# 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.