The document discusses Lagrange interpolation and divided differences. It explains that the Lagrange interpolation polynomial can be written in terms of divided differences, where the coefficients are divided differences of the function values. Divided differences are defined recursively, and a pattern is identified to write them in terms of the function values at nodes. An example divided difference table is given for a set of data points.

1. Adv. Engg. Mathematics
MTH-812 Divided Differences Interpolations
Dr. Yasir Ali (yali@ceme.nust.edu.pk)
DBS&H, CEME-NUST
December 4, 2017
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

2. Interpolation Problem
Given 1 The (n + 1) nodes: x0, x1, ..., xn
2 The functional values f0, f1, ..., fn at these nodes
3 An Intermediate (nontabulated) point: xI
Predict fI : the value at x = xI.
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

3. Suppose that Pn(x) is the nth Lagrange polynomial that agrees with
the function f at the distinct numbers x0, x1, ..., xn.
P(x) = Ln,0(x)f(x0) + Ln,1(x)f(x1) + · · · +, Ln,n(x)f(xn)
=
n
k=0
f(xk)Ln,k(x), where Ln,k(x) =
n
i=0
i=k
x − xi
xk − xi
,
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

4. Suppose that Pn(x) is the nth Lagrange polynomial that agrees with
the function f at the distinct numbers x0, x1, ..., xn.
P(x) = Ln,0(x)f(x0) + Ln,1(x)f(x1) + · · · +, Ln,n(x)f(xn)
=
n
k=0
f(xk)Ln,k(x), where Ln,k(x) =
n
i=0
i=k
x − xi
xk − xi
,
Although this polynomial is unique, there are alternate algebraic
representations that are useful in certain situations. The divided
differences of f with respect to x0, x1, ..., xn are used to express Pn(x)
in the form
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

5. Suppose that Pn(x) is the nth Lagrange polynomial that agrees with
the function f at the distinct numbers x0, x1, ..., xn.
P(x) = Ln,0(x)f(x0) + Ln,1(x)f(x1) + · · · +, Ln,n(x)f(xn)
=
n
k=0
f(xk)Ln,k(x), where Ln,k(x) =
n
i=0
i=k
x − xi
xk − xi
,
Although this polynomial is unique, there are alternate algebraic
representations that are useful in certain situations. The divided
differences of f with respect to x0, x1, ..., xn are used to express Pn(x)
in the form
Pn(x) = a0+a1(x−x0)+a2(x−x0)(x−x1)+ · · ·+an(x−x0) · · · (x−xn−1)
(1)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

6. Suppose that Pn(x) is the nth Lagrange polynomial that agrees with
the function f at the distinct numbers x0, x1, ..., xn.
P(x) = Ln,0(x)f(x0) + Ln,1(x)f(x1) + · · · +, Ln,n(x)f(xn)
=
n
k=0
f(xk)Ln,k(x), where Ln,k(x) =
n
i=0
i=k
x − xi
xk − xi
,
Although this polynomial is unique, there are alternate algebraic
representations that are useful in certain situations. The divided
differences of f with respect to x0, x1, ..., xn are used to express Pn(x)
in the form
Pn(x) = a0+a1(x−x0)+a2(x−x0)(x−x1)+ · · ·+an(x−x0) · · · (x−xn−1)
(1)
for appropriate constants
a0, a1, ..., an
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

7. For
Pn(x) = a0+a1(x−x0)+a2(x−x0)(x−x1)+ · · ·+an(x−x0) · · · (x−xn−1)
Pn(x) at x0 leaves
a0 = Pn(x0) = f0 (2)
Pn(x) at x1 leaves
Pn(x1) = a0 + a1(x1 − x0) we know
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

8. For
Pn(x) = a0+a1(x−x0)+a2(x−x0)(x−x1)+ · · ·+an(x−x0) · · · (x−xn−1)
Pn(x) at x0 leaves
a0 = Pn(x0) = f0 (2)
Pn(x) at x1 leaves
Pn(x1) = a0 + a1(x1 − x0) we know Pn(x1) = f1
a1 =
f1 − a0
x1 − x0
using (2), we get
a1 =
f1 − f0
x1 − x0
(3)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

