Numerical Methods

Chapter 1
The Calculus of Finite Difference
The Operators E, ∆ and ∇
Shift Operator, E: Independent
The Shift Operator is denoted by E and defined by 𝐸𝑓 𝑥 = 𝑓(𝑥 + ℎ) . variable x is
known as
The Operator 𝐸 − 1 is the inverse operator of the operator E and defined as, Argument.
𝐸 − 1𝑓 𝑥 = 𝑓 𝑥 + −1 ℎ = 𝑓 𝑥 − ℎ .
Dependent
Forward or Descending Difference Operator, ∆: Va r i a b l e y i s
The forward difference is defined as ∆𝑓 𝑥 = 𝑓 𝑥 + ℎ − 𝑓(𝑥). known as Entry.

Backward or Ascending Difference Operator, ∇:

The backward difference operator is defined as 𝛻𝑓 𝑥 = 𝑓 𝑥 − 𝑓(𝑥 − ℎ).
 Between the
Relation between Operators
Operators
(a) 𝐸 ≡ 1 + ∆ or ∆ ≡ 𝐸 − 1
We have ∆𝑓 𝑥 = 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 𝐸𝑓 𝑥 − 𝑓 𝑥 = 𝐸 − 1 𝑓 𝑥 .
Thus ∆𝑓 𝑥 = 𝐸 − 1 𝑓(𝑥), for any function 𝑓(𝑥).
∴ ∆≡ 𝐸 −1
or 𝐸 ≡ 1 + ∆

(b) 𝛻 ≡ 1 − 𝐸 − 1 or 𝐸 − 1 ≡ 1 − 𝛻
We have 𝛻𝑓 𝑥 = 𝑓 𝑥 − 𝑓 𝑥 − ℎ = 𝑓 𝑥 − 𝐸 − 1𝑓 𝑥 = 1 − 𝐸 − 1 𝑓 𝑥
Thus 𝛻𝑓 𝑥 = 1 − 𝐸 − 1 𝑓(𝑥), for any function 𝑓(𝑥).
∴ 𝛻 ≡ 1 − 𝐸 −1
or 𝐸 − 1 ≡ 1 − 𝛻

(c) 𝐸𝛻 ≡ 𝛻𝐸 ≡ ∆
We have, 𝐸𝛻 𝑓 𝑥 = 𝐸 𝛻𝑓 𝑥 =𝐸 𝑓 𝑥 −𝑓 𝑥−ℎ = 𝐸𝑓 𝑥 − 𝐸𝑓 𝑥 − ℎ = 𝑓 𝑥 + ℎ − 𝑓 𝑥 = ∆𝑓 𝑥 … (1)
Also, 𝛻𝐸 𝑓 𝑥 = 𝛻 𝐸𝑓 𝑥 = 𝛻𝑓 𝑥 + ℎ = 𝑓 𝑥 + ℎ − 𝑓 𝑥 = ∆𝑓 𝑥 … (2)
From (1) and (2), we have
𝐸𝛻 ≡ ∆ and 𝛻𝐸 ≡ ∆.
Thus 𝐸𝛻 ≡ 𝛻𝐸 ≡ ∆.
Forward Difference Table
First differences
Argument x Entry 𝒚 = 𝒇(𝒙) Second differences ∆𝟐𝒇(𝒙)
∆𝒇(𝒙)

𝒂 𝑓(𝑎)
𝑓 𝑎 + ℎ − 𝑓 𝑎 = ∆𝑓(𝑎)
∆𝑓 𝑎 + ℎ − ∆𝑓 𝑎
𝒂+𝒉 𝑓(𝑎 + ℎ)
𝑓 𝑎 + 2ℎ − 𝑓 𝑎 + ℎ = ∆2𝑓(𝑎)
= ∆𝑓(𝑎 + ℎ)
∆𝑓 𝑎 + 2ℎ − ∆𝑓 𝑎 + ℎ
𝒂 + 𝟐𝒉 𝑓(𝑎 + 2ℎ)
= ∆2𝑓(𝑎 + ℎ)
𝑓 𝑎 + 3ℎ − 𝑓 𝑎 + 2ℎ
= ∆𝑓(𝑎 + 2ℎ) ∆𝑓 𝑎 + 3ℎ
𝒂 + 𝟑𝒉 𝑓(𝑎 + 3ℎ) − ∆𝑓 𝑎 + 2ℎ
𝑓 𝑎 + 4ℎ − 𝑓 𝑎 + 3ℎ = ∆2𝑓(𝑎 + 2ℎ)
= ∆𝑓(𝑎 + 3ℎ)
𝒂 + 𝟒𝒉 𝑓(𝑎 + 4ℎ)
Backward Difference Table
First differences
Argument x Entry 𝒚 = 𝒇(𝒙) Second differences ∆𝟐𝒇(𝒙)
∆𝒇(𝒙)

𝒂 𝑓(𝑎)
𝑓 𝑎+ℎ −𝑓 𝑎
= 𝛻𝑓(𝑎 + h)
𝛻𝑓 𝑎 + 2ℎ − 𝛻𝑓 𝑎 + h
𝒂+𝒉 𝑓(𝑎 + ℎ)
𝑓 𝑎 + 2ℎ − 𝑓 𝑎 + ℎ = 𝛻2𝑓(𝑎 + 2h)
= 𝛻𝑓(𝑎 + 2ℎ)
𝛻𝑓 𝑎 + 3ℎ − 𝛻𝑓 𝑎 + 2ℎ
𝒂 + 𝟐𝒉 𝑓(𝑎 + 2ℎ)
= 𝛻2𝑓(𝑎 + 3ℎ)
𝑓 𝑎 + 3ℎ − 𝑓 𝑎 + 2ℎ
= 𝛻𝑓(𝑎 + 3ℎ)
𝛻𝑓 𝑎 + 4ℎ − 𝛻𝑓 𝑎 + 3ℎ
𝒂 + 𝟑𝒉 𝑓(𝑎 + 3ℎ)
= 𝛻2𝑓(𝑎 + 4ℎ)
𝑓 𝑎 + 4ℎ − 𝑓 𝑎 + 3ℎ
= 𝛻𝑓(𝑎 + 4ℎ)
𝒂 + 𝟒𝒉 𝑓(𝑎 + 4ℎ)
Example
Prove that,
(a) 𝑓 4 = 𝑓 3 + ∆𝑓 2 + ∆2𝑓 1 + ∆3𝑓(1)
(b) 𝑓 4 = 𝑓 0 + 4∆𝑓 0 + 6∆2𝑓 −1 + 10∆3𝑓(−1)

Sol: (a) We have,

𝑓(4) − 𝑓(3) = ∆𝑓(3)
= ∆[𝑓(2) + ∆𝑓(2)] {∵ ∆𝑓 2 = 𝑓 3 − 𝑓 2 }
= ∆𝑓(2) + ∆2𝑓(2)
= ∆𝑓(2) + ∆2[𝑓(1) + ∆𝑓(1)] { ∵ ∆𝑓 1 = 𝑓 2 − 𝑓 1 }
= ∆𝑓(2) + ∆2𝑓(1) + ∆ 3𝑓(1).
∴ 𝑓 4 = 𝑓 3 + ∆𝑓 2 + ∆2𝑓 1 + ∆3𝑓 1 .
Sol: (b) We have,
𝑓 4 = 𝐸5𝑓 −1 = 1 + ∆ 5f(−1)
= 1 + 5𝐶1 ∆ + 5C2 ∆2 + 5C3∆3 f −1 , taking upto 3rd differences
= 𝑓 −1 + 5∆𝑓 −1 + 10∆2𝑓 −1 + 10∆3𝑓(−1)
= 𝑓 −1 + ∆𝑓 −1 + 4 ∆𝑓 −1 + ∆2𝑓 −1 + 6∆2𝑓 −1 + 10∆3𝑓(−1)
= 𝑓 −1 + ∆𝑓 −1 + 4∆ 𝑓 −1 + ∆𝑓 −1 + 6∆2𝑓 −1 + 10∆3𝑓 −1
= 𝑓 0 + 4∆𝑓 0 + 6∆2𝑓 −1 + 10∆3𝑓 −1 .

[∵ 𝑓 −1 + ∆𝑓 −1 = 𝑓 −1 + ∆𝑓 0 − 𝑓 −1 = 𝑓 0 ]
 Chapter 2
Interpolation with Equal Intervals
What is Interpolation?
The technique or method of estimating unknown values from given set of
observation is known as interpolation.

