Numerical analysis - Appunti + esami svolti

Analisi dei risultati

Come possiamo vedere dai risultati, il metodo di Newton converge a α con una sola iterazione.

Nella parte 2, i valori di α sono 10 e 1.6129 10 .

T (x) è consistente se T (α ) = α e T (α ) = α .

T (α ) = 10 = α

T (α ) = 1.6129 10 = α

Il metodo è consistente.

7. 2 α1′ −1·|T = 2.5163 10 < 1(α )| =1 2−3 α 12 α2′ 0|T ·(α )| = = 8.0922 10 > 12 2−3 α 2

Il metodo è convergente solo per α .

10. −1 0· ̸ ·T (α ) = 3.6104 10 = α T (α ) = 1.6129 10 = α2 1 1 2 2 23

Il metodo è consistente solo per α .

12. −αe 1′ 0√ ·|T = 3.9741 10 > 1(α )| =1 −α−3 2 e 1−αe 2 −1′ √ ·|T = 1.2358 10 < 1(α )| =2

−α−3 2 e 2

The method is convergent only for α .215. −1 0−3.6104 · −1.6129 · ̸T (α ) = 10 = α T (α ) = 10 = α2 1 1 2 2 2

The method is consistent only for α .117. −αe 1′ 0√|T − ·(α )| = = 3.9741 10 > 11 −α−3 2 e 1−αe 2 −1′ √ ·|T − = 1.2358 10 < 1(α )| =2 −α−3 2 e 2

The method is convergent only for α .2 4

Exam 07-09-2020

Lorenzo Sostegni

1 Numerical Linear Algebra

1.1 Part 1

1. (k+1) (k)x = B x + g
2. −1−B = I α P Aα ̸ −3. To satisfy this condition we need det(A ) = 0 for i = 1, . . . , n 1.i
3. A generic linear iterative method is convergent if and only if ρ(B) < 1, where ρ(B) =|λ |max and λ is the i-th eigenvalue of matrix B.i i i

1.2 Part 2 −16true ·1. ρ(B) = 0.7083 < 1. The method is convergent.

iteration is convergent and we have:(k+1)-α x' = Φ (α)lim (k)-α xk→+∞ 2∈3. Let I denote a sufficiently small closed interval containing. Assume g C (I ) withα α' '(k+1) (k)̸ -g (α) = 0. Then the sequence x = Φ(x ), with Φ(x) = x g(x)/g (x), converges for(0) ∈every x I and we have:α '' ''(k+1)-α x Φ (α) g (α)lim = =(k) 2- 2 2 g (α)(α x )k→+∞ 34.2 Part 2 -2.83841. α = 0, α = and α = 2.8384.1 2 3 '|Φ2. In order to obtain convergence we need (x)| < 1.' '' '|Φ |Φ |Φ(α )| = 0.3333 (α )| = (α )| = 2.85801 2 3As we can see we have convergence only in α .1-16 -16-2.1758 · ·3. niter = 33, α = 10 and e = 2.1758 101,c c4Exam 18-01-2021Lorenzo Sostegni1 Numerical linear

is faster in converging to the exact solution than the Jacobi one.

1.2 Part 2

det(A) = 4

det(A) = 15

det(A) = 56

det(A) = 2091

1 2 3 4

-11 0 0 0

0 4 0 0

0 1 0 0

0 0 0 0

-0.25 -11 0 0

0 0 3.75 0

0 0 0 1

0 0 0 0

-0.2667 -10 1 0

0 0 0 3.7333

0 0 0 1

0 0 0 0

-0.2679 -10 0 1

0 0 0 0

0 0 3.7321 0

0 0 0 1

[L, U, P]=LU pivot(A);

1 d=det(L)*det(U);

2 1det(A) = 0

ρ = 0.4330 < 1. The method is convergent.

J10. ρ = 0.1875 < 1. The method is convergent.

J12. niterJ = 33, niterGS = 18.

- -

rel err J = 9.7617e 13, rel err GS = 3.9611e 13.