9. Pn(x) at x2 leaves
Pn(x2) = a0 + a1(x2 − x0) + a2(x2 − x0)(x2 − x1)
we know Pn(x2) = f2
a2 =
f2 − a0 − a1(x2 − x0)
(x2 − x0)(x2 − x1)
using (2) and (3), we get
a2 =
f2 − f0 − f1−f0
x1−x0
(x2 − x0)
(x2 − x0)(x2 − x1)
After simplification we get
a2 =
f2−f1
x2−x1
− f1−f0
x1−x0
x2 − x0
(4)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

10. Comparing a0, a1 and a2 can we get some pattern?
a0 = f0
a1 =
f1 − f0
x1 − x0
a2 =
f2−f1
x2−x1
− f1−f0
x1−x0
x2 − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

11. Comparing a0, a1 and a2 can we get some pattern?
a0
depends on x0
= f0
a1
depends on x0 and x1
=
f1 − f0
x1 − x0
a2
depends on x0 x1 and x2
=
f2−f1
x2−x1
− f1−f0
x1−x0
x2 − x0
If we denote a0 = f0 as
f[x0]. Then its easy to write
a1 as
a1 =
f[x1] − f[x0]
x1 − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

12. Comparing a0, a1 and a2 can we get some pattern?
a0
depends on x0
= f0
a1
depends on x0 and x1
=
f1 − f0
x1 − x0
a2
depends on x0 x1 and x2
=
f2−f1
x2−x1
− f1−f0
x1−x0
x2 − x0
If we denote a0 = f0 as
f[x0]. Then its easy to write
a1 as
a1 =
f[x1] − f[x0]
x1 − x0
If we denote a1 = f[x0, x1]. Then
a2 =
f[x2, x1] − f[x1, x0]
x2 − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

13. By looking at the pattern of a1 and a2 is it possible to write a3?
a1
f[x1,x0]
=
f[x1] − f[x0]
x1 − x0
a2
f[x2,x1,x0]
=
f[x2, x1] − f[x1, x0]
x2 − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

14. By looking at the pattern of a1 and a2 is it possible to write a3?
a1
f[x1,x0]
=
f[x1] − f[x0]
x1 − x0
a2
f[x2,x1,x0]
=
f[x2, x1] − f[x1, x0]
x2 − x0
a3
f[x3,x2,x1,x0]
=
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

15. By looking at the pattern of a1 and a2 is it possible to write a3?
a1
f[x1,x0]
=
f[x1] − f[x0]
x1 − x0
a2
f[x2,x1,x0]
=
f[x2, x1] − f[x1, x0]
x2 − x0
a3
f[x3,x2,x1,x0]
=
f[
First three
x3, x2, x1] − f[
Last three
x2, x1, x0]
x3 − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

16. By looking at the pattern of a1 and a2 is it possible to write a3?
a1
f[x1,x0]
=
f[x1] − f[x0]
x1 − x0
a2
f[x2,x1,x0]
=
f[x2, x1] − f[x1, x0]
x2 − x0
a3
f[x3,x2,x1,x0]
=
f[
First three
x3, x2, x1] − f[
Last three
x2, x1, x0]
x3 − x0
ak
f[x0,x1,··· ,xk]
=
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

17. By looking at the pattern of a1 and a2 is it possible to write a3?
a1
f[x1,x0]
=
f[x1] − f[x0]
x1 − x0
a2
f[x2,x1,x0]
=
f[x2, x1] − f[x1, x0]
x2 − x0
a3
f[x3,x2,x1,x0]
=
f[
First three
x3, x2, x1] − f[
Last three
x2, x1, x0]
x3 − x0
ak
f[x0,x1,··· ,xk]
=
f[
First k−1
... ] − f[
Last k−1
... ]
xk − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

