Aitken Interpolation, Summaries of Mathematics

Solve an equation using Aitken Interpolation. It is a Note with Examples.

Typology: Summaries

2022/2023

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Aitken Interpolation
After reading this topic, you should be able to:
1. derive the Aitken interpolation method,
2. apply the Aitken iteration formula
Overview
The Aitken iteration formula is similar to Newtonโ€™s formula. This formula is also successively
generating the higher degree interpolation polynomials. But, the most advantage of this formula is,
it can be easily programmed for a computer.
Assume that the points ๐‘ฅ๐‘ฅ0,๐‘ฅ๐‘ฅ1,๐‘ฅ๐‘ฅ2,โ€ฆ,๐‘ฅ๐‘ฅ๐‘›๐‘› and values ๐‘ฆ๐‘ฆ0=๐‘“๐‘“(๐‘ฅ๐‘ฅ0),๐‘ฆ๐‘ฆ1=๐‘“๐‘“(๐‘ฅ๐‘ฅ1),๐‘ฆ๐‘ฆ2=๐‘“๐‘“(๐‘ฅ๐‘ฅ2),โ€ฆ , ๐‘ฆ๐‘ฆ๐‘›๐‘›=
๐‘“๐‘“(๐‘ฅ๐‘ฅ๐‘›๐‘›), of the function ๐‘“๐‘“(๐‘ฅ๐‘ฅ) being interpolated are given. The interpolating polynomial ๐‘ƒ๐‘ƒ๐‘›๐‘›(๐‘ฅ๐‘ฅ) can be
evaluated by using the following ๐‘›๐‘›-iterative algorithm.
First Iteration
At the beginning, ๐‘›๐‘› polynomials ๐‘ƒ๐‘ƒ0๐‘—๐‘—(๐‘ฅ๐‘ฅ) of degree 1 are created in the following manner:
๐‘ƒ๐‘ƒ0๐‘—๐‘—(๐‘ฅ๐‘ฅ)=1
๐‘ฅ๐‘ฅ๐‘—๐‘—โˆ’๐‘ฅ๐‘ฅ0๏ฟฝ๐‘ฆ๐‘ฆ0๐‘ฅ๐‘ฅ0โˆ’๐‘ฅ๐‘ฅ
๐‘ฆ๐‘ฆ๐‘—๐‘—๐‘ฅ๐‘ฅ๐‘—๐‘—โˆ’๐‘ฅ๐‘ฅ๏ฟฝ,๐‘—๐‘—= 1, 2, 3, โ€ฆ , ๐‘›๐‘›.
That is, each of these polynomials passes through the points (๐‘ฅ๐‘ฅ0,๐‘ฆ๐‘ฆ0) and ๏ฟฝ๐‘ฅ๐‘ฅ๐‘—๐‘—,๐‘ฆ๐‘ฆ๐‘—๐‘—๏ฟฝ, ๐‘—๐‘—= 1, 2, 3, โ€ฆ , ๐‘›๐‘›.
Second Iteration
Polynomials ๐‘ƒ๐‘ƒ0๐‘—๐‘—(๐‘ฅ๐‘ฅ) evaluated above allow us to create the set of ๐‘›๐‘›โˆ’1 polynomials of degree 2:
๐‘ƒ๐‘ƒ01๐‘—๐‘—(๐‘ฅ๐‘ฅ)=1
๐‘ฅ๐‘ฅ๐‘—๐‘—โˆ’๐‘ฅ๐‘ฅ1๏ฟฝ๐‘ƒ๐‘ƒ01 ๐‘ฅ๐‘ฅ1โˆ’๐‘ฅ๐‘ฅ
๐‘ƒ๐‘ƒ0๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘—โˆ’๐‘ฅ๐‘ฅ๏ฟฝ,๐‘—๐‘—= 2, 3, 4, โ€ฆ , ๐‘›๐‘›.
The quadratic polynomial passing through the points (๐‘ฅ๐‘ฅ0,๐‘ฆ๐‘ฆ0),(๐‘ฅ๐‘ฅ1,๐‘ฆ๐‘ฆ1) and ๏ฟฝ๐‘ฅ๐‘ฅ๐‘—๐‘—,๐‘ฆ๐‘ฆ๐‘—๐‘—๏ฟฝ is determined.
Third Iteration
At the third stage of iteration:
๐‘ƒ๐‘ƒ012๐‘—๐‘—(๐‘ฅ๐‘ฅ)=1
๐‘ฅ๐‘ฅ๐‘›๐‘›โˆ’๐‘ฅ๐‘ฅ๐‘›๐‘›โˆ’1๏ฟฝ๐‘ƒ๐‘ƒ012 ๐‘ฅ๐‘ฅ2โˆ’๐‘ฅ๐‘ฅ
๐‘ƒ๐‘ƒ01๐‘—๐‘—๐‘ฅ๐‘ฅ๐‘—๐‘—โˆ’๐‘ฅ๐‘ฅ๏ฟฝ,๐‘—๐‘—= 3, 4, โ€ฆ , ๐‘›๐‘›.
Successive Iterations
In successive iterations the polynomials of higher degrees are evaluated analogously, as shown in
Table 1.
Last Iteration
Only one polynomial ๐‘ƒ๐‘ƒ0123โ€ฆ๐‘›๐‘›=๐‘ƒ๐‘ƒ๐‘›๐‘›(๐‘ฅ๐‘ฅ) of degree ๐‘›๐‘› is evaluated in the last (๐‘›๐‘›) iteration.
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Aitken Interpolation

