2014年1月计算方法期末考试题A

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上 海 海 事 大 学 试 卷

2013 — 2014 学年第一学期期末考试

《 计算方法 》（A 卷）

班级 学号 姓名 总分

1(18’)(a)Find all fixed points of 2x 39.0g(x)-=.

(b)To which of the fixed-points is Fixed-Point Iteration locally convergent, why?

(c)Compute 3 steps for the convergent fixed-point starting with the initial value 4.0x 0=, and find the convergent rate S.

2(18’)(a)The zero of the function A x x f -=3)( is the cubic root of a number A. Show that the Newton’s

Method to this function produce the iteration

21k 332x k

k x A

x +=

+ (b)Apply 3 steps of Newton iteration method to find the cubic root of 2 to 9 decimal places starting

with 10=x .

(c)Assume the error is 32-=i i x e , if after 3 steps the error 6310-=e , estimate 4e .

3(20’)(a)Find the LU factorization of the given matrices. Check by matrix multiplication.

⎥⎥

⎥⎥⎦

⎤⎢

⎢⎢

⎢⎣⎡=1-12044310120211-1A (b) Use the result of L and U to compute the second column of 1-A , then justify the result is right. (c) Compute the approximation number of operations involving the whole solution process.

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订

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4(14’) (a) Rearrange the equations to form a strictly diagonally dominant system. Apply 2 steps of the

Jacobi and Gauss-Seidel Methods from starting vector [0, 0, 0].

⎪⎩

⎪

⎨⎧-=+-=++=--2345128w v u w v u w v u (b) The exact solution of the system is ]49

57

,74,4961[--

=X , and the error 2-norm between the exact solution and the Jacobi second iteration is 3760878601.02

2

≈-Y X , estimate the same ones about

the Gauss-Seidel iteration. Is Gauss-Seidel method faster than Jacobi method?

5(20’)(a)Find the degree 2 interpolating polynomial )(2x P through the points

)0,(),1,2

(),0,0(ππ

.

(b) Calculate an approximation for )4

sin(π

.

(c) Using the interpolating error formula to give an error bound for the approximation in part (b). (d) Compare the actual error to your error bound.

6(10)Let )(x P is an polynomial and ⎰dx x P )( is the integral of )(x P . Write out the program with

MATLAB to evaluated the polynomial

⎰dx x P )(.(assume that the integral constant C=1)

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参考答案

1(18’) (a) since 2x 39.0g(x)-=, so 2x 39.0x -=,3.1,3.0x 21-==x

(b) -2x (x)g =',

convergent

,16.03.02(0.3)g <=⨯-=' divergent ,16.2)3.1(2(-1.3)g >=-⨯-=' (c) ,27636359.03371.039.0,

3371.023.039.0,23.04.039.04

.0232

2210=-==-==-==x x similarly

x x (d) S=0.6

2(18’) (a)

2

1

2

3232,32

32(x)f f(x)-x g(x)k

k k x x x so x

x +=+='=+

(b) ,

259933493.1263888889

,1,34

32321

3210=≈=+=

=x x sim ilarly

x x

(c) 12

2

3433107937.07937

.0)2(f 2)2(f (r)f 2(r)f M -⨯≈=∴≈'''='''=Me e

3(20’) (a)⎥⎥

⎥⎥⎦⎤⎢⎢⎢

⎢⎣⎡=10

10

01210010

0001L , ⎥⎥⎥⎥⎦

⎤⎢⎢⎢

⎢⎣⎡=10

021********

1

-1U , by checking, LU=A.