𝑥 𝑓(𝑥) 𝑥 𝑓(𝑥)
1971 1000 1975 670
1981 1025 1980 685
1991 1080 1987 750
2001 1120 1990 800
2011 1200 1994 915
Equal Interval Unequal Interval
 Methods
Newton formula for Forward Interpolation,
Here,
𝑢 𝑢 𝑢−1 2 𝑢 𝑢−1 𝑢−2 3 𝑎 = 𝐵𝑎𝑠𝑒
𝑓 𝑎 + ℎ𝑢 = 𝑓 𝑎 + ∆𝑓 𝑎 + ∆𝑓 𝑎 + ∆ 𝑓 𝑎 +⋯ ℎ = 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓
1! 2! 3!
𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙
Newton formula for Backward Interpolation, 𝑥−𝑎
𝑢=
𝑢 𝑢 𝑢+1 2 𝑢 𝑢+1 𝑢+2 3 ℎ
𝑓 𝑎 + ℎ𝑢 = 𝑓 𝑎 + 𝛻𝑓 𝑎 + 𝛻𝑓 𝑎 + 𝛻 𝑓 𝑎 +⋯
1! 2! 3!
𝑥 𝑓(𝑥)
Steps to do Math
1971 = 𝑎 1000
Step 1: Do the Difference Table 𝑥 = 1975
1987 1025
Step 2: Apply the required formula and Calculate
1991 1080
2001 1120
𝑥 = 2006
2011 = 𝑎 1200
Example
From the following table, estimate the number of students who obtained marks between 40 and 45. (Using
Newton’s Formula for forward interpolation)
Marks 30-40 40-50 50-60 60-70 70-80
No. of Students 31 42 51 35 31
Sol: First we prepare the cumulative frequency table, as given below:
Marks above 30 but less than No. of Students
𝒙 𝒇(𝒙)
40 31
50 73
60 124
70 159
80 190
Now we prepare the difference table for this data
𝒙 𝒇(𝒙) ∆𝒇(𝒙) ∆𝟐𝒇(𝒙) ∆𝟑𝒇(𝒙) ∆𝟒𝒇(𝒙)
40 31
42
50 73 9
51 -25
60 124 -16 37
35 12
70 159 -4
31
80 190
We shall find: 𝑓 45 = number of students with marks less than 45. Taking 𝑎 + 𝑢ℎ = 45, we get

1
40 + 𝑢 × 10 = 45 ⇒ 𝑢 =
2
Using Newton’s formula for forward interpolation, we get

1 1 1 ∆2𝑓(40) 1 1 3 ∆3𝑓(40) 1 1 3 5 ∆4𝑓(40)

𝑓 45 = 𝑓 40 + ∆𝑓 40 + − + − − + − − −
2 2 2 2! 2 2 2 3! 2 2 2 2 4!

1 1 1 1 1 1 3 1 1 1 3 5 1
= 31 + × 42 + − ×9+ − − −25 + − − − × 37
2 2 2 2! 2 2 2 3! 2 2 2 2 4!

= 47.868 𝑜𝑛 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 .
Hence the number of students with marks less than 45 is 47.868 i.e., 48.
But the number of students with marks less than 40 is 31.
Hence the required number of students getting marks between 40 and 45 is = 48 − 31 = 17.

Given,
𝒙 1 2 3 4 5 6 7 8
𝒇(𝒙) 1 8 27 64 125 216 343 512
Find 𝑓 7.5 .

Sol: The value to be interpolated lies at the end of the given observation i.e. near 𝑥 = 8. So in this case
Newton’s backward formula will be more suitable.
𝑥−𝑎 7.5−8
Here 𝑢= = = −0.5.
ℎ 1

The difference table is as below:

 𝒙 𝒇(𝒙) 𝛁𝐟(𝐱) 𝛁𝟐𝐟(𝐱) 𝛁𝟑𝐟(𝐱)
1 1
7
2 8 12
19 6
3 27 18
37 6
4 64 24
61 6
5 125 30
91 6
6 216 36
127 6
7 343 42
169
8 512

Since, 𝛻3f(x) is constant, so we can leave higher order differences.

By Newton’s backward interpolation formula,

𝑢 𝑢+1 2 𝑢 𝑢+1 𝑢+2 3

𝑓 𝑎 + 𝑢ℎ = 𝑓 𝑎 + 𝑢𝛻f a + 𝛻f a + 𝛻 f(a)
2! 3!
 −0.5 −0.5 + 1 2 −0.5 −0.5 + 1 −0.5 + 2 3
𝑓 7.5 = 𝑓 8 + −0.5 𝛻f 8 + 𝛻f 8 + 𝛻f 8
2 6

(−0.5)(0.5) (−0.5)(0.5)(1.5)
= 512 + −0.5 × 169 + × 42 + ×6
2 6
= 512 − 84.5 − 5.25 − 0.375
= 421.875.
 Chapter 3
Interpolation with Unequal Intervals
Methods
We will learn 2 methods for Interpolation with unequal interval of the argument.
1. Lagrange’s Method
2. Newton Divided Difference

Methods
The Lagrange’s Interpolation formula:
If, 𝑦 = 𝑓(𝑥) takes the values 𝑦0, 𝑦1, … , 𝑦𝑛 corresponding to 𝑥 = 𝑥0, 𝑥1, … , 𝑥𝑛 then,

𝑥 − 𝑥2 𝑥 − 𝑥3 … (𝑥 − 𝑥𝑛) 𝑥 − 𝑥1 𝑥 − 𝑥3 … (𝑥 − 𝑥𝑛) 𝑥 − 𝑥1 𝑥 − 𝑥2 … (𝑥 − 𝑥𝑛 − 1)
𝑓 𝑥 = 𝑦1 + 𝑦2 + ⋯ + 𝑦
𝑥1 − 𝑥2 𝑥1 − 𝑥3 … (𝑥1 − 𝑥𝑛) 𝑥2 − 𝑥1 𝑥2 − 𝑥3 … (𝑥2 − 𝑥𝑛) 𝑥𝑛 − 𝑥1 𝑥𝑛 − 𝑥2 … (𝑥𝑛 − 𝑥𝑛 − 1) 𝑛
Example
For the following table find the form of the function 𝑓(𝑥). (Using Lagrange’s Formula)
𝒙 0 1 2 5
𝒇(𝒙) 2 3 12 147
Sol: Here, 𝑥0 = 0, 𝑥1 = 1, 𝑥2 = 2, 𝑥3 = 5
By Lagrange’s formula, we have
𝑓 𝑥

𝑥 − 𝑥1 𝑥 − 𝑥2 𝑥 − 𝑥3 𝑥 − 𝑥0 𝑥 − 𝑥2 𝑥 − 𝑥3 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥 − 𝑥3 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥 − 𝑥2
= 𝑓 𝑥0 + 𝑓 𝑥1 + 𝑓 𝑥2 + 𝑓(𝑥3)
𝑥0 − 𝑥1 𝑥0 − 𝑥2 𝑥0 − 𝑥3 𝑥1 − 𝑥0 𝑥1 − 𝑥2 𝑥1 − 𝑥3 𝑥2 − 𝑥0 𝑥2 − 𝑥1 𝑥2 − 𝑥3 𝑥3 − 𝑥0 𝑥3 − 𝑥1 𝑥3 − 𝑥2
Substituting the values of 𝑥0, 𝑥1, 𝑥2, 𝑥3 in this, we get

(𝑥 − 1)(𝑥 − 2)(𝑥 − 5) (𝑥 − 0)(𝑥 − 2)(𝑥 − 5) (𝑥 − 0)(𝑥 − 1)(𝑥 − 5) (𝑥 − 0)(𝑥 − 1)(𝑥 − 2)

𝑓 𝑥 = ×2+ ×3+ × 12 + × 147
(0 − 1)(0 − 2)(0 − 5) 1 − 0 1 − 2 (1 − 5) (2 − 0)(2 − 1)(2 − 5) (5 − 0)(5 − 1)(5 − 2)

𝑓(𝑥) 1 3 2 49
𝑜𝑟, =− + − +
𝑥(𝑥 − 1)(𝑥 − 2)(𝑥 − 5) 5𝑥 4 𝑥 − 1 𝑥 − 2 20 𝑥 − 5

20𝑥3 + 20𝑥2 − 20𝑥 + 40

=
20𝑥(𝑥 − 1)(𝑥 − 2)(𝑥 − 5)
⇒ 𝑓 𝑥 = 𝑥3 + 𝑥2 − 𝑥 + 2.
 Methods
Newton Divided Difference:
𝑓 𝑥 = 𝑓 𝑥0 + 𝑥 − 𝑥0 ∆𝑓 𝑥0 + 𝑥 − 𝑥0 𝑥 − 𝑥1 ∆2𝑓 𝑥0 + 𝑥 − 𝑥0 𝑥 − 𝑥1 𝑥 − 𝑥2 ∆3𝑓 𝑥0 + ⋯

Divided Difference Table

First differences
Argument x Entry 𝒚 = 𝒇(𝒙) Second differences ∆𝟐𝒇(𝒙)
∆𝒇(𝒙)

𝒂 𝑓(𝑎)
𝑓 𝑎 + ℎ − 𝑓(𝑎) ∆𝑓 𝑎 + ℎ − ∆𝑓(𝑎)
= ∆𝑓(𝑎)
𝒂+𝒉 𝑓(𝑎 + ℎ) 𝑎+ℎ −𝑎 𝑎 + 2ℎ − 𝑎
𝑓 𝑎 + 2ℎ − 𝑓(𝑎 + ℎ) = ∆2𝑓(𝑎)
𝑎 + 2ℎ − (𝑎 + ℎ)
𝒂 + 𝟐𝒉 𝑓(𝑎 + 2ℎ) = ∆𝑓(𝑎 + ℎ)

Steps to do Math
Step 1: Do the Difference Table
Step 2: Apply the required formula and Calculate
Example
By means of Newton’s divided difference formula, find the values of 𝑓(8) and 𝑓(15) from the following table:

𝒙 4 5 7 10 11 13
𝒇(𝒙) 48 100 294 900 1210 2028

Sol: The divided difference table is:

