Unit-4
Numerical analysis
Q1) Evaluate-
A1)
We have-
Q2) Using Newton’s forward difference formula, find the sum
A2)
Putting
It follows that
Since is a fourth degree polynomial in n.
Further,
By Newton Forward Difference Method
Q3) Given find , by using Newton forward interpolation method.
A3)
Let , then
0.7071 | 0.7660 | - | 0.8192 | 0.8660 |
The table of forward finite difference is given below:
45
50
55
60 | 0.7071
0.7660
0.8192
0.8660 |
0.0589
0.0532
0.0468 |
-0.0057
-0.0064 |
-0.0007 |
By Newton forward difference method
Here initial value = 45, difference of interval h = 5 and the value to be calculated at x=52.
By Formula
Q4) Find the missing term in the following:
0 | 1 | 2 | 3 | 4 | |
1 | 3 | 9 | ? | 81 |
A4)
Let
First we construct the forward difference table:
0
1
2
3
4 | 1
3
9
81 |
2
6
|
4
|
|
Now,
Q5) Find from the following table:
0.20 | 0.22 | 0.24 | 0.26 | 0.28 | 0.30 | |
1.6596 | 1.6698 | 1.6804 | 1.6912 | 1.7024 | 1.7139 |
A5)
Consider the backward difference method
0.20
0.22
0.24
0.26
0.28
0.30 | 1.6596
1.6698
1.6804
1.6912
1.7024
1.7139 |
0.0102
0.0106
0.0108
0.0112
0.0115 |
0.0004
0.0002
0.0004
0.0003 |
-0.0002
0.0002
-0.0001 |
0.0004
-0.0003 |
-0.0007 |
Here
By Newton backward difference formula
Q6) The following table give the amount of a chemical dissolved in water:
Temp. | ||||||
Solubility | 19.97 | 21.51 | 22.47 | 23.52 | 24.65 | 25.89 |
Compute the amount dissolve at
A6)
Consider the following backward difference table:
Temp. x | Solubility y | |||||
10
15
20
25
30
35 | 19.97
21.51
22.47
23.52
24.65
25.89 |
1.54
0.96
1.05
1.13
1.24 |
-0.58
0.09
0.08
0.11 |
0.67
-0.01
0.03 |
-0.68
0.04 |
0.72 |
Here
By Newton Backward difference formula
Q7) The following are the marks obtained by 492 candidates in a certain examination
Marks | 0-40 | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 |
No. of candidates | 210 | 43 | 54 | 74 | 32 | 79 |
Find out the number of candidates:
a) Who secured more than 48 but not more than 50 marks?
b) Who secured less than 48 but not less than 45 marks?
A7)
Consider the forward difference table given below:
Marks upto x | No. of candidates y | |||||
40
45
50
55
60
65 | 210
210+43=253
253+54=307
307+74=381
381+32=413
413+79= 492 |
43
54
74
32
79 |
11
20
-42
47 |
9
-62
89 |
-71
151 |
222 |
Here
By Newton Forward Difference formula
f
a) No. of candidate secured more than 48 but not more than 50 marks
b) No. of candidate secured less than 48 but not less than 45 marks
Q8) In the table given below, the values are consecutive terms of a series of which 23.6 is the 6th term then find out the 1st and 10th term.
X | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
y | 4.8 | 8.4 | 14.5 | 23.6 | 36.2 | 52.8 | 73.9 |
A8)
The difference table will be as follows-
X | Y | ||||
3
4
5
6
7
8
9 | 4.8
8.4
14.5
23.6
36.2
52.8
73.9 |
3.6
6.1
9.1
12.6
16.6
21.1 |
2.5
3.0
3.5
4.0
4.5 |
0.5
0.5
0.5
0.5
|
0
0
0 |
To find the first term, we will use Newtons’s forward interpolation formula-
With
We get,
To find the 10th term-