mathbb{Z+}, \ [i-1]\in\mathbb{Z-}):\Delta^{k+j}(x_{i+1}^k)=\nabla^{k+j}(x_{i-1}^k)=0, \newline \ j>1, \ \textrm{since} \ \Delta C\equiv\delta C\equiv\nabla C\equiv0$
\item $\forall(f=x^n, \ n\in\mathbb{N}, \ k\leq n):\Delta ^k f=(-1)^{n-1}\cdot \nabla^k f$.
\item $\Delta f(x_{i+1})=|\nabla f(x_{i-1})|$
\item $\delta^2 f(x_0)=2\cdot(\delta x)^n, \ \forall(f(x_j)=x_j^n, \ n \ \textrm{mod2}=0)$
\item $\forall n \ \mathrm{mod}2=0:\delta^{2j+1}f(x_0)=0, \ j\in\mathbb{N}_0$ (see \hyperref[a1]{Appendix 1} for reference)
\item $\forall n \ \mathrm{mod}2=1:\delta^{2j}f(x_0)=0, \ j\in\mathbb{N}_1$
\end{enumerate}
Hereby, according to above properties, we can write the lemma (\hyperref[lem1]{3.1}) for enough large sets $\mathbb{Z}^+, \ \mathbb{Z}^-$ as
\begin{equation}\label{f12}
\frac{d^nx^n}{dx^n}=\frac{\Delta^n(x_i^n)}{\Delta x^n}=\frac{\delta^n(x_i^n)}{\delta x^n}=\frac{\nabla^n(x_{i}^n)}{\nabla x^n}=n!
\end{equation}
Or
\begin{equation}\label{f13}
\left(\frac{d}{dx}\right)^n x^n=E(x^n)=n!
\end{equation}
\section{Difference of polynomials}
Let be polynomial $P_n(x_g)$ defined as
\begin{equation}\label{14}
P_n(x_g)=\sum_{i=0}^{n}a_ix^{i}_g
\end{equation}
Finite differences of such kind polynomial, are $\Delta P_n(x_i)=P_n(x_{i+1})-P_n(x_i)$, $\nabla P_n(x_{i-1})=P_n(x_{i})-P_n(x_{i-1})$, $\delta P_n(x_i)=P_n\left(x_{i+\tfrac12}\right)-P_n\left(x_{i-\tfrac12}\right)$. Such way, according to the properties (\hyperref[f4]{1}, \hyperref[f4]{2}, \hyperref[f4]{3}) from section \hyperref[d1]{1}, high order finite differences of polynomials could be written as:
$$\Delta^kP_n(x_{i+1})=\Delta^k(a_0\cdot x_{i+1}^0+\cdots+a_n\cdot x_{i+1}^n)=\Delta^k(a_0\cdot x_{i+1}^0)+\cdots+\Delta^k(a_n\cdot x_{i+1}^n)$$
\begin{equation}\label{r1}
=a_0\cdot\Delta^k(x_{i+1}^0)+\cdots+a_n\cdot\Delta^k(x_{i+1}^n)
\end{equation}
Backward difference, respectively, is
$$\nabla^kP_n(x_{i-1})=\nabla^k(a_0\cdot x_{i-1}^0+\cdots+a_n\cdot x_{i-1}^n)=\nabla^k(a_0\cdot x_{i-1}^0)+\cdots+\nabla^k(a_n\cdot x_{i-1}^n)$$
\begin{equation}\label{r2}
=a_0\cdot\nabla^k(x_{i-1}^0)+\cdots+a_n\cdot\nabla^k(x_{i-1}^n)
\end{equation}
And central
$$\delta^kP_n(x_i)=\delta^k(a_0\cdot x_{i}^0+\cdots+a_n\cdot x_{i}^n)=\delta^k(a_0\cdot x_{i}^0)+\cdots+\delta^k(a_n\cdot x_{i}^n)$$
\begin{equation}\label{r3}
=a_0\cdot\delta^k(x_{i}^0)+\cdots+a_n\cdot\delta^k(x_{i}^n)
\end{equation}
\newline Above expressions hold for each build natural $n$-order polynomial.
