Culegere de probleme de Analiz˘a numeric˘a

116 Operatori liniari s¸i pozitivi
Bi,k(x) = ωi,k(x)Bi,k−1(x)+[1−ωi+1,k(x)]Bi+1,k−1(x)
ω0,2(x) = x−t0
= 0, ω0,1(x) = 0, ω1,2(x) = x, ω1,1(x) = 0
t2 −t0
ω2,2(x) = x
2 , ω2,1(x) = x, ω3,2(x) = x−1
, ω3,1(x) = x−1
2
ω4,2(x) = x−2, ω4,1(x) = x−2, ω5,2(x) = 0, ω5,1(x) = 0, ω6,1(x) = 0
B0,2(x) = (1−x)B1,1, B1,1(x) = (1−x)B2,0
B0,2(x) = (1−x) 2
2 (1−x) x ∈ [0,1)
B2,0(x) =
0 în rest
B1,2(x) = ω1,2B1,1 +(1−ω2,2)B2,1 = xB1,1 + 2−x
2 B2,1
B2,1(x) = ω2,1B0,2 +(1−ω3,1)B0,3 = xB2,0 +(2−x)B3,0
B1,2(x) = x(1−x)B2,0 + 2−x
2 xB2,0 + (2−x)2
B3,0
⎧
2
⎨ x
=
⎩
2− 3
2x x ∈ [0,1)
(x−2) 2
x ∈ [1,2) .
2
0 în rest
B2,2(x) = ω2,2B2,1 +(1−ω3,2)B3,1 = x
2 B2,1 + 3−x
2 B3,1
B3,1(x) = ω3,1B3,0 +(1−ω4,1)B4,0 = (x−1)B3,0 +(3−x)B4,0
B2,2 = x
2 xB2,0 + x(2−x)
B3,0 +
2
3−x
(x−1)B3,0 +
2
(3−x)2
⎧
2
⎪⎨
x
=
⎪⎩
2
x ∈ [0,1)
2
x(2−x) (3−x)(x−1)
+ x ∈ [1,2)
2 2
x ∈ [2,3)
(3−x) 2
2
B4,0 =
B3,2(x) = ω3,2B3,1 +(1−ω4,2);B4,1 = x−1
2 B3,1 +(3−x)B4,1
B4,1(x) = ω4,1B4,0 +(1−ω5,1);B5,0 = (x−2)B4,0
B3,2(x) = x−1
(x−1)B3,0 +
2
x−1
(3−x)B4,0 +(3−x)(x−2)B4,0 =
⎧
2
(x−1)
⎨
=
⎩
2
x ∈ [1,2)
2
(3−x)
x−1+2x−4 x ∈ [2,3)
2
0 în rest