Departamento de Ciencias Matematicas

Universidad de Puerto Rico, Mayaguez

MATE-6672

Asignacion 1

Paul Castillo

018324040

Paul Castillo, ID-018324040 1

Problema 1.

P-1.1 Jacobi. Notice that the usual theorem don’t apply since the matrix is neither strictlydiagonally dominant nor symmetric. The analysis reduces to the study of the spectral radius of theiteration matrix M−1N = I −M−1A:

I −M−1A =

1 0 00 1 00 0 1

− 1/5 0 0

0 1/2 00 0 1

5 −1 0−1 2 −1

0 −3/2 1

=

0 1/5 01/2 0 1/2

0 3/2 0

Then

p(λ) = det

−λ 1/5 01/2 −λ 1/2

0 3/2 −λ

= −λ(λ2 − 17/20)

It follows that ρ(M−1N) < 1 then Jacobi iteration converges for any initial value.

P-1.2 Forward Gauss-Seidel. In this case we have

M =

5 0 0−1 2 00 −3/2 1

M−1 =

1/5 0 01/10 1/2 03/20 3/4 1

.Then

I −M−1A =

1 0 00 1 00 0 1

− 1/5 0 0

1/10 1/2 03/20 3/4 1

5 −1 0−1 2 −1

0 −3/2 1

=

0 1/5 00 1/10 1/20 3/20 3/4

Using Gershgorin eigenvalue estimate, if λ ∈ σ(M−1N) then we have

|λ| ≤ 1/5 or |λ− 1/10| ≤ 1/2 or |λ− 3/4| ≤ 3/20

therefore−1/5 ≤ λ ≤ 1/5 or − 2/5 ≤ λ ≤ 3/5 or 3/5 ≤ λ ≤ 9/10

then ρ(M−1N) < 1, the iteration of Forward Gauss-Seidel converges for any initial value.

Problema 2.

P-2.1 Since we have 3 weights and nodes the maximum degree of exactness is 5. They shouldsatisfy te following non linear system of equations

ω0 + ω1 + ω2 = 2 (1)

ω0x0 + ω1x1 + ω2x2 = 0 (2)

ω0x20 + ω1x

21 + ω2x

22 = 2/3 (3)

ω0x30 + ω1x

31 + ω2x

32 = 0 (4)

ω0x40 + ω1x

41 + ω2x

42 = 2/5 (5)

ω0x50 + ω1x

51 + ω2x

52 = 0 (6)

2 Paul Castillo, ID-018324040

We can simplify this system by assuming a symmetric distributions of the nodes: x1 = 0 andx2 = −x0. Then the system reduces to

ω0 + ω1 + ω2 = 2 (7)

(ω0 − ω2)x0 = 0 (8)

(ω0 + ω2)x20 = 2/3 (9)

(ω0 − ω2)x30 = 0 (10)

(ω0 + ω2)x40 = 2/5 (11)

(ω0 − ω2)x50 = 0 (12)

From Eqn (??) we either haveω0 = ω2 or x0 = 0.

By symmetry x0 6= 0, then ω0 = ω2. And the systems now reduces to

2ω0 + ω1 = 2 (13)

ω0x20 = 1/3 (14)

ω0x40 = 1/5 (15)

Combining Eqns (??) and (??) we have

x20 = 3/5

since x0 < 0 we get x0 = −√

3/5 and ω0 = 5/9; then x2 = −x0 =√

3/5 and ω2 = ω0 = 5/9.Finally from Eqn (??) we get ω1 = 2− 2ω0 = 8/9. The quadrature reads

(ω0, x0) = (5/9,−√

3/5) (ω1, x1) = (8/9, 0) (ω2, x2) = (5/9,√

3/5).

Problema 3.

P-3.1 Aquı va la respuesta del primer inciso del problema 3.

P-3.2 Aquı va la respuesta del segundo inciso del problema 3.

P-3.3 Aquı va la respuesta del tercer inciso del problema 3.