\begin{lem}\label{lem2}
$\forall ([i+1]\in\mathbb{Z}^+, \ [i-1]\in\mathbb{Z}^-):\Delta^{k+j}(x_{i+1}^k)\equiv\nabla^{k+j}(x_{i-1}^k)\equiv0, \ j\geq1$
\end{lem}
\begin{proof}
According to lemma (\hyperref[lem1]{3.1}), the $n$-th difference of $n$-th power is constant, consequently, the constant rule (\hyperref[f4]{3}) holds $\Delta C=\delta C=\nabla C =0$.
\end{proof}
According to lemma (\hyperref[lem2]{4.5}) and properties (\hyperref[f4]{2}, \hyperref[f4]{3}), taking the limits of (\hyperref[r1]{4.2}, \hyperref[r2]{4.3}, \hyperref[r3]{4.4}), receive:
\begin{equation}\label{f15}
\Delta^{k \rightarrow n}P_n(x_{i+1})=\lim\limits_{k\to n}\left\{\Delta^k(a_0\cdot x_{i+1}^0)+\cdots+\Delta^k(a_n\cdot x_{i+1}^0)\right\}
\end{equation}
$$=\Delta^n(a_n\cdot x_{i+1}^n)=a_n\cdot \Delta^n(x_{i+1}^n)$$
\begin{equation}\label{f16}
\delta^{k \rightarrow n}P_n(x_i)=\lim\limits_{k\to n}\left\{\delta^k(a_0\cdot x_i^0)+\cdots+\delta^k(a_n\cdot x_i^n)\right\}
\end{equation}
$$=\delta^n(a_n\cdot x_i^n)=a_n\cdot\delta^n(x_i^n)$$
\begin{equation}\label{f17}
\nabla^{k \rightarrow n}P_n(x_{i-1})=\lim\limits_{k\to n}\left\{\nabla^k(a_0\cdot x_{i-1}^0)+\cdots+\nabla^k(a_n\cdot x_{i-1}^n)\right\}
\end{equation}
$$=\nabla^n(a_n\cdot x_{i-1}^n)=a_n\cdot\nabla^n(x_{i-1}^n)$$
Since the $n$-th difference of $n$-th power equals to $n!$, we have theorem.
\begin{thm}\label{th1}
Each $n$-order polynomial has the constant $n$-th finite (divided) difference and derivative, which equals each other and equal constant times $n!$, where $n$ is natural.
\end{thm}
\begin{proof}
According to limits (\hyperref[f15]{4.6}, \hyperref[f16]{4.7}, \hyperref[f17]{4.8}), we have $\Delta^nP_n(x_{i+1})=a_n\cdot\Delta^n(x_{i+1}^n), \ \newline \nabla^nP_n(x_{i-1})=a_n\cdot\nabla^n(x_{i-1}^n), \ \delta^nP_n(x_{i})=a_n\cdot\delta^n(x_{i}^n)$, going from lemma (\hyperref[lem1]{3.1}), the $n$-th difference of $n$-order polynomial equals to $k_n\cdot n!$, the properties (\hyperref[f11]{1}, \hyperref[f11]{2}, \hyperref[f11]{3}, \hyperref[f11]{4}) proofs that for enough large sets $\mathbb{Z^+}, \ \mathbb{Z^-}$ we have $\Delta^n(x^n_{i+1})\equiv\Delta^n(x^n_{i}), \ \delta^n(x^n_{i})\equiv\delta^n(x^n_{i+j}), \ \nabla^n(x^n_{i-1})\equiv\nabla^n(x^n_{i}), \ \min(\mathbb{Z^-})\leq n\leq\max(\mathbb{Z^+}).$ Therefore, we have equality
\begin{equation}\label{f18}
\frac{d^nP_n(x)}{dx^n}=\frac{\Delta^nP_n(x_i)}{(\Delta x)^n}=\frac{\delta^nP_n(x_{i})}{(\delta x)^n}=\frac{\nabla^nP_n(x_{i})}{(\nabla x)^n}=a_n\cdot n!
