matlab ex ercise 1 matrices &...

MATLAB Exercise 5 School of Mathematical Sciences Xiamen University http://gdjpkc.xmu.edu.cn

Ex51

MATLAB Exercise 5 – Symbol Computation 1. Compare the following commands, and show the data types and values of a1, a2, a3 and a4.

>> a1=1/4+1/6 >> a2=sym(1/4+1/6) >> a3=’1/4+1/6’ >> a4=’1/4+1/6’; eval(a4)

2. Input following commands, try to compare and analysis the results. 1) a) >> clear; c1= a+2*a; c1 2) a) >> clear; c3=’ a+2*a’; eval(c3)

b) >> clear; c4=’ a+2*a’; a=3; eval(c4) 3) a) >> clear; syms a; c5=’ a+2*a’; eval(c5)

b) >> clear; syms a; c6= a+2*a; eval(c6) c) >> clear; syms a; c7=sym('a+2*a'); subs(c7, a, 2) d) >> clear; syms a; c8=sym('a+2*a'); subs(c8, 2)

4) a) >> clear; c9=sym(‘a+2*a+b’); c9 b) >> clear; c10=sym('a+2*a+b'); subs(c10, 1) c) >> clear; c11=sym(‘a+2*a+b’); subs(c11, a, 1) d) >> clear; syms a; c12=sym('a+2*a+b'); subs(c12,a,1) e) >> clear; syms a b; c13=sym('a+2*a+b'); subs(c13,a,1, b,2) f) >> clear; syms a b; c14=sym('a+2*a+b'); subs(c14, [a,b], [1, sym('pi')])

3. Let 3 2 6 11 6 f x x x = − + − , ( 1)( 2)( 3) g x x x = − − − , [ ( 6) 11] 6 h x x x = − + − . Please

use function factor, horner, expand to prove the following conclusion: 1) f is the expanded form of g and h 2) g is the factor form of f 3) h is the nested form of f

4. Please simply the following functions by using simple, simplify and pretty respectively

1) 3 3 2 1 6 12 8 x x x

+ + + 2) 2 cos sin x x + −

3) 1 1 1 1 x x

+ + −

4) ln c a b e +

5) sin 2 sin cos

(1 cos 4 )(1 cos )(1 cos ) α α α

α α α + + − 6) cos 4 4cos 2 3 α α − +

5. Let s express the symbolic expression sin( ) y a x e + . Use subs function to compute the

following expressions, and simply the results.

1) ln sin( ) t a x e + 2) sin( ) 1 3

a π +

6. Set 2 2 2 3 p a xy xy x y = + + + . Please use collect function to


Ex52

1) collects all the coefficients with the same power of x 2) views p as a polynomial in its symbolic variable y

7. Use randn function to generate a random 2×2 matrix A. Substitute a matrix A into the

symbolic expression 2 3 2 5 x x − + .

8. Set symbolic expression x f e − = . Please compute the value of

1) (0) f

2) ( ) f eps

3) (1), (2), (3),..., (20) f f f f

9. Generate two symbolic matrices _ , _ a b u v

s a s b c d s t

= =

, compute

1) s_a+s_b 2) 2*s_a 3) s_a*s_b 4) s_a.*s_b 5) s_a\s_b 6) s_a/s_b 7) s_c=s_a+i*s_b 8) s_c’ 9) s_c .’ 10) det(s_a) 11) inv(s_a)

10. Let a be 2×2 matrix, b be a symbolic matrix 1

b x

=

. For the following coefficient

matrices a, solve the linear equations ax=b.

1) 1 2 3 4

a =

2) 1 2 3 6

a =

3) 1 2

a =

4) 1 2 3

a c

=

5)

c d a

e f

=

6) 1 2 3 2 3 4

a =

11. For the following matrix a, compute the eigenvalues and eigenvectors of a.

1) numeric matrix 1 2 3 4

a =

2) symbolic matrix 1 2 3 4

a =

3) symbolic matrix 1 3 4

x a =

4) symbolic matrix

s t a

u v

=

12. For the matrices as Ex11, compute their lu, schur, svd decompositions respectively.

Key to MATLAB Ex 1 - Matrices & Arrays.pdf

Key to MATLAB Exercise 1 School of Mathematical Sciences Xiamen University http://gdjpkc.xmu.edu.

Key to Ex11

Key to MATLAB Exersise 1 Matrices & Arrays 1. >> x0=10; v0=15; t=5; a=9.81; x=x0+v0*t+1/2*a*t^2 2. >> x=3;v=4;

1) >> log(x^2+v^2) 2) >> (x3)^(1/2)/(x2*v)^2 3) >> 4/3*pi*v^2 4) >> abs(sin(2*x))*exp(v) 5) >> (x5)^(1/2) or >> sqrt(x5) 6) >> x/(v4) Warning: Divide by zero. ans =Inf 7) >> x/(v4) Warning: Divide by zero. ans =Inf 8) >> eps 9) >> (x3)/(v4) Warning: Divide by zero. ans = NaN

3. ans = 2 ans =0.5000

4. 1) >> d=[23:3:2] >> numel(d,5) or >> d=[23,20,17,14,11,8,5,2] >> numel(d,5)

2) >> a=[1 2 3 4 2 4 6 8 3 6 9 12]；

or >> a = [1:4; 2:2:8; 3:3:12]; or >> a1 =[1:4]; a = [a1; 2*a1; 3*a1];

5. 1) >> size(a) 2) >> a(2,3)

>> b=a(:,[1,3]) 3) >> c=a([1,3,2],:) 4) >> x=a(:,end) 5) >> a(1,1)=0 6) omited 7) omited 8) >> a35=a; a35(:,5)=[1;1;1] or >> aa35=[a, [10; 20; 30]]

>> a44=a; a44(4,:)=[1 1 1 1] or >> a44=(a; 1 1 1 1)

9)14) omited


Key to Ex12

6. >> x=A\B ans x= 0.5000 1.0000 0.5000 >> A/B ??? Error using ==> mrdivide matrix dimensions must agree. A/B is equivalent to the expression 1 AB − , but here B is a vector, there is no invert of B.

7. 1) >> x=ones(6,1); A*x ans = 0

0 0 0 0 0

From A*x, we get the sum of every row of A, so we know that the determinant of A equal to 0, that is A must be singular. 2) omited

8. >> A = magic (8); sum(A(:,1:7)) ans =260 260 260 260 260 260 260

9. >> a=[1 2 1]; A=[a;a*2;a*3]; C=[1 11]'; D=[2 2 4]'; 1) >> A*C >> A*D The results show that though A≠0, AC=AD, but C is not equal to D. Therefore the proposition is not true. 2) >> A=round(10*rand(3)); B=round(10*rand(3)); >> inv(A+B) >> inv(A)+inv(B) The proposition is not true. 3) >> (A+B)^2 >> A^2+2*A*B+B^2 The proposition is not true. 4) >> A=round(10*rand(3)); B=A+A’; >> B == B’ ans = 1 1 1 1 1 1 1 1 1 We may try it more times. All results show the proposition is true.

10. omited

Key to MATLAB Ex 2 - Solving Linear Systems of Equations.pdf


Key to Ex21

Key to MATLAB Exercise 2 Solving Linear Systems of Equations

1. >> A=round(10*rand(5)) >> B=round(20*rand(5))10 a) >> det(A) ans =5972

>> det(A') ans = 5972 Yes b) >> det(A+B) ans = 36495

>> det(A)+det(B) ans =26384 No c) >> det(A*B) ans =121900464

>> det(A)*det(B) ans =121900464 Yes d) >> det(A')*det(B') ans =121900464

>> det(A'*B') ans =121900464 Yes e) >> det(inv(A)) ans = 1.6745e004

>> inv(det(A)) ans =1.6745e004 Yes f) >> det(A*inv(B)) ans =0.2926

>> det(A)*inv(det(B)) ans =0.2926 Yes 2. >> A=round(10*rand(6)).

a) >> A=round(10*rand(6)); B=A;B(2,:)=A(1,:);B(1,:)=A(2,:); >> det(A) ans = 4636 >> det(B) ans = 4636 Interchanging two rows of a matrix changes the sign of the determinant.

b) >> C=A; C(3, :)=4*A(3, :) >> det(C) ans =18544 >> det(A)*4 ans = 18544 Multiplying a single row of a matrix by a scalar has the effect of multiplying the value of the determinant by that scalar.

c) >> D=A; D(5, :)=A(5, :)+2*A(4, :) >> det(A) ans =4636 >> det(D) ans=4636 Adding a multiple of one row to another does not change the value of the determinant.

3. a) >> A=[1 2; 2 2]; b=[4;2]; x=A\b

x = 2 1>> A*x ans = 4 2 >> det(A) ans =6 Or>> A=[1 2; 2 2]; b=[4;2]; x=inv(A)*b Or


Key to Ex22

>> A=[1 2; 2 2]; b=[4;2]; c=[A b]; d=rref(c); x=d(:, end) Same does below. Omited.

b) >> A=[2 3; 5 1]; b=[1;4]; x=A\b x = 1 1 >> A*x ans = 1 4>> det(a) ans = 13

c) >> A=[4 2 1;3 1 2;11 3 0]; b=[2;10;8]; x=A\b Warning: Matrix is close to singular or badly scaled.

Results may be inaccurate. RCOND = 5.139921e018. x = 1.0e+016 * 0.4053 1.4862 1.3511

>> A*x ans =

0 8 8

>> det(a) ans =0 (it isn’t the solution, because the det(a)=0) d) >> A=[1 3 1; 2 1 1; 2 2 1]; b=[1;5;8]; x=A\b

x = 21 2

>> A*x ans =

1 58

>> det(A) ans =

3 4.

