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Numerical
Analysis
TOPIC
Interpolation
Interpolation is the process of estimating the
value of function for any intermediate value of
the variable with the help of its given set of
values .
Let us assume that the function y=f(x) is
known for certain values of x say a for
x0,x1,x2,………xn.
As f(x0),f(x1),……..f(xn). The process of finding
the value of f(x) corresponding to x=xi.
Where x0<xi<xn. With the help of given data is
called interpolation .
1.Finite Difference Operators
2.Newton’s Forward Difference
Interpolation Formula
3.Newton’s Backward Difference
Interpolation Formula
4.Lagrange’s Interpolation Formula
rth
forward difference
r 1
r 1
 yi   yi 1   yi
r
kth
backward difference
 yi  
k
k 1
yi  
k 1
i  n,(n  1),..., k
yi 1 ,
Thus
yx  yxh  yx  f ( x  h)  f ( x)
 yx  yxh  yx
2
Similarly
yx  yx  yxh  f ( x)  f ( x  h)
Shift operator, E
E f ( x)  f ( x  h)
E f ( x)  f ( x  nh)
n
E yx  yxnh
n
The inverse operator
is defined as
-1
E
1
E f ( x)  f ( x  h)
Similarly,
n
E f ( x)  f ( x  nh)
Average Operator,
1
 f ( x)   f
2
h

x 
2

1
  yx  ( h / 2)  yx ( h / 2) 
2

h 

f  x  
2 

Differential Operator, D
d

Df ( x) 
f ( x)  f ( x) 

dx

2
d
2
n

D f ( x)  2 f ( x)  f ( x)

dx
Important Results
  E 1
E 1
  1 E 
E
1
1 1/ 2
1/ 2
  (E  E )
2
Newton’s
Forward
Difference
Interpolation
Formula
E f ( x)  f ( x  ph).
p
f ( x0  ph)  E f ( x0 )  (1  ) f ( x0 )
p
p
p( p  1) 2 p( p  1)( p  2) 3

 1  p 
 
 
2!
3!


 f ( x0 )
f ( x0  ph)  f ( x0 )  pf ( x0 )
p ( p  1) 2
p ( p  1)( p  2) 3

 f ( x0 ) 
 f ( x0 )
2!
3!
p ( p  1) ( p  n  1) n
 
 f ( x0 )  Error
n!
This is known as Newton’s
forward difference formula for
interpolation, which gives the
value of f (x0 + ph) in terms of f (x0)
and its leading differences.
This formula is also known as
Newton-Gregory forward
difference interpolation formula.
Here p=(x-x0)/h.
An alternate expression is
p( p  1) 2
yx  y0  py0 
 y0
2!
p( p  1)( p  2) 3

 y0 
3!
p( p  1)( p  n  1) n

 y0  Error
n!
p ( p  1) 2
f ( x0  ph)  f ( x0 )  pf ( x0 ) 
 f ( x0 )
2
p ( p  1)( p  2) 3

 f ( x0 )
6
Here,
x  x0 x  0
p

x
h
1
f ( x0 )  6
 f ( x0 )  2
2
 f ( x0 )  6
3
x ( x  1)
f ( x )  3  6 x 
(2)
2
x ( x  1)( x  2)

(6)
6
f ( x)  x  2x  7 x  3,
3
2
The required cubic polynomial.
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),
f ( xn  ph)  E p f ( xn )  (E 1 ) p f ( xn )  (1 ) p f ( xn )
Binomial expansion yields,
p( p  1) 2 p( p  1)( p  2) 3

f ( xn  ph)  1  p 
 
 
2!
3!

p( p  1)( p  2) ( p  n  1) n


  Error  f ( xn )
n!