\end{equation}
Or, by means of definition (\hyperref[f10]{3.14}) one has
\begin{equation}\label{f19}
\left(\frac{d}{dx}\right)^nP_n(x)=E(P_n(x))=a_n\cdot E(x^n)
\end{equation}
\end{proof}
\begin{ppty}\label{pp1}
Let be a plot of $\nabla^kx_i^n(k), \ i\in\mathbb{Z}^-$ (see \hyperref[a1]{Appendix 1}, second line for reference)\\
\begin{center}
\begin{tikzpicture}[scale=1]
\begin{axis}[
xlabel = {$k\in[0;10]$},
ylabel = {$\nabla^kx_{-10}^{10}(k)$},
minor tick num = 1
]
\addplot[smooth, blue] table [x = b, y = a] {
a b
10000000000 0
-6513215599 1
4100173022 2
-2478397020 3
1425878520 4
-771309000 5
385363440 6
-172972800 7
66528000 8
-19958400 9
3628800 10
};
\end{axis}
\end{tikzpicture}
\end{center}
\begin{center}
\textrm{Figure 1. Plot of} $\nabla^kx_i^n(k), \ i\in\mathbb{Z}^-$
\end{center}
It's seen that each k-order backward difference (acc. to \hyperref[a1]{app 1}) of power $n$, such that $n\geq k$ could be well interpolated by means of general Harmonic oscillator equation
\begin{equation}\label{i1}
x=A_0e^{-\beta t}\sin(\omega t+\varphi_0)
\end{equation}
Particularizing \hyperref[i1]{4.13} to Figure 1, we get
\begin{equation}\label{i2}
\nabla^kx_i^n(j\leq k)=x^ne^{-\beta_{i} k}\sin(\omega_{i} k+\varphi_0)
\end{equation}
In the points of local minimum and maximum of $\int x^ne^{-\beta k}\sin(\omega k+\varphi_0)dk$ we have $\nabla^k x^n, \ k\in[1; \ n]\subset\mathbb{N}_1$. By means of (\hyperref[f11]{5}) we have relation with forward difference
\begin{equation}\label{g22}
\Delta^kx_i^n(k)=(-1)^{n-1}x^ne^{-\beta_{i} k}\sin(\omega_{i} k+\varphi_0)
\end{equation}
\end{ppty}
Property \hyperref[pp1]{4.12} as well holds for polynomials.
\section{Relation with Partial derivatives}
Let be partial finite differences defined as
\begin{equation}\label{f20}
\Delta f(u_1, \ u_2,\ldots, \ u_n)_{u_1}:=f(u_1+h, \ u_2,\ldots, \ u_n)-f(u_1, \ u_2,\ldots, \ u_n)
\end{equation}
\begin{equation}\label{f21}
\delta f(u_1, \ u_2,\ldots, \ u_n)_{u_1}:=f(u_1+h, \ u_2,\ldots, \ u_n)-f(u_1-h, \ u_2,\ldots, \ u_n)
\end{equation}
\begin{equation}\label{f22}
\nabla f(u_1, \ u_2,\ldots, \ u_n)_{u_1}:=f(u_1, \ u_2,\ldots, \ u_n)-f(u_1-h, \ u_2,\ldots, \ u_n)
\end{equation}
By means of mathematical induction, going from Lemma (\hyperref[lem1]{3.1}), we have equality between $n$-th partial derivative and $n$-th partial difference, while be taken of polynomial defined function or power function.
