Добавил:
Upload Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:
Lecture Notes on Solving Large Scale Eigenvalue Problems.pdf
Скачиваний:
49
Добавлен:
22.03.2016
Размер:
2 Мб
Скачать
☆

136 CHAPTER 7. SIMULTANEOUS VECTOR OR SUBSPACE ITERATIONS

has the order n = 1095. Subspace iteration is applied with five vectors as an inverse iteration to

L−1AL−T (Lx) = λ(Lx), B = LLT . (Cholesky factorization)

X(0) is chosen to be a random matrix. Here, we number the eigenvalues from small to big. The smallest six eigenvalues of (7.10) shifted by 0.01 to the right are

ˆ

= 0.01,

ˆ

= 0.066635,

λ1

λ4

ˆ

= 0.022690,

ˆ

= 0.126631,

λ2

λ5

ˆ

= 0.054385,

ˆ

= 0.147592.

λ3

λ6

and thus the ratios of the eigenvalues that determine the rate of convergence are

ˆ

ˆ

2

= 0.194,

ˆ

ˆ

2

= 0.277,

(λ1

/λ2)

 

(λ4

/λ5)

 

ˆ

ˆ

2

= 0.174,

ˆ

ˆ

2

= 0.736,

(λ2

/λ3)

 

(λ5

/λ6)

 

ˆ

ˆ

2

= 0.666.

 

 

 

 

(λ3

/λ4)

 

 

 

 

 

So, the numbers presented in Table 7.1 reflect quite accurately the predicted rates. The numbers in column 6 are a little too small, though.

The convergence criterion

max k(I − X(k)X(k) )x(ik−1)k ≤ ǫ = 10−5

1≤i≤p

was not satisfied after 50 iteration step.

7.3Accelerating subspace iteration

Subspace iteration potentially converges very slowly. It can be slow even it one starts with a subspace that contains all desired solutions! If, e.g., x(0)1 and x(0)2 are both elements in R([u1, u2]), the the vectors x(ik), i = 1, 2, . . ., still converge linearly towards u1 und u2 although they could be readily obtained from the 2 × 2 eigenvalue problem,

2

 

h

 

 

i

x1(0)

 

# A

(0)

(0)

 

"x(0)

 

x1

, x2

y = λy

The following theorem gives hope that the convergence rates can be improved if one proceeds in a suitable way.

Theorem 7.6 Let X Fn×p as in Theorem 7.2. Let ui, 1 ≤ i ≤ p, be the eigenvectors corresponding to the eigenvalues λ1, . . . , λp of A. Then we have

 

 

 

 

 

 

λi

 

k

min

 

 

 

 

 

 

 

sin

(ui, x)

≤ c λp+1

 

 

x R(X(k))

 

 

Proof. In the proof of Theorem 7.2 we have seen that

 

 

R(X(k)) = R U

S(k)

,

S(k) F(n−p)×p,

 

 

Ip

 

 

 

 

 

7.3. ACCELERATING SUBSPACE ITERATION

137

where

= sij λp+i

, 1 ≤ i ≤ n − p, 1 ≤ j ≤ p.

 

sij

 

(k)

 

λj

 

k

 

 

 

 

But we have

min

x R(X(k))

sin (ui, x) ≤ sin

 

ui, U S(k)

ei ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ip

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

uiui )U

 

 

0

 

 

 

 

 

 

 

/

 

 

 

Ip

 

ei

 

(I

−

 

 

.

 

 

 

 

 

 

 

 

 

S

(k)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s1i(λi/λp+1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

p,i(λi/λn)

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sn

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λi

 

 

 

 

 

 

 

 

 

 

(I

−

uiui ) ui +

 

 

sj−p,i

 

 

 

 

 

 

 

 

uj

 

 

 

 

 

 

λ

p+j

 

 

 

 

≤

 

 

 

 

 

 

j=p+1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

X

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λi

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−

 

2

λi

 

 

 

−

 

2

 

 

 

 

 

 

 

 

 

 

 

n

p

 

2k

 

 

 

 

k

 

n

 

 

p

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

uX

 

 

 

 

 

 

 

 

 

uX

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

 

 

 

 

t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= v

 

 

sji

2k

 

 

 

 

 

v

 

 

 

sji.

