Interpolation functions

1. Interpolation produces a
function that matches the given
data exactly. The function then
can be utilized to approximate
the data values at intermediate
points.

2. Interpolation may also be used
to produce a smooth graph of a
function for which values are
known only at discrete points,
either from measurements or
calculations.

3. Given data points
Obtain a function, P(x)
P(x) goes through the data points
Use P(x)
To estimate values at intermediate
points

4. Given data points:
At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8
Find the following:
At x = 4, y = ?

6. P(x) should satisfy the following conditions:
P(x = 2) = 3 and P(x = 5) = 8
P( x ) = 3 L0 ( x ) + 8 L1 ( x )
P(x) can satisfy the above conditions if
at x = x0 = 2, L0(x) = 1 and L1(x) = 0 and
at x = x1= 5, L0(x) = 0 and L1(x) = 1

7. At x = x0 = 2, L0(x) = 1 and L1(x) = 0 and
at x = x1= 5, L0(x) = 0 and L1(x) = 1
The conditions can be satisfied if L0(x) and
L1(x) are defined in the following way.
x−5
L0 ( x ) =
and
2−5
x − x1
L0 ( x ) =
x0 − x1
x−2
L1 ( x ) =
5−2
x − x0
and L1 ( x ) =
x1 − x0

8. P( x ) = 3 L0 ( x ) + 8 L1 ( x )
P( x ) = L0 ( x ) y0 + L1 ( x ) y1
Lagrange Interpolating Polynomial
P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )

9. P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
 x−5
 x−2
P( x ) = 
( 3 ) + 
( 8 )
 2−5
5−2
5x − 1
P( x ) =
3
5× 4 −1
P( 4) =
= 6.333
3

10. The Lagrange interpolating polynomial
passing through three given points; (x0, y0),
(x1, y1) and (x2, y2) is:
P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2
P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
+ L2 ( x ) f ( x2 )

11. ( x − x1 )( x − x2 )
L0 ( x ) =
( x0 − x1 )( x0 − x2 )
At x0, L0(x) becomes 1. At all
other given data points L0(x) is 0.

12. ( x − x0 )( x − x2 )
L1 ( x ) =
( x1 − x0 )( x1 − x2 )
At x1, L1(x) becomes 1. At all
other given data points L1(x) is 0.

13. ( x − x0 )( x − x1 )
L2 ( x ) =
( x2 − x0 )( x2 − x1 )
At x2, L2(x) becomes 1. At all
other given data points L2(x) is 0.

14. General form of the Lagrange Interpolating
Polynomial
P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2
+ ........... + Ln ( x ) yn
P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
+ L2 ( x ) f ( x2 ) + ........... + Ln ( x ) f ( xn )

15. ( x − x0 )( x − x1 )...( x − xk −1 )( x − xk +1 )...( x − xn )
Lk ( x ) =
( xk − x0 )( xk − x1 )...( xk − xk −1 )( x k − x k +1 )...( xk − xn )
( x − xi )
Lk ( x ) = ∏
i =0 ( x k − xi )
n
i ≠k

16. Numerator of
Lk ( x )
( x − x0 ) ( x − x1 ) ( x − x2 ) ×LL
× ( x − xk −1 ) ( x − xk +1 ) × LL
× ( x − xn −1 ) ( x − xn )

17. Denominator of
Lk ( x )
( xk − x0 ) ( xk − x1 ) ( xk − x2 ) ×LL
× ( xk − xk −1 ) ( xk − xk +1 ) × LL
× ( xk − xn −1 ) ( xk − xn )

18. Find the Lagrange Interpolating
Polynomial using the three given points.
( x0 , y0 ) = ( 2, 0.5)
( x1 , y1 ) = ( 2.5, 0.4)
( x2 , y2 ) = ( 4, 0.25)

19. ( x − x1 )( x − x2 )
L0 ( x ) =
( x0 − x1 )( x0 − x2 )
( x − 2.5)( x − 4)
L0 ( x ) =
( 2 − 2.5)( 2 − 4)
= x − 6.5 x + 10
2

20. ( x − x0 )( x − x2 )
L1 ( x ) =
( x1 − x0 )( x1 − x2 )
( x − 2)( x − 4)
L1 ( x ) =
( 2.5 − 2)( 2.5 − 4)
− x + 6x − 8
=
0.75
2

21. ( x − x0 )( x − x1 )
L2 ( x ) =
( x2 − x0 )( x2 − x1 )
( x − 2)( x − 2.5)
L2 ( x ) =
( 4 − 2)( 4 − 2.5)
x − 4. 5 x + 5
=
3
2

22. P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
+ L2 ( x ) f ( x2 )

23. (
)
P( x ) = x − 6.5 x + 10 ( 0.5 )
2
 − x + 6x − 8 
( 0.4 )
+


0.75


2
 x − 4 .5 x + 5 
( 0.25 )
+


3


2

24. P( x ) = 0.05 x − 0.425 x + 1.15
2
The three given points were taken from
the function
1
f ( x) =
x

25. 1
f ( 3) = = 0.333
3
An approximation can be obtained
from the Lagrange Interpolating
Polynomial as:
P( 3) = 0.05( 3) − 0.425( 3) + 1.15
= 0.325
2

26. Newton’s Interpolating Polynomials
Newton’s equation of a function that
passes through two points
( x0 , y 0 )
and
( x1 , y1 )
is
P( x ) = a 0 + a1 ( x − x0 )