𝒙 𝒇(𝒙) ∆𝒇(𝒙) ∆𝟐𝒇(𝒙) ∆𝟑𝒇(𝒙) ∆𝟒𝒇(𝒙)

4 48
100 − 48
= 52
5−4 97 − 52
5 100 = 15
294 − 100 7−4 21 − 15
= 97 =1
7−5 202 − 97 10 − 4
7 294 = 21 0
900 − 294 10 − 5 27 − 21
= 202 =1
10 − 7 310 − 202 11 − 5
10 900 = 27 0
1210 − 900 11 − 7 33 − 27
= 310 =1
11 − 10 409 − 310 13 − 7
11 1210 = 33
2028 − 1210 13 − 10
= 409
13 − 11
13 2028
Using Newton’s interpolation formula for unequal intervals we get,
𝑓 8 = 48 + 8 − 4 × 52 + 8 − 4 8 − 5 × 15 + 8 − 4 8 − 5 8 − 7 × 1
= 448.
and,
𝑓 15 = 48 + 15 − 4 × 52 + 15 − 4 15 − 5 × 15 + 15 − 4 15 − 5 15 − 7 × 1
= 3150.
 Chapter 4
Central Difference Interpolation Formulae
The Central Difference Interpolation Formulae are used for interpolating values of the function near the middle of the tabulated set.
The operator 𝛿 , called the central difference operator, is defined by the operator equation
Τ𝟐
Gauss Forward Central − 𝐄 −𝟏Τ𝟐 Interpolation Formula
𝛿 =𝐄 𝟏Difference
The first central difference 𝛿f(x) of f(x) is given by
Gauss Backward Central Difference 1 Interpolation
1 Formula
𝛿f(x) = f(x + ℎ) − f(x − ℎ)
Bessel’s Interpolation Formula 2 2

Stirling Interpolation Formula

• Some Central Difference Formulae for Equal Interval
• Gauss Forward Central Difference Interpolation Formula
• Gauss Backward Central Difference Interpolation Formula
• Bessel’s Interpolation Formula
• Stirling Interpolation Formula
 Gauss Forward Central Difference Interpolation Formula

𝒖(𝒖𝒖(𝒖
− 𝟏)
− 𝟏) 𝟐
𝟐
𝒖𝒖
++𝟏 𝟏
𝒖(𝒖𝒖(𝒖
− 𝟏) −𝟑𝟏)
𝟑𝒚
𝒚 𝒙 = 𝒚𝒚𝟎 +
𝒙 𝒖
= ∆𝒚
𝟎+ 𝟎
𝒚𝟎 + 𝒖 ∆𝒚 +
𝟐! ∆ ∆ 𝒚𝒚 −𝟏 +
−𝟏
+
𝟑! ∆
∆ 𝒚−𝟏 +
−𝟏 +
𝟐! 𝟑!
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
𝒖+𝟏 𝒖(𝒖 − 𝟏)(𝒖 ∆ 𝒚 + ∆ −
𝒚−𝟐𝟐)
𝟒! − 𝟐) −𝟐 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖
𝟓!
∆𝟒 𝒚−𝟐 + ∆𝟓 𝒚−𝟐
𝟒!
+ …………………………………………………………………………………
𝟓!
𝒙 −𝒙𝟎
Where 𝒖=
+ … … …𝒐𝒓,
𝒉
… 𝒙…=…𝒙 …+…𝒖𝒉… … … … … … … … … … … … … … … … … … … … … … …
𝟎
Here, 0< 𝒖 < 𝟏𝒖 = 𝒙 −𝒙𝟎
Where
𝒉
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
Here, 0< 𝒖 < 𝟏
 Example on Gauss Forward Formula
𝒖(𝒖 − 𝟏) 𝟐 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚𝟎 + 𝒖 ∆𝒚𝟎 + ∆ 𝒚−𝟏 + ∆ 𝒚−𝟏 +
𝟐! 𝟑!
Use Gauss forward formula to find the value of y when x = 3.75, given the following table :
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
x: 2.5 3.0 3.5 4.0∆ 𝒚−𝟐 +4.5 5.0 ∆ 𝒚−𝟐
𝟒! 𝟓!
y : 24.145 22.043 20.225 18.644 17.262 16.047
+ …………………………………………………………………………………
𝒙 −𝒙𝟎
Where 𝒖= 𝒉
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
Here, 0< 𝒖 < 𝟏
 Solution
𝒖(𝒖 − 𝟏) 𝟐 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚𝟎 + 𝒖 ∆𝒚𝟎 + ∆ 𝒚−𝟏 + ∆ 𝒚−𝟏 +
𝟐! 𝟑!
Let us take 3.5 as the origin and .5 as
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐)
the unit.
(𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐)
∆ 𝟒𝒚 + ∆ 𝟓𝒚
𝑥 −𝑥 0 −𝟐 −𝟐
We know 𝑢 = 𝟒! , Here x = 3.75 , 𝑥0 = 3.5, h = .5
𝟓!
ℎ
+ …… ……………………………………………………………………………
3.75−3.5
Where 𝒖 =∴ 𝒉 𝑢 = = .5
𝒙 −𝒙𝟎
.5
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
We0<require
Here, 𝒖 <𝟏 the value of y fot u=.5
The difference table is given below :
 x u yu ∆ yu ∆𝟐 yu ∆𝟑 yu ∆𝟒 yu ∆𝟓 yu
2.5 -2 24.145
-2.102
𝒖(𝒖 − 𝟏) 𝟐 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
3.o 𝒚
-1 𝒙 = 𝒚 + 𝒖 ∆𝒚
𝟎 22.043𝟎 + ∆ 𝒚.284
−𝟏 + ∆ 𝒚−𝟏 +
𝟐! 𝟑!
-1.818 -.047
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
3.5 0 20.225 ∆ 𝒚 +
−𝟐 .237 .009 ∆ 𝒚−𝟐
𝟒! 𝟓!
-1.581 -.038 -.003
+ …………………………………………………………………………………
Where4.0 𝒖 = 𝒉 1𝟎 18.644 .199 .006
𝒙 −𝒙

𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉 -1.382 -.032

0< 𝒖 < 𝟏 2
Here, 4.5 17.262 .167
-1.215
5.0 3 16.047
 Gauss’s Forward Formula is

𝑢(𝑢−1) 𝑢+1 𝑢(𝑢−1)

𝑦 𝑥 = 𝑦0 + 𝑢 ∆𝑦0 + ∆2 𝑦−1 + ∆3 𝑦−1 +
2! 3!
𝒖(𝒖 − 𝟏) 𝟐 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚
𝑢 + 1 𝑢(𝑢 − + 𝒖 ∆𝒚
𝟎 1)(𝑢 − +
𝟎 2) ∆ 𝒚
(𝑢−𝟏 + ∆ 𝒚−
+ 2) 𝑢 + 1𝟑!𝑢(𝑢 − 1)(𝑢 +
−𝟏2)
𝟐!
4
∆ 𝑦−2 + ∆5 𝑦−2
4! 5!
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐 + ∆ 𝒚
+ … … … … …𝟒!… … … … … … … … … … … … … … … … … 𝟓!… … … … … … … … …−𝟐

+ … … … … … … … … … … ….5
…….5…−…1… … … … … … ….5…+
…1… ….5….5
……−…
1…
𝒙 −𝒙𝟎
Where 𝒖∴ =
𝑦 𝑥 = 20.225 + .5 −1.581 + × .237 + −.038
𝒉 2 6
.5 +𝒐𝒓,
1 𝒙.5= .5
𝒙𝟎 −
+1𝒖𝒉 .5 − 2 (.5 + 2) .5 + 1 .5 .5 − 1 .5 − 2
+
Here, 0< 𝒖 < 𝟏 .009 +
24 120
(−.003)

= 20.225 − 0.7905 − 0.029625 + 0.002375 + 0.00210938 − 0.0000352

= 19.40 𝑎𝑝𝑝𝑟𝑜𝑥.
Hence the value of y for x=3.75 is 19.40 approximately.
 Gauss Backward Central Difference Interpolation Formula

𝒖(𝒖 − 𝟏) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚𝟎 + 𝒖 ∆𝒚𝟎𝑢(𝑢
+ + 1) ∆𝟐2𝒚−𝟏 + 𝑢 + 1 𝑢(𝑢 ∆−𝒚1)
−𝟏 + 3
𝑦 𝑥 = 𝑦0 + 𝑢 ∆𝑦−1 + 𝟐! ∆ 𝑦−1 + 𝟑! ∆ 𝑦−2 +
2! 3!
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐 + ∆ 𝒚−𝟐
(𝑢 + 2) 𝟒!
𝑢 + 1 𝑢(𝑢 − 1) 4 𝟓!
(𝑢 + 2) 𝑢 + 1 𝑢(𝑢 − 1)(𝑢 − 2)
∆ 𝑦−2 + ∆5 𝑦−3
+4!
……………………………………………………………… 5!… … … … … … …
𝒙 −𝒙𝟎
Where 𝒖= 𝒉
+ … … …𝒐𝒓,…𝒙…= …
𝒙𝟎 … ………………………………………………………………
+ 𝒖𝒉
Here, 0< 𝒖 < 𝟏 0 𝑥 −𝑥
Where 𝑢 =
ℎ
𝑜𝑟, 𝑥 = 𝑥0 + 𝑢ℎ
Here, -1< 𝑢 < 0
 Example on Gauss Backward Formula

Interpolate by means of𝒖(𝒖Gauss’s

𝒚 𝒙 = 𝒚𝟎 + 𝒖 ∆𝒚𝟎 +
− 𝟏) 𝟐 Backward
∆ 𝒚−𝟏 +
Formula
𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑 the
∆ 𝒚−𝟏 +
𝟐! 𝟑!
population for the year 1936, given the following table :
Year 𝒖 +: 𝟏1901 1911
𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 1921
∆𝟒 𝒚−𝟐 +1931 1941 1951
(𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐
𝟒! 𝟓!
Population : 12 15 20 27 39 52
(in thousands)+ …………………………………………………………………………………
𝒙 −𝒙
𝟎
Where 𝒖 = 𝒉
Solution: 𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
Here, 0< 𝒖 < 𝟏 We take 1931 as origin and 10 years as
the unit, tehn the population is to be estimated for u =
1936−1931
= .5
10
 𝒖(𝒖 − 𝟏) 𝟐 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚𝟎 + 𝒖 ∆𝒚𝟎 + ∆ 𝒚−𝟏 + ∆ 𝒚−𝟏 +
𝟐! 𝟑!

𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓

∆ 𝒚−𝟐 + ∆ 𝒚−𝟐
𝟒! 𝟓!

+ …………………………………………………………………………………
𝒙 −𝒙𝟎
Where 𝒖= 𝒉
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
Here, 0< 𝒖 < 𝟏
 Now Gauss’s Backward Formula is
𝑢(𝑢 + 1) 2 𝑢 + 1 𝑢(𝑢 − 1) 3
𝑦𝑢 = 𝑦0 + 𝑢 ∆𝑦−1 + ∆ 𝑦−1 + ∆ 𝑦−2 +
2! 3!

(𝑢 + 2)
𝒚 𝒙𝑢 +
= 1𝒚𝟎 𝑢(𝑢 − 1)+ 𝒖(𝒖
+ 𝒖 ∆𝒚 4
− 𝟏) (𝑢𝟐+ 2) 𝑢𝒖++1𝟏 𝑢(𝑢
∆ 𝒚 +
𝒖(𝒖−
− 1)(𝑢
𝟏) 𝟑− 2)
∆ 𝒚−𝟏 +∆5 𝑦
𝟎 ∆ 𝑦𝟐!
−2 +
−𝟏
𝟑! −3
4! 5!
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐)
+ … … … … … … … … … … … …∆…𝟒 𝒚…−𝟐…+… … … … … … … … … … … … … … … ∆…𝟓 𝒚−𝟐
𝟒! 𝟓!

+ … … ….5…+…1… (.5)
… … … … … … .5
…… +…1… .5
… …(.5
…… −…1)… … … … … … … … … …
∴ 𝑦 = 27 + .5 × 7 +
𝒙 −𝒙 ×5+ ×3
Where .5𝒖 = 𝒉 𝟎 2 6
(.5 + 2) .5 + 𝒐𝒓,
1 𝒙.5=(.5
𝒙𝟎 −
+ 1)
𝒖𝒉× (−7) + .5 + 2 .5 + 1 .5 .5 − 1 .5 − 2 × (−10)
+
Here, 0< 𝒖 < 𝟏 24 120

= 27+3.5+1.875-0.1875+0.2734-0.1172
= 32.3437 thousand
Thus the estimated population for 1936 is 32.3437 thousand
 Bessel’s Central difference Interpolation Formula

𝒖(𝒖 − 𝟏) 𝟐𝟑 𝒖 +𝟏𝟏 𝒖(𝒖 − 𝟏) 𝟑

𝒚𝟎 +𝒚𝟏 𝒚 𝒙 = 𝒚𝟏𝟎 + 𝒖 ∆𝒚𝟎𝒖+𝒖−𝟏 ∆𝟐 𝒚−𝟏+∆
∆ 𝒚𝒚−𝟏 + (𝒖− )𝒖 𝒖−𝟏 ∆ 𝒚 +
𝟑! ∆𝟑 𝒚−𝟏 −𝟏
𝟎 𝟐
y(x) = + 𝒖− ∆𝒚𝟎 + 𝟐! +
𝟐 𝟐 𝟐! 𝟐 𝟑!
𝟏
𝒖 𝒖𝟐 −𝟏 (𝒖−𝟐)
𝒖 +∆𝟒𝟏𝒚−𝟐𝒖(𝒖
+∆𝟒 𝒚
−−𝟏𝟏)(𝒖 (𝒖−
− 𝟐))𝒖 𝒖𝟐 −𝟏 (𝒖−𝟐)
(𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐)
+ + 𝟐 ∆𝟒 𝒚−𝟐 + ∆𝟓 𝒚−𝟐 + … … … … … ∆𝟓 𝒚−𝟐
𝟒! 𝟐 𝟒! 𝟓! 𝟓!

𝒙𝒖
−𝒙=
𝒙+−𝒙…𝟎
………………………………………………………………………………
Where
Where 𝒖= 𝟎
𝒉 𝒉
𝒐𝒓,𝒐𝒓,
𝒙 𝒙= = 𝒙𝒙𝟎𝟎++𝒖𝒉
𝒖𝒉
Here, 0< 𝒖 𝟏< 𝟏 𝟑
Here, < 𝒖 <
𝟒 𝟒
 Example on Bessel’s Formula
𝒖(𝒖 − 𝟏) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
Apply Bessel’s
𝒚 𝒙Formula ∆𝒚𝟎 𝑦+25 , given ∆𝟐 𝒚−𝟏 +
= 𝒚𝟎 +to𝒖find ∆ 𝒚−𝟏 +
𝟐! 𝟑!
𝑦20 = 24 , 𝑦24 = 32 , 𝑦28 = 35 , 𝑦32 = 40 .
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐 + ∆ 𝒚−𝟐
𝟒! 𝟓!
Solution:
+ …………………………………………………………………………………
𝒙 −𝒙𝟎 Take x=24 as the origin and 4 as the unit.
Where 𝒖= 𝒉 25−24
The value of y required
𝒐𝒓, 𝒙 = 𝒙will be for u=
𝟎 + 𝒖𝒉
= 0.25
4
Here, 0< 𝒖 < 𝟏
The difference table is given below :
 x u Yu ∆Yu ∆𝟐 Yu ∆𝟑 Yu
𝒖(𝒖 − 𝟏) 𝟐 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
20 𝒚 𝒙 = 𝒚 +
𝟎-1 𝒖 ∆𝒚 𝟎 + 24𝟐! ∆ 𝒚 −𝟏 + ∆ 𝒚−𝟏 +
𝟑!

𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖8+ 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓

∆ 𝒚−𝟐 + ∆ 𝒚−𝟐
𝟒! 𝟓!
24 0 32 -5
+ …………………………………………………………………………………
𝒙 −𝒙𝟎 3 7
Where 𝒖= 𝒉
28 𝒐𝒓, 𝒙 =1 𝒙𝟎 + 𝒖𝒉 35 2
Here, 0< 𝒖 < 𝟏
5

32 2 40
 The Bessel’s Formula is
1
𝑦0 +𝑦1 1 𝑢 𝑢−1 ∆2 𝑦−1 +∆3 𝑦0 (𝑢−2)𝑢 𝑢−1
𝑦𝑢 = + 𝑢 − ∆𝑦0 + + ∆3 𝑦−1
2 2 2! 2 3!
1
𝑢 𝑢2 −1 (𝑢−2) ∆4 𝑦−2 +∆4 𝑦−1 (𝑢−2)𝑢 𝑢2 −1 (𝑢−2)
+ + 𝒖(𝒖 − 𝟏) 𝟐 ∆5 𝑦𝒖−2++𝟏 … ……
𝒖(𝒖 …… 𝟑
− 𝟏)
4! 𝒚 𝒙 = 𝒚𝟎2 + 𝒖 ∆𝒚𝟎 + 5! ∆ 𝒚−𝟏 + ∆ 𝒚−𝟏 +
𝟐! 𝟑!
1 𝒖 + 𝟏 𝒖(𝒖 − 1𝟏)(𝒖1− 𝟐) 1 1 (𝒖 + 𝟐)−5
𝒖
+
+
2 𝒖(𝒖 1
𝟏 −
1 −
𝟏)(𝒖
1 𝟐)
∴ 𝑦0.25 = 35 + 32 + − × 3∆+
𝟒𝒚 +− 1 + − ∆𝟓 𝒚−𝟐
2 𝟒! 4 2 4−𝟐 4 4 𝟓! 4 2 4
1 7
−1
4 6𝒙 −𝒙𝟎 + … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …
Where 𝒖 = 𝒉
= 33.5 − .75 + 0.140625 + 0.0546875 = 32.945313
𝒐𝒓, 𝒙value
Thus the estimated = 𝒙𝟎 of
+ 𝒖𝒉
𝑦25 is 32.945313
Here, 0< 𝒖 < 𝟏
 Stirling Central difference Interpolation Formula

