UNIT III
(21MAB206T)
ACADEMIC YEAR 2023-2024 (ODD SEMESTER)
NUMERICAL METHODS AND ANALYISIS
TOPICS
 Numerical Differenetiation
 Numerical differentiation using Newton Forward difference formulae
 Numerical differentiation using Newton backward difference formulae
 Numerical integration
 Trapezoidal rule and their applications
 Simpson’s one-third rule and their application
 Simpson’s three-eighth rule and their application
NUMERICAL DIFFERENTIATION
Numerical differentiation using Newton
forward difference formulae
 Suppose we have (𝑛 + 1) ordered pairs (𝑥𝑖 , 𝑦𝑖 ), for 𝑖 = 0,1, … , 𝑛.
 And we want to find the derivative of 𝑦 = 𝑓(𝑥) passing through the (𝑛 + 1) points, at a
point nearer to the starting value 𝑥 = 𝑥0 .
 We use Newton forward difference interpolation formula to get the derivative.
 Newton’s forward difference interpolation formula is
𝑢(𝑢−1) 2
𝑢(𝑢−1)(𝑢−2) 3
Δ
𝑦
+
Δ 𝑦0
0
2!
3!
𝑥−𝑥
in 𝑥 and 𝑢 = ℎ 0
 𝑦 𝑥 = 𝑦 𝑥0 + 𝑢ℎ = 𝑦𝑢 = 𝑦0 + 𝑢Δ𝑦0 +
 Where 𝑦(𝑥) is a polynomial of degree 𝑛
+⋯
 Differentiating 𝑦(𝑥) with respect to 𝑥, we get

𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑢
= 𝑑𝑢 ⋅ 𝑑𝑥
 Also

𝑑𝑢
𝑑𝑥
=
𝑑
𝑥−𝑥0
ℎ
𝑑𝑥
1
=ℎ
 So

𝑑𝑦
𝑑𝑥
1
ℎ
= ⋅
𝑑𝑦
𝑑𝑢
 Again

𝑑𝑦
𝑑𝑢
= Δ𝑦0 +
2𝑢−1 2
Δ 𝑦0
2
+
3𝑢2 −6𝑢+2 3
Δ 𝑦0
6
+
4𝑢3 −18𝑢2 +22𝑢−6 4
Δ 𝑦0
24
+⋯
 Hence

𝑑𝑦
𝑑𝑥
=
1
{Δ𝑦0
ℎ
+
2𝑢−1 2
Δ 𝑦0
2
𝑑𝑦
+
3𝑢2 −6𝑢+2 3
Δ 𝑦0
6
+
4𝑢3 −18𝑢2 +22𝑢−6 4
Δ 𝑦0
24
+ ⋯}
 This is the value of 𝑑𝑥 at general 𝑥 which may be anywhere in the interval.
 Some other higher order derivatives of 𝑦(𝑥) with respect to 𝑥 are given as

𝑑2 𝑦
𝑑𝑥 2
=

𝑑3 𝑦
𝑑𝑥 3
= ℎ3 {Δ3 𝑦0 +
1
ℎ2
1
Δ2 𝑦0
+ 𝑢−1
Δ3 𝑦0
12𝑢−18 4
Δ 𝑦0
12
+
6𝑢2 −18𝑢+11 4
Δ 𝑦0
12
+ ⋯}
+⋯
 In special case 𝑥 = 𝑥0 , we have 𝑢 = 0. So

𝑑𝑦
𝑑𝑥 𝑥=𝑥0
=
𝑑𝑦
𝑑𝑥 𝑢=0
1
1
1
1
= ℎ {Δ𝑦0 − 2 Δ2 𝑦0 + 3 Δ3 𝑦0 − 4 Δ4 𝑦0 + ⋯ }
𝑑𝑦
 This is the value of 𝑑𝑥 at general 𝑥 which may be anywhere in the interval.
 Similarly,

