Problem 6.1#
We begin with the following Taylor expansions
\[\begin{align*}
f(x+\Delta x) = &f(x)
+\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{1}{3!}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{1}{4!}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4}
+\cdots \\
f(x-\Delta x) = &f(x)
-\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{1}{3!}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{1}{4!}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4}
-\cdots \\
f(x+2\Delta x) = &f(x)
+2\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(2\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{1}{3!}\,(2\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{1}{4!}\,(2\Delta x)^4\,\frac{d^4f(x)}{dx^4}
+\cdots \\
f(x-2\Delta x) = &f(x)
-2\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(2\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{1}{3!}\,(2\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{1}{4!}\,(2\Delta x)^4\,\frac{d^4f(x)}{dx^4}
-\cdots
\end{align*}\]
where each of these expansions has been truncated at \(\mathcal{O}(\Delta x)^5\). Now let us combine these expansions together as follows
\[\begin{align*}
f(x-2\Delta x) + 8 &f(x+\Delta x) - 8 f(x-\Delta x) - f(x+2\Delta x) = \\
&f(x)
-2\,\Delta x\,\frac{df(x)}{dx}
+ 2\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{8}{6}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{2}{3}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4}
\\
+ 8 &f(x)
+8 \Delta x\,\frac{df(x)}{dx}
+4\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+ \frac{8}{6}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{1}{3}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4}
\\
- 8 &f(x)
+8 \Delta x\,\frac{df(x)}{dx}
-4\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+ \frac{8}{6}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
-\frac{1}{3}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4}
\\
- &f(x)
-2\Delta x\,\frac{df(x)}{dx}
-2\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{8}{6}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
-\frac{2}{3}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4} \\
+& \mathcal{O}(\Delta x)^5 \quad .
\end{align*}\]
Evidently all terms on the right-hand side cancel out, except for the \(\frac{df(x)}{dx}\) terms, yielding
\[\begin{equation*}
12\,\Delta x\,\frac{df(x)}{dx} + \mathcal{O}(\Delta x)^5 = f(x-2\Delta x) + 8 f(x+\Delta x) - 8 f(x-\Delta x) - f(x+2\Delta x)
\end{equation*}\]
and therefore
\[\begin{equation*}
\frac{df(x)}{dx} = \frac{f(x-2\Delta x) + 8 f(x+\Delta x) - 8 f(x-\Delta x) - f(x+2\Delta x)}{12\,\Delta x} - \mathcal{O}(\Delta x)^4 \quad
\end{equation*}\]
approximates the first derivative of \(f(x)\) to 4th-order accuracy, noting that the division by \(\Delta x\) reduces the error from order \(\mathcal{O}(\Delta x)^5\) to \(\mathcal{O}(\Delta x)^4\).
Similarly, we may consider the Taylor expansions:
\[\begin{align*}
f(x+ \frac{3}{2}\Delta x) \ = \ f(x)
+ \frac{3}{2}\,\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(\frac{3}{2}\,\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{1}{3!}\,(\frac{3}{2}\,\Delta x)^3\,\frac{d^3f(x)}{dx^3} \\ +\: \frac{1}{4!}\,(\frac{3}{2}\,\Delta x)^4\,\frac{d^4f(x)}{dx^4}
+\mathcal{O}(\Delta x)^5
\end{align*}\]
\[\begin{align*}
f(x-\frac{3}{2}\Delta x) \ = \ f(x)
-\frac{3}{2}\,\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(\frac{3}{2}\,\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{1}{3!}\,(\frac{3}{2}\,\Delta x)^3\,\frac{d^3f(x)}{dx^3} \\ +\: \frac{1}{4!}\,(\frac{3}{2}\,\Delta x)^4\,\frac{d^3f(x)}{dx^4}
-\mathcal{O}(\Delta x)^5
\end{align*}\]
\[\begin{align*}
f(x+\frac{1}{2}\,\Delta x) \ = \ f(x)
+\frac{1}{2}\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(\frac{1}{2}\,\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{1}{3!}\,(\frac{1}{2}\,\Delta x)^3\,\frac{d^3f(x)}{dx^3}\\ +\: \frac{1}{4!}\,(\frac{1}{2}\,\Delta x)^4\,\frac{d^4f(x)}{dx^4}
+\mathcal{O}(\Delta x)^5
\end{align*}\]
\[\begin{align*}
f(x-\frac{1}{2}\,\Delta x) \ = \ f(x)
-\frac{1}{2}\,\Delta x\,\frac{df(x)}{dx}
+\frac{1}{2!}\,(\frac{1}{2}\,\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{1}{3!}\,(\frac{1}{2}\,\Delta x)^3\,\frac{d^3f(x)}{dx^3}\\ +\: \frac{1}{4!}\,(\frac{1}{2}\,\Delta x)^4\,\frac{d^4f(x)}{dx^4}
- \mathcal{O}(\Delta x)^5 \quad.
\end{align*}\]
Let us combine multiples of these expansions, each up to order \(\mathcal{O}(\Delta x)^5\), as follows:
\[\begin{align*}
- f(x+\frac{3}{2}\,\Delta x) \, + \, &27 f(x+\frac{1}{2}\,\Delta x) \, - \, 27 f(x-\frac{1}{2}\,\Delta x) \, + \, f(x-\frac{3}{2}\,\Delta x) = \\
- \, &f(x)
- \frac{3}{2}\Delta x\,\frac{df(x)}{dx}
-\frac{9}{8}\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{9}{16}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3} - \frac{27}{128}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4} \\
+\, 27\, &f(x)
+\frac{27}{2}\,\Delta x\,\frac{df(x)}{dx}
+\frac{27}{8}\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{9}{16}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3} + \frac{9}{128}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4} \\
- 27\, &f(x)
+\frac{27}{2}\,\Delta x\,\frac{df(x)}{dx}
-\frac{27}{8}\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{9}{16}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
-\frac{9}{128}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4} \\
+\, & f(x)
-\frac{3}{2}\,\Delta x\,\frac{df(x)}{dx}
+\frac{9}{8}\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{9}{16}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\frac{27}{128}\,(\Delta x)^4\,\frac{d^4f(x)}{dx^4}
\\
+\, & \mathcal{O}(\Delta x)^5\quad
\end{align*}\]
Again, we observe that all right-hand-side terms of \(f(x)\), \(\frac{d^2f(x)}{dx^2}\) and \(\frac{d^3f(x)}{dx^3}\) cancel out, leaving only terms of \(\frac{df(x)}{dx}\) such that
\[\begin{equation*}
- f(x+\frac{3}{2}\,\Delta x) + 27 f(x+\frac{1}{2}\,\Delta x) - 27 f(x-\frac{1}{2}\,\Delta x) + f(x-\frac{3}{2}\,\Delta x) = 24 \Delta x\,\frac{df(x)}{dx} +\mathcal{O}(\Delta x)^5 \quad,
\end{equation*}\]
and so dividing by \(24\,\Delta x\) yields, as required
\[\begin{equation*}
\frac{df(x)}{dx}+\mathcal{O}(\Delta x)^4 = \frac{1}{\Delta x} \left[ -\frac{1}{24} f(x+\frac{3}{2}\,\Delta x) + \frac{9}{8} f(x+\frac{1}{2}\,\Delta x) - \frac{9}{8} f(x-\frac{1}{2}\,\Delta x) + \frac{1}{24} f(x-\frac{3}{2}\,\Delta x) \right] \quad.
\end{equation*}\]