数值分析第4版答案

导读:第四章数值积分与数值微分,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????0.60397

数值分析第4版答案

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?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?7.4961692t

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,?,7,N?8

????1,????e2615?04?4i,????e????e

37??2i ??i,i?3?4,k 0 1 2 3 4 5 6 7 xk 4 3 2 1 0 1 2 3 3A1 4 4 4 2? 4 0 4 ?2? A2 8 4 0 4 8 22 0 ?22 Cj 16 4?22 0 4?22 0 4?22 0 4?22

23,用辗转相除法将R22(x)?解

R22(x)??3??3?x?923x?6xx?6x?6223x?6xx?6x?622化为连分式。

12x?18x?6x?6123?4x??322

?3?12x?4.50.75x?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

x3sinx?x?3!?x55!?x77!??

得C0?0, C1?1,C2?0,1

C3????,3!61C4?0,C5?15!?1120,

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?0k?1又?ak?则

?Cj?0jbk?j?Ck(k?0,1,2,3)

a0?C0?0a1?C0b1?C1?0a2?C0b2?C1b1?0a3?C0b3?C1b2?C2b1?C3??760

R33(x)?x??1??760120a0?a1x?a2x?a3x1?b1x?b2x?b3xxx32323

260x?7x60?3x3325。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

x2e?1?x?x2!?x33!??

得 C0?1,C1?1,C2?C3?12!13!??1 ,216,从而

?C2b1?C3

?12b1?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

R21(x)?1??23a0?a1x?a2x1?b1x16x22x?13x2

1??6?4x?x6?2x

(2)?S??(x)?f??(x)?S??(x)?dxab??baS??(x)d?f?(x)?S?(x)?ba??S??(x)?f?(x)?S?(x)???f?(x)?S?(x)?d[S??(x)]ab?S??(b)?f?(b)?S?(b)??S??(a)?f?(a)?S?(a)???S??(b)?f?(b)?S?(b)??S??(a)?f?(a)?S?(a)???S??(b)?f?(b)?S?(b)??S??(a)?f?(a)?S?(a)???S??(b)?f?(b)?S?(b)??S??(a)?f?(a)?S?(a)??baS???(x)?f?(x)?S?(x)?dxS???(xk?xk?12xk?xk?12)??xk?1xk

n?1?k?0n?1?f?(x)?S?(x)?dxxk?1xk?k?0S???()??f?(x)?S?(x)?

第四章 数值积分与数值微分

1.确定下列求积公式中的特定参数,使其代数精度尽量高,并指明所构造出的求积公式所具有的代数精度: (1)?h?h2h?2h1?1hf(x)dx?A?1f(?h)?A0f(0)?A1f(h);f(x)dx?A?1f(?h)?A0f(0)?A1f(h);(2)?(3)?

f(x)dx?[f(?1)?2f(x1)?3f(x2)]/3;2(4)?f(x)dx?h[f(0)?f(h)]/2?ah[f?(0)?f?(h)];0解:

求解求积公式的代数精度时,应根据代数精度的定义,即求积公式对于次数不超过m的多项式均能准确地成立,但对于m+1次多项式就不准确成立,进行验证性求解。 (1)若(1)?h?hf(x)dx?A?1f(?h)?A0f(0)?A1f(h)

令f(x)?1,则

2h?A?1?A0?A1

令f(x)?x,则

0??A?1h?A1h

令f(x)?x2,则

23h?hA?1?hA1

322从而解得 4?A??03h?1??A1?h

3?1?A?h??13?令f(x)?x3,则

?h?hf(x)dx??h?hxdx?0

3A?1f(?h)?A0f(0)?A1f(h)?0

故?h?hf(x)dx?A?1f(?h)?A0f(0)?A1f(h)成立。

令f(x)?x4,则

h?hh?h4?f(x)dx??xdx?25h5A?1f(?h)?A0f(0)?A1f(h)?23

h5故此时,

?h?hf(x)dx?A?1f(?h)?A0f(0)?A1f(h)

h?h故?f(x)dx?A?1f(?h)?A0f(0)?A1f(h)

具有3次代数精度。 (2)若?2h?2hf(x)dx?A?1f(?h)?A0f(0)?A1f(h)

令f(x)?1,则

4h?A?1?A0?A1

令f(x)?x,则

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