>> A=[1 2 3; 2 3 4; 5 4 6]; b=[0;0;1]; >> format rational >> A1=det([b A(:, 2:3)]); A2=det([A(:, 1) b A(:, 3)]); A3=det([A(:, 1:2) b]); ADet=det(A); >> x=[A1/ADet; A2/ADet; A3/ADet]


Key to Ex23

x = 1/3 2/3 1/3

5. >> AdjOfA=det(A)*inv(A) AdjOfA =

2 0 1 8 9 2 7 6 1

6. a) >> A=[2 1 3; 4 5 1; 2 1 4]; b=[0;8;2]; >> x=A\b x =

42 2

>> c=null(A,'r') c =

Empty matrix: 3by0 so x is the solution >> rref([A b])

42 2

b) >> A=[2 1 3; 4 5 1; 2 4 4]; b=[0;8;8]; >> c=null(A,'r') c =

8/3 7/3 1

>> x0=A\b Warning: Matrix is singular to working precision. x0 =

4/3 8/3 0

>> syms k >> x=x0+c*k x =

4/3+8/3*k1 8/37/3*k1

k1 >> d=rref([A b])


Key to Ex24

d= 1 0 8/3 4/3 0 1 7/3 8/3 0 0 0 0

>> x=d(:, end)+d(: end1)*k >> x=d(:, end)+d(:, end1)*k1 x =

4/38/3*k 8/3+7/3*k

0 7.

>> format short >> A=[2 1 3; 4 5 1; 2 2 10]; d=rref(A) d =

1.0000 0 2.6667 0 1.0000 2.3333 0 0 0

>> i=rank(A) i=

2 >> x=[d(:, i+1); 1] x =

2.6667 2.3333 1.0000

>> A*x ans = 1.0e014 * 0.0444 0.1776

0 8. Omitted 9. Omitted 10.

>> A=[1 2 1; 2 4 2; 2 1 1]; B=[3 1 2; 1 2 2; 3 1 4]; C=A+i*B 1).C =

1.0000 + 3.0000i 2.0000 + 1.0000i 1.0000 + 2.0000i 2.0000 + 1.0000i 4.0000 + 2.0000i 2.0000 + 2.0000i 2.0000 + 3.0000i 1.0000 + 1.0000i 1.0000 + 4.0000i

>> C' ans =

1.0000 3.0000i 2.0000 1.0000i 2.0000 3.0000i 2.0000 1.0000i 4.0000 2.0000i 1.0000 1.0000i 1.0000 2.0000i 2.0000 2.0000i 1.0000 4.0000i


Key to Ex25

>> C.' ans =

1.0000 + 3.0000i 2.0000 + 1.0000i 2.0000 + 3.0000i 2.0000 + 1.0000i 4.0000 + 2.0000i 1.0000 + 1.0000i 1.0000 + 2.0000i 2.0000 + 2.0000i 1.0000 + 4.0000i

The first matrix is the complex conjugate transpose of C while the second one is the transpose of C. Because the elements of C are complex, the results are different.

2). >> A A =

1 2 1 2 4 2 2 1 1

>> A' ans =

1 2 2 2 4 1 1 2 1

>> A.' ans =

1 2 2 2 4 1 1 2 1

The first matrix is equal to the second one. 3). >> inv(A) Warning: Matrix is singular to working precision. ans =

Inf Inf Inf Inf Inf Inf Inf Inf Inf

>> pinv(A) ans =

0.0727 0.1455 0.6364 0.1273 0.2545 0.3636 0.0182 0.0364 0.0909

The first matrix is not equal to the second one. 4). >> inv(B) ans =

0.6000 0.2000 0.2000 0.2000 0.6000 0.4000 0.5000 0 0.5000

>> pinv(B) ans =

0.6000 0.2000 0.2000 0.2000 0.6000 0.4000


Key to Ex26

0.5000 0.0000 0.5000 Because B is not singular, inv(B) is equal to pinv(B).

11. Omitted

Key to MATLAB Ex 3 - Linear Space.pdf


Key to Ex31

Key to MATLAB Exercise 3 – Linear Space 1.

1) >> a1=[1;0;0]; a2=[0;1;0]; a3=[0;0;1]; A=[a1,a2,a3] A =

1 0 0 0 1 0 0 0 1

>> rank(A) ans = 3 {a1,a2,a3} is spanning set for 3 1 R × , because the rank of A equals to 3, therefore. Or>> det(A) ans =

1 {a1,a2,a3} is spanning set for 3 1 R × , for the determinant of A is not zero. 2) >> a1=[1;0;0]; a2=[0;1;1]; a3=[1;0;1]; a4=[1;2;3]; A=[a1,a2,a3,a4]; rref(A) ans =

1 0 0 0 0 1 0 2 0 0 1 1

The result shows that {a1 a2 a3} is spanning set for 3 1 R × . Or>> A1=A(:, 1:3) A1 =

1 0 1 0 1 0 0 1 1

>> rank(A1) ans =

3 >> rank(A) ans =

3 The result shows that {a1 a2 a3} is spanning set for 3 1 R × . 3) >> a1=[2;1;2]; a2=[3;2;2]; a3=[2;2;0]; A=[a1,a2,a3] A =

2 3 2 1 2 2 2 2 0

>> rref(A)


Key to Ex32

ans = 1 0 2 0 1 2 0 0 0

They are not spanning sets for 3 1 R × , because there are less than 3 independent vectors. Or>> det(A) ans =

0 They are not spanning sets for 3 1 R × , because the matrix A, which is formed by the corresponding vectors, is singular. 4) >> A=[2 1 4; 1 1 2; 2 2 4 ]; >> rref(A)

ans = 1 1 2 0 0 0 0 0 0

Or>> rank(A)

ans = 1

They are not spanning sets for 3 1 R × . 2.

1) >> x1=[1;2;3]; x2=[3;4;2]; x=[2;6;6]; y=[9;2;5]; >> A1=[x1,x2,x]; >> rref(A1) ans =

1 0 0 0 1 0 0 0 1

1 2 Span( , ) x x x ∉ , for the vectors x1, x2, x are independent.

Or>> rank(A1) ans =

3 >> rank([x1, x2]) ans =

2

1 2 Span( , ) x x x ∉ , for the rank of { x1, x2} is not equal to that of {x1, x2, x}.

2)


Key to Ex33

>> rref([x1,x2,y]) ans =

1 0 3 0 1 2 0 0 0

Result implied that 1 2 Span( , ) y x x ∈ .

3. 1) >> x1=[1; 0; 0]; x2=[0; 0; 1]; x3=[1; 0; 1]; A=[x1, x2, x3]; r=rank(A); >> m2=size(A, 2); % calculate the column size of A >> if r==m2

disp (‘They are spanning sets for P3’); elsedisp (‘They are not spanning sets for P3’) end

It is not spanning set for P3. 2) >> x1=[1; 0; 0]; x2=[2; 1; 0]; x3=[0; 1; 0]; x4=[1; 0; 1]; rank([x1, x2, x3, x4]) ans=

3

{ } 2 2, 2, , 2 1 x x x − + is spanning set for P3. 3) >> x1=[1; 1; 0]; x2=[2; 1; 0]; x3=[3; 0; 1]; rref([x1, x2, x3]) ans =

1 0 0 0 1 0 0 0 1

It means that { } 2 1, 2, 3 x x x + − + is spanning set for P3. 4.

1) >> a1=[1 1 1]'; a2=[0 1 1]'; a3=[1 0 1]'; rank([a1, a2, a3]) ans =

3 They are linearly independent. 2) >> a1=[2 1 2]'; a2=[2 2 0]'; A=[a1, a2]; >> rank(A)==size(A, 2) %check whether the rank of A is equal to the column size of A. ans =

1 % 1 means “true”, 0 means “false” They are linearly independent. 3)


Key to Ex34

>> a1=[2 1 2]'; a2=[2 1 2]'; a3=[4 2 4]'; A=[a1,a2,a3]; rank(A)==size(A, 2) ans =

0 They are linearly dependent.

5. For example 2) >> x1=[1; 0; 0]; x2=[2; 1; 0]; x3=[0; 1; 0]; x4=[1; 0; 1]; rref([x1, x2, x3, x4]) ans =

1 0 2 0 0 1 1 0 0 0 0 1

Span{ } 2 2, 2, , 2 1 x x x − + =Span{ } 2 2, 2, 2 1 x x − + . 6.

>> x1=[1 2 3]'; x2=[3 4 2]'; x3=[0 10 11]'; 1) >> rank([x1,x2,x3]) ans =

2 Or>> rank(x1, x2, x3)==size([x1, x2, x3], 2) ans =

0

1 2 3 , , x x x are linearly dependent.

2) >> rank([x1 ,x2]) ans =

2

1 x and 2 x are linearly independent.

3)

The rank of vector set { 1 2 3 , , x x x } shows the dimension is 2.

4) They are the 2dimension subspace of 3 1 R × .

7. >> x1=[1 2 3]'; x2=[3 4 2]'; x3=[0 10 11]'; x4=[2 7 3]'; x5=[1 3 2]'; rref([x1,x2,x3,x4,x5]) ans =

1.0000 0 3.0000 0 0.4194 0 1.0000 1.0000 0 0.6452 0 0 0 1.0000 0.6774

{x1, x2, x4} forms a basis for R 3 . 8.


Key to Ex35

>> x1=[3;2]; x2=[1;3]; A=[x1 x2]; x=[10;7]; c=inv(A)*x c =

3.2857 0.1429

9. >> x1=[2; 4]; x2=[1; 3]; B=[x1, x2]; >> Transition_matrix=B Transition_matrix =

2 1 4 3

>> x=[10; 7]; >> coordinates =inv(Transition_matrix) *x coordinates =

11.5000 13.0000

10. >> X=[x1, x2]; Y=[y1, y2]; >> Transition_matrix =X\Y Transition_matrix =

2.5000 2.5000 4.0000 2.0000

11. 1) >> A=[1 2 3;2 3 6;9 4 6] A =

1 2 3 2 3 6 9 4 6

>> x1=A(1,:); x2=A(2,:); x3=A(3,:); >> y1=A(:,1); y2=A(:,2); y3=A(:,3); >> rref(A) ans =

1 0 0 0 1 0 0 0 1

So a basis for the row space is {x1, x2, x3}, a basis for the column space is {y1, y2, y3}. The nullspace is {0}. 2) >> x1=A(1,:); x2=A(2,:); x3=A(3,:); y1=A(:,1); y2=A(:,2); y3=A(:,3); y4=A(:,4); >> rref(A) ans =

1 0 1 0 0 1 1 0 0 0 0 1


Key to Ex36

So a basis for the row space is {x1, x2, x3}; a basis for the column space is {y1, y2, y4} and a basis for the null space is [1 1 1 0]’. 3) >> A=[1 2 1 0;3 2 5 6;2 0 6 6 ]; >> x1=A(1,:); x2=A(2,:); x3=A(3,:); y1=A(:,1); y2=A(:,2); y3=A(:,3); y4=A(:,4); >> rref(A) ans =

1.0000 0 0 0.7500 0 1.0000 0 0 0 0 1.0000 0.7500

So a basis for the row space is {x1, x2, x3}; a basis for the column space is {y1, y2, y3}. Notice that [ 0.7500 0 0.7500 1]’ is a solution of Ax=0 and rank(A)=3, then a basis for the null space is [ 0.7500 0 0.7500 1]’.