\begin{thm}\label{th3}
For each $n$-th natural power of many variables the $n$-th partial divided differences and $n$-th partial derivatives equal each other.
\begin{proof}
Let be function $Z=f(u_1, \ u_2,\ldots, \ u_n)=(u_1, \ u_2, \ldots, \ u_n)^n$, where dots mean the general relations, i.e multiplication and summation between variables. We denote the equality operator of partial difference as $E(F(u_1, \ u_2, \ \ldots, \ u_n))_{u_k}$, where $u_k$ is variable of taken difference. On this basis
\begin{equation}\label{f23}
\frac{\partial^n Z}{\partial u_k^n}=\frac{\Delta^nZ_{u_k}}{\Delta u_k^n}=\frac{\delta^nZ_{u_k}}{\delta u_k^n}=\frac{\nabla^nZ_{u_k}}{\nabla u_k^n}=A\cdot n!
\end{equation}
Or, using equality operator
\begin{equation}\label{f24}
\frac{\partial^n Z}{\partial u_k^n}=E(Z)_{u_k}=A\cdot n!
\end{equation}
where $A$ is free constant, depending of relations between variables and $0\leq k \leq n$.
\end{proof}
\end{thm}
\begin{ppty}\label{p1}
Let be partial differences of the function $f(u_1, \ \cdots \ ,u_k)=u_1^n\pm u_2^n\pm \ \cdots \ \pm u_k^n, \ n\in\mathbb{N},$ $\Delta f(u_1, \ \cdots \ ,u_k)_{M}, \ \delta f(u_1, \ \cdots \ ,u_k)_{M}, \ \nabla f(u_1, \ \cdots \ ,u_k)_{M}$, where $M$ - complete set of variables, i.e $M=\{u_i\}_i^k$ the $n$-th partial differences of each variables are
\begin{equation}\label{f25}
\Delta^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3,\ldots,u_k}=\pm k\cdot n!\cdot(\Delta u_1)^n\cdots (\Delta u_k)^n
\end{equation}
\begin{equation}\label{f26}
\delta^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3, \ \ldots,u_k}=\pm k\cdot n!\cdot(\delta u_1)^n\cdots (\delta u_k)^n
\end{equation}
\begin{equation}\label{f27}
\nabla^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3, \ \ldots,u_k}=\pm k\cdot n!\cdot(\nabla u_1)^n\cdots (\nabla u_k)^n
\end{equation}
$$\forall k\in\mathbb{Z}:\max(\mathbb{Z^+})>n>\min(\mathbb{Z^-}), \ (\delta u_1)\equiv(\delta u_2)\equiv\ldots\equiv(\delta u_k),$$
$$(\nabla u_1)\equiv(\nabla u_2)\equiv\ldots\equiv(\nabla u_k), \ (\Delta u_1)\equiv(\Delta u_2)\equiv\ldots\equiv(\Delta u_k)$$
$\mathrm{Otherwise}$
\begin{equation}\label{y1}
\Delta^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3, \ \ldots,u_k}=n!\cdot\sum_{i=1}^{k}(\Delta u_i)^n
\end{equation}
\begin{equation}\label{y2}
\delta^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3, \ \ldots,u_k}=n!\cdot\sum_{i=1}^{k}(\delta u_i)^n
\end{equation}
\begin{equation}\label{y1}
\nabla^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3, \ \ldots,u_k}=n!\cdot\sum_{i=1}^{k}(\nabla u_i)^n
\end{equation}
Note that here the partial differences of non-single variable defined as
$$\Delta^n f(u_1^,\ldots,u_k)_{M}=\Delta^{n-1} f(u_1+h,\ldots,u_k+h)_{M}-\Delta^{n-1} f(u_1,\ldots,u_k)_{M}$$
$$\delta^n f(u_1,\ldots,u_k)_{M}=\delta^{n-1} f(u_1+h,\ldots,u_k+h)_{M}-\delta^{n-1} f(u_1-h,\ldots,u_k-h)_{M}$$
