李庆扬数值分析第五版习题复习资料清华大学出版社

观察所给数据的特点，采用方程

y?ae,(a,b?0)

两边同时取对数，则

?btblny?lna?

t取??span?1,??,S?lny,x?? 则S?a?bx

**??1?t?1t?02?11,?12?0.062321,(?0,?1)??0.603975,(?0,f)??87.674095,(?1,f)?5.032489,则法方程组为

2211?0.603975??a*???87.674095??????b*????5.032489? ?0.6039750.062321??????从而解得

*??a??7.5587812 ?*??b?7.4961692因此

a?ea?5.2151048b?b?7.4961692?y?5.2151048e?**

7.4961692t

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,L,7,N?8

?0??4?1,????e2615?i4?,

????e????e

37?i2?3?i4???i,,45 / 65

k 0 1 2 3 4 5 6 7 xk 4 3 2 1 0 1 2 3 A1 4 4 4 2? 4 0 4 ?2?3 A2 8 4 0 4 8 22 0 ?22 Cj 16 4?22 0 4?22 0 4?22 0 4?22 3x2?6x23，用辗转相除法将R22(x)?2化为连分式。

x?6x?6解

3x2?6xR22(x)?2x?6x?612x?18?3?2x?6x?612?3? 39x??42x?32120.75?3??x?4.5x?1.524。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

x3x5x7sinx?x????L

3!5!7!得C0?0,

C1?1,C2?0,11

C3????,3!6C4?0,C5?11?, 5!12046 / 65

C6?0,

从而

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

??10?1??6??b??0????0?1??60??3??b??2???1?? ???01?b?120?1???1??6120????0???从而解得

??b3?0??b?1 ?220??b1?0k?1又Qak??Cjbk?j?Ck(k?0,1,2,3)j?0则

a0?C0?0a1?C0b1?C1?0a2?C0b2?C1b1?0

a73?C0b3?C1b2?C2b1?C3??60故

47 / 65

a0?a1x?a2x2?a3x3R33(x)?1?b1x?b2x2?b3x373x60?11?x22060x?7x3?60?3x3x?x

25。求f(x)?e在x?0处的(2,1)阶帕德逼近R21(x)。 解：

由f(x)?e在x?0处的泰勒展开为

xx2x3e?1?x???L

2!3!x得

C0?1,C1?1,11 ?,2!211C3??,3!6C2?从而

?C2b1?C3

即

11?b1? 26解得

1b1??

3又Qak?则

?Cbj?0k?1jk?j?Ck(k?0,1,2)

a0?C0?1

23 1a2?C1b1?C2?6a1?C0b1?C1?48 / 65