f ( xn  ph)  f ( xn )  pf ( xn )
p( p  1) 2

 f ( xn )
2!
p( p  1)( p  2) 3

 f ( xn ) 
3!
p( p  1)( p  2) ( p  n  1) n

 f ( xn )  Error
n!
This formula is known as
Newton’s backward
interpolation formula. This
formula is also known as
Newton’s-Gregory backward
difference interpolation
formula.
p( p  1) 2
yx  yn  pyn 
 yn
2!
p( p  1)( p  2) 3

 yn 
3!
p( p  1)( p  2) ( p  n  1) n

 yn  Error
n!
p
x  xn
h
Example:For the following table of
values, estimate f (7.5).
Difference Table
p ( p  1) 2
y x  yn  pyn 
 yn
2!
p ( p  1)( p  2) 3

 yn
3!
In this problem,
x  xn 7.5  8.0
p

 0.5
h
1
yn  169,  yn  42,  yn  6
2
3
(0.5)(0.5)
y7.5  512  (0.5)(169) 
(42)
2
(0.5)(0.5)(1.5)

(6)
6
 512  84.5  5.25  0.375
 421.875
-3
1979  1982
p
 1.5
2
and
yn  5,  yn  1,
2
 yn  2,  yn  5
3
4
Newton’s interpolation formula gives
Therefore,
DERIVATION:Let y = f (x) be a function which
takes the values,
y0 , y1 ,…yn
corresponding to x0 , x1, …xn . Since
there are (n + 1) values of y
corresponding to (n + 1) values of x,
we can represent the function f (x)
by a polynomial of degree n.
or in the form
Here, the coefficients ak are
so chosen as to satisfy this
equation by the (n + 1) pairs
(xi, yi). Thus we get
Therefore,
y1
a1 
( x1  x0 )( x1  x2 )
ai 
( xi  x0 )( xi  x1 )
( x1  xn )
yi
( xi  xi 1 )( xi  xi 1 )
( xi  xn )
and
yn
an 
( xn  x0 )( xn  x1 )
( xn  xn 1 )
y  f ( x) 
( x  x1 )( x  x2 ) ( x  xn )
( x  x0 )( x  x2 ) ( x  xn )
y0 
y1 
( x0  x1 )( x0  x2 ) ( x0  xn )
( x1  x0 )( x1  x2 ) ( x1  xn )
( x  x0 )( x  x1 ) ( x  xi 1 )( x  xi 1 ) ( x  xn )

yi 
( xi  x0 )( xi  x1 ) ( xi  xi 1 )( xi  xi 1 ) ( xi  xn )
( x  x0 )( x  x1 )( x  x2 ) ( x  xn1 )

yn
( xn  x0 )( xn  x1 )( xn  x2 ) ( xn  xn 1 )
The Lagrange’s formula for
interpolation
We can easily observe that,
and
Thus introducing Kronecker
delta notation
Example:Find Lagrange’s
interpolation polynomial
fitting the points
y(1) = -3, y(3) = 0,
y(4) = 30, y(6) = 132.
Hence find y(5).
Using Lagrange’s interpolation
formula, we have
On simplification, we get
which is required Lagrange’s
interpolation polynomial.
Now, y(5) = 75.
Solution:-
Using Lagrange’s formula,
( x  x1 )( x  x2 )
f ( x) 
f ( x0 )
( x0  x1 )( x0  x2 )
( x  x0 )( x  x2 )

f ( x1 ) 
( x1  x0 )( x1  x2 )
( x  x0 )( x  x1 )
f ( x2 )
( x2  x0 )( x2  x1 )
Therefore
1.
Evaluate : (x.log x)
2.
Evaluate : (sin2x.cos4x)
3.
Estimate the missing term:
x
0
1
2
3
4
Y
1
3
9
_
81
4.Derive newton forward formula for
interpolation.
5. Derive newton backward formula
for interpolation.
1. Evaluate (x2+sinx) , the
interval of difference being .
2. Find the lowest degree
polynomial which satisfies the
following table:
X
0
1
2
3
4
5
F(x)
0
3
8
15
24
35
3.The population of a town is as
follow:
YEAR
1961
1971
1981
1991
2001
2011
POPULATION
20
24
35
40
42
51