Đang tải... (xem toàn văn)
Thus, the order of convergence is 1. b) Give a modification of Newton’s method so that the order of convergence is 2.. The result is true for all i. The theorem is proved.. Proof.. We ha[r]
(1)FINAL PROJECT IN
Iterative Solution of Nonlinear Equations in Several Variables 2012
The Minh Tran
1) Let f(x)=x3ex
a) Write down Newton’s method for this function What is the order of convergence?
- Solution :
+ We will use the Newton’s method with the formula : ( )
) ( '
n n n
n
x f
x f x
x + = − for the function x
e x x
f( )=
We have : xn
n
n x e
x
f( )= ⇒ '( )=3 + = ( n +3)
x n x n x n
n x e x e x e x
x
f n n n So
( )
3 2 )
3 ( )
(
2
3 '
1
+ + = + −
= −
=
+
n n n n
x x n n
n n n
n
x x x x
e x
e x x
x f
x f x x
n n
+ To continue, we apply the formula p n n n
x x
α α
− −
+ ∞ →
1
lim to find the order of convergence, where
0 ) (α =
f ; α =0 We compute that :
p n p
n
n n n
p n n
n n n p n n n p n n
n x x
x x
x x
x x
x x x
x
3 2 lim
1 3 2 lim
lim
lim 1
2
1
+ + =
+ + =
= −
−
+ ∞ → ∞
→ + ∞ → +
∞
→ α
α
We assume that xn →α =0 so :
p n n n p n p
n
n n n
p n n
n x
x x
x
x x
x x
3 lim 3
2 lim
lim 1
2
∞ → +
∞ → +
∞
→ + =
+ =
− − α
α
Only if p =1 then 3
lim = > ∞
→ p
n n n x
x
, which is nonzero and positive
Thus, the order of convergence is
b) Give a modification of Newton’s method so that the order of convergence is
- Solution
(2)such that c x
x f
k x→ ( − ) =
) ( lim
0 α
We have : lim 3
0 = ≠
→ x
e
x x
x . Clearly, we see that the root α =0 has multiplicity for
x
e x x
f( )=
Hence, we can have a modification of Newton’s method to become quadratic convergence is : ( )
3 )
3 ( 3
) (
2
3 '
1
+ = + −
= −
=
+
n n n
x x n n
n n n
n
x x x
e x
e x x
x f
x f N x x
n n
Where N = is the multiplicity of root
αof f(x)=x3ex
Finally, we can observe that p n n
n
n x x
x 1
3 lim
2
+
∞
→ converges to a nonzero constant whenever p =
2 Given a linear system Ax=b where A is SDD
a) Describe Jacobi method applied to this system and prove a convergence theorem + Describe Jacobi method applied to this system
- Solution :
The system AX = b or
n n nn n
n
n n
b x a x
a x a
b x a x
a x a
= +
+
= +
+
. . . . . . . . .
. .
. . . . . . . .
2 1
1
2 12 11
We can rewrite ( assumption that aii ≠0, i=1 n) as :
( )
( 1 , 1)
1
12 11
. . . . . . . . 1
. . .