18. Newton’s Divided Difference
Hence, the interpolating polynomial (1) may be expressed as
Pn(x) = f[x0] + f[x1, x0](x − x0) + a2(x − x0)(x − x1)+
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
f[xi] = f(xi) zeroth divided difference
f[xi+1, xi] =
f[xi+1] − f[xi]
xi+1 − xi
1st divided difference
f[xi+2, xi+1, xi] =
f[xi+2, xi+1] − f[xi+1, xi]
xi+2 − xi
2nd divided difference
f[x0, x1, · · · , xk] =
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

19. Newton’s Divided Difference
Hence, the interpolating polynomial (1) may be expressed as
Pn(x) = f[x0] + f[x1, x0](x − x0) + a2(x − x0)(x − x1)+
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
f[xi] = f(xi) zeroth divided difference
f[xi+1, xi] =
f[xi+1] − f[xi]
xi+1 − xi
1st divided difference
f[xi+2, xi+1, xi] =
f[xi+2, xi+1] − f[xi+1, xi]
xi+2 − xi
2nd divided difference
f[x0, x1, · · · , xk] =
f[xk, xk−1 · · · , x1] − f[xk−1, xk−2, · · · , x0]
xk − x0
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

20. Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

21. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
♥
1 1.3 0.6201
2 1.6 0.4554
3 1.9 0.2818
4 2.2 0.1103
♥
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

22. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
♥
1 1.3 0.6201
2 1.6 0.4554
3 1.9 0.2818
4 2.2 0.1103
♥ The First Divided Difference involving x0 and x1 is
f[x0, x1] =
f[x1] − f[x0]
x1 − x0
=
0.6201 − 0.7651
1.3 − 1.0
= −0.4833.
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

23. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
♥
1 1.3 0.6201
2 1.6 0.4554
3 1.9 0.2818
4 2.2 0.1103
♥ The First Divided Difference involving x0 and x1 is
f[x0, x1] =
f[x1] − f[x0]
x1 − x0
=
0.6201 − 0.7651
1.3 − 1.0
= −0.4833.
Remaining First Divided Differences are as follows
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

24. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 ♠
-0.549
2 1.6 0.4554
-0.5787
3 1.9 0.2818
-0.5717
4 2.2 0.1103
♠
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

25. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 ♠
-0.549
2 1.6 0.4554
-0.5787
3 1.9 0.2818
-0.5717
4 2.2 0.1103
♠The Second Divided Difference involving x0, x1 and x2 is
f[x2, x1, x0] =
f[x2, x1] − f[x1, x0]
x2 − x0
=
−0.549 − (−0.4833)
1.6 − 1.0
= −0.1094
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

26. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 ♠
-0.549
2 1.6 0.4554
-0.5787
3 1.9 0.2818
-0.5717
4 2.2 0.1103
♠The Second Divided Difference involving x0, x1 and x2 is
f[x2, x1, x0] =
f[x2, x1] − f[x1, x0]
x2 − x0
=
−0.549 − (−0.4833)
1.6 − 1.0
= −0.1094
Remaining Second Divided Difference are as follows
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

27. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 ♣
2 1.6 0.4554 -0.0494
-0.5787
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
♣
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

28. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 ♣
2 1.6 0.4554 -0.0494
-0.5787
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
♣The Third Divided Difference f[x3, x2, x1, x0]
=
f[x3, x2, x1] − f[x2, x1, x0]
x3 − x0
=
−0.1094 − (−0.494)
1.9 − 1.0
= −0.0667
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

29. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 ♣
2 1.6 0.4554 -0.0494
-0.5787
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
♣The Third Divided Difference f[x3, x2, x1, x0]
=
f[x3, x2, x1] − f[x2, x1, x0]
x3 − x0
=
−0.1094 − (−0.494)
1.9 − 1.0
= −0.0667
Remaining Third Divided Differences are as follows
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

30. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 -0.0667
2 1.6 0.4554 -0.0494
-0.5787 0.0679
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

31. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 -0.0667
2 1.6 0.4554 -0.0494
-0.5787 0.0679
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
The Fourth Divided Difference f[x4, x3, x2, x1, x0]
=
f[x4, x3, x2, x1] − f[x3, x2, x1, x0]
x3 − x0
=
−0.0679 − (−0.0667)
2.2 − 1.0
= −0.001
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

32. Complete the Divided Difference table for the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 -0.0667
2 1.6 0.4554 -0.0494
-0.5787 0.0679
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
The Fourth Divided Difference f[x4, x3, x2, x1, x0]
=
f[x4, x3, x2, x1] − f[x3, x2, x1, x0]
x3 − x0
=
−0.0679 − (−0.0667)
2.2 − 1.0
= −0.001
This would be last divided difference in this case.
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

33. The coefficients of the Newton forward divided-difference form of the
interpolating polynomial are along the diagonal in the table. This
polynomial is
P4(x) = 0.7651 − 0.4833(x − 1) − 1094(x − 1)(x − 1.3)+
−0667(x − 1)(x − 1.3)(x − 1.6) − 0.001(x − 1)(x − 1.3)(x − 1.6)(x − 1.9)
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 -0.0667
2 1.6 0.4554 -0.0494
-0.5787 0.0679
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

34. The coefficients of the Newton forward divided-difference form of the
interpolating polynomial are along the diagonal in the table. This
polynomial is
P4(x) = 0.7651 − 0.4833(x − 1) − 1094(x − 1)(x − 1.3)+
−0667(x − 1)(x − 1.3)(x − 1.6) − 0.001(x − 1)(x − 1.3)(x − 1.6)(x − 1.9)
i xi f[xi] f[xi, xi−1] f[xi, xi−1, xi−2] f[xi, xi−1, xi−2, xi−3]
0 1.0 0.7651
-0.4833
1 1.3 0.6201 -0.1094
-0.549 -0.0667
2 1.6 0.4554 -0.0494
-0.5787 0.0679
3 1.9 0.2818 0.0117
-0.5717
4 2.2 0.1103
f[x4, x3, x2, x1, x0] = −0.001
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

35. Newton’s divided-difference formula can be expressed in a
simplified form when the nodes are arranged consecutively
with equal spacing. In this case, we introduce the notation
xi+1 − xi = h for each i = 0, 1, · · · , n − 1
Let x = x0 + sh, s ∈ R, then we can write
x − xi = (x0 + sh) − (x0 + ih)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

36. Newton’s divided-difference formula can be expressed in a
simplified form when the nodes are arranged consecutively
with equal spacing. In this case, we introduce the notation
xi+1 − xi = h for each i = 0, 1, · · · , n − 1
Let x = x0 + sh, s ∈ R, then we can write
x − xi = (x0 + sh) − (x0 + ih) ⇒ x − xi = (s − i)h
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

37. Using x − xi = (s − i)h in
Pn(x) = f[x0] + f[x1, x0](x − x0) + a2(x − x0)(x − x1)+
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

38. Using x − xi = (s − i)h in
Pn(x) = f[x0] + f[x1, x0](x − x0) + a2(x − x0)(x − x1)+
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain
Pn(x) = Pn(x0 + sh) = f[x0] + shf[x1, x0] + s(s − 1)h2
f[x2, x1, x0]+
s(s − 1)(s − 2)h3
f[x3, x2, x1, x0] + · · · +
+s(s − 1)(s − 2) · · · (s − n + 1)hn
f[xn, xn−1, · · · , x0]
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

39. Using x − xi = (s − i)h in
Pn(x) = f[x0] + f[x1, x0](x − x0) + a2(x − x0)(x − x1)+
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain
Pn(x) = Pn(x0 + sh) = f[x0] + shf[x1, x0] + s(s − 1)h2
f[x2, x1, x0]+
s(s − 1)(s − 2)h3
f[x3, x2, x1, x0] + · · · +
+s(s − 1)(s − 2) · · · (s − n + 1)hn
f[xn, xn−1, · · · , x0]
Note that
x − x0 = sh
x − x1 = (s − 1)h & x − x0 = sh ⇒
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