 

 

 

 

 

 

 

 

 

 

 

uj=1

 

λp+j

≤ λp+1 uj=1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

These considerations lead to the idea to complement Algorithm 7.1 by a so-called Rayleigh-Ritz step. Here we give an ‘inverted algorithm’ to compute the smallest eigenvalues and corresponding eigenvectors of a symmetric/Hermitian matrix. For the corresponding nonsymmetric algorithm see [1].

Algorithm 7.2 Subspace or simultaneous inverse iteration combined with Rayleigh-Ritz step

1:Let X Fn×p with X X = Ip:

2:Set X(0) := X.

3:for k = 1, 2, . . . do

4:Solve AZ(k) := X(k−1)

5:

Q(k)R(k) := Z(k)

 

 

 

/* QR factorization of Z(k) (or modified Gram–Schmidt) */

6:

ˆ (k)

:= Q

(k)

 

AQ

(k)

,

 

 

 

 

 

 

H

 

 

 

 

 

 

 

 

 

 

7:

ˆ (k)

=: F

(k)

Θ

(k)

F

(k)

ˆ (k)

 

Fp

×

p

*/

H

 

 

 

 

/* Spectral decomposition of H

 

 

8:X(k) = Q(k)F (k).

9:end for

Remark 7.1. The columns x(ik) of X(k) are called Ritz vectors and the eigenvalues ϑ(1k) ≤ · · · ≤ ϑ(pk) in the diagonal of Θ are called Ritz values. According to the RayleighRitz principle 2.19 we have

λi ≤ ϑ(ik) 1 ≤ i ≤ p, k > 0.

138 CHAPTER 7. SIMULTANEOUS VECTOR OR SUBSPACE ITERATIONS

ˆ

The solution of the full eigenvalue problems H(k)y = ϑy is solved by the symmetric QR algorithm.

ˆ

The computation of the matrix H(k) is expensive as matrix-vector products have to be executed. The following considerations simplify matters. We write X(k) in the form

 

X(k) = Z(k)G(k),

G(k) Fp×pnonsingular

Because X(k) must have orthonormal columns we must have

 

(7.11)

G(k) Z(k) Z(k)G(k) = Ip.

 

Furthermore, the columns of Z(k)G(k) are the Ritz vectors in R(A−kX) of A2,

 

G(k) Z(k) A2Z(k)G(k) =

(k)−2,

 

where

(k) is a diagonal matrix. Using the definition of Z(k)

in Algorithm 7.2 we see that

 

G(k) X(k−1) X(k−1)G(k) = G(k) G(k) =

(k)−2,

and that Y (k) := G(k) (k) is orthogonal. Substituting into (7.11) gives

 

Y (k) Z(k) Z(k)Y (k) =

(k)2.

 

So, the columns of Y (k) are the normalized eigenvectors of H(k) := Z(k) Z(k).

Thus we obtain a second variant of the inverse subspace iteration with Rayleigh-Ritz step.

Algorithm 7.3 Subspace or simultaneous inverse vector iteration combined with Rayleigh-Ritz step, version 2

1:Let X Fn×p with X X = Ip.

2:Set X(0) := X.

3:for k = 1, 2, . . . do

4:AZ(k) := X(k−1);

5:

H(k) := Z(k) Z(k) /* = X(k−1) A−2X(k−1) */

6:

H(k) =: Y (k) (k)2Y (k)

/* Spectral decomposition of H(k) */

7:

X(k) = Z(k)Y (k) (k)−1

/* = Z(k)G(k) */

8:

end for

 

 

 

 

Remark 7.2. An alternative to Algorithm 7.3 is the subroutine ritzit, that has been programmed by Rutishauser [4] in ALGOL, see also [2, p.293].

