Overview on Aitken Algorithm to spend up the rate of convergence of sequences that converges linearly.

1. Overview of Aitken Method
Numerical Analysis
Isaac Amornortey Yowetu
NIMS-Ghana
June 28, 2020

2. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Outline
Background of Aitken ∆2
Method
Derivation of Aitken’s ∆2
Method
Working With the Aitken’s Algorithm
Finding root with Fixed-Point Iteration
Fixed-Point Iteration with Aitken Method
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

3. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Background of Aitken ∆2
Method
This method is used to accelerate the rate of convergence
of sequences generated from an iterative method.
It is most useful for accelerating the convergence of
sequences that converges linearly.
Aitken ∆2
Method is said to be named after Alexander
Aitken (1926).
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

4. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Derivation of Aitken’s ∆2
Method
suppose xn − α = δn and ∃λ ∈ R, such that
δn+1 = λδn (1)
δn+2 = λδn+1 (2)
We divide eqn(1) by eqn(2)
δn+1
δn+2
=
λδn
λδn+1
(3)
λδ2
n+1 = λδn · δn+2 (4)
δ2
n+1 = δn · δn+2 (5)
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

5. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Derivation Continues...
Substitute δn = xn − α
(xn+1 − α)2
= (xn − α) · (xn+2 − α) (6)
x2
n+1 − 2α · xn+1 + α2
= xn · xn+2 − α · xn+2 − α · xn + α2
(7)
x2
n+1 − 2α · xn+1 = xn · xn+2 − α · xn+2 − α · xn
(8)
α · xn+2 − 2α · xn+1 + α · xn = xn · xn+2 − x2
n+1 (9)
α(xn+2 − 2xn+1 + xn) = xn · xn+2 − x2
n+1 (10)
α =
xn · xn+2 − x2
n+1
xn+2 − 2xn+1 + xn
(11)
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

6. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Derivation Continues....
α =
xn(xn+2 − 2xn+1 + xn) − xn(−2xn+1 + xn) − x2
n+1
xn+2 − 2xn+1 + xn
(12)
α =
xn(xn+2 − 2xn+1 + xn) + 2xnxn+1 − x2
n − x2
n+1
xn+2 − 2xn+1 + xn
(13)
α =
xn(xn+2 − 2xn+1 + xn) − (x2
n+1 − 2xnxn+1 + x2
n )
xn+2 − 2xn+1 + xn
(14)
α =
xn(xn+2 − 2xn+1 + xn) − (xn+1 − xn)2
xn+2 − 2xn+1 + xn
(15)
α = xn −
(xn+1 − xn)2
xn+2 − 2xn+1 + xn
(16)
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

7. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Working With the Aitken’s ∆2
Algorithm
Step 1. Get x0
Step 2. Solve for
x1 = g(x0) and x2 = g(x1)
Step 3. Solve for x3 using Aitken’s method to accelerate the
convergence
x3 = x0 −
(x1 − x0)2
x2 − 2x1 + x0
Step 4. Set x3 = x0 and start the iteration again.
Step 5. Continue till convergence is obtained.
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

8. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Finding root with Fixed-Point Iteration
Example 1
Find a root of xex
= 1 on I = [0, 1] using fixed-point iteration.
Considering our f (x) = xex
− 1 = 0
Solution
We could express f (x) as x = e−x
y1 = x and y2 = g(x) = e−x
g(x) = e−x
(17)
g (x) = −e−x
(18)
choose x0 = 0.5 ∈ [0, 1]
|g (0.5)| = | − e−0.5
| = 0.6065 < 1. Hence, g(x) is good
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

9. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Solution Continue...
x1 = g(x0) = e−0.5
= 0.6065 (19)
x2 = g(x1) = 0.5452 (20)
x3 = g(x2) = 0.5797 (21)
x4 = g(x3) = 0.5601 (22)
x5 = g(x4) = 0.5712 (23)
x6 = g(x5) = 0.5649 (24)
x7 = g(x6) = 0.5684 (25)
x8 = g(x7) = 0.5664 (26)
x9 = g(x8) = 0.5676 (27)
After successive iterations, the approximated value converges
x∗
= 0.5671
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

10. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Finding root with Fixed-Point Iteration & Aitken
Example 2
Find a root of xex
= 1 on I = [0, 1] using fixed-point iteration
with Aitken’s method.
Considering our f (x) = xex
− 1 = 0
Solution
From eqn(17), g(x) = e−x
Using x0 = 0.5, x1 = 0.6065, x2 = 0.5452
x3 = 0.5 −
(0.6065 − 0.5)2
0.5452 − 2(0.6065) + 0.5
= 0.5676 (28)
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

11. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
Solution Continues...
x4 = g(x3) = e−0.5676
= 0.5669 (29)
x5 = g(x4) = 0.5672 (30)
Using x3 = 0.5676, x4 = 0.5669, x5 = 0.5672
x6 = 0.5676 −
(0.5669 − 0.5676)2
0.5672 − 2(0.5669) + 0.5676
= 0.5671 (31)
After successive iterations, the approximated value converges
x∗
= 0.5671
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method

12. Background of Aitken ∆2
Method Derivation of Aitken’s ∆2
Method Working With the Aitken’s Algorithm Finding ro
End
THANK YOU
Isaac Amornortey Yowetu NIMS-Ghana
Overview of Aitken Method