After reading this topic, you should be able to:

  1. derive the Aitken interpolation method,
  2. apply the Aitken iteration formula

Overview

The Aitken iteration formula is similar to Newtonโ€™s formula. This formula is also successively generating the higher degree interpolation polynomials. But, the most advantage of this formula is,

it can be easily programmed for a computer.

Assume that the points ๐‘ฅ๐‘ฅ 0 , ๐‘ฅ๐‘ฅ 1 , ๐‘ฅ๐‘ฅ 2 , โ€ฆ , ๐‘ฅ๐‘ฅ๐‘›๐‘› and values ๐‘ฆ๐‘ฆ 0 = ๐‘“๐‘“(๐‘ฅ๐‘ฅ 0 ), ๐‘ฆ๐‘ฆ 1 = ๐‘“๐‘“(๐‘ฅ๐‘ฅ 1 ), ๐‘ฆ๐‘ฆ 2 = ๐‘“๐‘“(๐‘ฅ๐‘ฅ 2 ), โ€ฆ , ๐‘ฆ๐‘ฆ๐‘›๐‘› =

๐‘“๐‘“(๐‘ฅ๐‘ฅ๐‘›๐‘›), of the function ๐‘“๐‘“(๐‘ฅ๐‘ฅ) being interpolated are given. The interpolating polynomial ๐‘ƒ๐‘ƒ๐‘›๐‘›(๐‘ฅ๐‘ฅ) can be evaluated by using the following ๐‘›๐‘›-iterative algorithm.

First Iteration

At the beginning, ๐‘›๐‘› polynomials ๐‘ƒ๐‘ƒ0๐‘—๐‘—(๐‘ฅ๐‘ฅ) of degree 1 are created in the following manner:

๐‘ฆ๐‘ฆ๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ ,^ ๐‘—๐‘—^ = 1, 2, 3, โ€ฆ ,^ ๐‘›๐‘›.

That is, each of these polynomials passes through the points (๐‘ฅ๐‘ฅ 0 , ๐‘ฆ๐‘ฆ 0 ) and ๏ฟฝ๐‘ฅ๐‘ฅ๐‘—๐‘—, ๐‘ฆ๐‘ฆ๐‘—๐‘— ๏ฟฝ, ๐‘—๐‘— = 1, 2, 3, โ€ฆ , ๐‘›๐‘›.

Second Iteration

Polynomials ๐‘ƒ๐‘ƒ0๐‘—๐‘—(๐‘ฅ๐‘ฅ) evaluated above allow us to create the set of ๐‘›๐‘› โˆ’ 1 polynomials of degree 2:

๐‘ƒ๐‘ƒ0๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ ,^ ๐‘—๐‘—^ = 2, 3, 4, โ€ฆ ,^ ๐‘›๐‘›.

The quadratic polynomial passing through the points (๐‘ฅ๐‘ฅ 0 , ๐‘ฆ๐‘ฆ 0 ), (๐‘ฅ๐‘ฅ 1 , ๐‘ฆ๐‘ฆ 1 )^ and ๏ฟฝ๐‘ฅ๐‘ฅ๐‘—๐‘—, ๐‘ฆ๐‘ฆ๐‘—๐‘— ๏ฟฝ is determined.

Third Iteration

At the third stage of iteration:

๐‘ƒ๐‘ƒ01๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— โˆ’ ๐‘ฅ๐‘ฅ ๏ฟฝ^ ,^ ๐‘—๐‘—^ = 3, 4, โ€ฆ ,^ ๐‘›๐‘›.

Successive Iterations

In successive iterations the polynomials of higher degrees are evaluated analogously, as shown in

Table 1.

Last Iteration

Only one polynomial ๐‘ƒ๐‘ƒ 0123 โ€ฆ๐‘›๐‘› = ๐‘ƒ๐‘ƒ๐‘›๐‘›(๐‘ฅ๐‘ฅ) of degree ๐‘›๐‘› is evaluated in the last (๐‘›๐‘›) iteration.