We are now going to show that the Ritz vectors converge to the eigenvectors, as Theorem 7.6 lets us hope. First we prove

Lemma 7.7 ([2, p.222]) Let y be a unit vector and ϑ F. Let λ be the eigenvalue of A closest to ϑ and let u be the corresponding eigenvector. Let

γ :=

min

 

λ

(A)

−

ϑ

|

 

λi(A)=λ|

i

 

 

 

 

 

 

6

 

 

 

 

 

 

 

and let ψ = (y, u). Then

 

 

 

 

 

 

 

 

 

 

sin ψ

≤

kr(y)k

:=

kAy − ϑyk

,

 

 

γ

 

 

 

γ

 

 

where r(y, ϑ) = Ay − ϑy plays the role of a residual.

7.3. ACCELERATING SUBSPACE ITERATION

139

Proof. We write y = u cos ψ + v sin ψ with kvk = 1. Then

r(y, ϑ) = Ay − ϑy = (A − ϑI)u cos ψ + (A − ϑI)v sin ψ,

= (λ − ϑ)u cos ψ + (A − ϑI)v sin ψ.

Because u (A − ϑI)v = 0, Pythagoras’ theorem implies

kr(y, ϑ)k2 = (λ − ϑ)2 cos2 ψ + k(A − ϑI)vk2 sin2 ψ ≥ γ2kvk2 sin2 ψ.

Theorem 7.8 ([2, p.298]) Let the assumptions of Theorem 7.2 be satisfied.

Let xj(k) =

 

 

 

 

 

(k)

 

 

I

ei

X(k)ej be the j-th Ritz vector as computed be Algorithm 7.3, and let yi

= U

S(k)

(cf. the proof of Theorem 7.2). Then the following inequality holds

 

 

 

 

 

 

 

λi

k

 

 

 

 

 

(k)

(k)

) ≤ c

,

1 ≤ i ≤ p.

 

 

 

 

sin (xi

, yi

 

 

 

 

 

λp+1

 

 

 

 

Proof. The columns of U

Ip

form a basis of R(X(k)). Therefore, we can write

S(k)

(k)

Ip

ti,

ti Fp.

xi = U

S(k)

Instead of the special eigenvalue problem

X(k−1) A−2X(k−1)y = H(k)y = µ−2y

in the orthonormal ‘basis’ X(k) we consider the equivalent eigenvalue problem

(7.12)

hIp, S(k) i UA−2U

Ip

t = µ−2 hIp, S(k) i

Ip

t.

S(k)

S(k)

Let (µ, t) be an eigenpair of (7.12). Then we have

 

0 = hIp, S(k) i UA−2U

S(k)

t − µ−2 hIp, S(k) i S(k)

 

 

 

 

 

 

 

Ip

 

 

 

 

 

 

 

 

 

Ip

 

= Λ1−2 + S(k) Λ2−2S(k) t − µ−2 Ip + S(k) S(k) t,

 

(7.13)

= (Λ1−2 − µ−2I) + S(k) (Λ2−2 − µ−2I)S(k) t

 

 

 

 

1

−

 

 

 

 

2

k

2

 

−

 

 

k

 

=

(Λ−2

 

 

µ−2I) + ΛkS(0) Λ−k

(Λ−2

 

 

µ−2I)Λ−kS(0)

Λk

 

= (Λ1−2 − µ−2I) +

1

 

 

 

 

 

 

 

 

 

2

1

 

λp+1 Λ1 Hk

λp+1 Λ1 ! t

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

1

 

 

 

with

 

 

 

 

= λ2k S(0) Λ−k

 

 

 

 

 

 

 

 

 

 

 

 

H

k

(Λ−2

−

µ−2I)Λ−kS(0).

 

 

 

 

 

p+1

2

 

2

 

 

 

 

2

 

 

As the largest eigenvalue of Λ−2 1 is 1/λp+1, Hk is bounded,

t

t

kHkk ≤ c1 k > 0.

140 CHAPTER 7. SIMULTANEOUS VECTOR OR SUBSPACE ITERATIONS

Thus,

 

 

 

 

 

 

 

→∞

 

 

k

k

 

 

 

 

 

1

 

 

1

 

 

 

 

 

 

 

Λ1

 

Hk

Λ1

t

−−−→

0.

