Finite Difference Approximation
$$
\frac{\partial f_i}{\partial x} \;=\; \frac{f_{j+1} - f_{j-1}}{2\Delta x} \;-\; \frac{\partial^3 f_j}{\partial x^3} \frac{\Delta x^2}{6} \;+\; \cdots
$$
The first error term contains $\Delta x^2$ instead of $\Delta x$,
showing that if we divide $\Delta x$ by two, the error will go down by a factor of four.
Since the error is directly related to the second power of $\Delta x$,
it is a second-order approximation to the first derivative
The second-order approximation can also be derived by taking the average of the first-order approximations using $f_{j+1}$ and $f_{j-1}$, and it is presumably intuitive that the average would give us a better approximation
\begin{align}
f_{j+1} + f_{j-1}
&= \Bigg[ f_j + \frac{\partial f_j}{\partial x}\Delta x
+ \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
+ \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+ \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^4}{24} + \cdots \Bigg] \\
&\quad + \Bigg[ f_j - \frac{\partial f_j}{\partial x}\Delta x
+ \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
- \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+ \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^4}{24} + \cdots \Bigg] \\[6pt]
&= 2 f_j
+ \cancel{\frac{\partial f_j}{\partial x}\Delta x} - \cancel{\frac{\partial f_j}{\partial x}\Delta x}
+ \frac{\partial^2 f_j}{\partial x^2} \Delta x^2
+ \cancel{\frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}} - \cancel{\frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}}
+ \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^4}{12} + \cdots \\
\implies \frac{\partial^2 f_j}{\partial x^2}
&= \frac{f_{j+1} - 2 f_j + f_{j-1}}{\Delta x^2}
- \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^2}{12} + \cdots
\end{align}
By using larger number of points we can derive approximations for higher derivatives and
with higher-order error terms
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