Finite Difference Approximation
Consider the function $f(x)$, given at points $x_{j-1}, x_j, x_{j+1},$ and so on. Denote $f_{j-1} = f(x_{j-1})$, $f_j = f(x_j)$, and $f_{j+1} = f(x_{j+1})$.
The values at different points can be related to each other by a Taylor series
$$
f_{j+1} = f_j + \frac{\partial f_j}{\partial x}\Delta x
+ \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
+ \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+ \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^4}{24} + \cdots
$$ where the derivatives are taken at $x_j$
Give an expression for the first derivative
$$
\frac{\partial f_j}{\partial x} = \frac{f_{j+1} - f_j}{\Delta x}
- \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x}{2}
- \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^2}{6}
- \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^3}{24} + \cdots
$$
As $\Delta x \to 0$, all the terms on the right-hand side go to zero, except for the first one.
Since the error is directly related to the first power of $\Delta x$, it is a first-order approximation to the first derivative
Instead of writing $f_{j+1}$ as a Taylor series expansion of $f_j$ to get an approximation for the derivative at $x_j$, we could just as well have written $f_{j-1}$ as a Taylor series expansion of $f_j$, by replacing $\Delta x$ by $-\Delta x$
$$
f_{j-1} = f_j - \frac{\partial f_j}{\partial x}\Delta x + \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
- \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+ \frac{\partial^4 f_j}{\partial x^4}\frac{\Delta x^4}{24} + \cdots
$$
\[
\begin{aligned}
f_{j+1}-f_{j-1}
&=\Bigg(
f_j + \frac{\partial f_j}{\partial x}\,\Delta x
+ \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
+ \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+ \cdots
\Bigg) \\
&\quad-
\Bigg(
f_j - \frac{\partial f_j}{\partial x}\,\Delta x
+ \frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
- \frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+ \cdots
\Bigg) \\
&= \cancel{f_j}-\cancel{f_j}
+\;2\,\frac{\partial f_j}{\partial x}\,\Delta x
+\;\cancel{\frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}
-\frac{\partial^2 f_j}{\partial x^2}\frac{\Delta x^2}{2}}
+\;2\,\frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{6}
+\;\cdots \\
&= 2\Delta x\,\frac{\partial f_j}{\partial x}
+\frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^3}{3}
+\cdots\\
\Rightarrow
\frac{\partial f_j}{\partial x}
&= \frac{f_{j+1}-f_{j-1}}{2\Delta x}
-\frac{\partial^3 f_j}{\partial x^3}\frac{\Delta x^2}{6}
+\cdots
\end{aligned}
\]
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