12. ² >> A10=rand(10)；

>> b10=rand(10, 1)； >> tic; >> x1=inv(A10)*b10; >> toc Elapsed time is 28.611000 seconds.

² >> tic; >> [l,u]=lu(A10); >> y1=inv(l)*b10; >> x2=inv(u)*y1; >> toc Elapsed time is 50.012000 seconds.

² >> tic; >> [q,r]=qr(A10); >> y1=q'*b; >> y1=q'*b10; >> x1=inv(r)*y1; >> toc Elapsed time is 105.983000 seconds.

Key to MATLAB Ex 4 - Eigenvalue.pdf

Key to MATLAB Exercise 4 School of Mathematical Sciences Xiamen University http://gdjpkc.xmu.edu.cn

Key to Ex41

Key to MATLAB Exercise 4 – Eigenvalue 1.

1) >> A=[3 2;3 2]; [v,d]=eig(A) v =

0.8944 0.3162 0.4472 0.9487

d = 4 0 0 3

>> det(v) ans =

0.9899 It isn't defective, because the determinant of v is not zero. Key to 2) to 9) are omitted.

2． >> A=[2 0 0;0 4 0;1 0 2]; B=[2 0 0;0 4 0;3 6 2]; >> eig(A) ans =

2 2 4

>> eig(B) ans =

2 2 4

Notice As 2 is the repeated eigenvalue, we can’t judge whether A is similar to B only by their eigenvalues. That is, we may say A is similar to B if A and B have the same Jordan normal forms. >> [XA, JordanA] = jordan(A) XA =

0 0 1 1 0 0 0 1 0

JordanA = 4 0 0 0 2 1 0 0 2

>> [XB, JordanB] = jordan(B) XB =

0 0 2 1 0 0


Key to Ex42

3 6 3 JordanB =

4 0 0 0 2 1 0 0 2

A is similar to B because JordanA is equal to JordanB. Or>> [VA, DA]=eig(A) VA =

0 0.0000 0 0 0 1.0000

1.0000 1.0000 0 DA =

2 0 0 0 2 0 0 0 4

>> [VB, DB]=eig(B) VB =

0 0.0000 0 0 0 0.3162

1.0000 1.0000 0.9487 DB =

2 0 0 0 2 0 0 0 4

>> rank(VA)==rank(VB) ans =

1 2, 2, 4 are the eigenvalues of A and B, and the rank of VA equals to that of VB, which means

that A has the same Jordan normal form as B. Therefore A is similar to B. 3．

>> A=rand(4); >> P=poly(A) P = 1.0000 1.5430 0.0898 0.0310 0.1878 >> poly2sym(P) ans = x^41543/1000*x^3+449/5000*x^231/1000*x+939/5000 >> roots(P) ans =

1.4313 0.5970 0.2427 + 0.4011i 0.2427 0.4011i


Key to Ex43

It means that the characteristic polynomial is 4 3 2 1.543 0.0898 0.031 0.1878 x x x x − + − +

and the eigenvalues (or characteristic roots ) of A are 1.4313, 0.5970. 0.2427+0.4011i and 0.2427 0.4011i. 4．

1) >> A=[0 1;1 0]; [X1, D]=eig(A) X1=

0.7071 0.7071 0.7071 0.7071

D = 1 0 0 1

>> rank(X1)==size(X1, 1) ans =

1

The answer is 0.7071 0.7071

1 0.7071 0.7071

X −

=

, 1 0 0 1

D −

=

. Here X1 is an invertible matrix.

We may check X1 by other functions such as det, rref and so on. Or>> [X2 S]=schur(A) X2 =

0.7071 0.7071 0.7071 0.7071

S = 1 0 0 1

>> X2*S*inv(X2) ans =

0 1.0000 1.0000 0

Therefore X2 is the answer, too. The key to 2) to 6) is omitted.

5． 1) >> A=[2 1;2 1]；[v, d]=eig(A)；B= v*sqrt(d)* inv(v) B =

1.5119 0.3780 0.7559 1.1339

>> B*B ans =

2.0000 1.0000 2.0000 1.0000


Key to Ex44

The key to 2) is omitted. 6．

>> format rational; A=[0 1 1;1 2 0;1 0 3]; >> [U, T]=schur(A) U = 755/833 1661/6691 1573/4601 1573/4601 755/833 1661/6691 1661/6691 1573/4601 755/833 T = 655/1006 0 0

0 2665/1172 0 0 0 2874/851

7． >> A=[1 1;2 3;1 0]; >> [u, s, v]=svd(A) u =

0.3583 0.2312 0.9045 0.9208 0.2474 0.3015 0.1541 0.9409 0.3015

s = 3.9090 0

0 0.8485 0 0

v = 0.6022 0.7983 0.7983 0.6022

>> SingularValuesOfA = diag(s) SingularValuesOfA =

3.9090 0.8485 >> RootOfAA=roots(poly(AA)) RootOfAA =

15.2801 0.7199

>> abs(SingularValuesOfA.^2RootOfAA) > A=[4 3 12; 17 11 0; 1 12 3]; PolyA=poly(A); roots(PolyA) ans =


Key to Ex45

16.8781 2.4391 +10.6951i 2.4391 10.6951i

>> eig(A) ans = 16.8781

2.4391 +10.6951i 2.4391 10.6951i

>> schur(A) ans = 16.8781 2.1204 12.4708

0 2.4391 7.7919 0 14.6799 2.4391

>> svd(A) ans =

21.8740 14.6390

6.3427 >> svd(A)^2 ans = 478.4705 214.3003 40.2293

>> eig(A'*A) ans =

40.2293 214.3003 478.4705

The above show that we may use functions eig and roots to calculate the eigenvalues of a matrix. The result of function schur in real, that is, it will not show the eigenvalues of A if there are complex eigenvalues. And svd provides the values which are the square roots of the eigenvalues of A’A. 9．

>> A1=rand(4); B1=A1+A1'; % generate a symmetric matrix B1 >> A2=10*rand(4); B2=A2*A2'; % generate a symmetric matrix B2 We omitted the following for it is similar to that of Ex10.

10. >> P=fix(10*ran(3)); [Q r]=qr(P); >> D1=diag([0, 1, 2]); D2=diag(1:3); Jordan1=[1 0 0; 0 2 1; 0 0 2]; >> SingularMatrix=P*D1; >> NotSingularMatrix=P*D2; >> DiagonalMatrix=P’ *P; % any symmetric matrix is diagonal. >> NotDiagonalMatrix=Q’*Jordan1*Q;

11. Omitted

Key to MATLAB Ex 5 - Symbol Computation.pdf


Key to Ex51

Key to MATLAB Exercise 5 – Symbol Computation 1.

>> a1=1/4+1/6 % double >> a2=sym(1/4+1/6) % sym >> a3=’1/4+1/6’ % char >> a4=’1/4+1/6’; eval(a4) % char

2. 1) a)

>> clear; c1=a+2*a; c1 ??? Undefined function or variable 'a'. b) >> clear; c2=‘a+2*a’; c2 c2 = a+2*a Conclusion. It is illegal in the statement of “c1=a+2*a” of a) where ‘a’ is not defined before we assign data to variable c1. But “c2=a+2*a” of b) is right, due to here ‘a’ being a character.

2) a) >> clear; c3='a+2*a'; eval(c3) ??? Error using ==> eval Undefined function or variable 'a'. b) >> clear; c4=’ a+2*a’; a=3; eval(c4) ans =

9 3) a)

>> clear; syms a; c5=’ a+2*a’; eval(c5) ans = 3*a b) >> clear; syms a; c6= a+2*a; eval(c6) ans = 3*a c) >> clear; syms a; c7=sym('a+2*a'); subs(c7, a, 2) ans =

6 d) >> clear; syms a; c8=sym('a+2*a'); subs(c8, 2) ans =

6 4) a)

>> clear; c9=sym(‘a+2*a+b’); c9


Key to Ex52

c8 = a+2*a+b b) >> clear; c10=sym('a+2*a+b'); subs(c10, 1) ans = 3*a+1 c) >> clear; c11=sym(‘a+2*a+b’); subs(c11, a, 1) ??? Undefined function or variable 'a'. d) >> clear; syms a; c12=sym('a+2*a+b'); subs(c12,a,1) ans = 3+b e) >> clear; syms a b; c13=sym('a+2*a+b'); subs(c13,a,1, b,2) ??? Error using ==> sym.subs Too many input arguments. f) ans = 3+pi

3. 1)

>> syms x; f=x^36*x^2+11*x6; g=(x1)*(x2)*(x3); h=x*[x*(x6)+11]6; >> expand(g) ans = x^36*x^2+11*x6 >> expand(h) ans = x^36*x^2+11*x6

2) >> factor(f) ans = (x1)*(x2)*(x3)

3) >> horner(f) ans = x*(x*(x6)+11)6

4. 1) >> syms x; f=[x^(3)+6*x^(2)+12*x^(1)+8]^(1/3) f = (1/x^3+6/x^2+12/x+8)^(1/3) >> simple(f) simplify:


Key to Ex53

((2*x+1)^3/x^3)^(1/3) radsimp: (2*x+1)/x combine(trig): ((1+6*x+12*x^2+8*x^3)/x^3)^(1/3) factor: ((2*x+1)^3/x^3)^(1/3) expand: (1/x^3+6/x^2+12/x+8)^(1/3) combine: (1/x^3+6/x^2+12/x+8)^(1/3) convert(exp): (1/x^3+6/x^2+12/x+8)^(1/3) convert(sincos): (1/x^3+6/x^2+12/x+8)^(1/3) convert(tan): (1/x^3+6/x^2+12/x+8)^(1/3) collect(x): (1/x^3+6/x^2+12/x+8)^(1/3) mwcos2sin: (1/x^3+6/x^2+12/x+8)^(1/3) ans = (2*x+1)/x >> simplify(f) ans = ((2*x+1)^3/x^3)^(1/3) >> pretty(f)

/ 1 6 12 \1/3 | + + + 8| | 3 2 x | \ x x /

Omitted 2) to 6). 5.

