1. INTERPOLATION
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) 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
The conditions can be satisfied if L0(x) and L1(x) are defined in the following way.

8. Lagrange Interpolating Polynomial

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

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

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

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

14. General form of the Lagrange Interpolating Polynomial

16. Numerator of

17. Denominator of

18. Find the Lagrange Interpolating Polynomial using the three given points.

24. The three given points were taken from the function

25. An approximation can be obtained from the Lagrange Interpolating Polynomial as:

28. It is suspected that the high amounts of tannin in mature oak leave inhibit the growth of the winter moth larvae that extensively damage these trees in certain years. The following table lists the average weight of two samples of larvae at times in the first 28 days after birth. The first sample was reared on young oak leaves, whereas the second sample was reared on mature leaves from the same tree.
Use Lagrange’s interpolation to approximate the average weight curve for each sample.
Find an approximate maximum average weight for each sample by determining the maximum of the interpolating polynomial.
Day 6 10 13 17 20 28
Sample 1 average weight(mg)
6.67
17.33
42.67
37.33
30.10
29.31
28.74
Sample 2 average weight(mg)
16.11
18.89
15.00
10.56
9.44
8.89

29. Newton’s Interpolating Polynomials
Newton’s equation of a function that passes through two points
and is

30. Set
Set

31. Newton’s equation of a function that passes through three points
and is

32. To find
, set

34. Newton’s equation of a function that passes through four points can be written by adding a fourth term .

35. The fourth term will vanish at all three previous points and, therefore, leaving all three previous coefficients intact.

36. Divided differences and the coefficients
The divided difference of a function,
with respect to
is denoted as
It is called as zeroth divided difference and is simply the value of the function, at

37. The divided difference of a function,
called as the first divided difference, is denoted
with respect to
and

38. The divided difference of a function,
called as the second divided difference, is denoted as
with respect to
and ,

39. The third divided difference with respect to,
and

40. The coefficients of Newton’s interpolating polynomial are:
and so on.

42. Example
Find Newton’s interpolating polynomial to approximate a function whose 5 data points are given below.
2.0
2.3
2.6
2.9
3.2

43. 2.0 1
2.3 2
2.6 3
2.9 4
3.2

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

46. P(x) can now be used to estimate the value of the function f(x) say at x = 2.8.

49. Complete the divided difference table for the following data and also construct the interpolating polynomial that uses all this data. x
f(x)
1.0
1.3
1.6
1.9
2.2