𝑑2 𝑦
𝑑𝑥 2 𝑥=𝑥0
= ℎ2 Δ2 𝑦0 − Δ3 𝑦0 + 12 Δ4 𝑦0 + ⋯

𝑑3 𝑦
𝑑𝑥 3 𝑥=𝑥0
= ℎ3 {Δ3 𝑦0 − 2 Δ4 𝑦0 + ⋯ }
1
1
11
3
Numerical differentiation using Newton
backward difference formulae
 Suppose we have (𝑛 + 1) ordered pairs (𝑥𝑖 , 𝑦𝑖 ), for 𝑖 = 0,1, … , 𝑛.
 And we want to find the derivative of 𝑦 = 𝑓(𝑥) passing through the (𝑛 + 1) points, at a
point nearer to the end value 𝑥 = 𝑥𝑛 .
 We use Newton backward difference interpolation formula to get the derivative.
 Newton’s backward difference interpolation formula is
𝑣(𝑣+1) 2
𝑣(𝑣+1)(𝑣+2) 3
𝛻
𝑦
+
𝛻 𝑦𝑛
𝑛
2!
3!
𝑥−𝑥
in 𝑥 and v= ℎ 𝑛
 𝑦 𝑥 = 𝑦 𝑥𝑛 + 𝑣ℎ = 𝑦𝑣 = 𝑦𝑛 + v𝛻𝑦𝑛 +
 Where 𝑦(𝑥) is a polynomial of degree 𝑛
+⋯
 Differentiating 𝑦(𝑥) with respect to 𝑥, we get

𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑣
𝑑𝑣
⋅ 𝑑𝑥
 Also

𝑑𝑣
𝑑𝑥
=
𝑑
𝑥−𝑥𝑛
ℎ
𝑑𝑥
1
=ℎ
 So

𝑑𝑦
𝑑𝑥
1
ℎ
= ⋅
𝑑𝑦
𝑑𝑣
 Again

𝑑𝑦
𝑑𝑣
= 𝛻𝑦𝑛 +
2𝑣+1 2
𝛻 𝑦𝑛
2
+
3𝑣 2 +6𝑣+2 3
𝛻 𝑦𝑛
6
+
4𝑣 3 +18𝑣 2 +22𝑣+6 4
𝛻 𝑦𝑛
24
+⋯
 Hence

𝑑𝑦
𝑑𝑥
=
1
{𝛻𝑦𝑛
ℎ
2𝑣+1
+ 2 𝛻 2 𝑦𝑛
𝑑𝑦
+
3𝑣 2 +6𝑣+2 3
𝛻 𝑦𝑛
6
+
4𝑣 3 +18𝑣 2 +22𝑣+6 4
𝛻 𝑦𝑛
24
+ ⋯}
 This is the value of 𝑑𝑥 at general 𝑥 which may be anywhere in the interval.
 Some other higher order derivatives of 𝑦(𝑥) with respect to 𝑥 are given as

𝑑2 𝑦
𝑑𝑥 2
=

𝑑3 𝑦
𝑑𝑥 3
= ℎ3 {𝛻 3 𝑦𝑛 +
1
ℎ2
1
𝛻 2 𝑦𝑛
+ 𝑣+1
Δ3 𝑦𝑛
12𝑣+18 4
𝛻 𝑦𝑛
12
+
6𝑣 2 +18𝑣+11 4
𝛻 𝑦𝑛
12
+ ⋯}
+⋯
 In special case 𝑥 = 𝑥𝑛 , we have v= 0. So

𝑑𝑦
𝑑𝑥 𝑥=𝑥𝑛
=
𝑑𝑦
𝑑𝑥 𝑣=0
1
1
1
1
= ℎ {𝛻𝑦𝑛 + 2 𝛻 2 𝑦𝑛 + 3 𝛻 3 𝑦𝑛 + 4 𝛻 4 𝑦𝑛 + ⋯ }
𝑑𝑦
 This is the value of 𝑑𝑥 at general 𝑥 which may be anywhere in the interval.
 Similarly,

𝑑2 𝑦
𝑑𝑥 2 𝑥=𝑥𝑛
= ℎ2 𝛻 2 𝑦𝑛 + 𝛻 3 𝑦𝑛 + 12 𝛻 4 𝑦𝑛 + ⋯

𝑑3 𝑦
𝑑𝑥 3 𝑥=𝑥𝑛
= ℎ3 {𝛻 3 𝑦𝑛 + 2 𝛻 4 𝑦𝑛 + ⋯ }
1
1
11
3
Examples
 Example 1:- Find the first two derivative of 𝑥
below:
50
𝒙
𝑦= 𝑥
1
3
51
3.6840 3.7084
1
3
at 𝑥 = 50 and 𝑥 = 56 given the data
52
53
54
55
56
3.7325
3.7563
3.7798
3.8030
3.8259
 Solution:- Since we require 𝑓 ′ (𝑥) at 𝑥 = 50, we use Newton’s forward difference formula and to get
𝑓 ′ 𝑥 at 𝑥 = 56, we use Newton’s backward difference formula.
𝑥
𝑦
50
3.6840
Δ𝑦
Δ2 𝑦
Δ3 𝑦
0.0244
51
3.7084
−0.0003
0.0241
52
3.7325
0
−0.0003
0.0238
53
3.7563
0
−0.0003
0.0235
54
3.7798
0
−0.0003
0.0232
55
3.8030
−0.0003
0.0229
56
3.8259
0
 By the differentiation of 𝑦(𝑥) derived from Newton’s forward interpolation formula, we have

