Newtons Divided Difference Formulation

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

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.

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
,

=
n
k=0
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

=
n
k=0
n
i=0
i=k
x − xi
xk − xi
,
in the form
Pn(x) = a0+a1(x−x0)+a2(x−x0)(x−x1)+ · · ·+an(x−x0) · · · (x−xn−1)
(1)

=
n
k=0
n
i=0
i=k
x − xi
xk − xi
,
in the form
(1)
for appropriate constants
a0, a1, ..., an

For
Pn(x) at x0 leaves
a0 = Pn(x0) = f0 (2)
Pn(x) at x1 leaves
Pn(x1) = a0 + a1(x1 − x0) we know

For
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)

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)

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

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

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

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

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]
=

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

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]
=

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

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] =

Newton’s Divided Difference
Hence, the interpolating polynomial (1) may be expressed as
· · · + 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

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
♥

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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.

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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
♠

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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
♣

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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

x 1.0 1.3 1.6 1.9 2.2
f(x) 0.7651 0.6201 0.4554 0.2818 0.1103
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.

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)
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 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)
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

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)

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

Using x − xi = (s − i)h in
· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain

· · · + 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]

· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain
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 ⇒

· · · + an(x − x0)(x − x1) · · · (x − xn),
where
ak = f[xk, xk−1, · · · , x1, x0] for k = 0, 1, · · · , n.
We obtain
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

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!

Pn(x0 + sh) = f[x0] +
n
k=1
s(s − 1) · · · (s − k + 1)hk
f[xk, xk−1, · · · , x0]
s
k
=
s(s − 1) · · · (s − k + 1)
k!
⇒ s(s−1) · · · (s−k +1) = k!
s
k
.

Pn(x0 + sh) = f[x0] +
n
k=1
s(s − 1) · · · (s − k + 1)hk
f[xk, xk−1, · · · , x0]
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]

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))) =

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)

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)

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).

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

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,

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).

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.

Error in Newton’s Forwards Formula
εn =
hn+1
(n + 1)!
s(s − 1) · · · (s − n)f(n+1)
(t)

εn =
hn+1
(n + 1)!
s(s − 1) · · · (s − n)f(n+1)
(t)
n = 3 f(t) = cosh t, f4
(t) = cosh(t)

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.

Newtons Divided Difference Formulation

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Newtons Divided Difference Formulation