现代数值数学和计算习题答案(同济版)

时间:2025-07-13

1.

22

π

113

:

,

22

113=3.14159292035398···,π=3.14159265358979···.

,

22

355

:

22π

≤5×10 4

5×10 4

355π

≤9×10 8,

9×10 8

2.

100cm50cm,

0.15cm2.

:

xcm

2

50±

x

2 2

0.15,

x=0.002. x 2

,3.

x

x

√n

n.

:

Taylor

,

√n

x+

1

x

n2

n

ξ2

(x x)2,

1

x

ξ

x

x

√ξ

n

xx

=1

n 1

x

ξ

2

(x x)2,

4.MatLab

1.7977e+308,

:[1e 8,4e+3],[1e+308,1.5e+308],[ 1.4e+308,1.4e+308].(

)

a+b

:

2

a

2

(

b

[1.0e 308,1.01e 308]

)

5.

[a,b]

n

x0=a,

xk=xk 1+h,k=1,···,n

xk=a+kh,

k=0,···,n

h=b a

a,

xk+1=xk+h

8.888888888888888e-0011.777777777777778e+0002.666666666666667e+0003.555555555555555e+0004.444444444444445e+0005.333333333333334e+0006.222222222222223e+0007.111111111111113e+0008.000000000000002e+000

A

sort

A

MatLab

A

(

)

n=size(A,1);fori=1:n-1,

forj=i+1:n,

ifabs(A(i,i))<abs(A(j,j)),

A(:,[i,j])=A(:,[j,i]);A([i,j],:)=A([j,i],:);endendend

A

7.

x=(x1,x2,...,xn)

Euclide1

x =

nx2i

i=1

2

2

=

,

MatLab

x = x(1:n 1) ,x(n) ,

x=(x1,x2)

1

x1,

1,x2

functions=nrmv(x)x=abs(x(x~=0));n=length(x);ifn==0,

s=0;elseifn==1,

s=abs(x);else

scale=0;ssq=1;forj=1:n,

if(scale<x(j)),

ssq=1+ssq*(scale/x(j))^2;scale=x(j);else

ssq=ssq+(x(j)/scale)^2;endend

s=scale*sqrt(ssq);end

BLAS

dnrm2

http://www.77cn.com.cn/blas/dnrm2.f.

8.

ax2√

+bx+x=0

x= b±2a

x=2ca=0cb2= 0

4ac

.a

c10

6e+154

-4e+154

1

1

1

-4

-1e155

1e+155

5

MatLab

:

b2 4ac

1

c=0

b

(Vieta)

)

a=0

b=0

(

function[x,flag]=root2qe(a,b,c)

ifnargin==1,

c=a(3);b=a(2);a=a(1);end

ifa~=0,%Solve2-orderequation

M=max(abs([a,b,c]));%scalinga=a/M;b=b/M;c=c/M;s=sign(b);

ifs==0,s=1;end

x1=(-b-s*sqrt(b^2-4*a*c))/2/a;x2=c/(a*x1);%avoidlargeerrorx=[x1x2]’;

flag=’twosolutions’;elseifb~=0,%Solve1-orderequation

x=-c/b;

flag=’onesolution’;elseifc~=0,%nonzeroconstant

x=[];

flag=’noxisasolution’;else%equality0=0

x=[1];

flag=’allxaresolutions’;end

6

a

10

6e+154

1

1

-4

-1e155

c

-2.00000000000000

-4e+154

-1

1

2.00100000000007

1e+155

x2

0.33333333333333

0.0000100000

0.14005494464026

9.

x1=1

1 k4xk=

12,xk+1=2.25xk 0.5xk 1.

3

c14 2+c222=1

c1=4

3

1

4

3(

2p 1 52 ≤

fl(x)

x

0.64421768720.6518337710

2-9

cosx

sin0.705

cos0.702

:

sin0.705

L1(x)=

x 0.70

=0.76160838x+0.1110918212

8

0.70 0.71×0.6442176872

x=0.705

sin0.705≈L1(0.705)=0.6480257291.

cos0.702

L1(x)=

x 0.70

= 0.64803113x+1.2184639782

0.70 0.71×0.7648421872

x=0.702

2.

cos0.702≈L1(0.702)=0.76354612494.

(

2-10).

0.4

0.6

sinx

0.47943

0.64422

0.6 0.5

×0.56464+x 0.6(0.57891(0.5 00.+

.6)5)××(0(0..557891 0.7)

0.7)

×0.47943(0.7 0.5)×(0.7 0.6)×0.64422

0.54714.

9

x0=0.4,x1=0.5,x3=0.6,

sin0.57891

sin0.57891≈

=

L2(0.57891)

(0.57891 0.5)×(0.57891 0.6)(0.5 0.4)×(0.5 0.6)(0.57891 0.4)×(0.57891 0.5)

×0.47943

+

xf(x)

19

432-11

(1)(2)

f(4.2)

:

(1)

f(xk)1

8

9

14

23

-10

3

128

259

3

-2.75

-8

8.5

34.5

1.875

(2)

N(x)=1+8x+3x(x 1) 2.75x(x 1)(x 2)

+1.875x(x 1)(x 2)(x 4),

x=4.2

f(4.2)≈N(4.2)=4.696.

4.

p(x)=x4 x3+x2 x+1

-2p(x)

(

2-12).2

05

1

61

x

31

-112-13

111

3

:

2-12

2-13

p(x) q(x)

x=0,±1,±2

p(x) q(x)=Ax(x 1)(x+1)(x 2)(x+2),

x=3

A=31

120

x(x 1)(x+1)(x 2)(x+2).

5.

(

2-14)

-2y

4

16

2

-4

:

x= 1,0,1,2

L(x)=4+7(x+1) x(x+1) x(x 1)(x+1).

x= 2,3

y=L(x),

x= 2,3

y=L(x),

y=1, 4,

:

f(xk)-2

3

-1

7

5

1

-3

2

-17

3

-1-1-1

21!

k=0

,f(x)=e x,f(21)(x)= e x,ξ

20

(x xk),

2

|f(21)(ξ)|=|e ξ|<1.

12

x0,x1,···,x20

k=10,

max|(x xk)(x x20 k)|≤1;

x

k=10

,

|x x10|≤1.

|R20(x)|<

1

9,0≤t≤9.

R(x)=R(x0+th)=

f(10)(ξ)

x10

|f(10)(ξ)|<9!.

,

R¯(t) <

1

10·910

=

1

2

Mh.8

f(x)=sin(x),

h

13

1

2!

(x x0)(x x1),

x

x0

x1

,

1

|(x x0)(x x1)|≤

12!

8

.

f(x)=sin(x),

1

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