11(x)=(x – x0)(x – x1)(x – x2)······(x – x10)
11(x)
5/18
插值函数L10(x)取 切比雪夫结点插值
插值函数L10(x)取 等距结点插值
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
-5 -4 -3 -2 -1
0
1
2
3
4
5
2
1.5
1
0.5
0
-0.5

aaxx12

by1 by2

c c

z1 z2
ax3 by3 c z3
P(x, y)=l1(x, y)z1+l2(x, y)z2+l3(x,y)z3
(x，y)
l1(x, y) l2(x, y) l3(x, y)
(x1，y1) 1 0 0
(x2，y2) 0 1 0
x
x0
H(x)
y0
H’(x)
m0
x1
y1
m1
7/18
例2. 已知插值条件:
求3次插值函数.
x
0
1
H(x)
0
1
H’(x)
0
0
解:设 H ( x) a0 a1x a2 x2 a3 x3
得 a0=0, a1=0, 列出方程组 求解, 得
a2a2 2a33 a31 0
a2 = 3 , a3 = – 2 所以,有

F (4) ( ) f (4) ( ) C( x)(4!) 0

f (4) ( )
C(x)
4!
R( x) C( x)( x x0 )2 ( x x1 )2
f
(4) (
4!
)
[(
x

x0
)(
x

x1 )]2
11/18
分段线性插值
插值节点满足: x0<x1<······<xn 已知 yj=f (xj) ( j= 0,1,2,···,n)
2(n 1)
)
——切比雪夫结点
3/18
例1. 函数
1
f ( x) 1 x2 x∈[-5, 5]
取等距插值结点: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
f (x)
L10 ( x)
f
(11) ( n
11 !
)
11 (
x)
11(x)=(x+5)(x+4)(x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)(x-4)(x-5)
当 x∈[xj，xj+1]时, 两点Hermite插值
Hh(x)

(1
2
x xj x j1 x j
)(
x j1 x x j1 x j
)2
yj

(1
2
x j1 x x j1 x j
)(
x xj x j1 x j
)2
y j1

(x

xj
)(
x j1 x x j1 x j
-5 -4 -3 -2 -1
0
1
2
3
4
5
6/18
三次Hermite插值问题
已知节点x0和x1处的函数值及导数值
f ( x0 ) y0 f ( x1 ) y1 f ( x0 ) m0 f ( x1 ) m1 求三次插值函数
H(x)= a0 + a1x + a2x2 + a3x3 满足插值条件 H ( x j ) y j H ( x j ) m j (j = 0,1)
l4(u，v)= (1 – u) v
16/18
[u,v]=meshgrid(0:0.1:1)； L1=(1-u).*(1-v); surf(u,v,L1) figure L2=u.*(1-v); surf(u,v,L2) figure L3=u.*v; surf(u,v,Lu3) figure L4=(1-u).*v; surf(u,v,L4)
18/18
11(x)
4/18
在[-5, 5]区间上,取11个切比雪夫结点
(2k 1)
xk 5cos( 22 )
( k=10, 9, 8, ···, 1, 0 )
-4.9491 -4.5482 -3.7787 -2.7032 -1.4087 0.0000 1.4087
2.7032 3.7787 4.5482 4.9491
证明: 由插值条件知
R(x0)=R’(x0)=0, R取(xx1)异=R于’x(0x和1)=x01, 设
R( x) C( x)( x x0 )2 ( x x1 )2
利用 f(x) – H(x)=C(x)(x – x0)2(x – x1)2
构造辅助函数
F (t ) f (t ) H (t ) C( x)(t x0 )2 (t x1 )2
)2 m j

(x

x j1 )(
x xj x j1 x j
)2 m j1
( j= 0,1,2,···,n-1)
15/18
矩形区域上函数f(x, y) y2
的双线性插值
y1
P(x, y) = ax + by + cxy + d
x1
x2
插值条件: P(x1, y1) = z1, P(x2, y1) = z2, P(x2, y2) = z3, P(x1, y2) = z4
x∈[xj，xj+1]时, 线性插值函数
Lh( x)
x j1 x x j1 x j
yj
x xj x j1 x j
y j1
( j= 0,1,···,n-1)
······ ····
12/18
分段三次Hermite插值
取 a≤x0<x1<······<xn≤b,已知函数值和导数值 yj=f (xj), mj = f ’(xj) ( j= 0,1,2,···,n)
选取: x0, x1 ,······, xn , 使
max
a xb
|
n1
(
x)
|
min
结论: 切比雪夫多项式Tn+1(x)的全部零点。
2/18
f(x)∈C[–1, 1], 令 x = cos , 则有 [–1, 1] [0, ]
将g( ) = f(cos )展开成余弦级数
P(x, y) = z1(1 – u)(1 – v)+ z2 u(1 – v) + z3 u v + z4 (1 – u)v
其中
u x x1 x2 x1
v y y1 y2 y1
l1(u，v)= (1 – u)(1 – v)
l2(u，v)= u(1 – v)
l3(u，v)= u v
0(x)

(1

2
x x0 x1 x0
)(
x1 x x1 x0
)2
0(
x)

(
x

x0
)(
x1 x x1 x0
)2 .
1(
x)

(1

2
x1 x x1 x0
)(
x x1
x0 x0
)2
1(
x)

(
x

x1
)(
x x1
x0 x0
)2
x x0 x1
x x0 x1
g(
)

1 2
a0

n1
an
cos
n
n+1阶切比雪夫多项式: Tn+1=cos(n+1)
cos = x 代入得 Tn+1( x ) = cos((n+1) arccos x )
取 (n 1)arccos x (2k 1) ( k=0,1,···,n )
2
即
xk
cosBiblioteka (2k 1)10/18显然,F(t)有三个零点x0, x, x1,由Roll定理知,存
在F’(t)的两个零点t0,t1满足x0<t0<t1<x1,而x0和 x1也是F’(x)的零点,故F’(x)有四个相异零
反点复. 应用Roll定理,得F(4)(t)有一个零点设为ξ
F (t ) f (t ) H (t ) C( x)(t x0 )2 (t x1 )2
H(x) = 3x2– 2x3 = (3 – 2x)x2
1 0.8 0.6 0.4 0.2
0 -0.2 -0.4 -0.6
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
8/18
利用基函数表示Hermite插值
H ( x) y00 ( x) y11( x) m00 ( x) m11( x)
x=a sinφ cosθ y=a sinφ sinθ z = b cos φ
三角形区域线性插值
(x3, y3)
P(x, y)=a x + b y + c
(x1, y1)
(x2, y2)
插值条件: z1= P(x1, y1) z2 = P(x2, y2)
z3 = P(x3, y3 ) 拉格朗日方法
(x3，y3) 0
0
1
17/18
精品课件！
精品课件！