. . . . . . . . 1
− −
− −
− =
− −
− =
n n n n
n nn n
n n
x a x
a b a x
x a x
a b a x
(3) + − − − − − − − = − 22 11 1 , 22 22 23 22 21 11 11 12 nn n n nn n n nn n n n n a b a b a b x x x a a a a a a a a a a a a a a x x x
Or X = BX + d If we write the matrix A in the form A = L +D +U where = − 0 0 , 32 21 n n n
n a a
a a a
L ,
= − 0 0 1 12 n n n a a a U , = n n a a a D , 22 11 0 0
From the above part, it is to see that :
B =−D−1(L+U) , d = D−1b
With the Jacobi matrix B =−D−1(L+U) , the vector Jacobi d =D−1band X = BX + d We have :
X(k+1) =−D−1(L+U)X( )k +D−1b
( ) ( ) n i x a b a x n i j j k j ij i ii k
i , 1,...,
1 1 = − = ⇒ ∑ ≠ = +
+ To prove a convergence theorem
-Theorem : If A is strictly diagonally dominant then the Jacobi method converges for any guess x( )0
Proof :
Because A is strictly diagonally dominant (SDD) , we have :
< ⇔
>∑ ∑
≠
≠ i j ii
ij j i ij ii a a a a
(4)Here G = D−1(L+U)
We choose ∞.Then
( ) max 1
1
1 + = <
= ∑
≠ ≤ ≤ ∞ −
∞
i j ii
ij m
i a
a U
L D G
Thus, the Jacobi method converges for any guess x( )0
b) Describe Gauss-Seidel method applied to this system and prove a convergence theorem + Describe Gauss-Seidel method applied to this system
- Solution :
The system AX = b or
n n nn n
n
n n
b x a x
a x a
b x a x
a x a
= +
+
= +
+
. . . . . . . . .
. .
. . . . . . . .
2 1
1
2 12 11
We have :
(D−L)X =UX +b
⇒ = −
− ⇒
=b (D L U)X b
AX ⇒ X =(D−L) (−1 UX+b)
⇒ X(k+1) =(D−L)−1(UX( )k +b)
Where A = D – L –U and D, L, and U represent the diagonal, lower triangular, and upper triangular parts
( ) ( ) ( )
n i
x a x
a b
a x
n i j
k j ij n
i j
k j ij i
ii k
i , 1,...,
1 1
1
=
− −
=
⇒ ∑ ∑
> <
+ +
(1) +To prove a convergence theorem
- Theorem : The Gauss-Seidel method for Ax=b is convergent if A is strictly diagonally dominant
Proof :
From AX=b ⇒ (D-L-U)X=b ⇒(D−L)X =UX +b Therefore : (D−L)(X(i+1) −X)=U(X(i+1)−X)
(2)
(5)Apply from (2), we have :
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( 1)
1
− − −
− −
+ =
⇒
+ =
⇒
= −
m m
m
m m
m
m m
Ue D Le D e
Ue Le
De
Ue e
L D
With the above process, and ∑( ) ∑( )
+ = −
= −
− = −
= =
n i j
ij i
j ij ii
a U
a L
a D
1
1
; ;
1
we have :
( ) ( ) ( ) ( ) ( )
− +
− =
⇒ ∑ ∑
+ =
− −
=
n i j
m j ij i
j
m j ij ii
m
i a e a e
a e
1
1
1
1
For i =1 :
( ) ∑( ) ( ) =
−
− =
⇒ n
j
m j j m
e a a
e
2
1
11
1
( ) ( )
( )
( ) ∞ −
= ∞
− =
−
≤ ≤ ≤
⇒
∑ ∑
1
2 11
1
1
11
1 1
m j
n j
j m
j n j
m j j m
e r
a a
e
e a a
e
where 1
2 11
1 = ∑ <
=
n j
j
a a
r by SDD and set ( )i
n i r
r ≤ ≤
= max
Now, we have for i≥2 and assume that ( ) ( )
∞ −
≤ m1
m
j r e
e
( ) ( ) ( )
( ) ( )
( ) ( ) ( )
∞ − ∞
−
≠ = ∞
−
+ = ∞ − −
= ∞ −
+ =
− −
=
≤ =
<
+ =
+ ≤
⇒
∑
∑ ∑
∑ ∑
1
1
1
1
1
1
1
1
1 . 1 1
m m
i n
i j j
ij ii
m
n i j
ij m
i j
ij m
ii
n i j
m j ij i
j
m j ij ii
m i
e r e
r a
a e
a e
a e
r a
e a e
a a
(6)The result is true for all i
( ) ( )
∞ − ≤
≤ ≤
1
max im m
n
i e r e
( ) ( )
∞ − ∞ ≤
⇒ m m
e r e
Thus ( )
∞ ∞ ≤
0
e r
e m m and so ( ) →0 →∞( <1)
∞ asm r
em
Or e( )m = x( )m −x→0as m→∞ as required The theorem is proved
( We also can apply this way to prove for the part a) )
3) Let A∈Rn×n
a) State and prove a QR- decomposition theorem
-Theorem : Suppose that A is an n×m matrix with linearly independent columns then A can be factored as,
A = QR
where Q is an n×m matrix with orthonormal columns and R is an invertible m×m upper triangular matrix
Proof :
Suppose that the columns of A are given by c1,c2,... ,cm We use the Gram- Schmidt process on these vectors and we have a set of orthonormal vectors u1,u2,...,um We can write
A will have columns A=[c1|c2 |...|cm]
Q will be a matrix with orthonormal columns Q=[u1|u2 |...|um]
We can write each ci as a linear combination of u1,u2,...,um in the following linear system :
m m m m
m m
m m
m m
u u c u
u c u u c c
u u c u
u c u u c c
u u c u
u c u u c c
, .