>>clear; syms a x y; s=a*sin(x)+exp(y) s = a*sin(x)+exp(y)

1) >> syms t >> subs(s, y, log(t)) ans = a*sin(x)+ t

2) >> subs(s,{x,y},{sym('pi/3'),0})


Key to Ex54

ans = 1/2*a*3^(1/2)+1

6. 1) >> clear; syms x y; p=a+x*y+2*x*y^2+3*x^2*y p = a+x*y+2*x*y^2+3*x^2*y >> collect(p, x) ans = 3*x^2*y+(y+2*y^2)*x+a

2) >> collect(a+x*y+2*x*y^2+3*x^2*y, y) ans = 2*x*y^2+(x+3*x^2)*y+a

7. >> A=randn(2); syms x; s=3*x^2+52*x; >> polyvalm(sym2poly(s), A) ans =

5.2187 4.6712 8.7431 13.0285

Notice: subs(s, A) cann’t be used here, because it equals to 3.*A.*A+5.*[1, 1; 1, 1]2.*A. 8.

>> clear; syms x; f=exp(x) f = exp(x)

1) >> subs(f, 0) ans =

1 2) >> subs(f, eps) ans =

1.0000 3) >> a=[1:1:20]; format short; subs(f, a) ans = Columns 1 through 9 0.3679 0.1353 0.0498 0.0183 0.0067 0.0025 0.0009

0.0003 0.0001 Columns 10 through 18 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 0.0000 Columns 19 through 20 0.0000 0.0000


Key to Ex55

9. >> syms a b c d u v s t; s_a=[a b;c d]; s_b=[u v;s t];

1) >> s_a+s_b ans = [ a+u, b+v] [ c+s, d+t]

2) >> 2 * s_a ans = [ 2*a, 2*b] [ 2*c, 2*d]

3) >> s_a * s_b ans = [ a*u+b*s, a*v+b*t] [ c*u+d*s, c*v+d*t]

4) >> s_a .* s_b ans = [ a*u, b*v] [ c*s, d*t]

5) >> s_a\s_b ans = [ (b*s+u*d)/(a*dc*b), (b*tv*d)/(a*dc*b)] [ 1/(a*dc*b)*(c*ua*s), (a*tc*v)/(a*dc*b)]

6) >> s_a/s_b ans = [ (s*b+a*t)/(u*tv*s), (v*a+u*b)/(u*tv*s)] [ (s*dc*t)/(u*tv*s), (u*dv*c)/(u*tv*s)]

7) >> s_c=s_a+i*s_b s_c = [ a+i*u, b+i*v] [ c+i*s, d+i*t]

8) >> s_c' ans = [ conj(a+i*u), conj(c+i*s)] [ conj(b+i*v), conj(d+i*t)

9) >> s_c.'


Key to Ex56

ans = [ a+i*u, c+i*s] [ b+i*v, d+i*t]

10) >> det(s_a) ans = a*dc*b

11) >> inv(s_a) ans = [ d/(a*dc*b), b/(a*dc*b)] [ c/(a*dc*b), a/(a*dc*b)]

10. 1) >> syms x ; b=[1;x]; >> a=[1 2; 3 4]; syms y; y=inv(a)*b y =

2+x 3/21/2*x

Omitted 2) to 6). 11.

1) >> a=[1 2;3 4]; [v,d]=eig(a) v =

0.8246 0.4160 0.5658 0.9094

d = 0.3723 0

0 5.3723 2) >> a=[sym(1) sym(2); sym(3) sym(4)] a = [ 1, 2] [ 3, 4] >> [v, d]=eig(a) ans = v = [ 1, 1] [ 3/4+1/4*33^(1/2), 3/41/4*33^(1/2)] d = [ 5/2+1/2*33^(1/2), 0] [ 0, 5/21/2*33^(1/2)]

3) >> syms x ; a=[sym(1) sym(x);sym(3) sym(4)]


Key to Ex57

a = [ 1, x] [ 3, 4] >> [v, d]=eig(a) v = [ 1/2+1/6*(9+12*x)^(1/2), 1/21/6*(9+12*x)^(1/2)] [ 1, 1] d = [ 5/2+1/2*(9+12*x)^(1/2), 0] [ 0, 5/21/2*(9+12*x)^(1/2)]

4) >> syms s t u v; a=[s t;u v]; [v,d]=eig(a) v = [ (1/2*s+1/2*v1/2*(s^22*s*v+v^2+4*t*u)^(1/2))/u, (1/2*s+1/2*v+1/2*(s^22*s*v+v^2+4*t*u)^(1/2))/u] [ 1, 1] d = [ 1/2*s+1/2*v+1/2*(s^22*s*v+v^2+4*t*u)^(1/2), 0] [ 0, 1/2*s+1/2*v1/2*(s^22*s*v+v^2+4*t*u)^(1/2)]

12. For example 2) >> [u,t]=schur(a) ??? Function 'schur' is not defined for values of class 'sym'. Error in ==> schur at 30 [varargout{1:nargout}] = builtin('schur', varargin{:});

>> [l,u]=lu(a) ??? Function 'lu' is not defined for values of class 'sym'. Error in ==> lu at 54 [varargout{1:nargout}] = builtin('lu', varargin{:});

>> [u,v,t]=svd(a) u = [ .40455358483375693164244872262782, .91451429567730445267917697381021] [ .91451429567730445267917697381021, .40455358483375693164244872262780] v = [ 5.4649857042190426504511884932842, 0] [ 0, .36596619062625782042296438426142] t =[ .57604843676632079133109858194273, .81741556047036327308865238846391] [ .81741556047036327308865238846391, .57604843676632079133109858194273]

MATLAB Ex 1 - Matrices & Arrays.pdf

MATLAB Exercise 1 School of Mathematical Sciences Xiamen University http://gdjpkc.xmu.edu

Ex11

MATLAB Exercise 1 Matrices & Arrays 1. The distance traveled by a ball falling in the air is given by the equation

2 0 0

1 2

x x v t at = + +

Use MATLAB to calculate the position of the ball at time t=5s if 0 10 x m = , 0 15 / v m s = , and

2 9.81 / sec a m = −

2. Suppose that 3 x = , 4 v = . Use MATLAB to evaluate the following expression:

1) 2 2 log( ) x v + 2) 2 3

( 2 ) x

x v −

− 3) 2

4 3 v π 4) | sin 2 | v x e

5) 5 x − 6) 4 x

v − 7)

4 x

v − −

8) eps 9) 3 4

x v

− −

3. 4/2 4\2

4. 1) Try to input a vector d = [23, 20, 17, 14, 11, 8, 5, 2] in different ways. Use numel to count the number of elements in d.

2) Please input matrix

1 2 3 4 2 4 6 8 3 6 9 12

a =

at least in three different methods.

5. 1) Compute the size of a. 1) Show the value of a(2,3)? Obtain a subarray b which is composed by the 1 st and 3 rd

columns of a. 2) Obtain an new matrix c by exchanging the 2 nd and 3 rd rows of a. 3) Obtain a vector x which is the last column of a. (end ) 4) Replace the value of a(1,1) with 0. 5) Input

a ( 10 ) a ( 10 ) = 20 a ( 10 )

and observe what happen. 6) Input

a ( 2, : ) a ( :, 3 ) a (:, :) a (:, 2 : 3) a (:) a (2 : 3)

7) Add a new column to a to form a 3×5 matrix a35, add a new row to a as its 1 st row to form a 4×4 matrix a44.

8) Input a ( 20 ) = 100; a and observe what happen.


Ex12

9) Input d = a’; d and compare the value of a with d. 10) Input e = a ([1 3]; : ) 11) *Input help format to study “format”

Display the value of pi in short / long format. Display 0.5 in short e / rat format

12) *In help windows seach fprintf to study “fprintf” Display the value of pi as an integer / exponential format / at the second new line

13) Input a 3×5 matrix newmatrix, compare the results after the following commands a .* newmatrix a * newmatirx a .+ newmatrix a + newmatrix a ^ newmatrix a .^newmatrix

6. Try to solve for x in the equation Ax = B where

1 2 1 1 2 3 2 , 1 1 0 1 0

A B = = −

. Compare the

command A\B and A/B, try to explain the result. 7. Set A = round (10*rand(6)). Let us change the 6 th column A so as to make the matrix

singular (奇异). B = A’; A (:, 6) = sum (B (1 : 5, : ) )’ 1) Set x = ones (6,1) and compute A*x. Explain why do we know A must be singular. 2) Set C = round ( 10*rand (6 ) ). Check whether AC=CA, why?

8. Set A = magic (8). Compute its sum of elements on each row, column and diagonal respectively.

9. Examine the following statements by MATLAB, and show whether they are true. 1) If A≠0 and AC = AD, then C = D;

2) If A and B are nonsingular (可逆), then 1 1 1 ( ) A B A B − − − + = +

3) For any 3×3 matrices A and B, 2 2 2 ( ) 2 A B A AB B + = + +

4) A+A’ is a symmetric matrix. (对称) 10. * Some test

test1 = linspace (1, 11, 6) test2 = linspace (1, 11, 5) num_test1 = numel (test1) [ i ] = find (test1 > 4) test1 ( find (test1 ==3)) = 0 test = [test1; 1:2:11] test (:, 2) = [ ] Magic = magic (5) Magic_diag = diag (Magic, 0) Magic_diag = diag (Magic, 1) Magic_diag = diag (Magic, 1)


Ex13

help clc clc help clear clear test test1 clc

MATLAB Ex 2 - Solving Linear Systems of Equations.pdf

MATLAB Exercise 2 School of Mathematical Sciences Xiamen University http://gdjpkc.xmu.edu.