40. Using x − xi = (s − i)h in
Pn(x) = f[x0] + f[x1, x0](x − x0) + a2(x − x0)(x − x1)+
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain
Pn(x) = Pn(x0 + sh) = f[x0] + shf[x1, x0] + s(s − 1)h2
f[x2, x1, x0]+
s(s − 1)(s − 2)h3
f[x3, x2, x1, x0] + · · · +
+s(s − 1)(s − 2) · · · (s − n + 1)hn
f[xn, xn−1, · · · , x0]
Note that
x − x0 = sh
x − x1 = (s − 1)h & x − x0 = sh ⇒ (x − x1)(x − x0) = s(s − 1)h2
similarly (x − x0)(x − x1)(x − x2) = s(s − 1)(s − 2)h3
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

41. Equi-spaced Divided Difference
Pn(x0 + sh) = f[x0] +
n
k=1
s(s − 1) · · · (s − k + 1)hk
f[xk, xk−1, · · · , x0]
Using binomial-coefficient notation,
s
k
=
s(s − 1) · · · (s − k + 1)
k!
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

42. Equi-spaced Divided Difference
Pn(x0 + sh) = f[x0] +
n
k=1
s(s − 1) · · · (s − k + 1)hk
f[xk, xk−1, · · · , x0]
Using binomial-coefficient notation,
s
k
=
s(s − 1) · · · (s − k + 1)
k!
⇒ s(s−1) · · · (s−k +1) = k!
s
k
.
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

43. Equi-spaced Divided Difference
Pn(x0 + sh) = f[x0] +
n
k=1
s(s − 1) · · · (s − k + 1)hk
f[xk, xk−1, · · · , x0]
Using binomial-coefficient notation,
s
k
=
s(s − 1) · · · (s − k + 1)
k!
⇒ s(s−1) · · · (s−k +1) = k!
s
k
.
Thus we can express Pn(x) compactly as
Pn(x) = Pn(x0 + sh) = f[x0] +
n
k=1
k!
s
k
hk
f[xk, xk−1, · · · , x0]
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

44. Newton forward-difference formula
The Newton forward-difference formula, is constructed by making use
of the forward difference notation ∆. With this notation,
f[x1, x0] =
f[x1] − f[x0]
x1 − x0
=
1
h
(f(x1) − f(x0)) =
1
h
∆f(x0)
Similarly
f[x2, x1, x0] =
f[x2, x1] − f[x1, x0]
x2 − x0
=
1
2h
(f[x2, x1] − f[x1, x0])
=
1
2h
1
h
(∆f(x1) − ∆f(x0))
=
1
2h2
(∆(f(x1) − f(x0)))
=
1
2h2
(∆(∆f(x0))) =
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

45. Newton forward-difference formula
The Newton forward-difference formula, is constructed by making use
of the forward difference notation ∆. With this notation,
f[x1, x0] =
f[x1] − f[x0]
x1 − x0
=
1
h
(f(x1) − f(x0)) =
1
h
∆f(x0)
Similarly
f[x2, x1, x0] =
f[x2, x1] − f[x1, x0]
x2 − x0
=
1
2h
(f[x2, x1] − f[x1, x0])
=
1
2h
1
h
(∆f(x1) − ∆f(x0))
=
1
2h2
(∆(f(x1) − f(x0)))
=
1
2h2
(∆(∆f(x0))) =
1
2h2
∆2
f(x0)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

46. Newton forward-difference formula
Note that
f[x1, x0] =
1
h
∆f(x0)
f[x2, x1, x0] =
1
2h2
∆2
f(x0)
In general
f[xk, xk−1, · · · , x0] =
1
k!hk
∆k
f(x0)
Pn(x) = Pn(x0 + sh) = f[x0] +
n
k=1
k!
s
k
hk
f[xk, xk−1, · · · , x0]
Pn(x) = Pn(x0 + sh) = f(x0) +
n
k=1
s
k
∆k
f(x0)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