. . . . ,
, .
.
, .
. . . . ,
, .
, . . . . . ,
,
2
1
2
2 1 2
1
2 1 1
+ +
+ =
+ +
+ =
+ +
+ =
(7)
=
m m m
m
m m
u c u
c u c
u c u
c u c
u c u
c u c
R
, . . . . , ,
. .
. .
, . . . . , ,
, . . . . , ,
2
2
2
1
2 1
Now, we can observe that the product QR=A
[ ]
[ ]
A
c c
c
u c u
c u c
u c u
c u c
u c u
c u c
u u
u QR
m
m m m
m
m m
m
= =
=
|. . . | |
, . . . . , ,
. .
. .
, . . . . , ,
, . . . . , ,
| . . . | |
2
2
2
2
1
2 1
2
We continue to is to show that R is an invertible upper triangular matrix
First, recall the matrix
=
m m m
m
m m
u c u
c u c
u c u
c u c
u c u
c u c
R
, . . . . , ,
. .
. .
, . . . . , ,
, . . . . , ,
2
2
2
1
2 1
from Gram- Schmidt process we know that ukis orthogonal to c1,c2,...,ck−1 This mean that all the inner product below the main diagonal must be zero and they are all of the form ci,uj =0 with
j
i< We know from Special matrices property that a triangular matrix will be invertible if the main diagonal entries ci,ui are non-zero We have the general formula for ui from the Gram-Schmidt process
1
2
1 '
, .
. . ,
, − − − − −
−
= i i i i i i
i c c u u c u u c u u
u ' ; ' 0
'
≠ =
→ i
i i
i u
(8)1
2
1 '
1
2
1 '
, .
. . ,
,
, .
. . ,
,
− −
− −
+ +
+ +
=
+ +
+ +
=
⇒
i i i i
i i i
i i i i
i i i
u u c u
u c u u c u u
u u c u
u c u u c u c
Now, we can rewrite the formula using the properties of the inner product
i i i i i
i i i
i i i
i i i i i
i i i i
i
u u u c u
u u c u u u c u u u
u u u c u
u c u u c u u u
c
, ,
. . . ,
, ,
, ,
, ,
. . . ,
, ,
1
2
1 '
1
2
1 '
− − −
−
+ +
+ +
=
+ +
+ +
=
Because the ui are an orthonormal basis vectors and so we see that
0 ,
, 0
, 0
, i = > i i ≠ ⇒ i i = i' i i ≠
j u i j u u c u u u u
u
And from the above content ,we also have ci,uj =0 with i< j
Hence, R is an invertible upper triangular matrix and it is presented the form following
=
m m m m
u c
u c u
c
u c u
c u c
R
, . . 0 . . 0 0
. . .
.
. .
0 .
