Forward Differences 01

In this article, we shall study to prepare the forward difference table from the given data.

Example – 1:

If f(x) = x² + x + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5. Hence find Df(o), D²f(1) and D³f(2).

Solution:

Given f(x) = x² + x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0² + 0 + 1  = 0 + 0 + 1 = 1

f(1) = 1² + 1 + 1  = 1 + 1 + 1 = 3

f(2) = 2² + 2 + 1  = 4 + 2 + 1 = 7

f(3) = 3² + 3 + 1  = 9 + 3 + 1 = 13

f(4) = 4² + 4 + 1  = 16 + 4 + 1 = 21

f(5) = 5² + 5 + 1  = 25 + 5 + 1 = 31

The forward difference table is constructed as follows.

x	f(x)	Δf(x)	Δ2f(x)	Δ3f(x)
0	1
		3 – 1 = 2
1	3		4 – 2 = 2
		7 – 3 = 4		2 – 2 = 0
2	7		6 – 4 = 2
		13 – 7 = 6		2 – 2 = 0
3	13		8 – 6 = 2
		21 – 13 = 8		2 – 2 = 0
4	21		10 – 8 = 2
		31 – 21 = 10
5	31

From table we can see that Δ²f(x) = 2 = constant and Δ³f(x) = 0

Note:

• Given function is f(x) = x² + x + 1. The highest power is 2, which obviously means Δ²f(x) = constant and Δ³f(x) = 0.
• No need to show steps of calculations as shown in Δf(x),  Δ²f(x) and Δ³f(x) columns. You can write the values directly. To explain the concept for some problems steps are shown.

Example – 2:

If f(x) = x² – 3x + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5. Verify that the second differences are constant.

Solution:

Given f(x) = x² – 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0² – 3(0) + 1 = 0 – 0 + 1 = 1

f(1) = 1² – 3(1) + 1 = 1 – 3 + 1 = -1

f(2) = 2² – 3(2) + 1 = 4 – 6 + 1 = -1

f(3) = 3² – 3(3) + 1 = 9 – 9 + 1 = 1

f(4) = 4² – 3(4) + 1 =16 – 12 + 1 = 5

f(5) = 5² – 3(5) + 1 = 25 – 15 + 1 = 11

The forward difference table is constructed as follows.

x	f(x)	Δf(x)	Δ2f(x)
0	1
		-1 -1 = -2
1	-1		0 – (-2) = 2
		-1 – (-1) = 0
2	-1		2 – 0 = 2
		1 – (-1) = 2
3	1		4 – 2 = 2
		5 – 1 = 4
4	5		6 – 4 = 2
		11 – 5 = 6
5	11

From table we can see that second order differences i.e. Δ²f(x) = 2 = constant

Example – 3:

If f(x) = 2x² + 5, construct a forward difference table by taking x = 0, 2, 4, 6, 8 i.e. 0(2)8. Verify that the third differences are zero.

Solution:

Given f(x) = 2x² + 5 and x = 0, 2, 4, 6, 8 i.e. 0(2)8.

f(0) = 2(0)² + 5 = 0 + 5 = 5

f(2) = 2(2)² + 5 = 8 + 5 = 13

f(4) = 2(4)² + 5 = 32 + 5 = 37

f(6) = 2(6)² + 5 = 72 + 5 = 77

f(8) = 2(8)² + 5 = 128 + 5 = 133

The forward difference table is constructed as follows.

x	f(x)	Δf(x)	Δ2f(x)	Δ3f(x)
0	5
		8
2	13		18
		24		0
4	37		16
		40		0
6	77		16
		56
8	133

From table we can see that third order differences i.e.  Δ³f(x) = 0

Example – 4:

If f(x) = 2x³ – x² + 3x + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5. Verify that the third differences are constant.

