solution

1. Evaluation
1. From the following table, estimate the number of students who obtained marks between
40 and 45 using the Newton-Gregory forward interpolation formula.
Marks 30-40 40-50 50-60 60-70 70-80
No. of students 31 42 51 35 31
[3 marks]
Solution:
First, we prepare the cumulative frequency table, as follows:
Marks less than (X) 40 50 60 70 80
No. of students (Y) 31 73 124 159 190
Difference Table
X Y ∆𝑌0 ∆2
𝑌0 ∆3
𝑌0 ∆4
𝑌0
40 31
42
50 73 9
51 -25
60 124 -16 37
35 12
70 159 -4
31
80 190

2. We have 𝑋0 = 40, 𝑋 = 45, ℎ = 10, 𝑝 =
𝑋−𝑋0
ℎ
= 0.5
From the difference table 𝑌0 = 31, ∆𝑌0 = 42, ∆2
𝑌0 = 9, ∆3
𝑌0 = −25, ∆4
𝑌0 = 37
From Newton-Gregory Forward Interpolation formula:
𝑓(𝑥) = 𝑌0 + 𝑝∆𝑌0 +
𝑝(𝑝 − 1)
2!
∆2
𝑌0 +
𝑝(𝑝 − 1)(𝑝 − 2)
3!
∆3
𝑌0 +
𝑝(𝑝 − 1)(𝑝 − 2)(𝑝− 3)
4!
∆4
𝑌0
= 31 + 0.5(42) +
0.5(0.5−1)
2×1
(9) +
0.5(0.5−1)(0.5−2)
3×2×1
(−25)+
0.5(0.5−1)(0.5−2)(0.5−3)
4×3×2×1
(37)
= 31 + 21 − 1.125 − 1.5625 − 1.4453
𝑓(𝑥) = 47.8674 ≈ 48
The number of students with marks less than 45 is 48, but the number of students with marks less
than 40 is 31.
Hence the number of students getting marks between 40 and 45 is 48-31= 17

3. 2. Estimate f(1.13) from the following table using Newton-Gregory backward interpolation
formula.
x 1.00 1.05 1.10 1.15 1.20
f(x) 1.0000 1.2567625 1.531000 1.820875 2.128000
[3 marks]
Solution:
Difference Table
X Y ∇𝑌0 ∇2
𝑌0 ∇3
𝑌0 ∇4
𝑌0
1.00 1.0000
0.2567625
1.05 1.2567625 0.017475
0.2742375 -0.020625
1.10 1.531000 -0. 00315 0.0786
0.2710875 0.057975
1.15 1.8020875 0.054825
0.3259125
1.20 2.128000
We have 𝑋0 = 1.15, 𝑋 = 1.13, ℎ = 0.05, 𝑝 =
𝑋−𝑋0
ℎ
= −0.4
From the difference table 𝑌0 = 1.8020875, ∇𝑌0 = 0.2710875, ∇2
𝑌0 = −0.00315, ∇3
𝑌0 =
−0.020625,

4. From Newton-Gregory Backward Interpolation formula:
𝑓(𝑥) = 𝑌0 + 𝑝∇𝑌0 +
𝑝(𝑝 + 1)
2!
∇2
𝑌0 +
𝑝(𝑝 + 1)(𝑝 + 2)
3!
∇3
𝑌0
= 1.8020875+ (−0.4)(0.2710875)+
(−0.4)(−0.4 + 1)
2 × 1
(−0.00315)
+
(−0.4)(−0.4 + 1)(−0.4 + 2)
3 × 2 × 1
(−0.020625)
= 1.8020875− 0.108435+ 0.000378 + 0.00132
𝑓(𝑥) = 1.6953505

5. 3. If U0 = 14, U4 = 24, U8 = 32, U12 = 35 and U16 =40, apply Gauss’s forward
interpolation formula to find U9.
[3 marks]
Solution:
Difference Table
X Y ∆𝑈0 ∆2
𝑈−1 ∆3
𝑈−1 ∆4
𝑈−2
0 14
10
4 24 -2
8 -3
8 32 -5 10
3 7
12 35 2
5
16 40
We have 𝑋 = 9, 𝑋0 = 8, ℎ = 4, 𝑝 =
𝑋−𝑋0
ℎ
0.25
From the table, 𝑈0 = 32, ∆𝑈0 = 3, ∆2
𝑈−1 = −5, ∆3
𝑈−1 = 7, ∆4
𝑈−2 = 10
From the Gauss’s Forward Interpolation Formula,
𝑦 = 𝑦0 + 𝑝∆𝑦0 +
𝑝(𝑝−1)
2!
∆2
𝑦−1 +
(𝑝+1)𝑝(𝑝−1)
3!
Δ3
𝑦−1 +
(𝑝+1)𝑝(𝑝−1)(𝑝−2)
4!
∆4
𝑦−2

6. = 32 + (0.25)(3)+
0.25(0.25− 1)
2 × 1
(−5) +
(0.25 + 1)0.25(0.25− 1)
3 × 2 × 1
(7)
+
(0.25 + 1)0.25(0.25− 1)(0.25 − 2)
4 × 3 × 2 × 1
(10)
= 32 + 0.75 + 0.46875 − 0.27343 + 0.17090
𝑈9 = 33.11622

7. 4. Using Gauss’s backward interpolation formula, find the population of the town in the
year 1936, given that:
Year (x) 1901 1911 1921 1931 1941 1951
Population (y) 12 15 20 27 39 52
[3 marks]
Solution:
Difference Table
X Y ∆𝑌−1 ∆2
𝑌−1 ∆3
𝑌−2 ∆4
𝑌−2 ∆5
𝑌−3
1901 12
3
1911 15 2
5 0
1921 20 2 3
7 3 -10
1931 27 5 -7
12 -4
1941 39 1
13
1952 52
We have x = 1936, x0 =1931, h = 1911-1921 = 10, 𝑝 =
𝑥−𝑥0
ℎ
= 0.5
From the table, y0 = 27, ∆y-1 = 7, ∆2
y-1 = 5, ∆3
y-2 = 3, ∆4
y-2=-7, ∆5
𝑌−3 = −10

8. From the Gauss’s Backward Interpolation Formula,
𝑦 = 𝑦0 + 𝑝∆𝑦−1 +
(𝑝+1)𝑝
2!
∆2
𝑦−1 +
(𝑝+1)𝑝(𝑝−1)
3!
Δ3
𝑦−2 +
(𝑝+2)(𝑝+1)𝑝(𝑝−1)
4!
∆4
𝑦−2 +
(𝑝+2)(𝑝+1)𝑝(𝑝−1)(𝑝−2)
5!
∆5
𝑦−3
= 27 + (0.5)(7)+
(0.5 + 1)(0.5)
2 × 1
(5) +
(0.5 + 1)(0.5)(0.5 − 1)
3 × 2 × 1
(3)
+
(0.5 + 2)(0.5 + 1)(0.5)(0.5− 1)
4 × 3 × 2 × 1
(−7)
+
(0.5 + 2)(0.5 + 1)(0.5)(0.5− 1)(0.5 − 2)
5 × 4 × 3 × 2 × 1
(−10)
= 27 + 3.5 + 1.875 − 0.1875 + 0.2734 − 0.1172
𝑌 = 32.3437