This document discusses various methods of interpolation and numerical differentiation using divided differences and Newton's formulas. It introduces Lagrange interpolation for both equal and unequal intervals. Inverse interpolation and Newton's divided difference interpolation are also covered. Forward and backward difference formulas are presented for interpolation with equal intervals. Numerical differentiation can be performed by taking derivatives of the interpolation polynomial or using forward difference formulas to estimate derivatives at the data points.

1. INTERPOLATION

2. INTERPOLATION FOR UNEQUAL INTERVALS
 LAGRANGE’S INTERPOLATING
FORMULA
 Let y = f(x) be the given function
 Let 𝑦0, 𝑦1, … 𝑦𝑛 be the (𝑛 + 1) points of the
given function corresponding to 𝑥0, 𝑥1, … 𝑥 𝑛
 The polynomial 𝑦 = 𝑓 𝑥 can be written as
 𝑓 𝑥 =
𝑥−𝑥1 𝑥−𝑥2 ··· 𝑥−𝑥 𝑛
𝑥0−𝑥1 𝑥0−𝑥2 ··· 𝑥0−𝑥 𝑛
𝑦0 +
𝑥 − 𝑥0 𝑥 − 𝑥2 ··· 𝑥 − 𝑥 𝑛
𝑥1 − 𝑥0 𝑥1 − 𝑥2 ··· 𝑥1 − 𝑥 𝑛
𝑦1 +
𝑥 − 𝑥0 𝑥 − 𝑥1 ··· 𝑥 − 𝑥 𝑛
𝑥2 − 𝑥0 𝑥2 − 𝑥1 ··· 𝑥2 − 𝑥 𝑛
𝑦2
+ ⋯ +
𝑥−𝑥0 𝑥−𝑥1 ··· 𝑥−𝑥 𝑛−1
𝑥 𝑛−𝑥0 𝑥 𝑛−𝑥1 ··· 𝑥 𝑛−𝑥 𝑛−1
𝑦 𝑛

3. INVERSE INTERPOLATION
 It is the process of finding a value of x
for the corresponding value of y and
we use Lagrange’s interpolation
formula by taking the independent
variable as y and the dependent
variable as x. It is the inverse process
of direct interpolation in which we find
the values of y corresponding to a
value of x, not present in the table.

4. INVERSE INTERPOLATION BY LAGRANGE’S
 𝑥 =
𝑦−𝑦1 𝑦−𝑦2 ··· 𝑦−𝑦 𝑛
𝑦0−𝑦1 𝑦0−𝑦2 ··· 𝑦0−𝑦 𝑛
𝑥0 +
𝑦−𝑦0 𝑦−𝑦2 ··· 𝑦−𝑦 𝑛
𝑦1−𝑦0 𝑦1−𝑦2 ··· 𝑦1−𝑦 𝑛
𝑥1 +
⋯
+
𝑦−𝑦0 𝑦−𝑦1 ··· 𝑦−𝑦 𝑛−1
𝑦 𝑛−𝑦 𝑦 𝑛−𝑦1 ··· 𝑦 𝑛−𝑦 𝑛−1
𝑥 𝑛

5. USE OF LAGRANGIAN
INTERPOLATION
 It is a process of computing
intermediate values of a function from
a given set of tabular values of the
function.

6. DIVIDED DIFFERENCE
 Let 𝑦 = 𝑓(𝑥) be the given function
which takes the values
f(𝑥0), 𝑓(𝑥1) … 𝑓(𝑥 𝑛) corresponding to
the arguments 𝑥0, 𝑥1, … 𝑥 𝑛
respectively, where the intervals
𝑥1 – 𝑥0 , 𝑥2 − 𝑥1 , … 𝑥 𝑛 – 𝑥 𝑛−1 need not
equal

