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^{3}f(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 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^{3}f(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 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^{3}f(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 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.