, . . . . , 0
, . . . . , ,
2
2
1
2 1
b) Prove a uniqueness theorem of the decomposition for a proper A, e.g nonsinglar and so on - Theorem : Let A be a m×n matrix with linearly independent columns Thus, A admits a QR decomposition.Further such a decomposition is unique
Proof
We have proved the existence of QR decomposition of the matrix A as in the part a) Now we can prove the uniqueness of this decomposition
Indeed, from the matrices A, Q, R have the properties as in the part a) Let A=Q1R1 =Q2R2
where Q1TQ1 =Q2TQ2 =Id
(9)Then, we can a reduction on the matrices and see that :
( )
( )
2
2 2
1 1 1
R R
R Q Q R
A A
R Q Q R R R
t t t t
t t t
= = = =
Hence
( )
1 1 2
2 1
− −
= ⇒
=R R R R R R R
Rt t t t We see the this equation have the left hand side is a lower triangular matrix and the right hand side is an upper triangular matrix Hence, both of them must be diagonal
Let αi and βi 1≤i≤n are the diagonal entries of R1 and R2,respectively Then αi >0; βi >0 for every i and
n i
n i
i i
i i i i
≤ ≤ −
≤ ≤ =
1 ,
1 ,
β α
β α β α
Hence R R− =( )R− t R =Id
1 1
2 ⇒ R1 =R2 Since Q1R1 =Q2R2,it follows that Q1 =Q2 Thus, the decomposition is unique
4) Let { 2} , , 1t t
A= be there vectors(polynomial) and let (f,g) f(x)g(x)dx
1
1
∫
−
= be the inner product under consideration Use the Gram-Schimidt process to orthogonalize the set A, what is the resulting orthonormal set
-Solution :
We have a unit vector basic A={1,t,t2} Let A1 =1;A2 =t;A3 =t2 and (f,g) f(x)g(x)dx
1
1
∫
−
=
Compute :
1
1 = A =
q ,( , )
1
1
1 1
1 = ∫ = ∫ =
− −
dx dx
q q q
q
( )
( ) ( )
t
dt t t
t t q q q
q A A q
=
− = −
= −
=
⇒ ∫
−
1
1
1
1 2
2 1 , ,
(10)( )
3 ,
1
1
1 2
2 = ∫ = ∫ =
− −
dx t dx q q q
q
( )
( )
( )
( ) ( ) ( )
3
2
1
, 1 , ,
, ,
,
2
1
1
1 2
2
2 2
2
1
1 3
− =
− −
=
− −
= −
− −
= ⇒
∫ ∫
− −
t
dt t t dt t t
t t t t
t q q q
q A q
q q
q A A q
The resulting orthonormal set is
− 3 1 ,
, 1 t t2
We can check the inner product again as :
( )
− ⊥ ⊥ ⇒
=
− =
− =
− =
− =
= ∫ ∫ ∫
− −
−
3 1
0
1
1 ,
1
1 , ,
1
2
1
1 2
1
1 2
1
1
t t
dt t
t dt t
t t
t dt t t
5) Give three examples of isometry onR2.They should be a reflector, a rotation, and a composition of the two You should specifically write down the matrix for each case
- Solution :
Example : Rotation
Let P be the point (x,y) where x=rcosϕ and y=rsinϕ
Rotating with the angle θ from P(x,y) to P'(X,Y)
Rotation through θ about the origin
where ( )
( )
+ =
+ =
+ =
− =
− =
− =
θ θ
θ ϕ θ
ϕ θ
ϕ
θ θ
θ ϕ θ
ϕ θ
ϕ
sin cos
sin cos cos