47. Compute cosh 0.56 using Newton’s Forward Difference
Formula (with 4 values)
i xi fi ∆fi ∆2fi ∆3fi
0 .5 1.127626
0.057839
1 .6 1.185465 0.011865
0.069704 0.000697
2 .7 1.255169 0.012562
0.082266
3 .8 1.337435
Using
P3(x) = P3(x0 + sh) = f(x0) +
3
k=1
s = x−x0
h
k
∆k
f(x0).
With x = 0.56, h = 0.1 and x0 = 0.5, we get
P3(0.56) = f(x0)+
0.6
1
∆f(x0)+
0.6
2
∆2
f(x0)+
0.6
3
∆3
f(x0).
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

48. Compute cosh 0.56 using Newton’s Forward Difference
Formula (with 4 values)
i xi fi ∆fi ∆2fi ∆3fi
0 .5 1.127626
0.057839
1 .6 1.185465 0.011865
0.069704 0.000697
2 .7 1.255169 0.012562
0.082266
3 .8 1.337435
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

49. Compute cosh 0.56 using Newton’s Forward Difference
Formula (with 4 values)
i xi fi ∆fi ∆2fi ∆3fi
0 .5 1.127626
0.057839
1 .6 1.185465 0.011865
0.069704 0.000697
2 .7 1.255169 0.012562
0.082266
3 .8 1.337435
Using
x = 0.56,
h = 0.1
and x0 = 0.5,
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

50. Compute cosh 0.56 using Newton’s Forward Difference
Formula (with 4 values)
i xi fi ∆fi ∆2fi ∆3fi
0 .5 1.127626
0.057839
1 .6 1.185465 0.011865
0.069704 0.000697
2 .7 1.255169 0.012562
0.082266
3 .8 1.337435
Using
x = 0.56,
h = 0.1
and x0 = 0.5,
P3(0.56) = f(x0)+
0.6
1
∆f(x0)+
0.6
2
∆2
f(x0)+
0.6
3
∆3
f(x0).
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

51. Compute cosh 0.56 using Newton’s Forward Difference
Formula (with 4 values)
i xi fi ∆fi ∆2fi ∆3fi
0 .5 1.127626
0.057839
1 .6 1.185465 0.011865
0.069704 0.000697
2 .7 1.255169 0.012562
0.082266
3 .8 1.337435
Using
x = 0.56,
h = 0.1
and x0 = 0.5,
P3(0.56) = f(x0)+
0.6
1
∆f(x0)+
0.6
2
∆2
f(x0)+
0.6
3
∆3
f(x0).
cosh(0.56) ≈ 1.127626 + (0.6)0.057839 +
(0.6)(−0.4)
2
0.011865+
(0.6)(−0.4)(−1.4)
6
0.000697 = 1.160944.
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

52. Error in Newton’s Forwards Formula
εn =
hn+1
(n + 1)!
s(s − 1) · · · (s − n)f(n+1)
(t)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

53. Error in Newton’s Forwards Formula
εn =
hn+1
(n + 1)!
s(s − 1) · · · (s − n)f(n+1)
(t)
n = 3 f(t) = cosh t, f4
(t) = cosh(t)
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics

54. Error in Newton’s Forwards Formula
εn =
hn+1
(n + 1)!
s(s − 1) · · · (s − n)f(n+1)
(t)
n = 3 f(t) = cosh t, f4
(t) = cosh(t)
ε3 =
(0.1)4
4!
0.6(−0.4)(−1.4)(−2.4)f(4)
(t) = −0.00000336 cosh t
We do not know t, but we get an inequality by taking the largest and
smallest cosh t in that interval:
A cosh 0.8 ≤ ε3(x) ≤ A cosh 0.5 where A = −0.00000336
Since
f(x) = P3(x) + ε3(x)
P3(0.56) + A cosh 0.8 ≤ cosh(0.56) ≤ P3(0.56)A cosh 0.5
1.160939 ≤ cosh(0.56) ≤ 1.160941.
Dr. Yasir Ali (yali@ceme.nust.edu.pk) Adv. Engg. Mathematics