𝒖(𝒖 − 𝟏) 𝟐 + 𝒖 + 𝟏 𝒖(𝒖 −𝟐𝟏) 𝟑

𝒚 𝒙 = 𝒚𝟎 +(∆𝒚 + ∆𝒚𝟐!−𝟏 ) ∆𝟐𝒖
𝒖 ∆𝒚𝟎𝟎 + 𝒚−𝟏 𝒖(𝒖 − ∆𝟏) (∆
𝒚−𝟏 +𝟑 𝒚−𝟏 + ∆𝟑 𝒚−𝟐 )
𝒚 𝒙 = 𝒚𝟎 + 𝒖 + ∆𝟐 𝒚−𝟏 +𝟑!
𝟐 𝟐 𝟑! 𝟐
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐 + ∆ 𝒚−𝟐
𝟒! 𝟓!
𝒖𝟐 (𝒖𝟐 − 𝟏) 𝟒 𝒖𝟐 (𝒖𝟐 − 𝟏𝟐 )(𝒖𝟐 − 𝟐𝟐 ) (∆𝟓 𝒚−𝟐 + ∆𝟓 𝒚−𝟑 )
+ + … … … … …∆… …𝒚…−𝟐… +… … … … … … … …𝟓!
… … … … … … … … … … … … …𝟐…
𝒙 −𝒙𝟎
𝟒!
Where 𝒖= 𝒉
+ …𝒐𝒓,
…𝒙…=…
𝒙𝟎…
+…𝒖𝒉…
Here, 0< 𝒖 < 𝟏 𝒙 −𝒙𝟎
Where 𝒖=
𝒉
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
𝟏 𝟏
Here, − <𝒖 <
𝟒 𝟒
 Example on Stirling’s Formula
𝒖(𝒖 − 𝟏) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
Use Stirling’s
𝒚 𝒙 formula
= 𝒚𝟎 + 𝒖to 𝟎 + 𝑦35 , given
∆𝒚find ∆𝟐 𝒚−𝟏that,
+ ∆ 𝒚−𝟏 +
𝟐! 𝟑!
𝑦20 = 512 , 𝑦30 = 439, 𝑦40 = 346, 𝑦50 = 243 .
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐 + ∆ 𝒚−𝟐
𝟒! 𝟓!
Solution :
+ …………………………………………………………………………………
Where 𝒖 =
𝒙 −𝒙𝟎 We take 𝑥0 = 30, ℎ = 10.
𝒉
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉 35−30
Then we
Here, 0< 𝒖 < 𝟏
require the value of y for u= = .5 .
12
The difference table is given below :
 x u 𝒚𝒖 ∆𝒚𝒖 ∆ 𝟐 𝒚𝒖 ∆ 3 𝒚𝒖

20 -1 512
𝒖(𝒖 − 𝟏) 𝟐 -73 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚𝟎 + 𝒖 ∆𝒚𝟎 + ∆ 𝒚−𝟏 + ∆ 𝒚−𝟏 +
𝟐! 𝟑!
30 0 439 -20
𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟒 (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) 𝟓
∆ 𝒚−𝟐 + -93 10 ∆ 𝒚−𝟐
𝟒! 𝟓!
40 + …1… … … … … … …
346 -10
………………………………… …………………………
𝒙 −𝒙𝟎
Where 𝒖= 𝒉 -103
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉
Here, 0< 𝒖 50
<𝟏 2 243
 Stirling’s formula is
(∆𝑦0 + ∆𝑦−1 ) 𝑢2 2 𝑢(𝑢2 − 1) (∆3 𝑦−1 + ∆3 𝑦−2 )
𝑦𝑢 = 𝑦0 + 𝑢 +𝒖(𝒖 ∆ 𝑦−1 +
− 𝟏) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏) 𝟑
𝒚 𝒙 = 𝒚𝟎 +2𝒖 ∆𝒚𝟎 + 2 ∆𝟐 𝒚−𝟏 +3! 2 ∆ 𝒚−𝟏 +
𝟐! 𝟑!
𝑢2 (𝑢𝒖2 +−𝟏1)𝒖(𝒖 𝑢 2 (𝑢 2 − 12 )(𝑢 2 − 22 ) (∆5 𝑦 + ∆5𝑦 )
∆4 𝑦−−2𝟏)(𝒖
+ − 𝟐) ∆𝟒 𝒚−𝟐 + (𝒖 + 𝟐) 𝒖 + 𝟏 𝒖(𝒖 − 𝟏)(𝒖 − 𝟐) ∆𝟓 𝒚−𝟐
−2 −3
+
4! 𝟒! 5! 𝟓!2

+ ……… +… ……… ………… … … … … … … … … … … … … … … … … … … … … … … … … … …

𝒙 −𝒙𝟎
Where 𝒖= 𝒉 −93 − 73 .52
∴ 𝑦.5 = 439 + .5
𝒐𝒓, 𝒙 = 𝒙𝟎 + 𝒖𝒉2
+ (−20)
2
Here, 0< 𝒖 < 𝟏
= 439 − 41.5 − 2.5 = 395
Hence the estimated value of 𝑦35 is 395.
 Chapter 5
Numerical Differentiation
The process of finding the derivative or derivatives of a function at some value of the independent variable , when we
know the values of the function corresponding to the given values of the independent variable is called numerical
differentiation
X → 𝑌0
𝑋1 → 𝑌1
𝑋2 → 𝑌2
𝑋3 → 𝑌3
-------------- ----
- - - - - - - - -- - - - - ----
𝑋𝑛 → 𝑌𝑛

For , x=𝑋1 , y’(x) ,y’’(x)

Cases-
1. Equal Interval
2. Unequal Interval
 Newton Forward Interpolation Formula :-

p p−1 2 p p−1 p−2 3

y x = y0 + pΔy0 + Δ y0 + Δ y0 + ⋯
2! 4!
x−x0
Where, p =
h
x = x0 + ph
P P−1 P p−1 P−2
=> y x0 + ph = y0 + pΔy0 + Δ2 y0 + Δ3 y0 + ⋯
2! 4!

Differentiating with respect to ‘p’

d𝑦 2𝑃 − 1 2 3𝑃2 − 6𝑃 + 2 3
𝑦′ 𝑥 ℎ= ℎ = Δ𝑦0 + Δ 𝑦0 + Δ 𝑦0 + ⋯ ⋅
d𝑥 2! 3!
ⅆ𝑦 1 2𝑃−1 3𝑃2 −6𝑃+2
=> ⅆ𝑥
= 𝑦′ 𝑥 = ℎ
[Δ𝑦0 + 2!
Δ2 𝑦0 + 3!
Δ3 𝑦0 + ⋯ ⋅

Definitely,
𝑑2 𝑦 1
=> = 𝑦 ′′ 𝑥 = ℎ2 [Δ2 𝑦0 + (𝑝 − 1)Δ3 𝑦0 + ⋯]
ⅆ𝑥 2
Ex-1
Find the first , second & find derivatives of the function table blew at the point x = 1.5

x : 1.5 2.0 2.5 3.0 3.5 4.0

y=f(x) : 3.375 7.00 13.625 24.000 38.875 59.000

ΔΔ
Soln :

Since the derivatives are required at x = 1.5 , which is near the beginning of the table , therefore we shall
use . Newton’s forward formula. The difference table is given below .
X y 𝜟y 𝜟𝟐 y 𝜟𝟑 y 𝜟𝟒 y

1.5 3.375
3.625
2.0 7.00 3.00
6.625 .750
2.5 13.625 3.750 0
ΔΔ
10.375 .750
3.0 24.00 4.500 0
14.875 .750
3.5 38.875 5.250
20.125
4.0 59.00
Newton’s forward formula is

𝑓 𝑎 + 𝑥ℎ = 𝑓 𝑎 + 𝑥𝐶1 Δ f(a) + 𝑥𝐶2 Δ2 f(a) + 𝑥𝐶3 Δ3 𝑓 𝑎

𝑥(𝑥−1) 2 𝑥(𝑥−1)(𝑥−3)
=f(a)+x Δ f(a) + Δ f(a) + Δ3 𝑓 𝑎
2! 3!

Differentiating w.r.to twice then putting x=0 in the equations obtained _ we get

1 1
h𝑓 ′ 𝑎 = Δ f(a) - 2 Δ2 f(a) + 3 Δ3 𝑓(a)
ℎ2 𝑓 ′′ 𝑎 = Δ2 f(a) - Δ3 𝑓(a)
ℎ3 𝑓 ′′′ 𝑎 = Δ3 𝑓(a)

Putting a=1.5,h=0.5 & the values of various differences in these equations , we get

1 1 1
𝑓 ′ 1.5 = 0.5[3.025- 2 (3.00) + 3 (0.75)] = 4.75
1
𝑓 ′′ 1.5 = 0.25 3.00 − 0.750 = 9.00
1
𝑓 ′′′ 1.5 = 0.75 = 6.00
0.125

Note:- the function tabulated in the given table of this problem is , y=x 3 −2x+3
Hence,
ⅆy 2 ⅆ2 y
=3x -2, =6
ⅆx ⅆx2
Putting x = 1.5 we get,
ⅆy 2 ⅆ2 y
(ⅆx)1.5 = 3 × (1.5) −2 = 4.750 = 6.00
ⅆx2 x=1.5
 Chapter 6
Numerical Integration
The antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is,
in terms of known functions). Consequently, rather than evaluate definite integrals of these functions directly, we
resort to various techniques of numerical integration to approximate their values. In this section we explore several of
these techniques. In addition, we examine the process of estimating the error in using these techniques.

