Computation of normalized first derivative in discrete case I am reading a paper on recognition of online handwritten characters. One of the features proposed in the paper is "normalized first derivatives, $(\hat{x}'_t,\hat{y}'_t)$", which they have defined as follows:

$$
\hat{x}'_t=\frac{x'_t}{\sqrt{(x'_t)^2+(y'_t)^2}}\quad\quad
\hat{y}'_t=\frac{y'_t}{\sqrt{(x'_t)^2+(y'_t)^2}}\,,
$$
where

$$
x'_t=\frac{\sum_{i=1}^2i(x_{t+i}-x_{t-i})}{2\sum_{i=1}^{2}i^2}\quad\quad
y'_t=\frac{\sum_{i=1}^2i(y_{t+i}-y_{t-i})}{2\sum_{i=1}^{2}i^2}$$
I have couple of doubts with respect to these definitions:


If the derivatives are to be normalized, what is the significance of the denominators in the definition of $x'_t$ and $y'_t$?
Why is the difference scaled by $i$ in the numerator? What I understand is that this scaling gives more weight to the points farther from the central point.
For the first part of your question, the significance is just to guarantee that the vector $(\hat{x}'_t,\hat{y}'_t)$ has length&nbsp;$1$. This means that it tells us in which direction $(x_t,y_t)$ is changing, but not by how much. For example, $(\hat{x}'_t,\hat{y}'_t)=(1,0)$ tells us that $x_t$&nbsp;is increasing and $y_t$&nbsp;is constant; $(\hat{x}'_t,\hat{y}'_t) = (-\sqrt{2}/2, \sqrt{2}/2)$ tells us that $x_t$&nbsp;is decreasing, and $y_t$&nbsp;is increasing at the same rate.

The second part isn't something I know about, so I'll leave that for somebody else.