𝑑𝑦
𝑑𝑥 𝑥=𝑥0
𝑑𝑦
𝑑𝑥 𝑢=0
=
1
1
1
= ℎ Δ𝑦0 − 2 Δ2 𝑦0 + 3 Δ3 𝑦0 − ⋯
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=50
= 1 0.0244 − 2 −0.0003 + 3 (0)
1
1
1
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=50
= 0.02455
 Also,

𝑑2 𝑦
𝑑𝑥 2 𝑥=𝑥0
 ⇒
=
𝑑2 𝑦
𝑑𝑥 2 𝑥=50
𝑑2 𝑦
𝑑𝑥 2 𝑢=0
1
= ℎ2 Δ2 𝑦0 − Δ3 𝑦0 …
= 1 −0.0003 = −0.0003
 By the differentiation of 𝑦(𝑥) derived from Newton’s backward interpolation formula, we have

𝑑𝑦
𝑑𝑥 𝑥=𝑥𝑛
𝑑𝑦
𝑑𝑥 𝑣=0
=
1
1
1
= ℎ 𝛻𝑦𝑛 + 2 𝛻 2 𝑦𝑛 + 3 𝛻 3 𝑦𝑛 + ⋯
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=56
= 1 0.0229 + 2 −0.0003 + 3 (0)
1
1
1
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=56
= 0.02275
 Also,

𝑑2 𝑦
𝑑𝑥 2 𝑥=𝑥𝑛
 ⇒
=
𝑑2 𝑦
𝑑𝑥 2 𝑥=56
𝑑2 𝑦
𝑑𝑥 2 𝑣=0
1
= ℎ2 𝛻 2 𝑦𝑛 + 𝛻 3 𝑦𝑛 …
= 1 −0.0003 = −0.0003
Examples
 Example 2:- The population f a certain town is given below. Find the rate of growth of the
population in 1931,1941,1961 and 1971.
Year
𝒙
𝟏𝟗𝟑𝟏
1941
𝟏𝟗𝟓𝟏
𝟏𝟗𝟔𝟏
𝟏𝟗𝟕𝟏
Population
in thousands
𝑦
40.62
60.80
79.95
103.56
132.65
 Solution:- We use the same difference table for backward and forward differences.
𝑥
𝑦
𝟏𝟗𝟑𝟏
40.62
Δ𝑦
Δ2 𝑦
Δ3 𝑦
𝚫𝟒 𝒚
20.18
𝟏𝟗𝟒𝟏
60.80
−1.03
19.15
𝟏𝟗𝟓𝟏
79.95
5.49
4.46
23.61
𝟏𝟗𝟔𝟏
103.56
𝟏𝟗𝟕𝟏
132.65
1.02
5.48
29.09
−4.47
To get 𝑓 ′ 1931 and 𝑓′(1941), we use the forward interpolation formula.
 By the differentiation of 𝑦(𝑥) derived from Newton’s forward interpolation formula, we have

𝑑𝑦
𝑑𝑥
1
= ℎ Δ𝑦0 +
2𝑢−1 2
Δ 𝑦0
2
+
3𝑢2 −6𝑢+2 3
Δ 𝑦0
6
+
4𝑢3 −18𝑢2 +22𝑢−6 4
Δ 𝑦0
24
+⋯
 Here 𝑥0 = 1931 and 𝑥1 = 1941
 First we find the value of 𝑓 ′ 1931 . So we use the following formula

𝑑𝑦
𝑑𝑥 𝑥=𝑥0
=
𝑑𝑦
𝑑𝑥 𝑢=0
=
1
ℎ
1
2
1
3
Δ𝑦0 − Δ2 𝑦0 + Δ3 𝑦0 − ⋯
 So,
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=1931
= 10 20.18 − 2 −1.03 + 3 5.49 − 4 (−4.47)
1
1
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=1931
= 2.36425
 Also, if 𝑥 = 1941, then 𝑢 =
1
1941−1931
10
1
10
= 10 = 1.
 Putting 𝑢 = 1 in equation (𝐴), we get