𝑋0 =0 → 𝑌0
𝑏 𝑏 𝑋1 =𝑋0 + h → 𝑌1
න 𝑓(𝑥) dx = න 𝑦 𝑑𝑥 𝑋2 =𝑋0 + 2h → 𝑌2
𝑎 𝑎 𝑋3 =𝑋0 + 3h → 𝑌3
𝑏−𝑎 -------------- ----
Where h= ℎ , usually we take h=6
- - - - - - - - -- - - - - ----
𝑋𝑛 =𝑋0 + nh → 𝑌𝑛
 Using newton’s cates formula
3
1. Trapezoidal Rules (n=1) 3. Simpson’s rules (n=3)
8
1
2. Simpson’s rules (n=2) 4. waddle's rules (n=6)
3

1. Trapezoidal Rules (n=1

Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids
rather than using rectangles. This integration works by approximating the region under the graph of a function as a
trapezoid, and it calculates the area. This rule takes the average of the left and the right sum.

𝑏=𝑥𝑛
ℎ
න 𝑦𝑑𝑥 = [ 𝑦0 + 𝑦𝑛 + 2( 𝑦1 + 𝑦2 + 𝑦3 + ⋯ + 𝑦𝑛−1 ]
𝑎=𝑥0 2
 1
Simpson’s rules
3

Simpson’s 1/3rd rule is an extension of the trapezoidal rule in which the integrand is approximated by a second-
order polynomial. Simpson rule can be derived from the various way using Newton’s divided difference
polynomial, Lagrange polynomial and the method of coefficients. Simpson’s 1/3 rule is defined by:

𝑏=𝑥𝑛
ℎ
න 𝑦𝑑𝑥 = [ 𝑦 + 𝑦𝑛 + 2 𝑦1 + 𝑦3 + 𝑦5 + ⋯ + 𝑦𝑛−1 + 2 𝑦2 + 𝑦4 + 𝑦6 + ⋯ + 𝑦𝑛−2 ]
𝑎=𝑥0 3 0
 3
Simpson’s rules
8

Another method of numerical integration is called “Simpson’s 3/8 rule”. It is completely based on the cubic
interpolation rather than the quadratic interpolation. Simpson’s 3/8 or three-eight rule is given by:

𝑏=𝑥𝑛
3ℎ
න 𝑦𝑑𝑥 = [ 𝑦0 + 𝑦𝑛 + 3 𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 + 𝑦7 + ⋯ . . + 2 𝑦3 + 𝑦6 + 𝑦9 + ⋯ + 𝑦𝑛−3 ]
𝑎=𝑥0 8
waddle's rules

𝑏=𝑥𝑛
න 𝑦𝑑𝑥
𝑎=𝑥0
3ℎ
= [ 𝑦 + 𝑦𝑛 + 5 𝑦1 + 𝑦7 + 𝑦13 + ⋯ ) + (𝑦2 + 𝑦8 + 𝑦14 + ⋯ . .
10 0
+ 6 𝑦3 + 𝑦9 + 𝑦15 + ⋯ ) + (𝑦4 + 𝑦10 + 𝑦16 + ⋯ ) + 5(𝑦5 + 𝑦11 + 𝑦17 + ⋯ . . ]
 Example
Calculate the values of integral by
3
5.2 1. Trapezoidal Rules (n=1) 3. Simpson’s
8
න 𝑙𝑜𝑔𝑎 𝑥 𝑑𝑥 rules
4 1
2. Simpson’s rules 4. waddle's rules
3

Solution

Taking h= .2 divide the whole range of integration into (4,5.2) six equals parts. The values of log a for each
point of sub division are given below.
Sub Division Table

x y = ln X
𝑥0 = 4.0 𝑦0 = 1.3862944
𝑥0 + ℎ = 4.2 𝑦0 = 1.3862944
𝑥0 + 2ℎ = 4.4 𝑦0 = 1.3862944
𝑥0 + 3ℎ = 4.6 𝑦0 = 1.3862944
𝑥0 + 4ℎ = 4.8 𝑦0 = 1.3862944
𝑥0 + 5ℎ = 5.0 𝑦0 = 1.3862944
𝑥0 + 6ℎ = 5.2 𝑦0 = 1.3862944
 a) By Trapezoidal Rule , We have
5.2
ℎ
න log 𝑥 𝑑𝑥 = 𝑦 + 𝑦6 + 2 𝑦1 + 𝑦2 + 𝑦3 + 𝑦4 + 𝑦5
4 2 0

.2
= 3.04935 + 2 ∗ 7.6207991
2

= .1(18.276551)

= 1.8276551
1
b) By Simpson’s rules , We have
3
5.2
ℎ
න log 𝑥 𝑑𝑥 = 𝑦0 + 𝑦6 + 4 𝑦1 + 𝑦3 + 𝑦5 ) + 2(𝑦2 + 𝑦4
4 3

.2
= 3.034953 + 4 ∗ 4.50705787 + 2 ∗ 3.0502204
3

= .1(8.276551)

= 1.8278472
 3
a) By Simpson’s 8 rules , We have

5.2
3ℎ
න log 𝑥 𝑑𝑥 = 𝑦0 + 𝑦6 + 3 𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 ) + 2𝑦3
4 8

3∗(.2)
= 3.034953 + 3 6.0947428 + 2(1.5260563
8

.6
= (24.371294)
8

= 1.827847
a) By Weddle’s rules , We have
5.2
3ℎ
න log 𝑥 𝑑𝑥 = 𝑦0 + 𝑦6 + 5 𝑦1 + 𝑦5 ) + 𝑦2 + 𝑦4 + 6𝑦3
4 10

3∗(.2)
= 3.034953 + 5 3.044522 + 3.0502204 + 6(1.5260563)
10

.6
= 10 (30.464123)

= 1.827847
Actual Value of
5.2
න log 𝑥 𝑑𝑥 = [𝑥 log 𝑥 − 1 ]5.2
4
4

3∗(.2)
= 5.2(log 5.2 − 1) − 4(log 4 − 1)
10

= [ 3.3730249 – 1.5451774 ]

= 1.8278475

Hence , The errors due to different formula are

a) .0001924
b) .0000003
c) .0000005
d) .0000001
 Chapter 11
Solution of ordinary differential equation
Ordinary Differential equation

𝑑𝑦
Suppose first order differential equation is =
𝑑𝑥
𝑓 𝑥, 𝑦 ,given 𝑦(𝑥0) = 𝑦0.
we have to find the value of y from differential
equation for a given value of 𝑥,
when the values are given at 𝑥 = 𝑥0 , 𝑦 = 𝑦0
Picard’s Method

The Picard’s method is an iterative method and is primarily used for approximating
solutions to differential equations.

Working Rules:
𝑑𝑦
Suppose the first order differential equation is = 𝑓 𝑥, 𝑦 , Given y(x0)=y0…..(1)
𝑑𝑥
1.First approximation:
𝑥0
y1 = y0 + ‫( 𝑓 𝑥׬‬x0,y0)𝑑𝑥
2.Second approximation:
𝑥0
y 2=y0 + ‫(𝑓 𝑥׬‬x ,y1)𝑑𝑥
3.nth approximation:
𝑥0
yn = y0 + ‫( 𝑓 𝑥׬‬x, yn-1 )𝑑𝑥
The picard ’s method is failed when f 𝑥, 𝑦 is not integral for the given
condition.

Example:
𝑑𝑦
Given that, = x + y , and that y = 1when x = 0, determine the value of y.
𝑑𝑥
Solution:
𝑑𝑦
Given that, = x + y.
𝑑𝑥
 𝑥0
hence, 𝑦 = 𝑦0 +‫𝑓 𝑥׬‬ 𝑥 + 𝑦 𝑑𝑥.
𝑥0
where 𝑦0 = 0. That is, y =‫𝑓 𝑥׬‬ 𝑥+𝑦 ,
First approximation:
𝑥0
𝑦1 = 𝑦0 + ‫ 𝑥 𝑓 𝑥׬‬+ 𝑦 𝑑𝑥
𝑥2
= 1+x+
2
Second approximation:
𝑥0
𝑦2 = 𝑦0 + ‫𝑥 𝑓 𝑥׬‬, 𝑦1 𝑑𝑥
𝑥2 𝑥3
=1+x+ +
2 3
Third approximation:
𝑥0
𝑦3 =𝑦0 + ‫𝑥(𝑓 𝑥׬‬, 𝑦2 )𝑑𝑥
2 𝑥3 𝑥4
=1 + 2𝑥 + 𝑥 + +
3 24
Euler’s method:
The simplest numerical method for solving ordinary differential equation is
Euler’s method. This method is so crude that it is seldom used in practice;
however, its simplicity makes it useful for illustrative purpose.