$$\nabla^n f(u_1,\ldots,u_k)_{M}=\nabla^{n-1} f(u_1,\ldots,u_k)_{M}-\nabla^{n-1} f(u_1-h,\ldots,u_k-h)_{M}$$
\end{ppty}
Moreover, the $n$-th partial difference taken over enough large set $\mathbb{Z^+}$ and $\forall i: \Delta x_i=1$ has the next connection with single variable $n$-th derivative of $n$-th power
\begin{equation}\label{k1}
\Delta^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3,\ldots,u_k}=\sum_{i=1}^{k}\left(\frac{d}{du_i}\right)^nf(u_i)
\end{equation}
With partial derivative we have relation
\begin{equation}\label{k2}
\Delta^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)_{u_1, \ u_2, \ u_3,\ldots,u_k}=\sum_{i=1}^{k}\left(\frac{\partial}{\partial u_i}\right)^n f(u_1, \ u_2, \ u_3, \ \ldots,\ u_k)
\end{equation}
Multiplied (\hyperref[k1]{5.14}) and (\hyperref[k2]{5.15}) by coefficient, as defined, gives us relation with $n$-th partial polynomial.
\begin{thm}\label{th4}
For each non-single variable polynomial with order $n$ holds the equality between $k\leq n$-order partial differences and derivative.
\begin{proof}
Let be non-single variable polynomial
\begin{equation}\label{f28}
\mathcal{P}_n(u_n)=\sum_{i=1}^{n}M_i\cdot u_i^i
\end{equation}
Going from property (\hyperref[p1]{5.7}), the $k$-th partial differences of one variable are
\begin{equation}\label{f29}
\Delta^k\mathcal{P}_n(u_n)_{u_k}=M_k\cdot k!\cdot(\Delta u_k)^k, \ \delta^k\mathcal{P}_n(u_n)_{u_k}=M_k\cdot k!\cdot(\delta u_k)^k,
\end{equation}
$$\nabla^k\mathcal{P}_n(u_n)_{u_k}=M_k\cdot k!\cdot(\nabla u_k)^k$$
$\ 0\leq k\leq n$.
The $k$-th partial derivative:
\begin{equation}\label{f30}
\frac{\partial^k\mathcal{P}_n(u_n)}{\partial u_k^k}=M_k\cdot k!
\end{equation}
Hereby,
\begin{equation}\label{f31}
\frac{\partial^k\mathcal{P}_n(u_n)}{\partial u_k^k}=\frac{\Delta^k\mathcal{P}_n(u_n)_{u_k}}{\Delta u_k^k}=\frac{\delta^k\mathcal{P}_n(u_n)_{u_k}}{\delta u_k^k}=\frac{\nabla^k\mathcal{P}_n(u_n)_{u_k}}{\nabla u_k^k}
\end{equation}
Also could be denoted as
\begin{equation}\label{f32}
\frac{\partial^k\mathcal{P}_n(u_n)}{\partial u_k^k}=E(\mathcal{P}_n(u_n))_{u_k}=M_k\cdot k!, \ \ \ \ k\leq n
\end{equation}
And completes the proof.
\end{proof}
\end{thm}
\section{Relations between finite differences}
In this section are shown relations between central, backward and central finite differences, generally, they are
\begin{equation}\label{f9}
\delta_{\mathrm{div}} f(x):=\frac{f(x+\Delta x)-f(x-\Delta x)}{2\cdot\Delta x}\stackrel{\rm def}{=}\frac{1}{2}\left(\frac{f(x+\Delta x)}{\Delta x}-\frac{f(x-\nabla x)}{\nabla x}\right)
\end{equation}
$$
=\Bigg|\begin{aligned}
f(x+\Delta x)=\Delta f(x)+f(x)
\\
f(x-\nabla x)=f(x)-\nabla f(x)
\end{aligned}
\Bigg|=\frac{1}{2}\left(\frac{\Delta f(x)+f(x)}{\Delta x}-\frac{f(x)-\nabla f(x)}{\nabla x}\right)
$$\\
$$=\frac{1}{2}\left(\frac{\Delta f(x)+\nabla f(x)}{\Delta x\equiv\nabla x}\right)$$
where "div" means divided, i.e $\delta_{\mathrm{div}} f(x):=\delta f(x)/(2\cdot\Delta x)$.
