Using Runge-Kutta method with measurements of acceleration (what to do with half-steps?) I'd like to perform a short-time motion estimation based on measurements from an Inertial Measurement Unit.
If I use the Runge-Kutta method, I will need to compute the k values at half-time steps (Ref).
Except that I don't have measurements at half time-steps! I don't think interpolating sounds right. Does it? Should I rather "double" my time-step?
 A: The Runge-Kutta method assumes that $\frac{dy}{dt}$ is given as some function $f(t,y)$ of both $t$ and $y$. But if you are capturing time-series data then you only have one value of $f$ for each value of $t$. In other words there is no dependence on $y$ (or at least, if there is, you have no data from which to estimate this dependence). In this case, where $f$ is a function of $t$ only, Rung-Kutta reduces to Simpson's rule.
A: Based on your answer, @gandalf61, the following is my derivation which seems reasonable, hopefully:
$$ a = \frac{dv}{dt}$$
$$v_{n+1}=v_{n} + \int_{t_n}^{t_{n+1}}a(t)dt$$
I sample the acceleration at uniform intervals. According to your answer, I can approximate that integral using Simpson's rule.
$$\int_{t_n}^{t_{n+1}}a(t)dt \approx \int_{t_n}^{t_{n+1}}P(t)dt,$$
where $P(t)$ is a polynomial that approximates $a(t)$. I will use as $P(t)$ the parabola that passes through the points $a_{n-1}$, $a_{n}$ and $a_{n+1}$. According to wikipedia's article on Simpson's rule, the polynomial is:
$$P(t)=a_{n-1}\frac{(t-m)(t-b)}{(a-b)(a-b)} + a_{n}\frac{(t-a)(t-b)}{(m-a)(m-b)} + a_{n+1}\frac{(t-a)(t-m)}{(b-a)(b-m)},$$
Here I'm using Wikipedia's notation where unsubscripted $a$, $m$ and $b$ are the first point, midpoint, and endpoints of the parabola ($a=t_{n-1}$, $m=t_{n}$ and $b=t_{n+1}$).

Without any loss of generality, I set $a=-1$, $m=0$ and $b=1$, which yields:
$$P(t) = \left[\frac{a_{-1}}{2}-a_{0}+\frac{a_{1}}{2}\right]t^2 + \left[-\frac{a_{-1}}{2}+\frac{a_{1}}{2}\right]t+a_{0}$$
Integrating this yields:
$$\int_{0}^{1}P(t)dt = -\frac{1}{12}a_{-1} + \frac{2}{3}a_{0} + \frac{5}{12}a_{1}$$
Rescaling, factoring and going back to the original notation:
$$v_{n+1} \approx v_{n} + \frac{1}{3}\left[-\frac{1}{4}a_{n-1} + 2a_{n} + \frac{5}{4}a_{n+1}\right]\delta t ,$$
where $\delta t$ is the sampling time. I can use the same expression when integrating position:
$$x_{n+1} \approx x_{n} + \frac{1}{3}\left[-\frac{1}{4}v_{n-1} + 2v_{n} + \frac{5}{4}v_{n+1}\right]\delta t ,$$
I'm posting this as an answer but I'd appreciate if its validated by someone.