The entire calculation can be represented as in Table 1 below:

๐‘ฅ๐‘ฅ๐‘—๐‘— ๐‘ฆ๐‘ฆ๐‘—๐‘— ๐‘ƒ๐‘ƒ 0 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 01 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 012 ๐‘—๐‘— โ‹ฏ ๐‘ƒ๐‘ƒ 0123 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 0123 ,โ€ฆ,๐‘—๐‘—โˆ’ 1 ,๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— โˆ’ ๐‘ฅ๐‘ฅ ๐‘ฅ๐‘ฅ 0 ๐‘ฆ๐‘ฆ 0 ๐‘ฅ๐‘ฅ 0 โˆ’ ๐‘ฅ๐‘ฅ ๐‘ฅ๐‘ฅ 1 ๐‘ฆ๐‘ฆ 1 ๐‘ƒ๐‘ƒ 01 ๐‘ฅ๐‘ฅ 1 โˆ’ ๐‘ฅ๐‘ฅ ๐‘ฅ๐‘ฅ 2 ๐‘ฆ๐‘ฆ 2 ๐‘ƒ๐‘ƒ 02 ๐‘ƒ๐‘ƒ 012 ๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ ๐‘ฅ๐‘ฅ 3 ๐‘ฆ๐‘ฆ 3 ๐‘ƒ๐‘ƒ 03 ๐‘ƒ๐‘ƒ 013 ๐‘ƒ๐‘ƒ 0123 ๐‘ฅ๐‘ฅ 3 โˆ’ ๐‘ฅ๐‘ฅ ๐‘ฅ๐‘ฅ 4 ๐‘ฆ๐‘ฆ 4 ๐‘ƒ๐‘ƒ 04 ๐‘ƒ๐‘ƒ 014 ๐‘ƒ๐‘ƒ 0124 ๐‘ฅ๐‘ฅ 4 โˆ’ ๐‘ฅ๐‘ฅ โ‹ฏ โ‹ฏ โ‹ฏ โ‹ฏ โ‹ฏ โ‹ฏ ๐‘ƒ๐‘ƒ 0123 โ€ฆ,๐‘›๐‘›โˆ’ 2 ,๐‘›๐‘›โˆ’ 1 โ‹ฏ ๐‘ฅ๐‘ฅ๐‘›๐‘› ๐‘ฆ๐‘ฆ 6 ๐‘ƒ๐‘ƒ 0 ๐‘›๐‘› ๐‘ƒ๐‘ƒ 01 ๐‘›๐‘› ๐‘ƒ๐‘ƒ 012 ๐‘›๐‘› โ‹ฏ ๐‘ƒ๐‘ƒ 0123 โ€ฆ,๐‘›๐‘›โˆ’ 2 ,๐‘›๐‘› ๐‘ƒ๐‘ƒ 0123 โ€ฆ,๐‘›๐‘›โˆ’ 2 ,๐‘›๐‘›โˆ’ 1 ,๐‘›๐‘› ๐‘ฅ๐‘ฅ๐‘›๐‘› โˆ’ ๐‘ฅ๐‘ฅ

Table 1: Aitkenโ€™s Algorithm

Example 1

From the following table, find the value of log 10 301 using Aitken interpolation method.

๐‘ฅ๐‘ฅ 300 304 305 307 log 10 ๐‘ฅ๐‘ฅ 2.4771 2.4829 2.4843 2.

Solution

In this case, ๐‘ฅ๐‘ฅ = 301, the following table can be form easily.

๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— ๐‘ฆ๐‘ฆ๐‘—๐‘— ๐‘ƒ๐‘ƒ 0 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 01 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 012 ๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— โˆ’ ๐‘ฅ๐‘ฅ 0 300 2.4771 โˆ’ 1 1 304 2.4829 3 2 305 2.4843 4 3 307 2.4871 6

Now we calculate the first iteration:

๐‘ฆ๐‘ฆ 1 ๐‘ฅ๐‘ฅ 1 โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ =

๏ฟฝ2.47712.4829^ โˆ’ 31 ๏ฟฝ = 2.

๐‘ฆ๐‘ฆ 2 ๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ =

๏ฟฝ2.4771^ โˆ’^1

๐‘ฆ๐‘ฆ 3 ๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ =

๏ฟฝ2.4771^ โˆ’^1

After first iteration, the updated table is

๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— ๐‘ฆ๐‘ฆ๐‘—๐‘— ๐‘ƒ๐‘ƒ 0 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 01 ๐‘—๐‘— ๐‘ƒ๐‘ƒ 012 ๐‘—๐‘— ๐‘ฅ๐‘ฅ๐‘—๐‘— โˆ’ ๐‘ฅ๐‘ฅ 0 300 2.4771 โˆ’ 1 1 304 2.4829 2. 47855 3 2 305 2.4843 (^2). 47854 4 3 307 2.4871 2. 47853 6

The second iterations are determined as follows.

๐‘ƒ๐‘ƒ 02 ๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ =

๏ฟฝ2.47855^3

๐‘ƒ๐‘ƒ 03 ๐‘ฅ๐‘ฅ 3 โˆ’ ๐‘ฅ๐‘ฅ๏ฟฝ^ =

๏ฟฝ2.47855^3