 

 

 

 

 

 

 

 

λp+1

 

 

λp+1

!

 

 

 

 

 

 

 

 

 

 

 

λ

 

 

Therefore, in (7.13) we can interpret this expression as an perturbation of the diagonal matrix Λ−2 2 − µ−2I. For su ciently large k (that may depend on i) there is a µi that is close to λi, and a ti that is close to ei. We now assume that k is so big that

|µ−i 2 − λ−i 1| ≤ ρ := 12 min |λ−i 2 − λ−j 2|

λj =6 λi

such that µ−2

is closer to λ−2 than to any other λ−2

, j = i.

 

 

 

i

i

j

6

 

 

 

 

We now consider the orthonormal ‘basis’

 

 

 

 

 

 

B =

Ip

 

1/2

 

 

 

 

 

 

 

 

 

S(k) Ip + S(k) S(k) − .

 

1/2ti

If (µi, ti) is an eigenpair of (7.12) or (7.13), respectively, then

µi−2, Ip + S(k) S(k)

 

is an eigenpair of

 

 

 

 

 

(7.14)

 

B A−2Bt = µ−2t.

 

 

 

 

As, for su ciently large k, (λi−2, ei) is a good approximation of the eigenpair (µi−2, ti)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/2

Ip + S(k) S(k)

 

1/2ei

is a good approximation to the eigenpair

of (7.13), then also λi−2

,

 

 

µi−2, Ip + S(k) S(k)

 

 

ti of (7.14). We now apply Lemma 7.7 with

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

γ = ρ,

 

 

 

 

ϑ = λi−2,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

y = Ip + S(k) S(k) 1/2ei/ Ip + S(k) S(k) 1/2ei ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

u = Ip

+ S

 

 

 

S

 

 

 

 

 

 

ti/

 

Ip + S

 

 

S

 

 

 

 

ti

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(k)

 

 

(k)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(k)

 

 

 

(k)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Now we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/2ei

 

 

 

 

 

 

 

 

 

 

 

1/2ei

 

 

 

 

 

kr(y)k =

(B A−2 B − λi−2I) Ip + S(k) S(k)

 

 

 

Ip + S(k) S(k)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

−

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(k

 

 

 

 

 

 

≤

 

Ip + S

 

 

S

 

 

 

 

 

 

 

 

Ip, S UA− U

 

 

(k)

−

 

 

 

Ip + S

 

S

 

 

ei

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h

 

 

 

 

 

 

 

 

 

i

 

 

 

 

 

 

 

 

S

 

 

λi

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(k)

 

 

(k)

 

 

2

 

 

 

 

 

 

(k)

 

 

 

 

 

 

2

 

 

 

 

Ip

 

 

 

 

 

1

 

 

)

 

 

 

 

 

 

≤

 

Ip + S

 

 

S

 

 

 

 

 

 

 

 

 

Λ1−

 

 

 

−

λi−

 

I + λp−+1Λ1

 

Hk

λp−+1

Λ1

 

 

ei

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(k)

(k)

 

−1/2

 

k

 

 

 

2

 

 

 

 

 

2

 

 

 

 

 

 

1

 

 

 

 

k

 

 

 

1

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

≤

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λp

 

k

 

 

λi

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

p−+1

 

 

 

1

 

 

 

k

 

 

p−+1

1

 

 

i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λ

Λ

 

H

λ

 

e

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Λ

 

 

 

 

e

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λ−1

 

Λ

 

 

 

 

 

H

kk

λ−1

 

Λ

 

 

 

i

 

 

c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

≤

 

1

 

k

 

1

 

≤

1 λp+1

λp+1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Then,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Lemma 7.7 implies that

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ip + S(k) S(k)

 

ei ≤ ρ1

λp+1

.

sin (xi

 

, yi

 

) = sin

 

Ip + S(k) S(k)

 

 

 

ti,

 

 

 

(k)

 

(k)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

1/2

 

 

c

 

 

λi

k

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]