sin sin
sin cos
sin sin cos
cos cos
y x
rs r
r Y
y x
r r
(11)We can write down matrix form
−
=
⇒
−
=
θ θ
θ θ
θ θ
θ θ
θ
cos sin
sin cos
cos sin
sin cos
R y
x Y
X
Example : Reflector
Let P be the point (x,y) where x=rcosϕ and y=rsinϕ
From the above figure, we have computed two reflection angles are equal to θ −ϕ
2 Reflection in the line
2 tanθ
x y=
We can find the angle θ θ ϕ=θ −ϕ
− + = ∠
2 '
OX P
where ( )
( )
+ =
+ =
+ =
− =
− =
− =
θ θ
ϕ θ ϕ
θ ϕ
θ
θ θ
ϕ θ ϕ
θ ϕ
θ
cos sin
sin cos cos
sin sin
sin cos
sin sin cos
cos cos
y x
r r
r Y
y x
r r
r X
We can write down matrix form
− =
⇒
− =
θ θ
θ θ
θ θ
θ θ
θ
cos sin
sin cos
cos sin
sin cos
M y
x Y
X
Example :Composition of the two
Let A∈R2×2.Write the matrix as :
=
d c
b a A
(12)
) 3 ( 0
) 2 ( 1
) 1 ( 1 2
2
= +
= +
= +
cd ab
d b
c a
From the equation (1), we can write a=cosθ ,c=sinθ for some θ
From the equation (2), we have b=cosϕ ,d =sinϕ for some ϕ
From the equation (3), we see that cosθcosϕ +sinθ.sinϕ=0
( )
cosθ −ϕ =
Thus
=
+ =
− =
+ =
+ =
− =
+ =
=
+ =
+ =
θ θ
π θ
θ π θ
π ϕ
θ θ
π θ
θ π θ
π ϕ
cos 2
3 sin ,
sin 2
3 cos ,
2 3
cos 2
sin ,
sin 2
cos ,
2
d b
case which in
Or
d b
case which in
So
Finally, we have :
From
=
d c
b a
A and the values a, b, c, d are found Thus
−
−
=
θ θ
θ θ
θ θ
θ θ
cos sin
sin cos
cos sin
sin cos
or A
6 ) Find the general solution of the linear difference equation
0 4
4 n+2 − n+1 + n =
U U
U
-Solution :
The characteristic polynomial is :
( )
(2 1) 4
2
= − =
+ − =
ξ ξ ξ ξ ρ
The equation have double root
2 =ξ =
ξ So the general solution has the form :
U c c n
n n
n
+
=
2 1 2
1
2
(13) =
+ +
+
1
1
n n n
n
U U A U
U
where 2×2 ∈R
A Computer n
A using the associated Jordan decomposition Find its limit as
→
n Must the spectral radius of A be less than one?
-Solution :
From the equation
=
+ +
+
1
1
n n n
n
U U A U
U
n n
n n
n n
U U
U U
U U
4 1 0
4
4 +2 − +1+ = ⇒ +2 = +1 −
− =
− =
⇒
=
− =
+ +
+
+ +
+ +
+
+ +
1
1
2
1
1
2
1 4 1
1 0
4 1 4 1
n n n
n n n
n
n n n
n n n
n
U U U
U U
U U
U U A U
U U
U U
We set
^ ^
^ ^
1
1 ^
U A U U
A U
U U
U n n n n n
n
= ⇒ =
⇒
= + +
+
According to in the part a) Using the Jordan decomposition to compute n
A
The equation have the root
( )
( )
2
4 1
4
1
2
0
1 4
2
2
= ⇒ = + − =
− −
− = −
= ⇒
= − =
+ − =
λ λ
λ λ λ λ
λ λ
λ λ λ ρ
(14)We have the Jordan matrix :
=
2
1
J
We have AR = RJ with R=[r1 r2 ]
With eigenvalues :
2 =λ =
λ
To apply Jordan decomposition :
( )
( )
= ⇒ =
− = −
= ⇒ =
− = −
2
1
1
2
2 2
2
1
1
r r r I A r I A
r r
