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