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Solve an equation using Aitken Interpolation. It is a Note with Examples.
Typology: Summaries
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After reading this topic, you should be able to:
Overview
The Aitken iteration formula is similar to Newtonโs 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 ๐ฅ๐ฅ 0 , ๐ฅ๐ฅ 1 , ๐ฅ๐ฅ 2 , โฆ , ๐ฅ๐ฅ๐๐ and values ๐ฆ๐ฆ 0 = ๐๐(๐ฅ๐ฅ 0 ), ๐ฆ๐ฆ 1 = ๐๐(๐ฅ๐ฅ 1 ), ๐ฆ๐ฆ 2 = ๐๐(๐ฅ๐ฅ 2 ), โฆ , ๐ฆ๐ฆ๐๐ =
๐๐(๐ฅ๐ฅ๐๐), of the function ๐๐(๐ฅ๐ฅ) being interpolated are given. The interpolating polynomial ๐๐๐๐(๐ฅ๐ฅ) can be evaluated by using the following ๐๐-iterative algorithm.
First Iteration
At the beginning, ๐๐ polynomials ๐๐0๐๐(๐ฅ๐ฅ) of degree 1 are created in the following manner:
That is, each of these polynomials passes through the points (๐ฅ๐ฅ 0 , ๐ฆ๐ฆ 0 ) and ๏ฟฝ๐ฅ๐ฅ๐๐, ๐ฆ๐ฆ๐๐ ๏ฟฝ, ๐๐ = 1, 2, 3, โฆ , ๐๐.
Second Iteration
Polynomials ๐๐0๐๐(๐ฅ๐ฅ) evaluated above allow us to create the set of ๐๐ โ 1 polynomials of degree 2:
The quadratic polynomial passing through the points (๐ฅ๐ฅ 0 , ๐ฆ๐ฆ 0 ), (๐ฅ๐ฅ 1 , ๐ฆ๐ฆ 1 )^ and ๏ฟฝ๐ฅ๐ฅ๐๐, ๐ฆ๐ฆ๐๐ ๏ฟฝ is determined.
Third Iteration
At the third stage of iteration:
Successive Iterations
In successive iterations the polynomials of higher degrees are evaluated analogously, as shown in
Table 1.
Last Iteration
Only one polynomial ๐๐ 0123 โฆ๐๐ = ๐๐๐๐(๐ฅ๐ฅ) of degree ๐๐ is evaluated in the last (๐๐) iteration.
The entire calculation can be represented as in Table 1 below:
๐ฅ๐ฅ๐๐ ๐ฆ๐ฆ๐๐ ๐๐ 0 ๐๐ ๐๐ 01 ๐๐ ๐๐ 012 ๐๐ โฏ ๐๐ 0123 ๐๐ ๐๐ 0123 ,โฆ,๐๐โ 1 ,๐๐ ๐ฅ๐ฅ๐๐ โ ๐ฅ๐ฅ ๐ฅ๐ฅ 0 ๐ฆ๐ฆ 0 ๐ฅ๐ฅ 0 โ ๐ฅ๐ฅ ๐ฅ๐ฅ 1 ๐ฆ๐ฆ 1 ๐๐ 01 ๐ฅ๐ฅ 1 โ ๐ฅ๐ฅ ๐ฅ๐ฅ 2 ๐ฆ๐ฆ 2 ๐๐ 02 ๐๐ 012 ๐ฅ๐ฅ 2 โ ๐ฅ๐ฅ ๐ฅ๐ฅ 3 ๐ฆ๐ฆ 3 ๐๐ 03 ๐๐ 013 ๐๐ 0123 ๐ฅ๐ฅ 3 โ ๐ฅ๐ฅ ๐ฅ๐ฅ 4 ๐ฆ๐ฆ 4 ๐๐ 04 ๐๐ 014 ๐๐ 0124 ๐ฅ๐ฅ 4 โ ๐ฅ๐ฅ โฏ โฏ โฏ โฏ โฏ โฏ ๐๐ 0123 โฆ,๐๐โ 2 ,๐๐โ 1 โฏ ๐ฅ๐ฅ๐๐ ๐ฆ๐ฆ 6 ๐๐ 0 ๐๐ ๐๐ 01 ๐๐ ๐๐ 012 ๐๐ โฏ ๐๐ 0123 โฆ,๐๐โ 2 ,๐๐ ๐๐ 0123 โฆ,๐๐โ 2 ,๐๐โ 1 ,๐๐ ๐ฅ๐ฅ๐๐ โ ๐ฅ๐ฅ
Table 1: Aitkenโs Algorithm
Example 1
From the following table, find the value of log 10 301 using Aitken interpolation method.
๐ฅ๐ฅ 300 304 305 307 log 10 ๐ฅ๐ฅ 2.4771 2.4829 2.4843 2.
Solution
In this case, ๐ฅ๐ฅ = 301, the following table can be form easily.
๐๐ ๐ฅ๐ฅ๐๐ ๐ฆ๐ฆ๐๐ ๐๐ 0 ๐๐ ๐๐ 01 ๐๐ ๐๐ 012 ๐๐ ๐ฅ๐ฅ๐๐ โ ๐ฅ๐ฅ 0 300 2.4771 โ 1 1 304 2.4829 3 2 305 2.4843 4 3 307 2.4871 6
Now we calculate the first iteration:
After first iteration, the updated table is
๐๐ ๐ฅ๐ฅ๐๐ ๐ฆ๐ฆ๐๐ ๐๐ 0 ๐๐ ๐๐ 01 ๐๐ ๐๐ 012 ๐๐ ๐ฅ๐ฅ๐๐ โ ๐ฅ๐ฅ 0 300 2.4771 โ 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.