Ex21

MATLAB Exercise 2 Solving Linear Systems of Equations 1. Generate random 5×5 matrices with integer entries(元素) by setting

A=round(10*rand(5)) and B=round(20*rand(5))10 Use MATLAB to compute each of the following pairs of numbers. In each case check whether the first is equal to the second. a) |A| |A’| b) |A+B| |A|+|B| c) |AB| |A||B| d) |A’B’| |A’||B’|

e) 1 | | A − 1| | A

f) 1 | | AB − | | | | A B

2. Set A=round(10*rand(6)). In each of the following, use MATLAB to compute a second matrix as indicated. State how the second matrix is related to A and compute the determinants of both matrices. How are the determinants related? a) B=A; B(2, :) = A(1, :); B(1, :)=A(2, :) b) C=A; C(3, :)=4*A(3, :) c) D=A; D(5, :)=A(5, :)+2*A(4, :)

3. Use Cramer’s rule (det), backslash (\) operation, inv function, rref function respectively to solve the following systems. And check whether the results are the exact solutions of the corresponding equations. If not, try to show the reason.

a) 1 2

1 2

2 4 2 2 2 x x x x

+ = − =

b) 1 2

1 2

2 3 1 5 4 x x x x + = −

+ =

c) 1 2 3

1 2 3

1 2

4 2 2 3 2 10 11 3 8

x x x x x x x x

+ − = − + = + =

d) 1 2 3

1 2 3

1 2 3

3 1 2 5 2 2 8

x x x x x x x x x

+ + = + + = − + − = −

4. Let

1 2 3 2 3 4 5 4 6

A = , compute the third of column of 1 A − by using Cramer’s rule to solve

Ax=e3. The result should be displayed in rational format. 5. Let A be the matrix in Ex4. Compute the adj A (A的伴随矩阵) by using the inv and det

function. 6. Compute the general solution of the following linear systems by using backslash（\）, null

and rref function.

a) 1 2 3

1 2 3

1 2 3

2 3 0 4 5 8 2 4 2

x x x x x x x x x

+ − = + + = − − + =

b) 1 2 3

1 2 3

1 2 3

2 3 0 4 5 8 2 4 4 8

x x x x x x x x x

+ − = + + = + + =

7. Compute the general solutions of 1 2 3

1 2 3

1 2 3

2 3 0 4 5 0 2 2 10 0

x x x x x x x x x

+ − = + + = − − =

by using rref function.

8. *Use rand function to generate several 4×3, 3×1; 3×4, 4×1 matrices, compare the results


Ex22

of the corresponding linear systems computed by \, inv and pinv. 9. *Use rand to generate a linear system Ax=b. Try the display the result if r(A b)=r(A); else

display “There is no solution.” (if) 10. Set A=[1 2 1; 2 4 2; 2 1 1]; B=[3 1 2; 1 2 2; 3 1 4]; C=A+i*B. Use MATLAB to compute each

of the following pairs of matrices. In each case check whether the first is equal to the second. If they are equal then display ‘The first matrix is equal to the second one.’ Otherwise display ‘The first matrix is not equal to the second one.’ 1) C’ C.’ 2) A’ A.’ 3) inv(A) pinv(A) 4) inv(B) pinv(B)

11. Use help to study functions: trace, rank, conj, rref, rrefmovie, format

MATLAB Ex 3 - Linear Space.pdf


Ex31

MATLAB Exercise 3 – Linear Space

1. Which of the following are spanning sets for 3 1 R × ? Justify your answers by at least two different methods. (related function may be used: rank, rref, det)

1)

1 0 1 0 , 1 , 0 0 1 1

2)

1 0 1 1 0 , 1 , 0 , 2 0 1 1 3

3)

2 3 2 1 , 2 , 2 2 2 0

− −

4)

2 2 4 1 , 1 , 2 2 2 4

− − − −

2. Given

1 2

1 3 2 9 2 , 4 , 6 , 2 3 2 6 5

x x x y − −

= = = = −

1) Is 1 2 Span( , ) x x x ∈ ?

2) Is 1 2 Span( , ) y x x ∈ ?

(related function: rref, rank) 3. Which of the following are spanning sets for P3? Justify your answer.

1) { } 2 2 1, , 1 x x +

2) { } 2 2, 2, , 2 1 x x x − +

3) { } 2 1, 2, 3 x x x + − + 4. Which of the following collections of vectors are linearly independent?

1)

1 0 1 1 , 1 , 0 1 1 1

2)

2 2 1 , 2 2 0

−

3)

2 2 4 1 , 1 , 2 2 2 4

− − − −

5. For each of the sets of vectors in Exercise 4, describe geometrically the span of the given vectors. (i.e. point out the maximal linearly independent subset for each of the sets of vectors.)

6. Given

1 2 3

1 3 0 2 , 4 , 10 3 2 11

x x x −

= = =

1) Show that 1 2 3 , , x x x are linearly dependent.

2) Show that 1 x and 2 x are linearly independent.


Ex32

3) What is the dimension of Span( 1 2 3 , , x x x )?

4) Give a geometric description of Span( 1 2 3 , , x x x ).

7. The vectors

1 2 3 4 5

1 3 0 2 1 2 , 4 , 10 , 7 , 3 3 2 11 3 2

x x x x x − −

= = = = =

Span R 3 . Pare down the set 1 2 3 4 5 , , , , x x x x x to form a basis for R 3 .

8. Let 1 2 3 1 10 , ,

2 3 7 x x x = = =

. Find the coordinates (坐标) of x with respect to 1 2 , x x .

(i.e. A=(x1,x2), the coordinates of x with respect to 1 2 , x x is c=A 1 x).

9. Let 1 2 2 1 ,

4 3 x x = =

be a basis of R 2 . Find the transition matrix from the standard basis {e1,

e2} to the basis {x1, x2} and determine the coordinates of 10 7

x =

under the basis {x1, x2}.

10. Find the transition matrix corresponding to the change of basis from 1 2 [ , ] x x to 1 2 [ , ] y y ,

where 1 2 2 1 ,

4 3 x x = =

, 1 2

1 3 ,

2 4 y y

− = =

. (X 1 Y)

11. For each of the following matrices, find a basis for the row space（行空间）, a basis for the

column space（列空间）, and a basis for the null space (零空间={ | 0} x Ax = ).

（If A is an m×n matrix, the subspace of R 1×n by the row vectors of A is called the row space of A. The subspace of R m×1 by the column vectors of A is called the column space of A.）

1)

1 2 3 2 3 6 9 4 6

A = −

2)

1 2 1 0 3 2 5 6 2 2 0 4

A −

= −

3)

1 2 1 0 3 2 5 6 2 0 6 6

A −

=

12. *Generate 10×10, 100×100, 1000×1000 random matrices, and corresponding column vectors. Compare the time used in solving linear system Ax = b by different functions or operators such as inv, \, lu and qr. (Refer to Lecture 3)

MATLAB Ex 4 - Eigenvalue.pdf


Ex41

MATLAB Exercise 4 – Eigenvalue 1. Find the eigenvalues and the corresponding eigenspaces for each of the following matrices.

And judge whether it is a defective matrix.

1) 3 2 3 2

−

2) 6 4 3 1

− −

3) 3 1 1 1

−

4) 3 8 2 3

−

5) 1 1 2 3

−

6)

0 1 0 0 0 1 0 0 1

7)

1 1 1 0 2 1 0 0 1

8)

1 1 1 0 3 1 0 5 1

−

9)

4 5 1 1 0 1 0 1 1

− − −

2. Judge whether the following two matrix are similar?

2 0 0 2 0 0 0 4 0 , 1 4 0 1 0 2 3 6 2

A B = = − −

3. Use poly and roots function to compute the characteristic polynomial and characteristic roots of a random 4×4 matrix. According to the result show the the characteristic polynomial and characteristic roots of A in mathematics formula.

4. In each of the following, factor the matrix A into a product XDX 1 , where D is diagonal. (use two different methods)

1) 0 1 1 0

A =

2) 5 6 2 2

A = − − 3)

2 8 1 4

A −

= −

4)

1 0 0 2 1 3 1 1 1

A = − −

5)

2 2 1 0 1 2 0 0 1

A = −

6)

1 2 1 2 4 2 3 6 3

A −

= − −

5. For each of the following, find a matrix B such that B 2 = A.

1) 2 1 2 1

A = − − 2)

9 5 3 0 4 3 0 0 1

A −

=

6. Given

0 1 1 1 2 0 1 0 3

A −

= −

,find an orthogonal matrix U that diagonalizes A. Please display the

results in rational format. (schur)


Ex42

7. Let

1 1 2 3 1 0

A =

. Compute the singular values and the singular value decomposition of A.

Compare square of the singular values of A with the eigenvalues of A’A. Are they the same?

8. Let

4 3 12 17 11 0 1 12 3

A −

= − −

. Compute the eigenvalues of A by roots and eig functions

respectivelyl. Compute its eigenvalue decomposition, Schur decomposition and singular value decomposition respectively. Compare the results and show the differences.

9. Please generate 4 symmetric matrices and check the following proposition “If the eigenvalues

of a symmetric matrix A are 1 2 , ,..., n λ λ λ ,then the singular values of A are

1 2 | |,| |,..., | | n λ λ λ ”.

10. Please generate 10 matrices (some of them are singular 奇异, others are nonsingular 非奇异即可逆, others are diagonalized matrices and others are not diagonalized matrices) and calculate their eigenvalues and singular values. Then count the number of nonzero eigenvalues and singular values. Show the conclusion you guess.

11. *Compute the singular value decomposition of matrix

2 5 4 6 3 0 6 3 0 2 5 4

A

=

.

1) Use the singular value decomposition to find orthonormal bases for ( ') R A and ( ) N A

2) Use the singular value decomposition to find orthonormal bases for ( ) R A and ( ) N A′

MATLAB Ex 6 - Polynomial.pdf


Ex61

MATLAB Exercise 6 – Polynomial 1. Generate the following polynomials as row vectors

1) 5 3 4 f x x = + − 2) 3 f x =

3) ( 1)( 3) f x x x = − + 4) 1 2

3 4 x

x −

+

2. Represent polynomial ( 1)( 2)( 3) f x x x = − − − in the following forms

1) row vector 2) sym 3) pretty 4) horner

3. Evaluate the polynomial ( 1)( 2)( 3) f x x x = − − − at

1) point 4 2) point 1 to 10 4. Execute following codes compare the results and the difference between subs and polyvalm.

