#### Example – 1:

- If f(x) = x
^{2}+ 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), D^{2}f(1) and D^{3}f(2). **Solution:**

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

f(0) = 0^{2} + 0 + 1 = 0 + 0 + 1 = 1

f(1) = 1^{2} + 1 + 1 = 1 + 1 + 1 = 3

f(2) = 2^{2} + 2 + 1 = 4 + 2 + 1 = 7

f(3) = 3^{2} + 3 + 1 = 9 + 3 + 1 = 13

f(4) = 4^{2} + 4 + 1 = 16 + 4 + 1 = 21

f(5) = 5^{2} + 5 + 1 = 25 + 5 + 1 = 31

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(x) | Δ |

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 Δ^{2}f(x) = 2 = constant and Δ^{3}f(x) = 0

**Note:**

- Given function is f(x) = x
^{2}+ x + 1. The highest power is 2, which obviously means Δ^{2}f(x) = constant and Δ^{3}f(x) = 0. - No need to show steps of calculations as shown in Δf(x), Δ
^{2}f(x) and Δ^{3}f(x) columns. You can write the values directly. To explain the concept for some problems steps are shown.

#### Example – 2:

- If f(x) = x
^{2}– 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) = x^{2} – 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0^{2} – 3(0) + 1 = 0 – 0 + 1 = 1

f(1) = 1^{2} – 3(1) + 1 = 1 – 3 + 1 = -1

f(2) = 2^{2} – 3(2) + 1 = 4 – 6 + 1 = -1

f(3) = 3^{2} – 3(3) + 1 = 9 – 9 + 1 = 1

f(4) = 4^{2} – 3(4) + 1 =16 – 12 + 1 = 5

f(5) = 5^{2} – 3(5) + 1 = 25 – 15 + 1 = 11

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(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. Δ^{2}f(x) = 2 = constant

#### Example – 3:

- If f(x) = 2x
^{2}+ 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) = 2x^{2} + 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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) = 0

#### Example – 4:

- If f(x) = 2x
^{3}– x^{2}+ 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) = 2x^{3} – x^{2} + 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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) = 12 = constant

#### Example – 5:

- If f(x) = x
^{3}– 2x^{2}+ 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) = x^{3} – 2x^{2} + 1 and x = 0, 1, 2, 3, 4 i.e. 0(1)4.

f(0) = 0^{3} – 2(0)^{2} + 1 = 0 – 0 + 1 = 1

f(1) = 1^{3} – 2(1)^{2} + 1 = 1 – 2 + 1 = 0

f(2) = 2^{3} – 2(2)^{2} + 1 = 8 – 8 + 1 = 1

f(3) = 3^{3} – 2(3)^{2} + 1 = 27 – 18 + 1 = 10

f(4) = 4^{3} – 2(4)^{2} + 1 = 64 – 32 + 1 = 33

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(x) | Δ^{3}f(x) | Δ^{4}f(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. Δ^{4}f(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) | Δ^{2}f(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. Δ^{2}f(x) are constant.

- To find f(7), extra 2 (shown in red colour) is written in D
^{2}f(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 D
^{2}f(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) | Δ^{2}f(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) = x
^{2}+ 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), D^{2}f(1) and D^{3}f(2). **Solution:**

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

f(0) = 0^{2} + 0 + 1 = 0 + 0 + 1 = 1

f(1) = 1^{2} + 1 + 1 = 1 + 1 + 1 = 3

f(2) = 2^{2} + 2 + 1 = 4 + 2 + 1 = 7

f(3) = 3^{2} + 3 + 1 = 9 + 3 + 1 = 13

f(4) = 4^{2} + 4 + 1 = 16 + 4 + 1 = 21

f(5) = 5^{2} + 5 + 1 = 25 + 5 + 1 = 31

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(x) | Δ |

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 Δ^{2}f(x) = 2 = constant and Δ^{3}f(x) = 0

**Note:**

- Given function is f(x) = x
^{2}+ x + 1. The highest power is 2, which obviously means Δ^{2}f(x) = constant and Δ^{3}f(x) = 0. - No need to show steps of calculations as shown in Δf(x), Δ
^{2}f(x) and Δ^{3}f(x) columns. You can write the values directly. To explain the concept for some problems steps are shown.

#### Example – 2:

- If f(x) = x
^{2}– 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) = x^{2} – 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0^{2} – 3(0) + 1 = 0 – 0 + 1 = 1

f(1) = 1^{2} – 3(1) + 1 = 1 – 3 + 1 = -1

f(2) = 2^{2} – 3(2) + 1 = 4 – 6 + 1 = -1

f(3) = 3^{2} – 3(3) + 1 = 9 – 9 + 1 = 1

f(4) = 4^{2} – 3(4) + 1 =16 – 12 + 1 = 5

f(5) = 5^{2} – 3(5) + 1 = 25 – 15 + 1 = 11

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(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. Δ^{2}f(x) = 2 = constant

#### Example – 3:

- If f(x) = 2x
^{2}+ 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) = 2x^{2} + 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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) = 0