Hereby,
\begin{equation}\label{dd}
2\cdot\delta_{\mathrm{div}} f(x)\cdot\Delta x=\Delta f(x)+\nabla f(x)
\end{equation}
And so on. Let be $\Delta x\to 0$
\begin{equation}\label{dd2}
\lim_{\Delta x\to 0}2\cdot\delta_{\mathrm{div}} f(x)\cdot\Delta x=2\cdot df(x)
\end{equation}
Or
\begin{equation}\label{dd3}
2\cdot\lim_{\Delta x\to 0}\delta_{\mathrm{div}} f(x)=2\cdot \frac{df(x)}{dx}\longrightarrow\lim_{\Delta x\to 0}\delta_{\mathrm{div}} f(x)=\frac{df(x)}{dx}
\end{equation}
where $f(x)$ is power function, hence, the general relation between derivative and each kind finite difference is reached, as desired.
\section{The error of approximation}
The error of derivative approximation done by forward finite difference with respect to order $k\leq n$ could be calculated as follows
\begin{equation}\label{r4}
\left(\frac{\Delta}{\Delta x}\right)^k x^n-\left(\frac{d}{dx}\right)^kx^n=\mathcal{O}(x^{n-k})
\end{equation}
For $n$-order polynomial is
\begin{equation}\label{r5}
\left(\frac{\Delta}{\Delta x}\right)^k P_n(x)-\left(\frac{d}{dx}\right)^k P_n(x)=\mathcal{O}(x^{n-k})
\end{equation}
The partial, if $ m\leq k$
\begin{equation}\label{r6}
\left(\frac{\Delta}{\Delta u_k}\right)^m Z-\left(\frac{\partial}{\partial u_k}\right)^m Z=\mathcal{O}(u_k^{\ k-m})
\end{equation}
Where $\mathcal{O}$ - Landau-Bachmann symbol (see \cite{8}, \cite{9}).
\section{Summary}
In this section we summarize the obtained results in the previous chapters and establish the relationship between them. According to lemma (\hyperref[lem1]{3.1}), theorems (\hyperref[th1]{4.9}), (\hyperref[th3]{5.4}), (\hyperref[th4]{5.16}) we have concluded
\begin{equation}\label{f33}
\frac{d^nx^n}{dx^n}=E(x^n)=n!
\end{equation}
\begin{equation}\label{f34}
\frac{d^nP_n(x)}{dx^n}=E(P_n(x))=a_n\cdot E(x^n)
\end{equation}
\begin{equation}\label{f35}
\frac{\partial^n Z}{\partial u_k^n}=E(Z)_{u_k}=A\cdot n!
\end{equation}
\begin{equation}\label{f36}
\frac{\partial^k\mathcal{P}_n(u_n)}{\partial u_k^k}=E(\mathcal{P}_n(u_n))_{u_k}=M_k\cdot k!
\end{equation}
Generalizing these expressions, we can derive the general relations between ordinary, partial derivatives and finite (divided) differences
\begin{equation}\label{f37}
\underbrace{E(u^n)=E(P_n(u_g))=E(Z)_{u_k}=E(\mathcal{P}_{n+j}(u_{n+j}))_{u_n}}_{Y}
\end{equation}
\begin{equation}\label{f38}
\underbrace{\frac{d^nu^n}{du^n}=\frac{d^nP_n(u)}{du^{n}}=\frac{\partial^n Z}{\partial u_k^n}=\frac{\partial^n\mathcal{P}_{n+j}(u_{n+j})}{\partial u_n^n}}_{U}, \ j\geq0
\end{equation}
$$\forall(A, \ M_n, \ a_n)=1$$
I.e the equalities hold with precision to constant.
