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The following is a screenshot of the formula booklet I'll be able to use in an exam this week. I'm used to seeing the formulae for numerical differentiation in a different format though and I'm not sure how to interpret the ones in the formula booklet:

I'm used to seeing the formulae in the following format:

I want to know how the formulae in the book relate to the ones I'm used to using. I need to understand how to use the formula book versions of the formulae as these are the ones I will have access to in the exam. So, what do the $\Delta$, $\delta$ and $\mu$ symbols mean?

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3 Answers

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$\Delta$ represents the difference between 2 consecutive values.

We have $x_0,...,x_n$ with $x_k=x_0 + kh$, and we have the function values $f_0, ..., f_n$.

Then: $$\Delta f_0 = f_1-f_0, \quad \Delta f_1 = f_2-f_1, \quad ... \\ \Delta^2 f_0 = \Delta f_1 - \Delta f_0, \quad ...$$

Formulas 3 and 4 follow from an interpolating polynomial that is differentiated once respectively twice.

As yet I do not know where $\delta f_0$ and $\mu$ are coming from. My guess is that it's an alternative interpolation that is again differentiated once respectively twice.

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Speculation based on surrounding context:

$\Delta$ is a forward difference operator: $(\Delta f)(x)=f(x+h)-f(x)$.

$\delta$ is a backward difference operator: $(\delta f)(x)=f(x)-f(x-h)$.

$\mu$ is a forward shift operator: $(\mu f)(x)=f(x+h/2)$.

In this case property 3 is satisfied as one can check by expanding $\ln(I+\Delta)$, which is to say $hD$ where $D$ is the derivative operator, in powers of $\Delta$. The proof is a lot uglier, but property 4 is also satisfied by considering an expansion of $\ln(I+\Delta)^2$ in powers of $\Delta$. Note that this would still be satisfied by a backward difference as well.

I haven't checked carefully, but in this case $\mu \delta f(x)=f(x+h/2)-f(x-h/2)$, which is an approximation of $hf'(x)$ with error scaling like $h^3$, which is what you want. What perplexes me is the fact that $\mu$ only seems to appear to the first power, but it seems like higher powers would be required to "recenter" the high powers of $\delta$ if $\delta$ is indeed a one-sided difference.

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$\Delta$ is the Laplace operator:

It's the sum of all second-order derivatives of the function:

$$\text{if for e.g.}\quad f:\Bbb{R}^3 \to \Bbb{R},\ \ f = f(x,y,z)$$ $$\Delta f = \frac{d^2}{dx^2}f(x,y,z)+\frac{d^2}{dy^2}f(x,y,z)+\frac{d^2}{dz^2}f(x,y,z)$$ Generally, $$\text{if}\quad f:\Bbb{R}^n \to \Bbb{R},\ \ f = f(x_1,\dots,x_n)$$ $$\Delta f = \sum_{i=1}^n\frac{d^2}{dx_i^2}f(x_1,\dots,x_n)$$ This gives you back another $\Delta f : \Bbb{R}^n \to \Bbb{R}$ function.

You can also take the Laplace of Laplace-$f$: $$\Delta ^2 f = \Delta (\Delta f) = \sum_{i=1}^n\frac{d^2}{dx_i^2}\Delta f(x_1,\dots,x_n),$$ and the Laplace of the Laplace of Laplace-$f$: $$\Delta ^3 f = \Delta (\Delta^2 f) = \Delta (\Delta (\Delta f)) = \sum_{i=1}^n\frac{d^2}{dx_i^2}\Delta^2 f(x_1,\dots,x_n),$$ and so on.

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