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programming - How to define the Bernstein function like the built-in BernsteinBasis?


We know the Bernstein function defined as below:


Bn,i(u)=(ni)ui(1u)ni


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:


Bn,i(u)=n[Bn1,i1(u)Bn1,i(u)]



So I can utilize this equation to calculate the derivative of order k


B(k)n,i(u)=n(n1)(nk+1)[{Bnk,ik(u),,Bnk,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



enter image description here



Expand the expression


PiecewiseExpand[expr_] ^:=
expr /. Bernstein[n_, i_, u_Symbol] :>
Binomial[n, i] u^i (1 - u)^(n - i)



enter image description here





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}]



enter image description here



Obviously, the Mathematica utilizes a recursive method by observing the result.




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