I A r I A
λ λ
So [ ]
− =
− = ⇒ =
= −
2
0 2
1
1
2 1
2
1 r R
r R
We observe that the matrix A can presented by the matrices R, R−1,J
=
− =
2 1 0
1 2 1 ;
1 4
1 1
0
J A
Clearly,
1
2 1 4 1
0 2 1
2 1 0
1 2 1
2 1
0 2 1
4
1 1
0
−
=
−
=
−
= RJR
A
We have
1
1
1
1
−
− −
− −
=
⇒
= =
=
R RJ A
R RJ RJR
RJR A
RJR A
(15)
=
= ⇒
= = ⇒ =
=
⇒ −
3
2
2 2
1
0
1
2
0
1
1
1
1
λ λ λ λ λ λ
λ λ
λ λ λ λ λ λ λ λ
λ
J
J J
have We
R R
A
n n
− =
=
= ⇒
= − − −
−
2
0 2
1 2
1
2
1
, 1
1
R and R
where R
n R
A n
J Thus
n n n n
n n n n
λ λ λ
When n→0, we need to compute
=
− = =
= − − −
→
→ 0 1
0
0 1
0
0
0
1
2
1 lim
lim 1
0
0 R R R
n R
A
n n n n
n n
The eigenvalue of the matrix A is
2
4 1
4
1
2
= = ⇒ = + − =
− −
− =
−λI λ λ λ λ λ λ
A
The spectral radius of A be less than one
( )
2 max = < =
⇒ i
i
A λ
ρ
7) Recall that the rank of a matrix is equal to the number of linearly independent columns
Prove that A∈Rn×n has rank one if and only if there exist nonzero vectors u,v∈Rn such that T
uv
A= .To what extent is there flexibility in the choice of u and v ?
-Proof
(16)n i
for a
a a
a a
ai,1 =αi−1 1,1, i,2 =αi−1 1,2, . . . , i,n =αi−1 1,n , = 2,. . . ,
Consider the vectors u,v defined as : [ ] [ ]T n
n n
T
n C and u a a C
v= 1,α1, ,α −1 ∈ = 1,1, , 1, ∈
From the definition u,v satisfy the relation A=uvTand they are two non-zero vectors Otherwise, A would be the zero-matrix, contradiction with A is having rank
+ The representation is unique up to a constant if A=u1v1T =u2v2T then 2
2
1 su ,v s v
u = = −
b) Show that if T
uv I
G= − is nonsingular, then G−1 has the form T
uv
I−β Give a formula for β
-Solution :
We see that GG−1 =I
and can check
( )( )
( )
( T T ) T
T T T
T
T T T
T
T T
uv u v uv I
uv u v uv
uv I
uv uv uv
uv I
uv I uv I I GG
β β
β β
β β
β
− + − =
+ − −
=
+ − −
=
− −
= =
−1
Thus, If GG−1 = I
then
1
− = ⇒ = −
+
v u u
v
uvT β T β T
β
Suppose G =I−uvT is nonsingular , we will show that G−1 = I−βuvT
where
1
− =
v uT
β
G is nonsingular, suppose T =1
uv ⇒G2 =(I−uvT)(I −uvT)= I−2uvT +uvTuvT =I −uvT =G
Since G is nonsingular then = = − T ⇒ T =0
uv uv
I G
I This is a contradition
1 ≠
⇒uTv We have
( )( )
( )
I
uv uv I
uv v
u uv uv
I
uv v u
uv uv
v u
uv I
v u
uv I uv I
uv I uv I GG
T T
T T
T T
T T
T T
T T
T T T
T T
=
+ − =
− + − + − =
− + − − − =
− − −
=
− −
=
−
1
1
1
1 β
“ The mathematical methods in Applied Mathematics are wonderful ! “
(17)1 Lectures, U.S.A
2.D.S.Watkins, Fundamentals of Matrix Computations,Wiley, 3rd Edition,2010 3.R.LeVeque’s 2006 Lecture Notes