数值分析第4版答案

导读:观察所给数据的特点,采用方程?by?aet,(a,b?0)bt两边同时取对数,则lny?lna?1?t?1t取??span?1,??,S?lny,x????则S?a*?b*x?022?11,122?0.062321,(?0,?1)??0.603975,(?0,f)??87.674095,(?1,f)?5.032489,则法方程组为11????0.603975

数值分析第4版答案

观察所给数据的特点,采用方程

?b

y?aet,(a,b?0)

b

t两边同时取对数,则 lny?lna? 1?

t?1t取??span?1,??,S?lny,x??

??

则S?a*?b*x

?022?11,122?0.062321,

(?0,?1)??0.603975,

(?0,f)??87.674095,(?1,f)?5.032489,

则法方程组为

11?

???0.603975?0.603975??a??*0.062321???b*???87.674095???? ??5.032489???从而解得

*??a??7.5587812 ?*??b?7.4961692

因此

a?ea

**?5.2151048 b?b?7.4961692

?y?5.2151048e?7.4961692

t

22。给出一张记录{fk}?(4,3,2,1,0,1,2,3),用FFT算法求{ck}的离散谱。 解:

{fk}?(4,3,2,1,0,1,2,3),

则k?0,1,?,7,N?8

????1,

????e

2615?04?4i,????e

????e

37??2 ??i,i?3?4,

23,用辗转相除法将R22(x)?解

R22(x)?

?3?

?3?

x?9

23x?6xx?6x?6223x?6xx?6x?622化为连分式。 12x?18x?6x?6123?x?

?322 ?3?12

x?4.50.75x?1.5

24。求f(x)?sinx在x?0处的(3,3)阶帕德逼近R33(x)。 解:

由f(x)?sinx在x?0处的泰勒展开为

x3

sinx?x?3!?x5

5!?x7

7!??

得C0?0,

C1?1,

C2?0,

1 C3????,3!61

C4?0,

C5?15!?1

120,

C6?0,

从而

?C1b3?C2b2?C3b1?C4?C2b3?C3b2?C4b1?C5 ?C3b3?C4b2?C5b1?C6即

?

?1

?

??0?

?

??1

??60?161???06b???3????1?0b???2??120??b1????01???120??0??? ???从而解得

?b3?0?1? b??220?

??b1?0

k?1

又?ak?

则 ?Cj?0jbk?j?Ck(k?0,1,2,3)

a0?C0?0

a1?C0b1?C1?0

a2?C0b2?C1b1?0a3?C0b3?C1b2?C2b1?C3??760 故

R33(x)?x??

1??760120a0?a1x?a2x?a3x1?b1x?b2x?b3xxx32323 260x?7x60?3x33

25。求f(x)?ex在x?0处的(2,1)阶帕德逼近R21(x)。 解: 由f(x)?ex在x?0处的泰勒展开为

x2

e?1?x?x

2!?x3

3!??

得 C0?1,C1?1,C2?C3?12!13!??1 ,21

6,从而 ?C2b1?C3 即 ?1

2b1?16 解得 b1??13

k?1又?ak?则 ?Cj?0jbk?j?Ck(k?0,1,2) a0?C0?1 a1?C0b1?C1?a2?C1b1?C2?2316

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