Function $Z$ defined as $Z=f(u_1, \ u_2,\ldots, \ u_n)=(u_1, \ u_2,\ldots, \ u_n)^n$.
And finally
$$
Y=U
$$
with same limitations.
\section{Conclusion}
In this paper were established the equalities between ordinary and partial finite (divided) differences and derivatives of power function and polynomials, with order equal between each other.
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M. Hanif Chaudhry (2007). Open-Channel Flow. Springer. p. 369. ISBN 978-0-387-68648-6.
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Peter Olver (2013). \href{http://chaosbook.org/library/Olver14.pdf}{Introduction to Partial Differential Equations} . Springer Science and Business Media. p. 182.
\bibitem{4} \label{b3}
Weisstein, Eric W. \href{http://mathworld.wolfram.com/FiniteDifference.html}{"Finite Difference."} From MathWorld
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D. Gleich (2005), Finite Calculus: A Tutorial for Solving Nasty Sums p 6-7. \href{https://www.cs.purdue.edu/homes/dgleich/publications/Gleich 2005 - finite calculus.pdf}{online copy}
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Bakhvalov N. S. Numerical Methods: Analysis, Algebra, Ordinary Differential Equations p. 42, 1977.
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G. M. Fichtenholz (1968). Differential and integral calculus (Volume 1). p. 244.
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Paul Bachmann. Analytische Zahlentheorie, vol.2, Leipzig, Teubner 1894.
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\end{thebibliography}
\eject \pdfpagewidth=14in \pdfpageheight=7in
\clearpage
\section{Appendix 1. Difference table up to tenth power}\label{a1}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
$i$ & $x_i$ & $f$ & $(\Delta,\nabla)^{}f/\delta^{10}f$ & $(\Delta,\nabla)^{2}f/\delta^{9}f$ & $(\Delta,\nabla)^{3}f/\delta^{8}f$ & $(\Delta,\nabla)^{4}f/\delta^{7}f$ & $(\Delta,\nabla)^{5}f/\delta^{6}f$ & $(\Delta,\nabla)^{6}f/\delta^{5}f$ & $(\Delta,\nabla)^{7}f/\delta^{4}f$ & $(\Delta,\nabla)^{8}f/\delta^{3}f$ & $(\Delta,\nabla)^{9}f/\delta^{2}f$ & $(\Delta,\nabla)^{10}f/\delta^{}f$
\\
\hline
-10 & -10 & 10000000000 & -6513215599 & 4100173022 & -2478397020 & 1425878520 & -771309000 & 385363440 & -172972800 & 66528000 & -19958400 & 3628800
\\
-9 & -9 & 3486784401 & -2413042577 & 1621776002 & -1052518500 & 654569520 & -385945560 & 212390640 & -106444800 & 46569600 & -16329600 & \textbf{-8926258176}
\\
-8 & -8 & 1073741824 & -791266575 & 569257502 & -397948980 & 268623960 & -173554920 & 105945840 & -59875200 & 30240000 & \textbf{7912982528} & -3204309152
\\
-7 & -7 & 282475249 & -222009073 & 171308522 & -129325020 & 95069040 & -67609080 & 46070640 & -29635200 & \textbf{-6959124480} & 2931599528 & -1013275648
\\
