Interpolation and Curve Fitting - Numerical Methods in Engineering with MATLAB - page 113

Here are a few sample calculations used in arriving at the figures in the table:

y 3 − y 1 59 − ( − 1)

∇ y 3 =

x 1 =

= 10

x 3 − 4 − ( − 2)

∇ y 3 −∇ y 2 10 − 1

2 y 3 =

∇ = = 3

x 3 − x 2 4 − 1

2 y 6 2 y 3

∇ −∇ − 5 − 3

3 y 6 =

∇ =

4 =

1

x 6 − x 3 − 4 −

From the table we see that the last nonzero coefficient(last nonzero diagonalterm)
of Newton's polynomial is

3 y 3 , which is the coefficient of the cubicterm. Hence the

∇

polynomial is a cubic.

EXAMPLE 3.3
Given the datapoints

x 4 . 0 3 . 9 3 . 8 3 . 7

y − 0 . 06604 − 0 . 02724 0 . 01282 0 . 05383

determine the root of y ( x ) = 0 by Neville's method.

Solution This is an example of inverse interpolation , where the roles of x and y are
interchanged. Instead of computing y at a given x

, we are finding x thatcorresponds

to a given y (in thiscase, y 0). Employing the formatofTable 3.2 (with x and y

interchanged,ofcourse), weobtain

=

i y i P 0 [] = x i P 1 [ , ] P 2 [ ,, ] P 3 [ ,,, ]

1 − 0 . 06604 4 . 0 3 . 8298 3 . 8316 3 . 8317

2 − 0 . 02724 3 . 9 3 . 8320 3 . 8318

3 0 . 01282 3 . 8 3 . 8313

4 0 . 05383 3 . 7

The following are a couple of sample computations usedinthe table:

( y − y 2 ) P 0 [ y 1 ] + ( y 1 − y ) P 0 [ y 2 ]

P 1 [ y 1 , y 2 ] =

y 1 − y 2

(0 + 0 . 02724)(4 . 0) + ( − 0 . 06604 − 0)(3 . 9)

= = 3 . 8298

− 0 . 06604 + 0 . 02724

( y − y 4 ) P 1 [ y 2 , y 3 ] + ( y 2 − y ) P 1 [ y 3 , y 4 ]

P 2 [ y 2 , y 3 , y 4 ] =

y 2 − y 4

(0 − 0 . 05383)(3 . 8320) + ( − 0 . 02724 − 0)(3 . 8313)

= = 3 . 8318

− 0 . 02724 − 0 . 05383