Interpolation produces a
function that matches the given
data exactly. The function then
can be utilized to approximate
th...
Interpolation may also be used
to produce a smooth graph of a
function for which values are
known only at discrete points,...
Given data points
Obtain a function, P(x)
P(x) goes through the data points
Use P(x)
To estimate values at intermedia...
Given data points:
At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8
Find the following:
At x = 4, y = ?
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 sat...
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) ...
P( x ) = 3 L0 ( x ) + 8 L1 ( x )

P( x ) = L0 ( x ) y0 + L1 ( x ) y1
Lagrange Interpolating Polynomial

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

P( x ) = L0...
( 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.
( 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.
( 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.
General form of the Lagrange Interpolating
Polynomial

P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2
+ ........... + Ln...
( x − x0 )( x − x1 )...( x − xk −1 )( x − xk +1 )...( x − xn )
Lk ( x ) =
( xk − x0 )( xk − x1 )...( xk − xk −1 )( x k − x...
Numerator of

Lk ( x )

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

Lk ( x )

( xk − x0 ) ( xk − x1 ) ( xk − x2 ) ×LL
× ( xk − xk −1 ) ( xk − xk +1 ) × LL
× ( xk − xn −1 ) ( ...
Find the Lagrange Interpolating
Polynomial using the three given points.

( x0 , y0 ) = ( 2, 0.5)
( x1 , y1 ) = ( 2.5, 0.4...
( 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
( 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
=
0.7...
( 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...
P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
+ L2 ( x ) f ( x2 )
(

)

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 )
+


...
P( x ) = 0.05 x − 0.425 x + 1.15
2

The three given points were taken from
the function

1
f ( x) =
x
1
f ( 3) = = 0.333
3
An approximation can be obtained
from the Lagrange Interpolating
Polynomial as:

P( 3) = 0.05( 3) − 0...
Newton’s Interpolating Polynomials
Newton’s equation of a function that
passes through two points

( x0 , y 0 )

and

( x1...
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 )
y1 − y 0
a1 =
x1 − x0
Newton’s equation of a function that passes
through three points

( x0 , y 0 ) ( x1 , y1 )
is

and

...
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 ...
y2 − y1 y1 − y0
−
x2 − x1 x1 − x0
a2 =
x2 − x0
Newton’s equation of a function that passes
through four points can be written by adding
a fourth term .

P ( x ) = a0 + a...
P ( x ) = a0 + a1 ( x − x0 )

+ a2 ( x − x0 ) ( x − x1 )

+ a3 ( x − x0 ) ( x − x1 ) ( x − x2 )
The fourth term will vanis...
Divided differences and the coefficients
The divided difference of a function, f
with respect to

xi

is denoted as

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

x
...
The divided difference of a function, f
with respect to xi , i +1 and
i +2
called as the second divided difference, is
den...
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 [ x...
The coefficients of Newton’s interpolating
polynomial are:
a0 = f [ x0 ]

a1 = f [ x0 , x1 ]

a 2 = f [ x0 , x1 , x 2 ]
a3...
First
divided differences

Second
divided differences

Third
divided differences
Example
Find Newton’s interpolating polynomial to
approximate a function whose 5 data points are
given below.

x

f ( x)

...
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
-0.3...
The 5 coefficients of the Newton’s interpolating
polynomial are:

a0 = f [ x0 ] = 0.85467

a1 = f [ x0 , x1 ] = −0.32617

...
P ( x ) = a0 + a1 ( x − x0 )

+ a2 ( x − x0 ) ( x − x1 )

+ a3 ( x − x0 ) ( x − x1 ) ( x − x2 )

+ a4 ( x − x0 ) ( x − x1 ...
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...
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 ) ( ...
Interpolation functions
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Interpolation functions

  1. 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. 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. 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. 4. Given data points: At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8 Find the following: At x = 4, y = ?
  5. 5. 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
  6. 6. 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
  7. 7. 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 )
  8. 8. 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
  9. 9. 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 )
  10. 10. ( 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.
  11. 11. ( 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.
  12. 12. ( 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.
  13. 13. 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 )
  14. 14. ( 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
  15. 15. Numerator of Lk ( x ) ( x − x0 ) ( x − x1 ) ( x − x2 ) ×LL × ( x − xk −1 ) ( x − xk +1 ) × LL × ( x − xn −1 ) ( x − xn )
  16. 16. Denominator of Lk ( x ) ( xk − x0 ) ( xk − x1 ) ( xk − x2 ) ×LL × ( xk − xk −1 ) ( xk − xk +1 ) × LL × ( xk − xn −1 ) ( xk − xn )
  17. 17. 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)
  18. 18. ( 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
  19. 19. ( 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
  20. 20. ( 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
  21. 21. P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 )
  22. 22. ( ) 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
  23. 23. P( x ) = 0.05 x − 0.425 x + 1.15 2 The three given points were taken from the function 1 f ( x) = x
  24. 24. 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
  25. 25. 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 )
  26. 26. 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 )
  27. 27. 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 )
  28. 28. 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
  29. 29. y2 − y1 y1 − y0 − x2 − x1 x1 − x0 a2 = x2 − x0
  30. 30. 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 )
  31. 31. 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.
  32. 32. 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 )
  33. 33. 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
  34. 34. 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
  35. 35. 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
  36. 36. 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.
  37. 37. First divided differences Second divided differences Third divided differences
  38. 38. 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
  39. 39. 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
  40. 40. 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
  41. 41. 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 )
  42. 42. 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.
  43. 43. 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
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