Is a rolling $z$-score a good proxy for a derivative? Given a time series $\{V_t\}_{t=1}^{n}$, where $t \mapsto V_t \in \mathbb{R}$, I want to have some smoothed notion of the "derivative" of this time series. It was recommended that I look at 
$$\tilde{V}_t = \frac{V_t - \text{$k$-step moving average of } V_t}{\text{$k$-step std dev of $V_t$}}.$$ 
The $k$-step moving average of $V_t$ is $\frac{1}{k} \sum_{i=0}^{k-1}V_{t-i}$.
This quantity, empirically, seems to act something like derivative. (For a line it is constant, for a sine curve it is almost a cosine curve, etc.) Can someone explain why, mathematically, this would be a heuristic for a derivative?
Thanks.
 A: This formula is not a good analogue of the derivative. Consider
$$V_t = \alpha t$$
for some $\alpha \in \Bbb R$, and take $k=3$, for example. Then the $k$-step average is $\alpha(t-1)$ and the $k$-step standard deviation is $|\alpha|$, so we have
$$\tilde{V}_t = \frac{\alpha t - \alpha(t-1)}{|\alpha|} = \pm 1$$
depending on the sign of $\alpha$. In other words, the result is independent of $\alpha$: we see whether the sequence is increasing or decreasing, but not by how much, whereas the point of the derivative is to get the rate of change.
Worse yet, for a sequence like $V_t = t^2$, the rolling score is positive and very close to $1$ but actually decreases over time. The same holds for other powers of $t$ and other values of $k$. You mentioned that the score for a sine wave gives something like a cosine wave, but it actually gives something much closer to $\operatorname{sgn}(\cos t)$.
As a more reasonable alternative, why not do a polynomial interpolation and take the derivative of that?
