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KIRINYAGA UNIVERSITY

DEPARTMENT OF PURE AND APPLIED SCIENCES

COURSE CODE : SPM2321

COURSE TITLE: NUMERICL ANALYSIS I

LECTURER: ANTHONY KINYANJUI

MODULE 05: FORWARD AND BACKWARD FINITE DIFFERENCE
TABLES

OBJECTIVE: AT THE END OF LECTURE, A STUDENT WILL BE ABLE
TO CONSTRUCT FORWARD AND BACKWARD FINITE DIFFERENCE
TABLES
1

Construction of finite difference tables
Consider a tabulated function (𝑥𝑖 , 𝑓𝑖)) from the table
𝑥𝑖

𝑓𝑖

𝑥0

𝑓0

𝑥1

𝑓1

∆𝑓𝑖

∆2𝑓𝑖

∆3𝑓𝑖

∆4𝑓𝑖

∆5𝑓𝑖

∆𝑓0
∆2𝑓0

𝑥2

∆𝑓1

𝑓2

∆2𝑓1
𝑥3

∆4𝑓0

∆𝑓2

𝑓3

∆2𝑓2
𝑥4

∆5𝑓0

∆3𝑓1
∆4𝑓1

∆𝑓3

𝑓4

∆2𝑓3
𝑥5

∆3𝑓0

∆3𝑓2

∆𝑓4

𝑓5

The above table is called a diagram difference table
...
The difference ∆𝑓0∆2𝑓0 ∆2𝑓0 are called the leading
difference
...

𝑥

𝑓(𝑥)

∆𝑓(𝑥)

𝑥+ℎ

𝑓(𝑥 + ℎ)

∆𝑓(𝑥)

∆2𝑓(𝑥)

∆3𝑓(𝑥)

∆2𝑓(𝑥)
𝑥 + 2ℎ

𝑓(𝑥 + 2ℎ)

∆𝑓(𝑥 + ℎ)
2

∆3𝑓(𝑥)

𝑥 + 3ℎ

𝑓(𝑥 + 3ℎ)

∆𝑓(𝑥 + 2ℎ)

∆2𝑓(𝑥 + ℎ)

Solved problem
Construct a forward difference table from the following table
𝑥

0

10

20

30

𝑓(𝑥)

0

0
...
347

0
...
174
10

−0
...
174

−0
...
173
20

−0
...
347
0
...
518

Construct a difference table for 𝑓 (𝑥 ) = 𝑥 3 + 2𝑥 + 1 𝑓𝑜𝑟 𝑥 = 1,2,3,4,5
𝑥

𝑓(𝑥)

∆𝑓(𝑥)
3

∆2𝑓(𝑥)

∆3𝑓(𝑥)

1

4
9

2

12

13

6

21
3

18

34

6

39
4

24

73
63

5

136

Backward difference
Let 𝑦 = 𝑓(𝑥) be a function given the values 𝑦0 , 𝑦1 ,
...
𝑥𝑛 of the independent variable x
...
Thus we have
𝑦1 − 𝑦0 = ∇𝑦1
𝑦2 − 𝑦1 = ∇𝑦2
𝑦𝑛 − 𝑦𝑛−1 = ∇𝑦𝑛
Where ∇ is called the backward difference operator
...
Hence use the results to find:
(i)

∇2𝑓2 𝑎𝑡 𝑥 = 2

(ii)

∆2𝑓2 𝑎𝑡 𝑥 = 3

(iii)

δf3 at x = 4

(iv)

𝜇𝑓2 𝑎𝑡 𝑥0 = 4

(v)

𝛿 2 𝑓2 𝑎𝑡 𝑥0 = 2

Solution
5

𝑥

𝑓(𝑥)

∆𝑓

1

6

7

∆2 𝑓

∆3 𝑓

2
2

13

0

9
2

3

22

0

11
2

4

33(𝑓0)

0

13(∆𝑓0)
2

5

46

0

15
2

6

61

0

17
2

7

78

0

19
2

8

97

0

21
2

(i)

9

118

10

141

∆𝑓𝑖 = 𝑓𝑖+1 − 𝑓𝑖

23

∆𝑓2 = 𝑓3 − 𝑓2

6

∆2𝑓𝑖 = 𝑓𝑖+2 − 2𝑓𝑖+1 + 𝑓𝑖
= 𝑓𝑖+2 − 2𝑓𝑖+1 + 𝑓𝑖
= 𝑓4 − 2𝑓3 + 𝑓2
= 61 − 2(46) + 33
=2
𝑥

𝑓(𝑥)

1

6

∇𝑓

∇2 𝑓

∇3 𝑓

−7
2

13

2
−9

3

22

0
2

−11
4

33

0
2

−13
5

46

0
2

−15
6

61

0
2

−17
7

78

0
2

−19
8

97

0
2

−21
9

118

0
2

−23

7

10

141

𝛻𝑓2 𝑎𝑡 𝑥0 = 3
= −13
∇2 𝑓2 = 2
𝛿𝑓3 𝑎𝑡 𝑥0 = 4
𝛿𝑓𝑖 = 𝑓𝑖+1 − 𝑓𝑖−1
2

2

𝛿𝑓3 = 𝑓3
...
5
= 87
...
5
= 18
𝛿 2 𝑓2 𝑎𝑡 𝑥0 = 2
𝛿 2 𝑓𝑖 = 𝑓𝑖+1 − 2𝑓𝑖 + 𝑓𝑖−1
𝛿 2 𝑓2 = 𝑓3 − 2𝑓2 + 𝑓1
= 46 − 2(33) + 22
=2

8

1
𝜇𝑓𝑖 = (𝑓𝑖+1 − 𝑓𝑖−1 )
2
2
2
𝜇𝑓3=𝑓3
...
5
2

1
= (87
...
5 )
2
= 61
...
Tabulate 2𝑥 2 − 2𝑥 + 1 from 1 to 8 and hence use the value to find

(i)

∇𝑓2 𝑎𝑡 𝑥 = 3

(ii)

∆3 𝑓𝑖 𝑎𝑡 𝑥 = 2

(iii)

𝜇𝑓4 𝑎𝑡 𝑥 = 1

(iv)

𝛿 3 𝑓2 𝑎𝑡 𝑥 = 2

9



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