2 Approximation

2.1 Part 1

Q (f) = 0.5h

e (f) = 0h

Q (f) = 0.25h

e (f) = 0.0833h

2.2 Part 2

2. E f = 2.3809.n 23.

4. E f = 0.6804.

3 Ordinary Differential Equations

3.1 Part 1

From page 147 : -u = u + θ Δt f(t, u) + (1 - θ)

Δt f(t, u)ⁿ⁺¹ = nⁿ⁺¹ - y(t) y'(t), F k+1 ky = kΔt³

Δt f(t, u)ⁿ⁺¹ = nⁿ⁺¹ - y(t) y'(t), B k k-1y = kΔt⁴

Δt f(t, u)ⁿ⁺¹ = nⁿ⁺¹ - y(t) y'(t), C k+1 k-1y = k²Δt^33.2 Part 28. 12. max err F E = 0.0885 max err BE = 0.0864 max err CN = 0.00374 Nonlinear Equations

4.1 Part 11. From page 157. Assume that T satisfies:

• α = T(α)
• ∇T(I) Iα α
• 1 ∈ T C(I)α
• ′|T ≤(α)| K < 1

Then the fixed point iteration is convergent and we have:

(k+1)-α x ′lim = T(α)(k)-α xk→+∞

2. From page 159 Newton’s method formula: ′ -g(x) = g(x) + g'(x)(x - x)(k+1)(k)(k)(k+1)(k)Iteration function: g(x)-T(x) = x ′g'(x)

4.2 Part 22. -1 -1 -1 -1-9.2388 · -3.8268 · · ·α = 10 α = 10 α = 3.8268 10 α = 9.2388 101 2 3 4

5. The derivative is equal: x′

−T (x) = 3182 −2 (x ) 4

From point 1 we can check that: −1−9.2388 ·α = T (α ) = 101 1 ∈T (I ) = [−0.9672, 0.9672] [−1, 1]α′|T |0.5858|(α )| = = 0.5858 < 11T converges to α .1 16. −1 −13−9.2388 · −9.1549 ·α = 10 err = 1018. The derivative is equal: x′T (x) = 3182 −2 (x ) 4From point 1 we can check that: −1·α = T (α ) = 9.2388 104 4 ∈T (I ) = [−0.9672, 0.9672] [−1, 1]α′|T |0.5858|(α )| = = 0.5858 < 14T converges to α .2 49. −1 −13· ·α = 9.2388 10 err = 9.1549 104 5

Exam 01-02-2021

Lorenzo Sostegni

1 Numerical Linear Algebra

1.1 Part 1

|1 | ± |1 | − |1|+ θ| > 1| = 1 + θ| > 1| + = 2

Since the second condition is more stringent we take this system of equations.

2. Solving the previous equation we obtain:−2 ∨1 +

θ < 1 + θ > 2−3 ∨θ < = θ θ > 1 = θ1 23. Since the matrix is strictly diagonally dominant by row and it is tridiagonal it can be2shown that the relationship ρ(B ) = (ρ(B )) holds. Since the smaller is the spectralGS Jradius the faster is the convergence of the method, we can say that the Gauss-Seideliterative method converges faster than the Jacobi method.

1.1.1 Part 2

ρ = 0.6006.

J2. ρ = 0.3608.

GS4. niterJ = 51. −13·5. rel err J = 5.3842 10 .

7. niterGS = 27. −13·8. rel err GS = 6.1355 10 .

2 Approximation

2.1 Part 1

1. (Page 121). 1 (n+1)||E ≤ ||f || ||ω ||f (x)||n n+1∞,[a,b] ∞,[a,b] ∞,[a,b](n + 1)!(n+1)Where f is the derivative of the function and therefore depends only on the function− − −chosen. ω is the nodal polynomial defined as ω (x) = (x x ) (x x ) . . . (x x ).n+1 n+1 0 1 n2. Since the function is a polynomial and the degree

The formula for the interpolating polynomial is: