Problem 6.2#
Here we will use the Taylor expansions listed in Problem 6.1. As in Problem 6.1 we combine multiples of these Taylor series, such that
\[\begin{align*}
-f(x+2\Delta x) \,+\, &16\,f(x+\Delta x) \,-\, 30\,f(x) \,+\, 16\,f(x-\Delta x) \,-\, f(x-2\Delta x) = \\
- &f(x) -2\Delta x\,\frac{df(x)}{dx}
-2\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{8}{6}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
-\mathcal{O}(\Delta x)^4 \\
+ &16 f(x) + 16\,\Delta x\,\frac{df(x)}{dx}
+ 8\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{8}{3}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\mathcal{O}(\Delta x)^4 \\
- &30 f(x) \\
+ &16 f(x) -16\,\Delta x\,\frac{df(x)}{dx}
+8\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
-\frac{8}{3}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
+\mathcal{O}(\Delta x)^4 \\
- &f(x) +2\Delta x\,\frac{df(x)}{dx}
-2\,(\Delta x)^2\,\frac{d^2f(x)}{dx^2}
+\frac{8}{6}\,(\Delta x)^3\,\frac{d^3f(x)}{dx^3}
-\mathcal{O}(\Delta x)^4 \quad,
\end{align*}\]
which simplifies to
\[\begin{equation*}
-f(x+2\Delta x) + 16 f(x+\Delta x) - 30 f(x) + 16 f(x-\Delta x) - f(x-2\Delta x) = 12(\Delta x)^2 \frac{d^2f(x)}{dx^2} +\mathcal{O}(\Delta x)^4 \quad,
\end{equation*}\]
and therefore dividing by \(12\,(\Delta x)^2\) yields
\[\begin{align*}
\frac{d^2f(x)}{dx^2} &= \frac{1}{(\Delta x)^2 } \left[ -\frac{1}{12}\,f(x+2\Delta x) + \frac{4}{3}\,f(x+\Delta x) - \frac{5}{2}\,f(x) + \frac{4}{3}\,f(x-\Delta x) - \frac{1}{12}\,f(x-2\Delta x) \right] \\ &+ \frac{\mathcal{O}(\Delta x)^4}{(\Delta x)^2} \quad.
\end{align*}\]
It is important to note here that due to the division by \((\Delta x)^2\), the error is of order \(\frac{\mathcal{O}(\Delta x)^4}{(\Delta x)^2} = \mathcal{O}(\Delta x)^2\).