7. REPRESENTATION BY DIVIDED DIFFERENCE
TABLE
𝐴𝑟𝑔𝑢𝑚𝑒𝑛𝑡
𝑥
𝐸𝑛𝑡𝑟𝑦
𝑓(𝑥)
𝐹𝑖𝑟𝑠𝑡 𝐷. 𝐷
Δ1 𝑓(𝑥)
𝑆𝑒𝑐𝑜𝑛𝑑 𝐷. 𝐷
Δ1
2
𝑓(𝑥)
𝑇ℎ𝑖𝑟𝑑 𝐷. 𝐷
Δ1
3
𝑓(𝑥)
𝑥0
𝑥1
𝑥2
𝑥3
𝑥4
𝑓(𝑥0)
𝑓 𝑥1
𝑓 𝑥2
𝑓 𝑥3
𝑓(𝑥4)
𝑓 𝑥1 –𝑓 𝑥0
𝑥1 –𝑥0
= 𝑓 𝑥0, 𝑥1
𝑓 𝑥2 –𝑓 𝑥1
𝑥2 –𝑥1
= 𝑓 𝑥1, 𝑥2
𝑓 𝑥3 –𝑓 𝑥2
𝑥3 –𝑥2
= 𝑓 𝑥2, 𝑥3
𝑓 𝑥4 –𝑓 𝑥3
𝑥4 –𝑥3
= 𝑓 𝑥3, 𝑥4
𝑓 𝑥1,𝑥2 −𝑓 𝑥0,𝑥1
𝑥2−𝑥0
=
𝑓(𝑥0, 𝑥1, 𝑥2)
𝑓 𝑥2,𝑥3 −𝑓 𝑥1,𝑥2
𝑥3−𝑥1
=
𝑓(𝑥1, 𝑥2, 𝑥3)
𝑓 𝑥3,𝑥4 −𝑓 𝑥2,𝑥3
𝑥4−𝑥2
=
𝑓(𝑥2, 𝑥3, 𝑥4)
𝑓 𝑥1,𝑥2,𝑥3 −𝑓 𝑥0,𝑥1,𝑥2
x3−𝑥0
=
𝑓(𝑥0, 𝑥1, 𝑥2, 𝑥3)
𝑓 𝑥2,𝑥3,𝑥4 −𝑓 𝑥1,𝑥2,𝑥3
𝑥4−𝑥1
=
𝑓(𝑥1, 𝑥2, 𝑥3, 𝑥4)

8. PROPERTIES OF DIVIDED DIFFERENCES
 1. The divided differences are
symmetrical in all their arguments, that
is, the value of any difference is
independent of the order of the
arguments.
 2. The divided difference of the product
of a constant and a function is equal to
the product of the constant and the
divided difference of the function.
 3. The operator Δ1 is linear
 4. The 𝑛 𝑡ℎ
divided difference of a
polynomial of degree 𝑛 is a constant

9. NEWTON DIVIDED DIFFERENCE INTERPOLATION
 If 𝑓(𝑥) is a polynomial of degree𝑛, and
𝑓(𝑥0), 𝑓(𝑥1), … 𝑓(𝑥 𝑛) are the
corresponding values of arguments
𝑥0, 𝑥1, … . 𝑥 𝑛 respectively, not
necessarly equally spaced.
 Then 𝑓 𝑥 = 𝑓 𝑥0 + ( 𝑥 −

10. INTERPOLATION WITH EQUAL INTERVALS
 FORWARD DIFFERENCE
 If 𝑦0, 𝑦1, … 𝑦𝑛 are the values of 𝑦 = 𝑓(𝑥)
corresponding to the arguments 𝑥0, 𝑥1, … 𝑥 𝑛 ,
where 𝑥1 − 𝑥0 , 𝑥2 − 𝑥1, … 𝑥 𝑛 − 𝑥 𝑛−1 are equal
 𝑖. 𝑒 𝑥𝑖 = 𝑥0 + 𝑖ℎ, 𝑖 = 0,1,2 … 𝑛
 Define Δ𝑦0 = 𝑦1 − 𝑦0 , Δ𝑦1 = 𝑦2 − 𝑦1 … . Δ𝑦 𝑛−1 =
𝑦𝑛 − 𝑦 𝑛−1
 And Δ2
𝑦0 = Δ𝑦1 − Δ𝑦0 , Δ2
𝑦1 = Δ𝑦2 −
Δ𝑦1 … . Δ2
𝑦 𝑛−1 = Δ𝑦𝑛 − Δ𝑦 𝑛−1
 And so on
 Here Δ is called Newton’s forward difference
operator

11.  NEWTON’S FORWARD
DIFFERENCE FORMULA
 If 𝑦0, 𝑦1, … 𝑦𝑛 are the values of 𝑦 =
𝑓(𝑥) corresponding to the arguments
𝑥0, 𝑥1, … 𝑥 𝑛
 Then y = 𝑦0 + 𝑝∆𝑦0 +
𝑝 𝑝−1
2!
∆2
𝑦0 +
𝑝 𝑝−1 𝑝−2
3!
∆3
𝑦0 + ⋯
 Where 𝑝 =
𝑥−𝑥0
ℎ
.

