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 Part I: Ch. 1

# 1 Finite Difference Calculus

 ... Yes, Computational Physics is that old! (Newton, 1670) More recent (Simple fluid MD, 1970) More recent still (Liquid crystal MD, 2001) etc.

Use finite differences in place of infinitesimal differentials:
Given $f(x)$, let $x_{k} \equiv k\Delta x(k=1,2,...)$

Remarks:
- History: opposite route; from finite differences to differentials
- Kepler 2- and 3-body problems (chaos!)
- Difference calculus remains applicable for any number of bodies and any potential
- Price paid: only tabulated trajectory

Subsections

## 1.1 Definitions

Given equidistant table values $f_{k} \equiv f(x_{k})$, define

$\begin{eqnarray} \Delta f_{k} &\equiv& f_{k+1}-f_{k} \;\; {\rm Forward \;\;Difference} \\ \nabla f_{k} &\equiv& f_{k}-f_{k-1} \;\; {\rm Backward \;\;Difference} \\ \delta f_{k} &\equiv& f_{k+1/2}-f_{k-1/2} \;\;\; {\rm Central}\;\;{\rm Difference}\;\; {\rm (*)} \end{eqnarray}$

(*) Table values at $x_{k \pm 1/2}$ not given; please have patience! In an emergency, use the "central mean"
$\begin{eqnarray} \delta f_{k} \rightarrow \mu \delta f_{k} &\equiv& \frac{1}{2} \left[ \delta f_{k+1/2}+ \delta f_{k-1/2}\right] \\ &=& \frac{1}{2} \left[f_{k+1}-f_{k-1}\right] \end{eqnarray}$

which uses only table values.

Recursive definition:
$\begin{eqnarray} \Delta^{2} f_{k} &\equiv& \Delta f_{k+1}-\Delta f_{k} = f_{k+2}-2f_{k+1}+f_{k} \\ \nabla^{2} f_{k} &\equiv& \dots \;\;{\rm (Exercise!?)} \\ \delta^{2} f_{k} &\equiv& \delta f_{k+1/2}-\delta f_{k-1/2} \\ &=& f_{k+1}-2f_{k}+f_{k-1} \;\; {\rm \;\;(*)} \end{eqnarray}$

etcetera.

(*) Here is the reward for your patience!
vesely 2005-10-10

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