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Numerical Analysis –MTH603 **VU**

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**
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 *y*n, Newton’s backward

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

…, *x*0. Suppose, we wish to evaluate the function *f* (*x*) at (*x*n + *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 *x*n,
*x*n-1, …, *x*n-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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**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, 6*n 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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**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 50.1172*y *=

**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

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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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