# What does the delta notation in this formula mean?

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?

• Most likely it is something like $(\Delta f)(x)=f(x+h)-f(x),(\delta f)(x)=f(x)-f(x-h)$. Not sure what's going on with $\mu$ though.
– Ian
May 21, 2018 at 13:26
• hmm that's along the lines of what I assumed too. Thanks for your help May 21, 2018 at 13:27
• One useful formal identity is that if $D$ is the differentiation operator and $\tau_h$ is the forward translation operator then $e^{hD}=\tau_h$. As a result again formally you have $hD=\ln(\tau_h)$ and so $h^2 D^2=\ln(\tau_h)^2$. Truncating the power series expansions of these formulas (as expansions about the identity) give you finite difference methods. For example, $\ln(\tau_h)=\ln(I+\delta_h)=\delta_h-\delta_h^2/2+\delta_h^3/3-\dots$ which is exactly your third formula.
– Ian
May 21, 2018 at 13:29
• So your $\Delta$ is probably my $\delta_h$. Now the only question is what $\mu$ is; once that is clear it should be clear what $\delta$ is. Presumably $\mu$ is used to build centered differences.
– Ian
May 21, 2018 at 13:31
• I suspect that $(\mu \delta f)(x)$ is $f(x+h)-f(x-h)$ or something very similar, because it is second order (the error in the actual derivative behaves like $h^2$). Also because your little writeup mentions the centered difference.
– Ian
May 21, 2018 at 13:43

$\Delta$ 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.

• I think $\mu$ might be a pure shift: if $\delta_h$ is a backward difference and $\tau_h$ is a forward translation then $\tau_{h/2} \delta_h$ is now the centered difference, thus a third order approximation for $hf'$ as desired. My confusion is then why no additional powers of $\mu$ seem to be required, unless this is just sloppy notation for $(\mu \delta)^3,(\mu \delta)^5$ etc. Further powers of $\mu$ would definitely be required for the higher order corrections to the centered difference.
– Ian
May 21, 2018 at 15:14

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.

• This seems accurate. Thanks @Ian May 21, 2018 at 15:59

$\Delta$ is the Laplace operator: https://en.wikipedia.org/wiki/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.

• I don't think that is the $\Delta$ that is meant here...
– Ian
May 21, 2018 at 13:14
• @Ian Usually in mathematics $\Delta f$ always means the Laplace of the function. Unfortunately without more context, that's the best guess I can give. May 21, 2018 at 13:19
• @user43712 Unfortunately I don't. If you give us more context or link to page where we can find more information about the topic, then we might be able to figure it out. May 21, 2018 at 13:19
• I'm pretty sure these are each finite difference operators though it isn't apparent to me which is which.
– Ian
May 21, 2018 at 13:24
• @user43712 Maybe somewhere in the same document in an obsure place we can find a definition for $\delta$, $\mu$ and $\Delta$. Look through the document, maybe they're mentioned somewhere else earlier or later. May 21, 2018 at 13:27