Why? Here eye(2) means [1 0; 0 1] 1) >> syms x; mat = eye (2); sym_pol_a = x^2+1; subs(sym_pol_a, mat) 2) >> clear; mat = eye (2); pol_a = [1 0 1]; polyvalm(pol_a, mat) 3) replace the code pol_a = [1 0 1] with syms x; pol_a = x^2+1 in 2). How about the

result? Why? How to correct it if pol_a = x^2+1 should be used?

5. Let 2 ( ) 4 f x x x = + − , 2 ( ) 2 1 g x x = + . Compute

1) ( ) ( ) f x g x + 2) ( ) ( ) f x g x 3) ( ) f g 4) ( ) ( ) f x g x

6. Let 2 ( ) f x x = , 5 ( ) 3 1 g x x = + . Compute following formula by two different methods (i.e.

operations of matrix & operations of sym objects).

1) ( ) ( ) f x g x + 2) ( ) ( ) f x g x

7. Please display the result of Ex6 1) sym form and the terms ordered by descending powers 2) as row vector

8. Generate matrices which are 1) a symmetric positive definite matrix 2) a symmetric positive semidefinite matrix

and check the matrices by roots function and ‘ operation.

9. Compute the quotient and remainder of ( ) ( ) f x g x

and display the results in sym form, where

5 4 3 2 ( ) 2 3 2 3 f x x x x x x = + − − − + , 4 3 2 ( ) 5 6 g x x x x = + − − .

10. *Try to compute the greatest common factor (最大公因式) of ( ) f x and ( ) g x in Ex9.

MATLAB Ex 7 - Calculus.pdf


Ex71

MATLAB Exercise 7 – Calculus 1. Try to express y to be a compose function of x.

1) 2 1 1 , x y u u e − = + = ;

2) 2 2 1 , ln , x y u u v v e − = + = = ;

3) 2 1 , ln , sin , x y u u v v w w e − = + = = = .

2. Try to generate the converse function of y.

1) 2 1 ln sin y x = +

2) ln sin y x u = + , where u is as the independent variable

3) ln sin y x u = + , where x is as the independent variable

3. Let 4

x π = . Try to evaluate the value of sin x (express the result as 1/2*2^(1/2) ) and

sin(sin ) arc x (express the result as pi/4 ).

4. Compute the following limits, and simplify them.

1) 0

tan sin lim x

nx mx x → −

2) lim x y

x y

e e x y →

− −

3) 3

lim 2 100 x x

x →+∞ +

4) 3

lim sin x x x →−∞

5) tan 2

4

lim (tan ) x x

x π +

→

6) lim tan 2 x x

π − →

5. 1) Compute the limit 0

( ) lim n n

h

x h x h →

+ − , and simplify it.

2) Compute ( ) n x ′ by diff.

6. Compute the following derivatives

1) 0 x

dgdx =

for 3

2

5 ( ) 2 7 x g x x

− =

+ 2)

2

1 x

d g dxdy

=

for 3

2

5 ( , ) 2 7 x y y g x y x

− =

+

3) 1, 2 x y

dgdy = =

for 3

2

5 ( , ) 2 7 x y y g x y x

− =

+ 4) (5) f for sin sin 2 sin 3 f x x x =

7. Let 1 2 [ , ,..., ] n A a a a = is a vector with n elements, say [1,5,8, 2,6,3] A = − , how can we

generate a new vector

1) 1 2 2 3 1 [ , ,..., ] n n B a a a a a a − = − − − ?


Ex72

2) 1 2 3 2 3 4 2 1 [ 2 , 2 ,..., 2 ] n n n C a a a a a a a a a − − = − + − + − + .

8. Calculate the following calculus

1) 1 1 dx

x + ∫ 2) 1

0

1 1 dx

x + ∫ 3) 0 1 1

t dx

x + ∫

4) 2 sin

( 1) y dx

x y +∞

−∞ + ∫ 5) 2 sin

( 1) y dxdy

x y +∞ +∞

−∞ −∞ + ∫ ∫ .

9. Let 2 1 f x = + , compare it with the results of int(diff(f)) and diff(int(f)), respectively.

10. Compute the following summations

1) 3 1

n

k k

= ∑ 2) 2 1 1 1 k k

∞

= − ∑ 3) 2 2 1 1 k k

∞

= − ∑ 4) 2

1 k

k k x ∞

= ∑ 11. EvaluateTaylor series expansions of

1) 2 ( ) x f x e = at point 0 to the first 15 items;

2) 2 ( ) x f x e = at point 1 to the first 9 items;

3) 2 ( ) xy f x e = the first 5 items of Taylor series expansion responding to x.

12. *Compare the result (cos 2 ) b

a x x dx + ∫ with

( ) sin 2 ( ) 2 2

a b a b b a + + + − when b equals

to 10 a π + , 5 a π + , a π + , 1/2 a π + , 1 64 a π + , 1 256 a π + , respectively. What

conclusion you may reach?

13. *Examine integral meanvalue theorem, that is for any ( ) [ , ] f x C a b ∈ , there is a ( , ) a b ξ ∈ ,

such that ( ) ( )( ) b

a f x dx f b a ξ = − ∫ . For example, try to find out the (0,1) ξ ∈ , such that

1

2 2 0

1 1 ( 1) ( 1)

dx x ξ

= + + ∫ .

MATLAB Ex 8 - Graphics-Curve.pdf


Ex81

MATLAB Exercise 8 – Graphics Curve 1. Plot following 2D curve

1) 0.3 cos 2 , [0,6] x y e x x − = ∈ ; 2) 4sin , 4cos , [0, 2 ] x t y t t π = = ∈ ;

3) 4 4 2 2 8 10 16 0 x y x y + − − + = ; 4) 4cos3 ρ θ = .

2. Let , [0,10] 1 nx y x x

= ∈ +

.

1) Set 1,5,10 n = respectively, display the corresponding curve 1 2 3 , , c c c at the same axes;

2) Use different color to distinguish the curves; 3) Legend the curves; 4) Add grid on the figure above; 5) Label the xaxis with ‘X轴’ and yaxis with ‘函数值’.

3. Add title for 4) in Ex1 with ‘Plot of 4cos3 ρ θ = ’.

4. Let 4 4 2 2 2( ) x y x y b + − + = .

1) Plot 4 figures 1 2 3 4 , , , c c c c with b=1.0, 0.7, 1.0, 1.05;

2) Partition the figure into 4 subplots, plot 1 2 3 4 , , , c c c c into the subplots in turn.

5. Create 2 figures which plot the following 3D curves, respectively.

1) 2 3 0.5 , 0.1 , 9cos2 , [0,6] x t y t z t t = = = ∈ ;

2) 0.5sin , 0.1cos , 9cos 2 , [0,9] x t y t z t t = = = ∈ .

6. Plot polylines(折线) with following data on the same axes. Distinguish them by color and line shape and legend them.

1) [0,1, 4,5,7], [1,3,4,5,5.6] x y = = ; 2) [1, 2,3, 4,5], [2, 4,1,6,3] x y = = .

7. Let 1 ( )

5 4cos f x

x =

+ , 1 ( ) ( ) f x f x ′ = , 2 ( ) ( ) f x f x ′′ = ,

2 2 ( ) ( ) g x f x d x = ∫ ∫ .

1) Plot 1 2 ( ), ( ), ( ) f x f x f x respectively;

2) Observe the plots of f (x) and g(x) are the same? Break the Figure window into an 1by2 matrix of small axes, plot f (x) and g(x). 3) Let e be the error between f (x) and g(x). Display the function e and plot it. What do you find out?

8. *Execute following codes several times, observe the results. What they imply? >> syms x y; a=fix(10*randn(1, 6)); f=a(1)*x^2+a(2)*y^2+a(3)*x*y+a(4)*x+a(5)*y+a(6); >> ezplot(f, [10, 10])

9. *Plot a pentagram(五角星).

Key to MATLAB Ex 6 - Polynomial.pdf


Key to Ex61

Key to MATLAB Exercise 6 – Polynomial 1.

1) >> a=[1 0 0 0 3 4] a =

1 0 0 0 3 4 2) >> a2=[1 0 0 0] a2 =

1 0 0 0 3) >> r=[0 1 3]; poly(r) ans =

1 2 3 0 4) >> a=[1 2; 3 4]; poly(a) ans =

1 3 10 2.

1) >> a=[1 2 3]; poly(a) ans =

1 6 11 6 2) >> syms x; f=(x1)*(x2)*(x3) f = (x1)*(x2)*(x3) 3) >> pretty(f)

(x 1) (x 2) (x 3) 4) >> horner(f) ans = (x1)*(x2)*(x3)

3. 1) >> subs(f, 4) ans =

6 2) >> subs(f, 1:10) ans =

0 0 0 6 24 60 120 210 336 504


Key to Ex62

4. 1) >> syms x; mat = eye (2); sym_pol_a = x^2+1; subs(sym_pol_a, mat) ans =

2 1 1 2

It generate a matrix, the (i, j) element of which is obtained by substituting the x in

2 1 x + with the (i, j) element of mat, that is, 2 2 2 2 11 12 2 2 2 2 21 22

1 1 1 1 0 1 .

1 1 0 1 1 1 mat mat mat mat

+ + + + =

+ + + + 2) >> clear; mat = eye (2); pol_a = [1 0 1]; polyvalm(pol_a, mat) ans =

2 0 0 2

The answer is the result of 2 1 0 1 0

0 1 0 1

+

.

