数值分析第五版习题总结(2)

xi xi 1xi 1 xi

11

(x x)i

1 xi21 xi 12

(xi 1 x)

各节点间中点处的Ih(x)与f(x)的值为 当x 4.5时，f(x) 0.0471,Ih(x) 0.0486 当x 3.5时，f(x) 0.0755,Ih(x) 0.0794 当x 2.5时，f(x) 0.1379,Ih(x) 0.1500 当x 1.5时，f(x) 0.3077,Ih(x) 0.3500 当x 0.5时，f(x) 0.8000,Ih(x) 0.7500 误差

h2

maxf(x) Ih(x) maxf ( ) xi x xi 18 5 x 51

2

1 x 2x

f (x) ,22

(1 x)

又 f(x)

6x2 2

f (x)

(1 x2)324x 24x3

f (x)

(1 x2)4

令f (x) 0

得f (x)的驻点为x1,2 1和x3 0

1

f (x1,2) ,f (x3) 2

2

1

maxf(x) Ih(x) 5 x 54

19．求f(x) x4在[a,b]上分段埃尔米特插值，并估计误差。 解：

在[a,b]区间上，x0 a,xn b,hi xi 1 xi,i 0,1, ,n 1, 令h maxhi

0 i n 1

f(x) x4,f (x) 4x3

函数f(x)在区间[xi,xi 1]上的分段埃尔米特插值函数为

Ih(x) ( (

x xi 12x xi

)(1 2)f(xi)xi xi 1xi 1 xi

x xi2x xi 1)(1 2)f(xi 1)xi 1 xixi xi 1

x xi 12 ()(x xi)f (xi)xi xi 1 (

x xi2

)(x xi 1)f (xi 1)xi 1 xi

xi4

3(x xi 1)2(hi 2x 2xi)hixi 14

3(x xi)2(hi 2x 2xi 1)hi

4xi

(x xi 1)2(x xi)2hi

3

4xi 13

2(x xi)2(x xi 1)hi

误差为

f(x) Ih(x)

1(4)

f( )(x xi)2(x xi 1)2 4!1h maxf(4)( )(i)424a x b2

又 f(x) x4

f(4)(x) 4! 24

hi4h4

maxf(x) Ih(x) max a x b0 i n 11616

试求三次样条插值，并满足条件：

(1)S (0.25) 1.0000,S (0.53) 0.6868;

(2)S (0.25) S (0.53) 0.

解：

h0 x1 x0 0.05h1 x2 x1 0.09h2 x3 x2 0.06h3 x4 x3 0.08 j 1

hj 1hj 1 hj

, j

hjhj 1 hj

533

, 2 , 3 , 4 11457

1

924

, 2 , 3 , 0 11457

f(x1) f(x0)

f x0,x1 0.9540

x1 x0f x1,x2 0.8533f x2,x3 0.7717f x3,x4 0.7150

(1)S (x0) 1.0000,S (x4) 0.6868d0

6

(f x1,x2 f0 ) 5.5200h0

f x1,x2 f x0,x1

4.3157

h0 h1

d1 6

f x2,x3 f x1,x2

d2 6 3.2640

h1 h2

f x3,x4 f x2,x3 d3 6 2.4300

h2 h3d4

6

(f4 f x3,x4 ) 2.1150h3

由此得矩阵形式的方程组为

2 1 M0 5.5200

59 2 M1 4.3157 1414

32

2 M2 3.2640

55

34

2 M3 2.4300

77

1 2 M4 2.1150

求解此方程组得

M0 2.0278,M1 1.4643

M2 1.0313,M3 0.8070,M4 0.6539

三次样条表达式为 S(x) Mj (yj

(xj 1 x)3

6hj

2

Mj 1

(x xj)3

6hjMj 1hj

6

2

Mjhj

6

)

xj 1 xhj

(yj 1

)

x xjhj

(j 0,1, ,n 1)