27. P( x ) = a 0 + a1 ( x − x0 )
Set x = x
0
P ( x0 ) = y0 = a0
Set
x = x1
P ( x1 ) = y1 = a0 + a1 ( x1 − x0 )

28. y1 − y 0
a1 =
x1 − x0
Newton’s equation of a function that passes
through three points
( x0 , y 0 ) ( x1 , y1 )
is
and
( x2 , y2 )

29. P ( x ) = a0 + a1 ( x − x0 )
+ a2 ( x − x0 ) ( x − x1 )
a 2 , set x = x 2
P ( x2 ) = a0 + a1 ( x2 − x0 )
+ a2 ( x2 − x0 ) ( x2 − x1 )
To find

30. y2 − y1 y1 − y0
−
x2 − x1 x1 − x0
a2 =
x2 − x0

31. Newton’s equation of a function that passes
through four points can be written by adding
a fourth term .
P ( x ) = a0 + a1 ( x − x0 )
+ a2 ( x − x0 ) ( x − x1 )
+ a3 ( x − x0 ) ( x − x1 ) ( x − x2 )

32. P ( x ) = a0 + a1 ( x − x0 )
+ a2 ( x − x0 ) ( x − x1 )
+ a3 ( x − x0 ) ( x − x1 ) ( x − x2 )
The fourth term will vanish at all three
previous points and, therefore, leaving all
three previous coefficients intact.

33. Divided differences and the coefficients
The divided difference of a function, f
with respect to
xi
is denoted as
f [ xi ]
It is called as zeroth divided difference and is
simply the value of the function, f
at xi
f [ xi ] = f ( xi )

34. The divided difference of a function, f
with respect to xi and
i +1
called as the first divided difference, is denoted
x
f [ xi , xi +1 ]
f [ xi , xi +1 ]
f [ xi +1 ] − f [ xi ]
=
xi +1 − xi

35. The divided difference of a function, f
with respect to xi , i +1 and
i +2
called as the second divided difference, is
denoted as
x
x
f [ xi , xi +1 , xi + 2 ]
f [ xi , xi +1 , xi +2 ]
f [ xi +1 , xi +2 ] − f [ xi , xi +1 ]
=
xi + 2 − xi

36. The third divided difference with respect to
,
,
xi + 2 and i + 3
i
i +1
x
x
f [ xi , xi +1 , xi +2 , xi +3 ]
x
f [ xi +1 , xi + 2 , xi +3 ] − f [ xi , xi +1 , xi + 2 ]
=
xi +3 − xi

37. The coefficients of Newton’s interpolating
polynomial are:
a0 = f [ x0 ]
a1 = f [ x0 , x1 ]
a 2 = f [ x0 , x1 , x 2 ]
a3 = f [ x0 , x1 , x 2 , x3 ]
a 4 = f [ x0 , x1 , x 2 , x3 , x 4 ]
and so on.

38. First
divided differences
Second
divided differences
Third
divided differences

39. Example
Find Newton’s interpolating polynomial to
approximate a function whose 5 data points are
given below.
x
f ( x)
2.0
0.85467
2.3
0.75682
2.6
0.43126
2.9
0.22364
3.2
0.08567

40. i
0
xi
f [ xi ]
2.0
f [ xi −1 , xi ] f [ xi − 2 , xi −1 , xi ]
f [ xi −3 , , xi ]
f [ xi −4 , , xi ]
0.85467
-0.32617
1
2.3
0.75682
-1.26505
-1.08520
2
2.6
0.43126
2.13363
0.65522
-0.69207
3
2.9
0.22364
4
3.2
0.08567
-0.29808
0.38695
-0.45990
-2.02642

41. The 5 coefficients of the Newton’s interpolating
polynomial are:
a0 = f [ x0 ] = 0.85467
a1 = f [ x0 , x1 ] = −0.32617
a2 = f [ x0 , x1 , x2 ] = −1.26505
a3 = f [ x0 , x1 , x2 , x3 ] = 2.13363
a4 = f [ x0 , x1 , x2 , x3 , x4 ] = −2.02642

42. P ( x ) = a0 + a1 ( x − x0 )
+ a2 ( x − x0 ) ( x − x1 )
+ a3 ( x − x0 ) ( x − x1 ) ( x − x2 )
+ a4 ( x − x0 ) ( x − x1 ) ( x − x2 ) ( x − x3 )

43. P ( x ) = 0.85467 − 0.32617 ( x − 2.0 )
-1.26505 ( x − 2.0 ) ( x − 2.3 )
+ 2.13363 ( x − 2.0 ) ( x − 2.3) ( x − 2.6 )
−2.02642 ( x − 2.0 ) ( x − 2.3) ( x − 2.6 ) ( x − 2.9 )
P(x) can now be used to estimate the value of the
function f(x) say at x = 2.8.

44. P ( 2.8 ) = 0.85467 − 0.32617 ( 2.8 − 2.0 )
-1.26505 ( 2.8 − 2.0 ) ( 2.8 − 2.3 )
+ 2.13363 ( 2.8 − 2.0 ) ( 2.8 − 2.3 ) ( 2.8 − 2.6 )
−2.02642 ( 2.8 − 2.0 ) ( 2.8 − 2.3) ( 2.8 − 2.6 ) ( 2.8 − 2.9 )
f ( 2.8 ) ≈ P ( 2.8 ) = 0.275