Working rules:
𝑑𝑦
Suppose any ordinary differential equation is = 𝑓 𝑥, 𝑦 ,given 𝑦(𝑥0 ) = 𝑦0 .
𝑑𝑥
 𝑥−𝑥0
1.Taking ℎ = ; such that 𝑥𝑛 = 𝑥0 + 𝑛ℎ.
𝑛
2.First Approximation:
𝑦1 =𝑦0 +ℎ𝑓(𝑥0 , 𝑦0 ); at 𝑥1 = 𝑥0 + ℎ

3. Second Approximation:
𝑦2 = 𝑦1 +ℎ𝑓(𝑥1 , 𝑦1 ); at 𝑥2 = 𝑥0 + 2ℎ.

4.nth Approximation:
𝑦𝑛 = 𝑦𝑛−1 +ℎ𝑓(𝑥𝑛−1 ,𝑦𝑛−1 );
at 𝑥𝑛 = 𝑥0 + 𝑛ℎ
 𝑑𝑦 𝑦−𝑥
Example: Given = with initial condition 𝑦 = 1 at 𝑥 = 0;
𝑑𝑥 𝑦+1
find y for x = 0.1 by Euler’s method.

Solution:
We divide the interval (0, 0.1) in to 2 steps, i.e., we take 𝑛 = 2
0.1−0
𝑎𝑛𝑑 ℎ = = 0.05
2
here, 𝑥1 = 0 + 0.05 = 0.05
𝑥2 = 0 + 2 × 0.05 = 0.1
. The various calculations are arranged as follows:
First approximation:
𝑦1 = 𝑦0 +ℎ𝑓 𝑥0, 𝑦0
1−0
= 1 + 0.05( ); at 𝑥1 = 0.05
1+0
=1.05
Second approximation:
𝑦2 = 𝑦0 + ℎ𝑓 𝑥1 , 𝑦1
1.05−0.05
= 1 + 0.05( ); at 𝑥2 = 0.1
1.05+0.05
=1.045.

Note that: The iteration can be continue for further steps.

 Modified euler’s method:
𝑑𝑦
Suppose any ordinary differential equation = 𝑓(𝑥, 𝑦),given 𝑦(𝑥0 ) = 𝑦0 .
𝑑𝑥
Working Rules:
𝑥−𝑥0
1.Taking ℎ = ; such that 𝑥𝑛 = 𝑥0 + 𝑛ℎ.
𝑛
2.First Approximation:
𝑦1 = 𝑦0 +ℎ𝑓(𝑥0 , 𝑦0 ); at 𝑥1 = 𝑥0b+ ℎ
Modified method:
ℎ
𝑦1(1) = 𝑦0 + [𝑓(𝑥0 , 𝑦0 ) + 𝑓(𝑥0 + ℎ𝑦1 )]
2
ℎ
𝑦2(2) = 𝑦0 + [𝑓(𝑥0, 𝑦0 )+𝑓(𝑥0 + ℎ𝑦1(1) )]
2
We repeat this step, until two consecutive values of y agree.
3.second approximation:
𝑦2 = 𝑦1 + ℎ𝑓(𝑥1 , 𝑦1 ); 𝑎𝑡 𝑥2 = 𝑥0 + 2ℎ
Modified method:
ℎ
𝑦2(1) = 𝑦1 + [𝑓 𝑥1, 𝑦1 + 𝑓 𝑥2 + ℎ𝑦2 ]
2
ℎ
𝑦2(2) = 𝑦1 + [𝑓 𝑥1 𝑦1 + 𝑓 𝑥2 + ℎ𝑦2 1 ]
2
We repeat this step, until two consecutive values of 𝑦 agree.
 Taylor’s Series:
𝑑𝑦
Suppose the first order differential equation is = f(x, y) Given 𝑦(𝑥0 ) = 𝑦0 … . . (1)
𝑑𝑥
Working rules:
𝑥−𝑥0
1.Taking ℎ = ; at least 𝑛 = 1;
𝑛
hence 𝑥𝑛 = 𝑥0 + 𝑛ℎ
2. Differentiating equation1 successively we 𝑔𝑒𝑡, 𝑦’, 𝑦’’, 𝑦’’’ and so on.
3.First approximation
ℎ ℎ2
𝑦1 = 𝑦0 + 𝑦′ + 𝑦′′ +…….𝑎𝑡 𝑥1 = 𝑥0 + ℎ
1! 2!
4. Second approximation
ℎ ℎ2
𝑦2 = 𝑦1 + (𝑦 ′) + (𝑦1 ′′) +
… … … … 𝑎𝑡 𝑥2 = 𝑥0 + 2ℎ
1! 1 2!
Note that :If ( x,y) is somewhat complicated and the calculation of higher order derivation’s
become difficult the taylor’s series cannot use.
 Example:
Solve 𝑦 = 𝑥 𝑦, 𝑦 0 = 1 by Taylor’s series method. Hence find the
values of 𝑦 𝑎𝑡 𝑥 = 0.1 𝑎𝑛𝑑 𝑥 = 0.2.
Solution:
Differentiating successively,
we get 𝑦’ = 𝑥 + 𝑦 ; 𝑦(0) = 1
𝑦’’ = 1 + 𝑦’; 𝑦’’(0) = 2
𝑦’’’ = 𝑦’’; 𝑦’’’(0) = 2
𝑦’’’ = 𝑦’’’; 𝑦’’’ 0 = 2 etc
Taylor’s series is,
ℎ ℎ2
𝑦1 = 𝑦0 + 𝑦′ + 𝑦′′ +…….
1! 2!
 𝐻𝑒𝑟𝑒 𝑥0 = 0, 𝑦0 = 1

𝑥2
𝑦 0.1 = 1 + 𝑥 1 +
2! (2)
=1+0.1+(0.1)2
= 1.11
𝑥2
𝑦(0.2) = 1 + 𝑥(1) +
2! (2)
= 1 + 0.2 + 0.22
= 1.24
 Milne’s Method:
𝑑𝑦
Given = 𝑓(𝑥, 𝑦)and 𝑦 = 𝑦0 ,𝑥 = 𝑥0 ; to find an approximate value of 𝑦 for
𝑑𝑥
𝑥 = 𝑥0 + 𝑛ℎ by Milne’s method, we proceed as follows:

The value Type equation here. 𝑦0 = 𝑦( 𝑥0 ) being given, we compute 𝑦1

= 𝑦(𝑥0 + ℎ), 𝑦2 = 𝑦(𝑥0 + 2ℎ), 𝑦3 = 𝑦(𝑥0 + 3ℎ), by Picard’s or Taylor’s
series method.

Next we calculate, 𝑓0 = 𝑓( 𝑥0 , 𝑦0 ), 𝑓1 = 𝑓( 𝑥0 + ℎ, 𝑦1 ), = 𝑓(𝑥2 + 2ℎ, 𝑦2 ),

𝑓3 = 𝑓(𝑥0 + 3ℎ, 𝑦3 )
Then to find 𝑦4 = 𝑦( 𝑥0 + 4ℎ),
we substitute Newton’s forward interpolation formula,
𝑛(𝑛−1) 2 𝑛 𝑛−1 (𝑛−2)
𝑓(𝑥 , 𝑦) = 𝑓0 + 𝑛∆ 𝑓0 + ∆ 𝑓0 + ∆3 𝑓0 +…….
2! 6
In the relation
𝑥
𝑦4 = 𝑦0 + ‫ 𝑥׬‬0 + 4ℎ𝑓(𝑥, 𝑦)𝑑𝑥

putting the value of 𝑓(𝑥, 𝑦) and by calculating the equation, we get

4ℎ
𝑦4 = 𝑦0 + (2 𝑓1 − 𝑓2 + 2𝑓3 )
3
′
Which is called 𝑚𝑖𝑙𝑛𝑒 𝑠 predictor formula
Having found y4 , we obtain a first approximation to
𝑓4 = 𝑓(𝑥0 + 4ℎ, 𝑦4 )
Then a better value of y4 is found by Simpson’s rule as
ℎ
𝑦4 = 𝑦2 + (𝑓2 + 4𝑓3 + 𝑓4 ) ; 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑚𝑖𝑙𝑛𝑒’𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.
3
 Example:

Using Milne’s method find y(4.5) given 5xy , 𝑦2 = 2 given y(4) =1, y(4.1) =1.0049, y(4.2)
=1.0097, y(4.3) =1.0143; y(4.4) =1.0187.

