We know the Bernstein function defined as below:
$$B_{n,i}(u)=\binom n i u^i(1-u)^{n-i}$$
And we define $B_{n,i}(u)=0$ when $i<0 $ or $i>n$
In addition, the first derivative of $B_{n,i}(u)$ has the below relationship:
$$B'_{n,i}(u)=n\left[B_{n-1,i-1}(u)-B_{n-1,i}(u)\right]$$
So I can utilize this equation to calculate the derivative of order $k$
$$B^{(k)}_{n,i}(u)=n(n-1)\cdots (n-k+1)\\ \left[\{B_{n-k,i-k}(u),\cdots,B_{n-k,i}(u)\} .coefficient\right]$$
Here,$coefficient$ has the style (1,-3,3,1),(1, -4, 6, -4, 1),etc
Implementation
Bernstein[n_, i_, u_] /; i < 0 || i > n := 0
Bernstein[0, 0, u_] := 1
Bernstein[n_, i_, u_?NumericQ] := Binomial[n, i] u^i (1 - u)^(n - i)
The derivative of Bernstein
D[Bernstein[n_, i_, u_], {u_, k_}] ^:=
Module[{coeff, body},
coeff = Times @@ Array[n - # &, k, 0];
body =
Array[Bernstein[n - k, #, u] &, k + 1, i - k].
CoefficientList[(1 - u)^k, u];
coeff* body
]
(*=======================================*)
D[Bernstein[n_, i_, u_], u_] ^:= D[Bernstein[n, i, u], {u, 1}]
However, the Mathematica give me the warining information
Expand the expression
PiecewiseExpand[expr_] ^:=
expr /. Bernstein[n_, i_, u_Symbol] :>
Binomial[n, i] u^i (1 - u)^(n - i)
Tesing
Successful case
Bernstein[3, 2, .4](*0.288*)
D[Bernstein[3, 2, u], u]
3 (Bernstein[2, 1, u] - Bernstein[2, 2, u])
D[Bernstein[3, 2, u], {u, 2}]
6 (Bernstein[1, 0, u] - 2 Bernstein[1, 1, u])
Failture
D[Bernstein[3, 2, u], u] // PiecewiseExpand
no expantion >_<
Question
- How to fix the warining information about
UpSetDelayed
? - Is it possible to implement the derivative of $B_{n,i}(u)$ by rule-based solution? I have a trial, but failed.
My trial
D[Berns[n_, i_, u_], {u_, k_}] ^:=
Do[
Bernstein[n, i, u] /.
Bernstein[x_, y_, z_] :>
x (Bernstein[x - 1, y - 1, u] - Bernstein[x - 1, y, z]), {k}]
(*failture*)
Built-in function
D[BernsteinBasis[6, 3, u], {u, 3}]
Obviously, the Mathematica utilizes a recursive method by observing the result.
Comments
Post a Comment