#### Example – 4:

- If f(x) = 2x
^{3}– x^{2}+ 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) = 2x^{3} – x^{2} + 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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) = 12 = constant

#### Example – 5:

- If f(x) = x
^{3}– 2x^{2}+ 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) = x^{3} – 2x^{2} + 1 and x = 0, 1, 2, 3, 4 i.e. 0(1)4.

f(0) = 0^{3} – 2(0)^{2} + 1 = 0 – 0 + 1 = 1

f(1) = 1^{3} – 2(1)^{2} + 1 = 1 – 2 + 1 = 0

f(2) = 2^{3} – 2(2)^{2} + 1 = 8 – 8 + 1 = 1

f(3) = 3^{3} – 2(3)^{2} + 1 = 27 – 18 + 1 = 10

f(4) = 4^{3} – 2(4)^{2} + 1 = 64 – 32 + 1 = 33

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(x) | Δ^{3}f(x) | Δ^{4}f(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. Δ^{4}f(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) | Δ^{2}f(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. Δ^{2}f(x) are constant.

- To find f(7), extra 2 (shown in red colour) is written in D
^{2}f(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 D
^{2}f(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) | Δ^{2}f(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. D^{2}f(x) are constant.

To find f(6), extra 2 (shown in red colour) is written in D^{2}f(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 D^{2}f(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) | D^{2}f(x) | D^{3}f(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) = x
^{2}+ 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), D^{2}f(1) and D^{3}f(2). **Solution:**

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

f(0) = 0^{2} + 0 + 1 = 0 + 0 + 1 = 1

f(1) = 1^{2} + 1 + 1 = 1 + 1 + 1 = 3

f(2) = 2^{2} + 2 + 1 = 4 + 2 + 1 = 7

f(3) = 3^{2} + 3 + 1 = 9 + 3 + 1 = 13

f(4) = 4^{2} + 4 + 1 = 16 + 4 + 1 = 21

f(5) = 5^{2} + 5 + 1 = 25 + 5 + 1 = 31

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(x) | Δ |

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 Δ^{2}f(x) = 2 = constant and Δ^{3}f(x) = 0

**Note:**

- Given function is f(x) = x
^{2}+ x + 1. The highest power is 2, which obviously means Δ^{2}f(x) = constant and Δ^{3}f(x) = 0. - No need to show steps of calculations as shown in Δf(x), Δ
^{2}f(x) and Δ^{3}f(x) columns. You can write the values directly. To explain the concept for some problems steps are shown.

#### Example – 2:

- If f(x) = x
^{2}– 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) = x^{2} – 3x + 1 and x = 0, 1, 2, 3, 4, 5 i.e. 0(1)5.

f(0) = 0^{2} – 3(0) + 1 = 0 – 0 + 1 = 1

f(1) = 1^{2} – 3(1) + 1 = 1 – 3 + 1 = -1

f(2) = 2^{2} – 3(2) + 1 = 4 – 6 + 1 = -1

f(3) = 3^{2} – 3(3) + 1 = 9 – 9 + 1 = 1

f(4) = 4^{2} – 3(4) + 1 =16 – 12 + 1 = 5

f(5) = 5^{2} – 3(5) + 1 = 25 – 15 + 1 = 11

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(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. Δ^{2}f(x) = 2 = constant

#### Example – 3:

- If f(x) = 2x
^{2}+ 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) = 2x^{2} + 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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) = 0

#### Example – 4:

- If f(x) = 2x
^{3}– x^{2}+ 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) = 2x^{3} – x^{2} + 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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) = 12 = constant

#### Example – 5:

- If f(x) = x
^{3}– 2x^{2}+ 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) = x^{3} – 2x^{2} + 1 and x = 0, 1, 2, 3, 4 i.e. 0(1)4.

f(0) = 0^{3} – 2(0)^{2} + 1 = 0 – 0 + 1 = 1

f(1) = 1^{3} – 2(1)^{2} + 1 = 1 – 2 + 1 = 0

f(2) = 2^{3} – 2(2)^{2} + 1 = 8 – 8 + 1 = 1

f(3) = 3^{3} – 2(3)^{2} + 1 = 27 – 18 + 1 = 10

f(4) = 4^{3} – 2(4)^{2} + 1 = 64 – 32 + 1 = 33

The forward difference table is constructed as follows.

x | f(x) | Δf(x) | Δ^{2}f(x) | Δ^{3}f(x) | Δ^{4}f(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. Δ^{4}f(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) | Δ^{2}f(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. Δ^{2}f(x) are constant.

- To find f(7), extra 2 (shown in red colour) is written in Δ
^{2}f(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 Δ
^{2}f(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 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) | Δ^{2}f(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. Δ^{2}f(x) are constant.

- To find f(6), extra 2 (shown in red colour) is written in Δ
^{2}f(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 Δ
^{2}f(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) | Δ^{2}f(x) | Δ^{3}f(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. Δ^{3}f(x) are constant and equals to 6.

- To find f(5), extra 6 (shown in red colour) is written in Δ
^{3}f(x) column. The entry in Δ^{2}f(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.