12. FORWARD DIFFERENCE
TABLE
𝑥 𝑦 Δ Δ2
Δ3
Δ4
Δ5
𝑥0
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑦0
𝑦1
𝑦2
𝑦3
𝑦4
𝑦5
Δ𝑦0
Δ𝑦1
Δ𝑦2
Δ𝑦3
Δ𝑦4
Δ2
𝑦0
Δ2
𝑦1
Δ2
𝑦2
Δ2
𝑦3
Δ3
𝑦0
Δ3
𝑦1
Δ3
𝑦2
Δ4
𝑦0
Δ4
𝑦1
Δ5
𝑦0

13. BACKWARD DIFFERENCE
 The differences 𝑦1 − 𝑦0, 𝑦2 − 𝑦1, … . 𝑦𝑛 −
𝑦 𝑛−1 are called first backward differences
and denoted by
 𝛻𝑦1 = 𝑦1 − 𝑦0, 𝛻𝑦2 = 𝑦2 − 𝑦1, … .
𝛻𝑦𝑛 = 𝑦𝑛 − 𝑦 𝑛−1
 And 𝛻2
𝑦1 = 𝛻𝑦1 − 𝛻𝑦0,
𝛻2
𝑦2 = 𝛻𝑦2 − 𝛻𝑦1, … .
𝛻2
𝑦𝑛 = 𝛻𝑦𝑛 − 𝛻𝑦 𝑛−1
and so on

14. NEWTON’S BACKWARD DIFFERENCE FORMULA
 If 𝑦0, 𝑦1, … 𝑦𝑛 are the values of 𝑦 =
𝑓(𝑥) corresponding to the arguments
𝑥0, 𝑥1, … 𝑥 𝑛
 Then 𝑦 = 𝑦𝑛 + 𝑝𝛻𝑦𝑛 +
𝑝 𝑝+1
2!
𝛻2
𝑦𝑛 +
𝑝 𝑝+1 𝑝+2
3!
𝛻3
𝑦𝑛 + ⋯
 Where 𝑝 =
𝑥−𝑥 𝑛
ℎ
.

15. BACKWARD DIFFERENCE
TABLE
𝑥 𝑦 𝛻 𝛻2
𝛻3
𝛻4
𝛻5
𝑥0
𝑥1
𝑥2
𝑥3
𝑥4
𝑥5
𝑦0
𝑦1
𝑦2
𝑦3
𝑦4
𝑦5
𝛻𝑦1
𝛻𝑦2
𝛻𝑦3
𝛻𝑦4
𝛻𝑦5
𝛻2
𝑦2
𝛻2
𝑦3
𝛻2
𝑦4
𝛻2
𝑦5
𝛻3
𝑦3
𝛻3
𝑦4
𝛻3
𝑦5
𝛻4
𝑦4
𝛻4
𝑦5
𝛻5
𝑦5

16. NUMERICAL
DIFFERENTIATION

17. DERIVATIVES USING DIVIDED DIFFERENCE
 Procedure
 Step 1. By using Newton’s divided
difference formula find 𝑓(𝑥) in terms of
x
 Step 2. Find the derivatives of 𝑓(𝑥)

18. DERIVATIVE - USING NEWTON’S FORWARD DIFFERENCE
FORMULA
 The first derivative of y at 𝑥 = 𝑥0 + 𝑝ℎ (near beginning of the data)
is
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑝
𝑑𝑝
𝑑𝑥

𝑑𝑦
𝑑𝑥
=
1
ℎ
∆𝑦0 +
2𝑝−1
2!
∆2 𝑦0 +
3𝑝2−6𝑝+2
3!
∆3 𝑦0 + ⋯
 The second derivative of y at 𝑥 = 𝑥0 + 𝑝ℎ is
𝑑2 𝑦
𝑑𝑥2 =
𝑑2 𝑦
𝑑𝑝2
𝑑𝑝
𝑑𝑥
2

