Higher-order Approximations

The derivative we just considered is first-order accurate. If we cut \(\Delta x\) in half, our error drops by a factor of 2 (at least until roundoff error dominates).

Centered-difference approximation

Note

Higher-order here means better truncation error, not a higher (e.g. second) derivative.

Let’s look back at our difference approximation, but now consider Taylor expanding \(f(x_0 + \Delta x)\) and \(f(x_0 - \Delta x)\):

\[f(x_0 + \Delta x) = f(x_0) + \left . \frac{df}{dx} \right |_{x_0} \Delta x + \frac{1}{2} \left . \frac{d^2f}{dx^2} \right |_{x_0} \Delta x^2 + \mathcal{O}(\Delta x^3)\]

\[f(x_0 - \Delta x) = f(x_0) - \left . \frac{df}{dx} \right |_{x_0} \Delta x + \frac{1}{2} \left . \frac{d^2f}{dx^2} \right |_{x_0} \Delta x^2 - \mathcal{O}(\Delta x^3)\]

Now, if we subtract these, we get:

\[f(x_0 + \Delta x) - f(x_0 - \Delta x) = 2 \left . \frac{df}{dx} \right |_{x_0} \Delta x + \mathcal{O}(\Delta x^3)\]

Notice that the \(\Delta x^2\) term cancels out. Now solving for the first derivative, we have:

\[\left . \frac{df}{dx} \right |_{x_0} = \frac{f(x_0 + \Delta x) - f(x_0 - \Delta x)}{2 \Delta x} + \mathcal{O}(\Delta x^2)\]

We see that this is second-order accurate. This is sometimes called a centered-difference.