# Newton’s Backward Difference Interpolation Formla-Numerical Analysis-Lecture Handouts, Lecture notes for Mathematical Methods for Numerical Analysis and Optimization. Chennai Mathematical Institute

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Numerical Analysis –MTH603 VU

NEWTON’S BACKWARD DIFFERENCE INTERPOLATION FORMULA For interpolating the value of the function y = f (x) near the end of table of values, and to extrapolate value of the function a short distance forward from yn, Newton’s backward

interpolation formula is used Derivation Let y = f (x) be a function which takes on values f (xn), f (xn-h), f (xn-2h), …, f (x0) corresponding to equispaced values xn, xn-h, xn-2h,

…, x0. Suppose, we wish to evaluate the function f (x) at (xn + ph), where p is any real

number, then we have the shift operator E, such that 1( ) ( ) ( ) ( ) (1 ) ( )p p pn n n nf x ph E f x E f x f x − − −+ = = = −∇

Binomial expansion yields,

2 3( 1) ( 1)( 2)( ) 1 2! 3!

( 1)( 2) ( 1) Error ( ) !

n

n n

p p p p pf x ph p

p p p p n f x n

+ + ++ = + ∇ + ∇ + ∇ + + + + − + ∇ + 

That is

2 3( 1) ( 1)( 2)( ) ( ) ( ) ( ) ( ) 2! 3!

( 1)( 2) ( 1) ( ) Error !

n n n n n

n n

p p p p pf x ph f x p f x f x f x

p p p p n f x n

+ + + + = + ∇ + ∇ + ∇ +

+ + + − + ∇ +

This formula is known as Newton’s backward interpolation formula. This formula is also known as Newton’s-Gregory backward difference interpolation formula. If we retain (r + 1)terms, we obtain a polynomial of degree r agreeing with f (x) at xn, xn-1, …, xn-r. Alternatively, this formula can also be written as

2 3( 1) ( 1)( 2) 2! 3!

( 1)( 2) ( 1) Error !

x n n n n

n n

p p p p py y p y y y

p p p p n y n

+ + + = + ∇ + ∇ + ∇ +

+ + + − + ∇ +

Here nx xp h

=

Example For the following table of values, estimate f (7.5).

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Numerical Analysis –MTH603 VU

Solution The value to be interpolated is at the end of the table. Hence, it is appropriate to use Newton’s backward interpolation formula. Let us first construct the backward difference table for the given data Difference Table

Since the 4 th

and higher order differences are zero, the required Newton’s backward interpolation formula is

2

3

( 1) 2!

( 1)( 2) 3!

x n n n

n

p py y p y y

p p p y

+ = + ∇ + ∇

+ + + ∇

In this problem,

7.5 8.0 0.5 1

nx xp h − −

= = = −

2 3169, 42, 6n n ny y y∇ = ∇ = ∇ =

7.5 ( 0.5)(0.5)512 ( 0.5)(169) (42)

2 ( 0.5)(0.5)(1.5) (6)

6 512 84.5 5.25 0.375 421.875

y −= + − +

− +

= − − − =

Example The sales for the last five years is given in the table below. Estimate the sales for the year 1979

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Numerical Analysis –MTH603 VU

Solution Newton’s backward difference Table

In this example,

1979 1982 1.5 2

p −= = −

and

2

3 4

5, 1,

2, 5 n n

n n

y y y y

∇ = ∇ =

∇ = ∇ =

Newton’s interpolation formula gives

1979

( 1.5)( 0.5) ( 1.5)( 0.5)(0.5)57 ( 1.5)5 (1) (2) 2 6

( 1.5)( 0.5)(0.5)(1.5) (5) 24

y − − − −= + − + +

− − +

57 7.5 0.375 0.125 0.1172= − + + + Therefore, 1979 50.1172y =

Example Consider the following table of values

x 1 1.1 1.2 1.3 1.4 1.5 F(x) 2 2.1 2.3 2.7 3.5 4.5 Use Newton’s Backward Difference Formula to estimate the value of f(1.45) . Solution

x y=F(x) y∇ 2 y∇ 3 y∇ 4 y∇ 5 y∇ 1 2 1.1 2.1 0.1 1.2 2.3 0.2 0.1 1.3 2.7 0.4 0.2 0.1 1.4 3.5 0.8 0.4 0.2 0.1 1.5 4.5 1 0.2 -0.2 -0.4 -0.5 docsity.com

Numerical Analysis –MTH603 VU

1.45 1.5 0.5

0.1 nx xp

h − −

= = = − , ny∇ = 1 , 2

ny∇ = .2 , 3

ny∇ = - .2 , 4

ny∇ = -.4 ,

5 ny∇ = -.5 As we know that

2 3

4 5

( 1) ( 1)( 2) 2! 3!

( 1)( 2)( 3) ( 1)( 2)( 3)( 4) 4! 5!

x n n n n

n n

p p p p py y p y y y

p p p p p p p p py y

+ + + = + ∇ + ∇ + ∇

+ + + + + + + + ∇ + ∇

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

0.5 ( 0.5 1) 0.5 ( 0.5 1)( 0.5 2) 4.5 0.5 (1) (0.2) 0.2

2! 3! 0.5 ( 0.5 1)( 0.5 2)( 0.5 3) 0.5 ( 0.5 1)( 0.5 2)( 0.5 3)( 0.5 4)

0.4 0.5 4! 5!

xy − − + − − + − +

= + − + + −

− − + − + − + − − + − + − + − + + − + −

= 4.5 0.5 0.025 + 0.0125 + 0.015625+ 0.068359 4.07148 − −

=

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