Solution:
We have, (2 – 12)/5 × 4
Then the starting values of the Milne’s method are,
2−12
𝑥0 = 0, 𝑦0 = 1, 𝑓 = = 0.05
5×4
𝑥1 = 4.1, 𝑦1 = 1.0049, 𝑓1 = 0.0485
𝑥2 = 4.2, 𝑦2 = 1.0097, 𝑓2 = 0.0467
𝑥3 = 4.3, 𝑦3 = 1.0143, 𝑓3 = 0.0452
𝑥4 = 4.4, 𝑦4 = 1.0187, 𝑓4 = 0.0437
 Since 𝑦5 is required, we use the predictor are
4ℎ
𝑦5 = 𝑦1 + (2𝑓2 − 𝑓3 + 2𝑓4 )
3
𝑥 = 4.5;
4×0.1
𝑦5 = 1.0049y5’=y1+4h/3(2f2-f3+2f4’)
+ (2 × 2.067 − 0.045 + 2 × 0.0437)
3
= 1.0232
𝑦
𝑓5 =2 − 5
5×4.5
1.0232
=2− (1.023 2 )
5×4.5
= 0.0424
Now using the corrector
ℎ
𝑦5 = 𝑦3 + (𝑓3 + 4𝑓4 + 𝑓5 )
3
= 1.023 ; 𝑦(4.5)
,
 Runge-Kutta Method
𝑑𝑦
Consider the differential equation , = 𝑓(𝑥, 𝑦), 𝑦(𝑥0 ) = 𝑦0 … … (1)
𝑑𝑥
Working rule to solve (1) by 𝑅𝑢𝑛𝑔𝑒 ’𝑠 method:
Calculate successively
𝑘1 = ℎ𝑓(𝑥0 , 𝑦0 ),
ℎ 𝑘
𝑘2 = ℎ𝑓(𝑥0 + , 𝑦0 + 1 ),
2 2
ℎ 𝑘2
𝑘3 = ℎ𝑓( 𝑥0 + , 𝑦0 + ),
2 2
𝑘4 = ℎ𝑓(𝑥0 + ℎ, 𝑦0 + 𝑘3 ),
Finally compute ,
1
𝑘 = (𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4 )
6
 Example
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑅𝑢𝑛𝑔𝑒𝑘𝑢𝑡𝑡𝑎 𝑚𝑒𝑡ℎ𝑜𝑑 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑎𝑛 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 𝑤ℎ𝑒𝑛 𝑥 0.2 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡
𝑑𝑦
= 𝑥 + 𝑦 𝑎𝑛𝑑 𝑦 1 𝑤ℎ𝑒𝑛 𝑥 0.
𝑑𝑥
Solution:
Here 𝑥0 = 0, 𝑦0 = 1, ℎ = 0.2, 𝑓(𝑥0 , 𝑦0 ) = 1
𝑘1 = ℎ𝑓 𝑥0 , 𝑦0
= 0.21
= 0.2000,
ℎ 𝑘
𝑘2 =ℎ𝑓(𝑥0 + , 𝑦0 + 1 )
2 2
= 0.2 × 𝑓(0.1 , 1.1)
= 0.2400
ℎ 𝑘
𝑘3 = ℎ𝑓(𝑥0 + , 𝑦0 + 2)
2 2
= 0.2 × 𝑓 0.1 , 1.12
= 0.2440
𝑘4 = ℎ𝑓(𝑥0 + ℎ,𝑦0 + 𝑘3 )
𝑘4 = ℎ𝑓(𝑥0 + ℎ, 𝑦0 + 𝑘3 )
= 0.2 × 𝑓(0.2 , 1.244)
= 0.2888

1
𝑘= (𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4 )
6
1
= 0.2000 + 2 × 0.2400 + 2 × 0.2440 + 0.2888
6
1
= × 1.4568
6
= 0.2428
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 𝑖𝑠 1.2428.
 Chapter 12
Solution Algebraic and Transcendental Equations
Methods
Regula-Falsi Method:

Step 01: Given equation, 𝑓 𝑥 = 0 … … … . . (1)

Find 𝑥0 and 𝑥1 such that,
𝑓 𝑥0 < 0 𝑎𝑛𝑑 𝑓 𝑥1 > 0
𝑖. 𝑒. , 𝑓 𝑥0 × 𝑓 𝑥1 < 0
Root of equation (1) lies between 𝑥0 and 𝑥1 .
1st approximation root by Regula-Falsi Method:

𝑥0 𝑓 𝑥1 − 𝑥1 𝑓(𝑥0 )
𝑥2 = 𝑓 𝑥1 −𝑓(𝑥0 )
Step 02: Find 𝑓(𝑥2 ) and estimate its sign:
if 𝑓 𝑥2 < 0 then replace 𝑥0 = 𝑥2
if 𝑓(𝑥2 ) > 0 then replace 𝑥1 = 𝑥2

Step 03: Find 2nd approximation root by Regula-Falsi Method:

𝑥0 𝑓 𝑥1 − 𝑥1 𝑓(𝑥0 )
𝑥3 =
𝑓 𝑥1 −𝑓(𝑥0 )

Step 04: Find 𝑓 𝑥3 , 𝑓 𝑥4 … 𝑓(𝑥𝑛 ) and estimate until the required roots.

Example
Solve the equation 3𝑥 − cos 𝑥 − 1 = 0 by False Position Method.
Sol: Let, 𝑥0 = 0.60 𝑎𝑛𝑑, 𝑥1 = 0.61
𝑓 𝑥0 = 3 × 0.60 − cos 0.60 − 1 = −0.025
𝑓 𝑥1 = 3 × 0.61 − cos 0.61 − 1 = 0.010
𝑓 𝑥0 × 𝑓 𝑥1 < 0
Using Regula Falsi Method,
1st iteration:

𝑥0 𝑓 𝑥1 − 𝑥1 𝑓(𝑥0 )
𝑥2 = 𝑓 𝑥1 −𝑓(𝑥0 )

0.60×0.010 −0.61×(−0.025)
= 0.010 −(−0.025)

= 0.607
𝑓 𝑥2 = 3 × 0.607 − cos 0.607 − 1 = −0.00036
2nd iteration: Here, 𝑥0 = 𝑥2 = 0.607

𝑥0 𝑓 𝑥1 − 𝑥1 𝑓(𝑥0 )
𝑥3 =
𝑓 𝑥1 −𝑓(𝑥0 )

0.607×0.010 −0.61×(−0.00036)
𝑥3 =
0.010 −(−0.00036)

= 0.607
Since 𝑥2 = 𝑥3 , so the value of the root is 0.607.
 Methods
Newton-Raphson Method:

Step 01: Given the function 𝑓(𝑥), find the derivative of the function 𝑓′(𝑥).
Step 02: Find 𝑎, 𝑏 so that 𝑓(𝑎) × 𝑓(𝑏) < 0.
Step 03: Assume 𝑥0 = 𝑎.

𝑓(𝑥𝑛 )
Step 04: Find out 𝑥𝑛 = 𝑥𝑛 −
𝑓′(𝑥𝑛 )
Step 05: Find 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , … , 𝑥𝑛 until any two successive two values are equal.

Example
Solve the equation 3𝑥 − cos 𝑥 − 1 = 0 by Newton Raphson Method.
Sol: Here, 𝑓 𝑥 = 3𝑥 − cos 𝑥 − 1
𝑓 ′ 𝑥 = 3 + sin 𝑥

𝑓 𝑥𝑛 3𝑥𝑛 −cos 𝑥𝑛 −1
Now, 𝑥𝑛+1 = 𝑥𝑛 − 𝑓′ = 𝑥𝑛 −
𝑥𝑛 3+sin 𝑥𝑛
 Taking, 𝑥0 = 0.60, we get

3 × 0.60 − cos(0.60) − 1
𝑥1 = 0.60 − = 0.60701
3 + sin(0.60)

3 × 0.60701 − cos 0.60701 − 1

𝑥2 = 0.60701 − = 0.60710
3 + sin 0.60701
Since 𝑥2 = 𝑥3 , so the value of the root is 0.607.

Methods
Iteration Method:

Step 01: Given equation, 𝑓 𝑥 = 0 … … … . . (1)

 Chapter 14
Simultaneous Linear Algebraic Equations
 Chapter
16
Curve Fitting
Given a set of data (𝑥𝑖 , 𝑦𝑖 ) with 1<= i <=n ,curve fitting revolves around finding a math metical model can describe the
relationship y(x) such that prediction of the math metical model match , as closely as possible the given data.

X →Y
𝑋1 → 𝑌1
𝑋2 → 𝑌2
𝑋3 → 𝑌3
-------------- ----
- - - - - - - - -- - - - - ----
𝑋𝑛 → 𝑌𝑛

Types-
1. Line fitting ( y= a+bx)
2. Parabola fitting (y= a+bx+c𝑥 2 )
 Line fitting working rule

Step -1 : let (𝑥1 , 𝑦1 ) , (𝑥2 , 𝑦2 )…, (𝑥𝑛 , 𝑦𝑛 )

Step-2 : find σ 𝑦, , σ 𝑥𝑦, σ 𝑥 , σ 𝑥 2

Step – 3 : Construct Normal equation for a and b

σ 𝑦= na + bσ 𝑥
σ 𝑥𝑦 = 𝑎 σ 𝑥 + 𝑏 σ 𝑥 2

Step – 4: Solve equation in 1 and find a and b

Step-5 : put them into equation of line y=a + bx

Parabola curve fitting

y= a+bx+c𝑥 2
 v= a+bu+c𝑢2 ….(1)
 Putting the values of a,b,c int this equation
 We get ,
 Y-b=a+b(x-a) +c(𝑥 − 𝑎)2
 Example – Fit a straight line to the following data regarding x as the independent variable

x: 0 1 2 3 4

y:1 1.8 3.3 4.5 6.3

Sol. Let the straight line to be fitted to the given data be y= a+bx
then the normal equation are

σ 𝑦= na + bσ 𝑥
σ 𝑥𝑦 = 𝑎 σ 𝑥 + 𝑏 σ 𝑥 2
 X Y xy 𝑥2
0 1 0 0
1 1.8 1.8 1
2 3.3 6.6 4
3 4.5 13.5 9
4 6.3 25.2 16
Total = 10 16.9 47.1 30
ΔΔ

In this case m=5 , σ 𝑥 = 10 , σ 𝑦=16.9 , σ 𝑥𝑦 = 47.1 , σ 𝑥 2 =30

Substituting theses values in the normal equations , We have
16.9= 5a +10 b
47.1=10 a+30 b
And solving the above equations, we obtain a=.72 , b=1.33 ,
Hence the fitted line is y= 0.72 +1.33x