Solution:

Given f(x) = 2x³ – x² + 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 2(0)³ – (0)² + 3(0) + 1 = 0 – 0 + 0 + 1 = 1

f(1) = 2(1)³ – (1)² + 3(1) + 1 = 2 – 1 + 3 + 1 = 5

f(2) = 2(2)³ – (2)² + 3(2) + 1 = 16 – 4 + 6 + 1 = 19

f(3) = 2(3)³ – (3)² + 3(3) + 1 = 54 – 9 + 9 + 1 = 55

f(4) = 2(4)³ – (4)² + 3(4) + 1 = 128 – 16 + 12 + 1 = 125

f(5) = 2(5)³ – (5)² + 3(5) + 1 = 250 – 25 + 15 + 1 = 241

The forward difference table is constructed as follows.

x	f(x)	Δf(x)	Δ2f(x)	Δ3f(x)
0	1
		4
1	5		10
		14		12
2	19		22
		36		12
3	55		34
		70		12
4	125		46
		116
5	241

From table we can see that third order differences i.e.  Δ³f(x) = 12 = constant

Example – 5:

If f(x) = x³ – 2x² + 1, construct a forward difference table by taking x = 0, 1, 2, 3, 4 i.e. 0(1)4. Verify that the fourth differences are zero.

Solution:

Given f(x) = x³ – 2x² + 1 and x = 0, 1, 2, 3, 4 i.e. 0(1)4.

f(0) = 0³ – 2(0)² + 1 = 0 – 0 + 1 = 1

f(1) = 1³ – 2(1)² + 1 = 1 – 2 + 1 = 0

f(2) = 2³ – 2(2)² + 1 = 8 – 8 + 1 = 1

f(3) = 3³ – 2(3)² + 1 = 27 – 18 + 1 = 10

f(4) = 4³ – 2(4)² + 1 = 64 – 32 + 1 = 33

The forward difference table is constructed as follows.

x	f(x)	Δf(x)	Δ2f(x)	Δ3f(x)	Δ4f(x)
0	1
		-1
1	0		2
		1		6
2	1		8		0
		9		6
3	10		14
		23
4	33

From the table, we can see that fourth-order differences i.e. Δ⁴f(x) are zero.

Example – 6:

By constructing a forward difference table find the 7^th and 8^th terms of a sequence 8, 14, 22, 32, 44, 58,….

Solution:

Let f(1) = 8, f(2) = 14, f(3) = 22, f(4) = 32, f(5) = 44, f(6) = 58

We have to find f(7) and f(8)

We prepare following forward difference table.

x	f(x)	Δf(x)	Δ2f(x)
1	8
		6
2	14		2
		8
3	22		2
		10
4	32		2
		12
5	44		2
		14
6	58		2
		16
7	74		2
		18
8	92

We can see that the second differences i.e. Δ²f(x) are constant.

• To find f(7), extra 2 (shown in red colour) is written in D²f(x) column. The entry in Df(x) is 2 + 14 = 16 (shown in red colour) is added. The entry in f(x) is 16 + 58 = 74 (shown in red colour) is added. Thus f(7) = 74.
• To find f(8), extra 2 (shown in green colour) is written in D²f(x) column. The entry in Df(x) is 2 + 16 = 18 (shown in green colour) is added. The entry in f(x) is 18 + 74 = 92 (shown in green colour) is added. Thus f(8) = 92.

Thus 7 th and 8 th terms of series are 74 and 92 respectively.

Example – 7:

By constructing a forward difference table find the 6^th and 7^th terms of a sequence 6, 11, 18, 27, 38,….

Solution:

Let f(1) = 6, f(2) = 11, f(3) = 18, f(4) = 27, f(5) = 38

We have to find f(6) and f(7)

We prepare following forward difference table.

x	f(x)	Δf(x)	Δ2f(x)
1	6
		5
2	11		2
		7
3	18		2
		9
4	27		2
		11
5	38		2
		13
6	51		2
		15
7	66

We can see that the second differences are 2 i.e. constant