3) >> syms x; mat = eye (2); pol_a =x^2+1; polyvalm(pol_a, mat) ??? Inputs to polyvalm must be floats, namely single or double. Error in ==> polyvalm at 27

Y = diag(p(1) * ones(m,1,superiorfloat(p,X))); >> clear;syms x; mat = eye (2); pol_a =x^2+1; polyvalm(sym2poly(pol_a), mat) ans =

2 0 0 2

5. 1) >> clear; syms x; f=x^2+x4; g=2*x^2+1; f_plus_g_sym=f+g % in sym form f_plus_g_sym = 3*x^2+x3 Or>> clear; f0=[1 1 4]; g0=[2 0 1]; f_plus_g_vector=f0+g0 % in vector form f_plus_g_vector =

3 1 3 Or>> clear; f0=[1 1 4]; g0=[2 0 1]; f_plus_g_sym=poly2sym(f0)+poly2sym(g0) f_plus_g_sym = 3*x^2+x3 2) >> clear; syms x; f=x^2+x4; g=2*x^2+1; f_multiply_g_sym=f*g f_multiply_g_sym = (x^2+x4)*(2*x^2+1)


Key to Ex63

>> poly2sym(sym2poly(f_multiply_g_sym)) ans = 2*x^47*x^2+2*x^3+x4 Or>> clear; f0=[1 1 4]; g0=[2 0 1]; f_multiply_g_vector=conv(f0, g0) f_multiply_g_vector =

2 2 7 1 4 >> poly2sym(f_multiply_g_vector) ans = 2*x^47*x^2+2*x^3+x4 3) >> clear; syms x; f=x^2+x4; g=2*x^2+1; subs(f, g) ans = (2*x^2+1)^2+2*x^23 4) >> clear; f0=[1 1 4]; g0=[2 0 1]; [q, r]=deconv(f0, g0) q =

0.5000 r =

0 1.0000 4.5000 6.

1) >> clear; syms x; f=x^2; g=3*x^5+1; f_plus_g_sym=f+g f_plus_g_sym = x^2+3*x^5+1 Or>> clear; f0=[0 0 0 1 0 0]; g0=[3 0 0 0 0 1]; f_plus_g_vector=f0+g0 f_plus_g_vector =

3 0 0 1 0 1 >> f_plus_g_sym =poly2sym(f_plus_g_vector) f_plus_g_sym = x^2+3*x^5+1 2) >> clear; syms x; f=x^2; g=3*x^5+1; f_multiply_g_sym=f*g f_multiply_g_sym = x^2*(3*x^5+1) >> poly2sym(sym2poly(f_multiply_g_sym)) ans = 3*x^7+x^2 Or>> clear; f0=[0 0 0 1 0 0]; g0=[3 0 0 0 0 1]; f_multiply_g_vector=conv(f0, g0) f_multiply_g_vector =

0 0 0 3 0 0 0 0 1 0 0 >> f_multiply_g_sym =poly2sym(f_multiply_g_vector)


Key to Ex64

f_multiply_g_sym = 3*x^7+x^2

7. for example 6.2) 1) >> clear; syms x; f=x^2; g=3*x^5+1; f_multiply_g_sym=f*g f_multiply_g_sym = x^2*(3*x^5+1) >> poly2sym(sym2poly(f_multiply_g_sym)) ans = 3*x^7+x^2 Orclear; f0=[0 0 0 1 0 0]; g0=[3 0 0 0 0 1]; f_multiply_g_vector=conv(f0, g0) f_multiply_g_vector =

0 0 0 3 0 0 0 0 1 0 0 >> f_multiply_g_sym =poly2sym(f_multiply_g_vector) f_multiply_g_sym = 3*x^7+x^2 2) >> clear; syms x; f=x^2; g=3*x^5+1; f_multiply_g_sym=f*g f_multiply_g_sym = x^2*(3*x^5+1) >> sym2poly(f_multiply_g_sym) ans =

3 0 0 0 0 1 0 0 8.

1) >> A=diag(1:3); p=fix(10*rand(3)); B=p'*A*p; B =

100 71 171 71 135 175 171 175 329

>> B==B' ans =

1 1 1 1 1 1 1 1 1

>> roots(poly(B)) ans = 513.6451 49.4764 0.8785

2) >> A0=diag(0:2); p=fix(10*rand(3)); B0=p'*A*p; B =


Key to Ex65

144 152 44 152 164 50 44 50 17

>> B==B' ans =

1 1 1 1 1 1 1 1 1

>> roots(poly(B)) ans = 320.9116 4.0884 0.0000

9. >> f=[1 2 3 1 2 3]; g=[1 1 5 6 0]; >> [q,r]=deconv(f,g) q =

1 1 r =

0 0 1 10 4 3 >> poly2sym(q) ans = 1+x >> poly2sym(r) ans = x^3+10*x^2+4*x+3

10. Omitted.

Key to MATLAB Ex 7 - Calculus.pdf


Key to Ex71

Key to MATLAB Exercise 7 – Calculus 1.

1) >> clear; syms y1 u x; y1=(1+u^2)^(1/2); u_x= exp(x); >> y1= compose(y1,u_x,x) Or>> clear; syms x; u=exp(x); y1=(1+u^2)^(1/2) % no recommend y1 = (1+exp(x)^2)^(1/2) 2) >> clear; syms y2 u v x; y2=(1+u^2)^(1/2); u =log(v); v=exp(x); >> u_x=compose(u,v,x); y2=compose(y2,u_x,x) Or>> clear; syms y2 u v x; y2=(1+u^2)^(1/2); u=log(v); v=exp(x); >> y2=compose(y2, compose(u, v, x)) Or>> clear; syms x; v=exp(x); u=log(v); y2=(1+u^2)^(1/2) % no recommend y2 = (1+log(exp(x))^2)^(1/2) 3) >> clear; syms y u v w x; y=(1+u^2)^(1/2); u=log(v); v=sin(w); w=exp(x); >> v_x=compose(v,w,x); u_x=compose(u,v_x,x); y=compose(y,u_x,x) Or>> clear; syms y u v w x; y=(1+u^2)^(1/2); u=log(v); v=sin(w); w=exp(x); >> y=compose(y, compose(u, compose(v, w, x))) Or>> clear; syms x; w=exp(x); v=sin(w); u=log(v); y=(1+u^2)^(1/2) % no recommend y = (1+log(sin(exp(x)))^2)^(1/2)

2. 1) >> clear; syms x y; y=(1+(log(sin(x)))^2)^(1/2); >> finverse(y,x) Warning: finverse((1+log(sin(x))^2)^(1/2)) is not unique. > In sym.finverse at 43 ans = asin(exp((1+x^2)^(1/2))) 2) >> clear; syms x u y; y=(x+log(sin(u)))^(1/2); >> finverse(y,u) ans = asin(exp(x+u^2)) 3)


Key to Ex72

>> clear; syms x u y; y=(x+log(sin(u)))^(1/2); >> finverse(y,x) ans = log(sin(u))+x^2

3. >> syms y1=sin(x); y2=asin(sin(x)); >> x=sym(’pi/4’); subs(y1,x) ans = 1/2*2^(1/2) >> subs(y2,x) ans = 1/4*pi Or>> syms y1=sin(x); y2=asin(sin(x)); >> compose(y1,pi/4) ans = 1/2*2^(1/2) >> compose(y2,pi/4) ans = 1/4*pi

4. 1) >> clear; syms n m x; y1=(tan(n*x)sin(m*x))/x; >> limit(y1,0) Or>> clear; syms n m x; y1=(tan(n*x)sin(m*x))/x; >> limit(y1) ans = nm 2) >> clear; syms x y; y2=(exp(x)exp(y))/ (xy); >> limit(y2,x,y) ans = exp(y) 3) >> clear; syms x; y3=x^3/(2*x+100); >> limit(y3,x,+inf) ans = Inf 4) >> clear; syms x; y4=x^3/sin(x); >> limit(y4,x,inf) ans = NaN


Key to Ex73

5) >> clear; syms x; y5=(tan(x)^(tan(x))); >> limit(y5,x,pi/4,’right’) ans = 1 6) >> clear; syms x; y=tan(x/2); >> limit(y,x,pi,'left') ans = Inf

5. 1) >> clear; syms x n h; y=((x+h)^nx^n)/h; >> z=limit(y,h,0) z = x^n/x*n >> simplify(z)) ans = x^(1+n)*n 2) >> clear; syms x n ; y= x^n; >> diff(y, x) ans = x^n/x*n >> simplify(diff(y,x)) ans = x^(1+n)*n

6. 1) >> clear; syms x g; g=(x^35)/(2*x^2+7); >> subs(diff(g,x),0) Or>> clear; syms x g; g=(x^35)/(2*x^2+7); >> compose(diff(g,x),0) ans = 0 2) >> clear; syms x y g; g=(x^3*y5*y)/(2*x^2+7); >> g_xy=diff(diff(g,x),y) g_xy = 3*x^2/(2*x^2+7)4*(x^35)/(2*x^2+7)^2*x >> subs(g_xy, 1) ans =

0.5309 Or


Key to Ex74

>> clear; syms x y g; g=(x^3*y5*y)/(2*x^2+7); g_xy=diff(diff(g,x),y) >> compose(g_xy,1) ans = 43/81 3) >> clear; syms x y g; g=(x^3*y5*y)/(2*x^2+7); >> g_y=diff(g,y) g_y = (x^35)/(2*x^2+7) >> subs(g_y, [x y], [1 2]) ans = 4/9 4) >> clear; syms x f; f=sin(x)*sin(2*x)*sin(3*x); >> diff(f,5) ans = 1696*cos(x)*sin(2*x)*sin(3*x)+2192*sin(x)*cos(2*x)*sin(3*x)+2208*sin(x)*sin(2*x)*cos(3 *x)1680*cos(x)*cos(2*x)*cos(3*x)

7. 1) >> A=[1,5,8,2,6,3]; B=(1)*diff(A) B =

4 3 10 8 3 2) >> A=[1,5,8,2,6,3]; C=diff(A,2) C =

1 13 18 11 8.

