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In studying the temporal evolution of a system according to the deterministic model, we begin by considering a Taylor series expansion for the displacement $r$. First, we consider a positive variation $+\Delta t$, then a negative variation $-\Delta t$

$$ \vec{r}(t+\Delta t) = \vec{r}|_{t} + \left( \vec{v}|_t \right) \Delta t + \dfrac{1}{2!} \left( \vec{a}|_t \right) \Delta t^2 + \dfrac{1}{3!} \left( \dfrac{\mathrm{d^3}}{\mathrm{d^3}t} \vec{r}|_{t} \right) \Delta t^3 + \dfrac{1}{4!} \left( \dfrac{\mathrm{d^4}}{\mathrm{d^4}t} \vec{r}|_{t} \right) \Delta t^4 $$

$$ \vec{r}(t-\Delta t) = \vec{r}|_{t} - \left( \vec{v}|_t \right) \Delta t + \dfrac{1}{2!} \left( \vec{a}|_t \right) \Delta t^2 - \dfrac{1}{3!} \left( \dfrac{\mathrm{d^3}}{\mathrm{d^3}t} \vec{r}|_{t} \right) \Delta t^3 + \dfrac{1}{4!} \left( \dfrac{\mathrm{d^4}}{\mathrm{d^4}t} \vec{r}|_{t} \right) \Delta t^4 $$

Subtracting member from member, we get

$$ \vec{r}(t+\Delta t) - \vec{r}(t-\Delta t) = 2\vec{v}|_t \Delta t + 2 \left[ \dfrac{1}{3!} \left( \dfrac{\mathrm{d^3}}{\mathrm{d^3}t} \vec{r}|_{t} \right) \Delta t^3 \right] $$

Dividing everything by 2

$$ \dfrac{\vec{r}(t+\Delta t) - \vec{r}(t-\Delta t)}{2} = \vec{v}|_t \Delta t + \dfrac{1}{3!} \left( \dfrac{\mathrm{d^3}}{\mathrm{d^3}t} \vec{r}|_{t} \right) \Delta t^3 $$

Dividing everything by $\Delta t$

$$ \dfrac{\vec{r}(t+\Delta t) - \vec{r}(t-\Delta t)}{2\Delta t} = \vec{v}|_t + \dfrac{1}{3!} \left( \dfrac{\mathrm{d^3}}{\mathrm{d^3}t} \vec{r}|_{t} \right) \Delta t^2 $$ So
$$ \vec{v}|_t = \dfrac{\vec{r}(t+\Delta t) - \vec{r}(t-\Delta t)}{2 \Delta t} - O(t^2) $$

where

$$ O(t^2) = \dfrac{1}{3!} \left( \dfrac{\mathrm{d^3}}{\mathrm{d^3}t} \vec{r}|_{t} \right) \Delta t^2 $$

Why does every text report $O(t^2)$ with positive sign?

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    $\begingroup$ I am not sure, but usually when one writes $O(t^2)$ it means that those terms don't really contribute to the calculation so one says that they just are there without much care to the signs. Also, you can just assume $O(t^2)$ has a minus sign so that it gets a plus in the expression for $v$ $\endgroup$ Commented Oct 11, 2024 at 16:59
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    $\begingroup$ It's called asymptotic notation, which represents a term that is roughly on the order of its argument. In this case, it's meant to express that it's negligible. It should be $O(\Delta t^2)$ BTW. $\endgroup$ Commented Oct 12, 2024 at 12:08
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    $\begingroup$ Doubt? Do you suspect the algorithm is wrong? $\endgroup$ Commented Oct 12, 2024 at 18:11
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    $\begingroup$ @MagneticMonopsone Could be that OP uses a variant of English (Indian English or something like that) where "doubt" is mostly synonymous with "question". $\endgroup$ Commented Oct 13, 2024 at 9:46
  • $\begingroup$ What a shame then, that there was no way to write a title that clearly, precisely and unambiguously describes the issue. $\endgroup$ Commented Oct 14, 2024 at 14:46

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In the $O$ notation, the sign doesn't matter. In your case, it should be evident by noticing that the real sign of the first neglected term depends on the sign of the third derivative of position (which may have both signs).

Notice that the algorithm you describe is the so-called position form of Verlet's. Unless there is a compelling reason for using such a form, the velocity form $$ \begin{align} \vec r(t+\Delta t) &= \vec r(t) + \Delta t ~~\vec v(t) + \frac{\Delta t^2}{2} \vec a(t) + O(\Delta t^3)\\ \vec v(t+\Delta t) &= \vec v(t) + \frac{\Delta t}{2} ~\left( \vec a(t) + \vec a(t + \Delta t) \right)+ O(\Delta t^3) \end{align} $$ is more convenient and provides a better local error for velocity.

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