Forward Difference

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

Example – 6:

  • By constructing a forward difference table find the 7th and 8th 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 6th and 7th 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 i.e.

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

Example – 6:

  • By constructing a forward difference table find the 7th and 8th 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 6th and 7th 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 i.e. D2f(x) are constant.

To find f(6), extra 2 (shown in red colour) is written in D2f(x) column. The entry in Df(x) is 2 + 11 = 13 (shown in red colour) is added. The entry in f(x) is 13 + 38 = 51 (shown in red colour) is added. Thus f(6) = 51.

To find f(7), extra 2 (shown in green colour) is written in D2f(x) column. The entry in Df(x) is 2 + 13 = 15 (shown in green colour) is added. The entry in f(x) is 15 + 51 = 66 (shown in green colour) is added. Thus f(7) = 66.

Thus 6 th and 7 th terms of series are 51 and 66 respectively.

 

Example – 8:

Estimate f(5) from the following table.

x 0 1 2 3 4
f(x) 3 2 7 24 59

 

 

 

 

 

 

 

We prepare following forward difference table.

x f(x) Df(x) D2f(x) D3f(x)
0 3
-1
1 2 6
5 6
2 7 12
17 6
3 24 18
35 6
4 59 24
59
5 118

We can see that the third  differences i.e.

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

Example – 6:

  • By constructing a forward difference table find the 7th and 8th 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 Δ2f(x) column. The entry in Δf(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 Δ2f(x) column. The entry in Δf(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 6th and 7th 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 i.e. Δ2f(x) are constant.

  • To find f(6), extra 2 (shown in red colour) is written in Δ2f(x) column. The entry in Δf(x) is 2 + 11 = 13 (shown in red colour) is added. The entry in f(x) is 13 + 38 = 51 (shown in red colour) is added. Thus f(6) = 51.
  • To find f(7), extra 2 (shown in green colour) is written in Δ2f(x) column. The entry in Δf(x) is 2 + 13 = 15 (shown in green colour) is added. The entry in f(x) is 15 + 51 = 66 (shown in green colour) is added. Thus f(7) = 66.

Thus 6 th and 7 th terms of series are 51 and 66 respectively.

Example – 8:

  • Estimate f(5) from the following table.
x 0 1 2 3 4
f(x) 3 2 7 24 59

We prepare following forward difference table.

x f(x) Δf(x) Δ2f(x) Δ3f(x)
0 3
-1
1 2 6
5 6
2 7 12
17 6
3 24 18
35 6
4 59 24
59
5 118

We can see that the third  differences i.e. Δ3f(x) are constant and equals to 6.

  • To find f(5), extra 6 (shown in red colour) is written in Δ3f(x) column. The entry in Δ2f(x) is 6 + 18 = 24 (shown in red colour) is added. The entry in Δf(x) is 24 + 35 = 59 (shown in red colour) is added. The entry in f(x) is 59 + 59 = 118 (shown in red colour) Thus f(5) = 118.

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