Interesting how the Bernstein basis is identical to the binomial distribution. How are the two related intuitively? I can't seem to make sense of it...
motoole2
@billiam I found this interactive demo (also mentioned in the subsequent slide) provides an intuitive explanation of Bezier curves, and can be used to develop the Bernstein basis as well.
The Bezier curve can be generated using a relatively simple algorithm known as De Casteljau's algorithm. The goal is to compute a point on the Bezier curve for a specific value x \in [0,1]. This is done recursively, by computing a new set of "control points" linearly interpolated from the previous set. The recurrence relation looks as follows:
p^{k+1}_i = (1-x) * p^k_i + x * p^k_{i+1}
Here, p^0_i and p^0_{i+1} are the original control points, and this recurrence relation is repeated up until only one "control point" remains.
So how is this related to the Bernstein basis? Well, one can expand this expression out for different degree Bezier curves:
... and so on. The resulting expression is a linear combination of control points p_0 through p_n, and the weight associated with each control point is given by the Bernstein basis function above. I hope this helps to clear up where the Bernstein basis comes from!
billiam
Wow your explanation above made me understand Bezier curves in a new light, thanks!
My observation was from a probabilistic perspective. Say we have a random variable $X$ that follows the binomial distribution with parameters $n \in \mathbb{N}$ and $p \in [0, 1]$ (probability of success is $p$ and failure is $1 - p$). Then the probability of getting $k$ successes in $n$ tries is given by
$$\operatorname{Pr}(X = k) = {{n} \choose {k}} p^k (1 - p)^{n - k}$$
$\textbf{That's literally identical to the Bernstein basis polynomials.}$
Here's an interesting demonstration of the two in action: https://www.wolframcloud.com/objects/demonstrations/BinomialDistributionsAreBernsteinVectors-source.nb.
I'm just gonna chalk it up to a bizarre coincidence since I didn't find anything that links the two explicitly on the internet.
Interesting how the Bernstein basis is identical to the binomial distribution. How are the two related intuitively? I can't seem to make sense of it...
@billiam I found this interactive demo (also mentioned in the subsequent slide) provides an intuitive explanation of Bezier curves, and can be used to develop the Bernstein basis as well.
The Bezier curve can be generated using a relatively simple algorithm known as De Casteljau's algorithm. The goal is to compute a point on the Bezier curve for a specific value
x \in [0,1]. This is done recursively, by computing a new set of "control points" linearly interpolated from the previous set. The recurrence relation looks as follows:p^{k+1}_i = (1-x) * p^k_i + x * p^k_{i+1}Here,
p^0_iandp^0_{i+1}are the original control points, and this recurrence relation is repeated up until only one "control point" remains.So how is this related to the Bernstein basis? Well, one can expand this expression out for different degree Bezier curves:
n = 1: (1-x)*p_0 + x*p_1n = 2: (1-x)*((1-x)*p_0 + x*p_1) + x*((1-x)*p_1 + x*p_2) = (1-x)^2 * p_0 + 2 * (1-x)*x * p_1 + x^2 * p_2... and so on. The resulting expression is a linear combination of control points
p_0throughp_n, and the weight associated with each control point is given by the Bernstein basis function above. I hope this helps to clear up where the Bernstein basis comes from!Wow your explanation above made me understand Bezier curves in a new light, thanks!
My observation was from a probabilistic perspective. Say we have a random variable $X$ that follows the binomial distribution with parameters $n \in \mathbb{N}$ and $p \in [0, 1]$ (probability of success is $p$ and failure is $1 - p$). Then the probability of getting $k$ successes in $n$ tries is given by $$\operatorname{Pr}(X = k) = {{n} \choose {k}} p^k (1 - p)^{n - k}$$
$\textbf{That's literally identical to the Bernstein basis polynomials.}$
Here's an interesting demonstration of the two in action: https://www.wolframcloud.com/objects/demonstrations/BinomialDistributionsAreBernsteinVectors-source.nb.
I'm just gonna chalk it up to a bizarre coincidence since I didn't find anything that links the two explicitly on the internet.