1) >> clear; syms x y; y=1/(x+1); >> int(y) ans = log(x+1) 2) >> clear; syms x y; y=1/(x+1); >> int(y,0,1) ans = log(2) 3) >> clear; syms x t y; y=1/(x+1); >> int(y,0,t) ans = log(t+1) 4)


Key to Ex75

>> clear; syms x y f; f=sin(y)/(x^2*y+1); >> int(f,x,inf,inf) ans = pi*sin(y)/y^(1/2) 5) >> clear; syms x y f; f=sin(y)/(x^2*y+1); >> int(int(f,x,inf,inf),inf,inf) ans = 0

9. >> clear; syms x f; f=x^2+1; >> f_int_diff=int(diff(f)) f _int_diff= x^2 >> f_diff_int=diff(int(f)) f_diff_int = x^2+1 diff(int(f)) is not equal to int(diff(f)), the difference between them is the constant item C.

10. 1) >> clear; syms k n; >> simplify(symsum(k^3,1,n)) ans = 1/4*n^4+1/2*n^3+1/4*n^2 2) >> clear; syms k; >> symsum(1/(k^21),1,inf) ans = sum(1/(k^21),k = 1 .. Inf) 3) >> clear; syms k; >> symsum(1/(k^21),2,inf) ans = 3/4 4) >> clear; syms k x f; f=k^2*x^k; >> symsum(f,k,1,inf) ans = x*(x+1)/(x1)^3

11. 1) >> clear; syms x f; f=exp(2*x); >> taylor(f,9) ans =


Key to Ex76

1+2*x+2*x^2+4/3*x^3+2/3*x^4+4/15*x^5+4/45*x^6+8/315*x^7+2/315*x^8 2) >> clear; syms x f; f=exp(2*x); >> taylor(f,9,1) ans = exp(2)+2*exp(2)*(x+1)+2*exp(2)*(x+1)^2+4/3*exp(2)*(x+1)^3+2/3*exp(2)*(x+1)^4+4/ 15*exp(2)*(x+1)^5+4/45*exp(2)*(x+1)^6+8/315*exp(2)*(x+1)^7+2/315*exp(2)*(x+1)^8 3) >> clear; syms x f; f=exp(2*x*y); >> taylor(f,x) ans = 1+2*x*y+2*y^2*x^2+4/3*y^3*x^3+2/3*y^4*x^4+4/15*y^5*x^5

12. >> clear; format short e; syms a x; f=cos(x)+2*x; f_int=int(f); >> b=[a+10*pi, a+5*pi, a+pi, a+1/2*pi, a+1/16*pi, a+1/1024*pi]; >> for k=1:6 y(k)=int(f,a,b(k)); y_app(k)=(subs(f_int, b(k))subs(f_int, a))*(b(k)a); end >> y, y_app y = [ 20*a*pi+100*pi^2, 2*sin(a)+10*a*pi+25*pi^2, 2*sin(a)+2*a*pi+pi^2, cos(a)+a*pi+1/4*pi^2sin(a), sin(a+1/16*pi)+1/8*a*pi+1/256*pi^2sin(a), sin(a+1/1024*pi)+1/512*a*pi+1/1048576*pi^2sin(a)] y_app = [ 10*((a+10*pi)^2a^2)*pi, 5*(2*sin(a)+(a+5*pi)^2a^2)*pi, (2*sin(a)+(a+pi)^2a^2)*pi, 1/2*(cos(a)+(a+1/2*pi)^2sin(a)a^2)*pi, 1/16*(sin(a+1/16*pi)+(a+1/16*pi)^2sin(a)a^2)*pi, 1/1024*(sin(a+1/1024*pi)+(a+1/1024*pi)^2sin(a)a^2)*pi] >> y_0=subs(y, 0); y_app_0=subs(y_app, 0); error_y=y_0y_app_0 error_y = 3.0019e+004 3.6290e+003 2.1137e+001 1.9792e+000 1.8777e001 3.0679e003

Conclusion: While b closes to a, ( ) sin 2 ( )

2 2 a b a b b a + + + −

approaches

to (cos 2 ) b

a x x dx + ∫ .

13. >> clear; format long e; syms x xi; f=1/(x+1)^2; >> f_int=int(f, 0, 1); xi= subs(finverse(ff_int), 0) xi =

4.142135623730949e001 % xi is in (0, 1)


Key to Ex77

>> abs(subs(f, xi)f_int) ans =

0 % 1

2 2 0

1 1 ( 1) ( 1)

dx x ξ

= + + ∫

Key to MATLAB Ex 8 - Graphics-Curve.pdf


Key to Ex81

Key to MATLAB Exercise 8 – Graphics Curve 1.

1) >> clear; x=linspace(0,6); y=exp(0.3.*x).*cos(2.*x); >> plot(x,y) 2) >> clear; t=linspace(0,2*pi); x=4.*sin(t); y=4.*cos(t); >> plot(x,y)

3) >> clear; syms x y; f=x^4+y^48*x^210*y^2+16; >> ezplot(f) 4) >> clear; theta=2*pi:0.1:2*pi; rho=4*cos(2*theta); >> polar(theta,rho)

2. >> clear; syms x y n; x=linspace(0,10); y=n.*x./(1+x); >> y1=subs(y,n,1); y2=subs(y,n,5); y3=subs(y,n,10);


Key to Ex82

>> plot(x,y1,x,y2,'r',x,y3,'g') >> legend('y1','y2','y3') >> grid on >> xlabel('X轴'); ylabel('函数值')

3. >> title('Plot of \rho=4cos3\theta')


Key to Ex83

4. 1) >> clear; syms x y b f; f=x^4+y^42*(x^2+y^2)b; >> f1=subs(f,b,0.1); f2=subs(f,b,0.7); f3=subs(f,b,1.0); f4=subs(f,b,1.05); >> ezplot(f1); >> ezplot(f2); >> ezplot(f3); >> ezplot(f4); 2) >> subplot(2,2,1); ezplot(f1) >> subplot(2,2,2); ezplot(f2) >> subplot(2,2,3); ezplot(f3) >> subplot(2,2,4); ezplot(f4)

5. >> figure(1); syms x t y; t=linspace(0,6); x=0.5*t.^2; y=0.1*t.^3; z=9*cos(2*t); >> plot3(x,y,z,'m*') >> figure(2); subplot; syms x t y; t=linspace(0,9); x=0.5*sin(t); y=0.1*cos(t); z=9*cos(2*t); >> plot3(x,y,z,'m*')

6. >> figure; x=[0,1,4,5,7]; y=[1,3,4,5, 5.6]; plot(x,y,'m*');


Key to Ex84

>> hold on; x1=[1,2,3,4,5];y1=[2,4,3,6,4]; plot(x1,y1,':cs'); legend('Group 1', 'Group 2');

7. >> x = sym('x'); f = 1/(5+4*cos(x)); ezplot(f)

>> f1 = diff(f); ezplot(f1)


Key to Ex85

>> f2 = diff(f,2); ezplot(f2);

>> g = int(int(f2)); ezplot(g);

At first glance, the plots for f and g look the same. Look carefully, however, at their formulas and their ranges on the yaxis. >> subplot(1,2,1); ezplot(f) >> subplot(1,2,2); ezplot(g)


Key to Ex86

e is the difference between f and g. It has a complicated formula, but its graph looks like a constant. >> e = f g >> subplot(1,1,1); ezplot(e) e = 1/(5+4*cos(x))+8/(tan(1/2*x)^2+9)

To show that the difference really is a constant, simplify the equation. This comfirms that the difference between them really is a constant. >> e = simple(e) ezplot(e) e = 1

8. They are curves of second order, such as elipse(椭圆), hyperbola (双曲线), parabola (抛物线) and so on.

9. >> phi=[pi/2:4*pi/5:4*pi, pi/2]; B=exp(i*phi); x1=real(B);y1=imag(B); >> plot(x1,y1,'r'); axis square; title('五角星')

MATLAB Ex 9 - Graphics-Surface.pdf


Ex91

Key to MATLAB Exercise 9 – Graphics Surface

1. Plot 2 2 2 2 0.15( ) 0.15( ) 0.3 0.03 x y x y z e e − + + = − on the region of D: 4 4 x − ≤ ≤ , 5 5 y − ≤ ≤ ,

by1) plot3 2) mesh 3) surf 4) meshz 5) surfc

2. Plot 2 2 2 2 sin( ) / z x y x y = + + on the region of D: 9 9, 9 9 x y − ≤ ≤ − ≤ ≤ , with

title, x, y, and z label, grid on, displaying multiple plots per figure 1) plot3 2) mesh 3) surf

3. Display a graph which is cos , sin , 2 8 ;0 , , 4 x u v y u v z u v u v π = = = + ≤ ≤ in three

figures by 1) plot3 2) mesh 3) surf

4. Plot the 3D curve 2 3 0.5 , 0.1 , 9cos2 , [0,6] x t y t z t t = = = ∈ .

5. Plot an upper semisphere. 上半球 6. Plot a polygon and fill it with blue color. 7. *Plot a pentagram. 五角星

8. *Displaying nonuniform data on a surface of cos , sin , 2 8 x u v y u v z u v = = = + .

Key to MATLAB Ex 9 - Graphics-Surface.pdf


Key to Ex91

Key to MATLAB Exercise 9 – Graphics Surface 1.>> clear; clf; x=linspace(4,4,20); y=linspace(5,5,20); >> [X, Y]=meshgrid(x,y); Z=[0.3*exp(0.15*(X.^2+Y.^2)) 0.3*exp(0.15*(X.^2+Y.^2))]; 1)>> plot3(X,Y,Z); grid on;

2) >> mesh(X,Y,Z)

3) >> surf(X,Y,Z)


Key to Ex92

4) >> meshz(X,Y,Z)

5) >> surfc(X,Y,Z)

2. >> clear; clf; >> x=[9:9];y=[9:9]; [X,Y]=meshgrid(x,y); >> XY= (X.^2+Y.^2).^(1/2)+eps; Z=sin(XY)./XY; 1) >> subplot(1,3,1); plot3(X, Y, Z); >> title('Display by plots function'); xlabel('x'); ylabel('y'); zlabel('z'); grid on 2) >> subplot(1,3,2); mesh(X,Y,Z); >> title('Display by mesh function'); xlabel('x'); ylabel('y'); zlabel('z'); grid on 3) >> subplot(1,3,3); surf(X, Y, Z); >> title('Display by surf function'); xlabel('x'); ylabel('y'); zlabel('z'); grid on

Key to MATLAB Exercise 9 School of Mathematical Sciences Xiamen University http://gd

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