$B(n, t) = \displaystyle \sum_{i=0}^n {n \choose i} (1-t)^{n-i}t^iP_i$ for points $Pi$, then $$\begin{align} B^{'}(n, t) &= \sum{i=0}^n {n \choose i}\left[i(1-t)^{n-i}t^{i-1} -(n-i)(1-t)^{n-i-1}t^i \right]Pi \ &= \underbrace{\sum{i=0}^n {n \choose i}i(1-t)^{n-i}t^{i-1}Pi}{1^{st}\text{ term vanishes}} - \underbrace{\sum_{i=0}^n {n \choose i}(n-i)(1-t)^{n-i-1}t^iPi}{n^{th}\text{ term vanishes}}\

&= \sum_{i=1}^n {n \choose i}i(1-t)^{n-i}t^{i-1}Pi - \sum{i=0}^{n-1} {n \choose i}(n-i)(1-t)^{n-i-1}t^iPi\ &= \sum{i=0}^{n-1} {n \choose i+1}(i+1)(1-t)^{n-i-1}t^iP{i+1} - \sum{i=0}^{n-1} {n \choose i}(n-i)(1-t)^{n-i-1}t^iPi\ &= n\sum{i=0}^{n-1} {n-1 \choose i}(1-t)^{(n-1)-i}t^iP{i+1} - n\sum{i=0}^{n-1} {n-1 \choose i}(1-t)^{(n-1)-i}t^iPi\ &= n\sum{i=0}^{n-1} {n-1 \choose i}(1-t)^{(n-1)-i}t^i(P_{i+1} - P_i) \end{align}$$

which is equal to $nB(n-1, t)$ with points $\left(P_{i+1} - P_i\right)$

$B(n, t) = \displaystyle \sum_{i=0}^n {n \choose i} (1-t)^{n-i}t^iP_i$ for points $Pi$, then $$\begin{align} B^{'}(n, t) &= \sum{i=0}^n {n \choose i}\left[i(1-t)^{n-i}t^{i-1} -(n-i)(1-t)^{n-i-1}t^i \right]Pi \ &= \underbrace{\sum{i=0}^n {n \choose i}i(1-t)^{n-i}t^{i-1}Pi}{1^{st}\text{ term vanishes}} - \underbrace{\sum_{i=0}^n {n \choose i}(n-i)(1-t)^{n-i-1}t^iPi}{n^{th}\text{ term vanishes}}\

&= \sum_{i=1}^n {n \choose i}i(1-t)^{n-i}t^{i-1}Pi - \sum{i=0}^{n-1} {n \choose i}(n-i)(1-t)^{n-i-1}t^iPi\ &= \sum{i=0}^{n-1} {n \choose i+1}(i+1)(1-t)^{n-i-1}t^iP{i+1} - \sum{i=0}^{n-1} {n \choose i}(n-i)(1-t)^{n-i-1}t^iPi\ &= n\sum{i=0}^{n-1} {n-1 \choose i}(1-t)^{(n-1)-i}t^iP{i+1} - n\sum{i=0}^{n-1} {n-1 \choose i}(1-t)^{(n-1)-i}t^iPi\ &= n\sum{i=0}^{n-1} {n-1 \choose i}(1-t)^{(n-1)-i}t^i(P_{i+1} - P_i) \end{align}$$

which is equal to $nB(n-1, t)$ with points $\left(P_{i+1} - P_i\right)$

a bezier curve of 3rd order.

a bezier curve of 2th order

An important thing to bear in mind is that this algorithm is intended to run just 1 time by t value, it return control points for two sub curves cutted off at t percent.

Below are listed just the coefficients, refer to wikipedia page.

Also bezier curves are just convex combinations of points.

$$B^{n}(t) = P_x = \sum_{i=0}^n {n \choose i} (1-t)^{n-i}t^iP_i$$

where $P_i$ are the control points and the extreme points.

Just the previous annotation, with the only difference that here, the guy is splitting the point by coordinates, and my equation use the entire point.

This is rather amaizing, what the plot below suggest is that. if we center every data with zero mean. we can with high porbability get groups that perhaps not perfectly as the plot below, we'll get groups the like in the plot. Then assuming all of these groups share the same mean(which imply that by the plot, they can be traited as random samples, which mean that the share the same mean, by the best estimator for the mean). This implies that the properties listed below meet automatically if we apart from assuming that the mean is a linear function of x, we as well assume that these groups are random samples.