𝑑2 𝑦
𝑑𝑥2 =
1
ℎ2 ∆2
𝑦0 + (𝑝 − 1)∆3
𝑦0 +
6𝑝2−18𝑝+11
12
∆4
𝑦0 + ⋯
 The third derivative y at 𝑥 = 𝑥0 + 𝑝ℎ

𝑑3 𝑦
𝑑𝑥3 =
1
ℎ3 ∆3
𝑦0 +
12𝑝−18
12
∆4
𝑦0+. .
 For tabular values, at 𝑥 = 𝑥0, (𝑝 = 0)

𝑑𝑦
𝑑𝑥
=
1
ℎ
∆𝑦0 −
∆2 𝑦0
2
+
∆3 𝑦0
3
−
∆4 𝑦0
4
+ ⋯

𝑑2 𝑦
𝑑𝑥2 =
1
ℎ2 ∆2
𝑦0 − ∆3
𝑦0 +
11
12
∆4
𝑦0 − ⋯

𝑑3 𝑦
𝑑𝑥3 =
1
ℎ3 ∆3 𝑦0 −
3
2
∆4 𝑦0+. .

19. DERIVATIVE - USING NEWTON’S BACKWARD DIFFERENCE
FORMULA
 The first derivative of y at 𝑥 = 𝑥 𝑛 + 𝑝ℎ (near end of the data) is
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑝
𝑑𝑝
𝑑𝑥

𝑑𝑦
𝑑𝑥
=
1
ℎ
𝛻𝑦 𝑛 +
2𝑝+1
2!
𝛻2 𝑦 𝑛 +
3𝑝2+6𝑝+2
3!
𝛻3 𝑦 𝑛 + ⋯
 The second derivative of y at 𝑥 = 𝑥 𝑛 + 𝑝ℎ is
𝑑2 𝑦
𝑑𝑥2 =
𝑑2 𝑦
𝑑𝑝2
𝑑𝑝
𝑑𝑥
2

𝑑2 𝑦
𝑑𝑥2 =
1
ℎ2 𝛻2
𝑦 𝑛 + (𝑝 + 1)𝛻3
𝑦 𝑛 +
6𝑝2+18𝑝+11
12
𝛻4
𝑦0 + ⋯
 The third derivative of y at 𝑥 = 𝑥 𝑛 + 𝑝ℎ

𝑑3 𝑦
𝑑𝑥3 =
1
ℎ3 𝛻3
𝑦 𝑛 +
12𝑝+18
12
𝛻4
𝑦0+. .
 For tabular values, at 𝑥 = 𝑥 𝑛, (𝑝 = 0)

𝑑𝑦
𝑑𝑥
=
1
ℎ
𝛻𝑦 𝑛 +
𝛻2 𝑦 𝑛
2
+
𝛻3 𝑦 𝑛
3
+ ⋯

𝑑2 𝑦
𝑑𝑥2 =
1
ℎ2 𝛻2
𝑦 𝑛 + 𝛻3
𝑦 𝑛 +
11
12
𝛻4
𝑦0 + ⋯

𝑑3 𝑦
𝑑𝑥3 =
1
ℎ3 𝛻3 𝑦 𝑛 +
3
2
𝛻4 𝑦0+. .

20. NUMERICAL INTEGRATION
SINGLE (LINEAR) INTEGRATION
1. TRAPEZOIDAL RULE
2. SIMPSON’S
𝟏
𝟑
RULE

21. TRAPEZOIDAL RULE

𝑥0
𝑥1
𝑓(𝑥) 𝑑𝑥 =
ℎ
2
𝑦0 + 𝑦𝑛 + 2( 𝑦1 + 𝑦2 +

22. SIMPSON’S
𝟏
𝟑
𝒓𝒅 RULE

𝑥0
𝑥1
𝑓(𝑥) 𝑑𝑥 =
ℎ
3
𝑦0 + 𝑦𝑛 + 4 𝑦1 + 𝑦3 + 𝑦5 + ⋯ + 2(𝑦2 + 𝑦4 + 𝑦6+. . )
=
ℎ
3
𝑓𝑖𝑟𝑠𝑡 + 𝑙𝑎𝑠𝑡 + 4 𝑠𝑢𝑚 𝑜𝑓 𝑜𝑑𝑑 𝑡𝑒𝑟𝑚𝑠 +
2 𝑠𝑢𝑚 𝑜𝑓 𝑒𝑣𝑒𝑛 𝑡𝑒𝑟𝑚𝑠 ]
 ℎ =
𝑥1−𝑥0
𝑛
, 𝑛 − 𝑖𝑠 𝑡ℎ𝑒 𝒆𝒗𝒆𝒏 𝑛𝑜 𝑜𝑓 𝑠𝑢𝑏 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠 𝑖𝑛 𝑥0 𝑥1
 Condition for applying the Simpson’s rule – Number
of subintervals must be even

