We know the Bernstein function defined as below:
Bn,i(u)=(ni)ui(1−u)n−i
And we define Bn,i(u)=0 when i<0 or i>n
In addition, the first derivative of Bn,i(u) has the below relationship:
B′n,i(u)=n[Bn−1,i−1(u)−Bn−1,i(u)]
So I can utilize this equation to calculate the derivative of order k
B(k)n,i(u)=n(n−1)⋯(n−k+1)[{Bn−k,i−k(u),⋯,Bn−k,i(u)}.coefficient]
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 Bn,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.
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