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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:

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

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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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    $\begingroup$ 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. $\endgroup$
    – Ian
    May 21, 2018 at 13:26
  • $\begingroup$ hmm that's along the lines of what I assumed too. Thanks for your help $\endgroup$
    – user43712
    May 21, 2018 at 13:27
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    $\begingroup$ 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. $\endgroup$
    – Ian
    May 21, 2018 at 13:29
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    $\begingroup$ 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. $\endgroup$
    – Ian
    May 21, 2018 at 13:31
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    $\begingroup$ 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. $\endgroup$
    – Ian
    May 21, 2018 at 13:43

3 Answers 3

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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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    $\begingroup$ 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. $\endgroup$
    – Ian
    May 21, 2018 at 15:14
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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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  • $\begingroup$ This seems accurate. Thanks @Ian $\endgroup$
    – user43712
    May 21, 2018 at 15:59
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$\Delta$ 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.

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    $\begingroup$ I don't think that is the $\Delta$ that is meant here... $\endgroup$
    – Ian
    May 21, 2018 at 13:14
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    $\begingroup$ @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. $\endgroup$
    – Daniel P
    May 21, 2018 at 13:19
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    $\begingroup$ @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. $\endgroup$
    – Daniel P
    May 21, 2018 at 13:19
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    $\begingroup$ I'm pretty sure these are each finite difference operators though it isn't apparent to me which is which. $\endgroup$
    – Ian
    May 21, 2018 at 13:24
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    $\begingroup$ @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. $\endgroup$
    – Daniel P
    May 21, 2018 at 13:27

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