23. NUMERICAL DOUBLE
INTEGRATION
 1. 𝐓𝐑𝐀𝐏𝐄𝐙𝐎𝐈𝐃𝐀𝐋 𝐑𝐔𝐋𝐄
 2. 𝐒𝐈𝐌𝐏𝐒𝐎𝐍’𝐒 𝐑𝐔𝐋𝐄

24. TRAPEZOIDAL RULE

𝑎
𝑏
𝑐
𝑑
𝑓 𝑥, 𝑦 𝑑𝑥𝑑𝑦 =
ℎ𝑘
4
𝑓1 + 𝑓3 + 𝑓7 + 𝑓9 +
2 𝑓2 + 𝑓4 + 𝑓6 + 𝑓8 + 4𝑓5]
 =
ℎ𝑘
4
𝑆𝑢𝑚 𝑜𝑓 𝑐𝑜𝑟𝑛𝑒𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 +
2 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 +
4(𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓(𝑥, 𝑦)]

25. y x 𝑥0 𝑥1 𝑥2 𝑥3
𝑦0 𝑓(𝑥0, 𝑦0) 𝑓(𝑥1, 𝑦0) 𝑓(𝑥2, 𝑦0) 𝑓(𝑥3, 𝑦0)
𝑦1 𝑓(𝑥0, 𝑦1) 𝑓(𝑥1, 𝑦1) 𝑓(𝑥2, 𝑦1) 𝑓(𝑥3, 𝑦1)
𝑦2 𝑓(𝑥0, 𝑦2) 𝑓(𝑥1, 𝑦2) 𝑓(𝑥2, 𝑦2) 𝑓(𝑥3, 𝑦2)
𝑦3 𝑓(𝑥0, 𝑦3) 𝑓(𝑥1, 𝑦3) 𝑓(𝑥2, 𝑦3) 𝑓(𝑥3, 𝑦3)
For example, if x takes the values 𝑥0, 𝑥1, 𝑥2, 𝑥3 and y takes values
𝑦0, 𝑦1, 𝑦2, 𝑦3
Red indicates – corner values
Blue indicates – boundary values
Black indicates – interior values

26. 𝑺𝑰𝑴𝑷𝑺𝑶𝑵’𝑺 𝑹𝑼𝑳𝑬

𝑎
𝑏
𝑐
𝑑
𝑓 𝑥, 𝑦 𝑑𝑥𝑑𝑦 =
ℎ𝑘
9
𝑓1 + 𝑓3 + 𝑓7 + 𝑓9 +
2 𝑓2 + 𝑓4 + 𝑓6 + 𝑓8 + 16𝑓5]
 =
ℎ𝑘
9
𝑆𝑢𝑚 𝑜𝑓 𝑐𝑜𝑟𝑛𝑒𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 +
4 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓 𝑥, 𝑦 +
16(𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑓(𝑥, 𝑦)]

27. Errors in Trapezoidal rule of numerical
integration.
 When evaluating 𝑎
𝑏
𝑓 𝑥 𝑑𝑥, the error
in the trapezoidal rule is
 <
𝑏−𝑎 2
12
ℎ2
𝑀, where ℎ =
𝑏−𝑎
𝑛
, n is the
number of subintervals of (a, b),
 And 𝑀 = 𝑚𝑎𝑥{|𝑦0
′′
|, |𝑦1
′′
|,· · ·
, |𝑦 𝑛−1
′′
|}, 𝑦𝑟′′ = 𝑓′′(𝑥 𝑟)
 Error in Trapezoidal rule is of order
𝒉 𝟐

28. Errors in Simpson’s rule of numerical
integration
 The error in Simpson’s rule is <
𝑏−𝑎
180
ℎ4
𝑀
 where ℎ =
𝑏−𝑎
2𝑛
, 2n is the number of
subintervals of (a, b),
 𝑀 = 𝑚𝑎𝑥 𝑦0
1
, 𝑦1
4
,· · ·