-6 & -6 & 60466176 & -50700551 & 41983502 & -34255980 & 27459960 & -21538440 & 16435440 & \textbf{6063636480} & -2668596480 & 953858048 & -272709624
\\
-5 & -5 & 9765625 & -8717049 & 7727522 & -6796020 & 5921520 & -5103000 & \textbf{-5225472000} & 2415240960 & -895488000 & 263003048 & -59417600
\\
-4 & -4 & 1048576 & -989527 & 931502 & -874500 & 818520 & \textbf{4443586560} & -2171473920 & 838164480 & -253355520 & 58370048 & -9706576
\\
-3 & -3 & 59049 & -58025 & 57002 & -55980 & \textbf{-3715891200} & 1937295360 & -781885440 & 243767040 & -57323520 & 9647528 & -1047552
\\
-2 & -2 & 1024 & -1023 & 1022 & \textbf{3096576000} & -1703116800 & 727695360 & -234178560 & 56279040 & -9588480 & 1046528 & -59048
\\
-1 & -1 & 1 & -1 & \textbf{-1857945600} & 1703116800 & -619315200 & 234178560 & -54190080 & 9588480 & -1044480 & 59048 & -1024
\\
0 & 0 & 0 & \textbf{3715891200} & 0 & 1238630400 & 0 & 108380160 & 0 & 2088960 & 0 & 2048 & 0
\\
1 & 1 & 1 & 1 & \textbf{1857945600} & 1703116800 & 619315200 & 234178560 & 54190080 & 9588480 & 1044480 & 59048 & 1024
\\
2 & 2 & 1024 & 1023 & 1022 & \textbf{3096576000} & 1703116800 & 727695360 & 234178560 & 56279040 & 9588480 & 1046528 & 59048
\\
3 & 3 & 59049 & 58025 & 57002 & 55980 & \textbf{3715891200} & 1937295360 & 781885440 & 243767040 & 57323520 & 9647528 & 1047552
\\
4 & 4 & 1048576 & 989527 & 931502 & 874500 & 818520 & \textbf{4443586560} & 2171473920 & 838164480 & 253355520 & 58370048 & 9706576
\\
5 & 5 & 9765625 & 8717049 & 7727522 & 6796020 & 5921520 & 5103000 & \textbf{5225472000} & 2415240960 & 895488000 & 263003048 & 59417600
\\
6 & 6 & 60466176 & 50700551 & 41983502 & 34255980 & 27459960 & 21538440 & 16435440 & \textbf{6063636480} & 2668596480 & 953858048 & 272709624
\\
7 & 7 & 282475249 & 222009073 & 171308522 & 129325020 & 95069040 & 67609080 & 46070640 & 29635200 & \textbf{6959124480} & 2931599528 & 1013275648
\\
8 & 8 & 1073741824 & 791266575 & 569257502 & 397948980 & 268623960 & 173554920 & 105945840 & 59875200 & 30240000 & \textbf{7912982528} & 3204309152
\\
9 & 9 & 3486784401 & 2413042577 & 1621776002 & 1052518500 & 654569520 & 385945560 & 212390640 & 106444800 & 46569600 & 16329600 & \textbf{8926258176}
\\
10 & 10 & 10000000000 & 6513215599 & 4100173022 & 2478397020 & 1425878520 & 771309000 & 385363440 & 172972800 & 66528000 & 19958400 & 3628800
\\
\hline
\end{tabular}
\end{center}
\begin{tabbing}
$Note \ that \ central \ differences \ divided \ by \ \textbf{bold} \ typeset. \ The \ table \ shows \ example \ in \ case \max\mathbb{Z}^+=10, \ \min\mathbb{Z}^-=-10, \ \Delta x=1, \ n=10$ \kill
% \> for next tab, \\ for new line...
$\textrm{Note \ that \ central \ differences \ divided \ by \ \textbf{bold} \ typeset \ and\ kept\ in\ the\ middle\ of\ table.\ \ The \ table \ shows \ example \ in \ case} \ \max\mathbb{Z}^+=10, \ \min\mathbb{Z}^-=-10, \ \Delta x=1, \ n=10$
\end{tabbing}
\end{document}