将M0,M1,M2,M3,M4代入得

6.7593(0.30 x)3 4.8810(x 0.25)3 10.0169(0.30 x) 10.9662(x 0.25)

x 0.25,0.30

2.7117(0.39 x)3 1.9098(x 0.30)3 6.1075(0.39 x) 6.9544(x 0.30) x 0.30,0.39 S(x) 33

2.8647(0.45 x) 2.2422(x 0.39) 10.4186(0.45 x) 10.9662(x 0.39) x 0.39,0.45

1.6817(0.53 x)3 1.3623(x 0.45)3 8.3958(0.53 x) 9.1087(x 0.45) x 0.45,0.53

(2)S (x0) 0,S (x4) 0

d0 2f0 0,d1 4.3157,d2 3.2640

d 2.4300,d

34 2f4 0

0 4 0

由此得矩阵开工的方程组为

M0 M4 0 290

14

32 2 M1 4.3157

55 M

2 3.2640 M3 2.4300

0372

求解此方程组，得

M0 0,M1 1.8809

M2 0.8616,M3 1.0304,M4 0

又 三次样条表达式为

S(x) M(xj 1 x)3

(x xj)3

j6h Mj 1

j

6hjM2

(yM2

jhj

j 1 xj 1hj

j

6

)

xh (yj 1

6

)

x x

jj

hj

将M0,M1,M2,M3,M4代入得

6.2697(x 0.25)3 10(0.3 x) 10.9697(x 0.25)

x 0.25,0.30

3.4831(0.39 x)3 1.5956(x 0.3)3 6.1138(0.39 x) 6.9518(x 0.30) x 0.30,0.39 S(x) 33

2.3933(0.45 x) 2.8622(x 0.39) 10.4186(0.45 x) 11.1903(x 0.39) x 0.39,0.45

2.1467(0.53 x)3 8.3987(0.53 x) 9.1(x 0.45) x 0.45,0.53

第三章 函数逼近与曲线拟合

2.当f(x) x时，求证Bn(f,x) x 证明：

若f(x) x，则

k

Bn(f,x) f()Pk(x)

nk 0

k n

xk(1 x)n kk 0n k n

kn(n 1) (n k 1)k

x(1 x)n k

k!k 0n

n

(n 1) [(n 1) (k 1) 1]k

x(1 x)n k

(k 1)!k 1

n

n

n 1 kn k

x(1 x)k 1 k 1 n

n 1 k 1(n 1) (k 1)

x x(1 x)k 1 k 1

n

x[x (1 x)]n 1 x

4。计算下列函数f(x)关于C[0,1]的f

,f1与f2：

(1)f(x) (x 1)3,x [0,1](2)f(x) x

1,2

(3)f(x) xm(1 x)n,m与n为正整数，

(4)f(x) (x 1)10e x

解：

(1)若f(x) (x 1)3,x [0,1]，则

f (x) 3(x 1)2 0

f(x) (x 1)3在(0,1)内单调递增

f

maxf(x)

0 x 1

max f(0),f(1) max 0,1

1f

maxf(x)

0 x 1

max f(0),f(1) max 0,1

1

f

2

( (1 x)dx)

1

6

12

1

172 [(1 x)]

07

1

,x 0,1 ，则

212

(2)若f(x) x

ff

maxf(x)

0 x 110

f(x)dx1

1

1 2

1(x )dx

22

1 4

f

21

( f(x)dx)

1

2

12

121

[ (x )dx]2

02

(3)若f(x) xm(1 x)n,m与n为正整数

当x 0,1 时,f(x) 0

f (x) mxm 1(1 x)n xmn(1 x)n 1( 1)

n mm 1n 1

x(1 x)m(1 x)

mm

)时,f (x) 0 n m

m

)内单调递减 f(x)在(0,

n mm

,1)时,f (x) 0 当x (

n m

m

,1)内单调递减。

f(x)在(

n m

当x (0,

x (

m

,1)f (x) 0n m

f maxf(x)

0 x 1

m

max f(0),f()

n m

mm

nn

(m n)m n

f

f(x)dx1

010

1

xm(1 x)ndx

2(sin2t)m(1 sin2t)ndsin2t