𝑑𝑦
𝑑𝑥 𝑥=1941
=
𝑑𝑦
𝑑𝑥 𝑢=1
1
1
1
1
= 10 20.18 + 2 −1.03 − 6 5.49 + 12 (−4.47) = 1.83775.
(𝐴)
To get 𝑓 ′ 1961 and 𝑓′(1971), we use the backward interpolation formula.
 By the differentiation of 𝑦(𝑥) derived from Newton’s backward interpolation formula, we have

𝑑𝑦
𝑑𝑥
1
= ℎ 𝛻𝑦𝑛 +
2𝑣+1 2
𝛻 𝑦𝑛
2
+
3𝑣 2 +6𝑣+2 3
𝛻 𝑦𝑛
6
+
4𝑣 3 +18𝑣 2 +22𝑣+6 4
𝛻 𝑦𝑛
24
+⋯
 Here 𝑥𝑛 = 1971 and 𝑥𝑛−1 = 1961
 First we find the value of 𝑓 ′ 1931 . So we use the following formula

𝑑𝑦
𝑑𝑥 𝑥=𝑥𝑛
=
𝑑𝑦
𝑑𝑥 𝑣=0
=
1
ℎ
1
2
1
3
𝛻𝑦𝑛 + 𝛻 2 𝑦𝑛 + 𝛻 3 𝑦𝑛 + ⋯
 So,
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=1971
= 10 29.09 + 2 5.48 + 3 1.02 + 4 (−4.47)
1
1
 ⇒
𝑑𝑦
𝑑𝑥 𝑥=1971
= 3.10525
 Also, if 𝑥 = 1961, then 𝑣 =
1
1961−1971
10
=
1
−10
10
= −1.
 Putting 𝑣 = 1 in equation (𝐵), we get

𝑑𝑦
𝑑𝑥 𝑥=1961
=
𝑑𝑦
𝑑𝑥 𝑣=1
1
1
1
1
= 10 29.09 − 2 5.48 − 6 1.02 − 12 (−4.47) = 2.65525.
(𝐵)
Examples
 Example 3:- The table given below reveals the velocity 𝑣 of a body at a time 𝑡 specified.
Find its acceleration at 𝑡 = 1.1.
𝒕
𝟏. 𝟎
1.1
𝟏. 𝟐
𝟏. 𝟑
𝟏. 𝟒
𝑣
43.1
47.7
52.1
56.4
60.8
 Solution:-𝑣 is dependent on time 𝑡, i.e. 𝑣 = 𝑣(𝑡). We require acceleration =
𝑑𝑣
.
𝑑𝑡
 Therefore, we have to find 𝑣 ′ 1.1 .
 That is, it is a problem of numerical differentiation.
𝒕
𝒗
𝟏. 𝟎
43.1
Δ𝐯
Δ2 𝐯
Δ3 𝐯
𝚫𝟒 𝒗
4.6
𝟏. 𝟏
47.7
−0.2
4.4
𝟏. 𝟐
52.1
0.1
−0.1
4.3
𝟏. 𝟑
56.4
𝟏. 𝟒
60.8
0.2
0.1
4.4
0.1
𝑑𝑣
As 𝑑𝑡 at 𝑡 = 1.1 is require, (nearer to the beginning value), we use forward interpolation formula.
 By the differentiation of v(𝑡) derived from Newton’s forward interpolation formula, we have

𝑑𝑣
𝑑𝑡
=
1
ℎ
Δ𝑣0 +
 Here 𝑢 =

𝑑𝑣
𝑑𝑡 𝑡=1.1
𝑡−𝑡0
ℎ
=
2𝑢−1 2
Δ 𝑣0
2
=
1.1−1
0.1
𝑑𝑣
𝑑𝑡 𝑢=1
+
3𝑢2 −6𝑢+2 3
Δ 𝑣0
6
+
4𝑢3 −18𝑢2 +22𝑢−6 4
Δ 𝑣0
24
+⋯
= 1.
1
1
1
1
= 0.1 4.6 + 2 −0.2 − 6 0.1 + 12 (0.1) = 44.917.
(𝐴)
NUMERICAL INTEGRATION
In this section we will be using
Trapezodial rule
Simpson’s one – third rule
Simpson’s three-eighth rule
Trapezoidal Rule
Simpson’s One-third Rule
Simpson’s Three-eighth Rule
Problems
THE END