This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Newton-Gregory Forward Interpolation Formula”.
1. Newton- Gregory Forward interpolation formula can be used _____________
a) only for equally spaced intervals
b) only for unequally spaced intervals
c) for both equally and unequally spaced intervals
d) for unequally intervals
View Answer
Explanation: Newton – Gregory Forward Interpolation formula is given by
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! + …..
This formula is obtained by the Newton’s Divided difference formula by substituting the intervals as h. This is done because we assume the intervals to be constant, that is, equally spaced.
2. Find n for the following data if f(0.2) is asked.
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| f(x) | 176 | 185 | 194 | 203 | 212 | 220 | 229 |
a) 0.4
b) 0.2
c) 1
d) 0.1
View Answer
Explanation: The formula is
x = x0 + nh.
Here x0 is 0 as 0 is the first element and h is 1.
Since in the question it is given that we have to find f(0.2), x= 0.2.
So, substituting the values in the formula we get,
0.2 = 0 + n(1) .
Hence, n= 0.2.
3. Find n for the following data if f(1.8) is asked.
| x | 0 | 0.5 | 1 | 1.5 | 2 |
| f(x) | 0.3989 | 0.3521 | 0.2420 | 0.1295 | 0.0540 |
a) 2.4
b) 3.4
c) 2.6
d) 3.6
View Answer
Explanation: Here, x0 is 0, h is 0.5, x is 1.8.
Substituting the values in the formula
x = x0 + nh,
1.8 = 0 + n(0.5)
n = 3.6.
4. Find the polynomial for the following data.
| x | 4 | 6 | 8 | 10 |
| f(x) | 1 | 3 | 8 | 16 |
a) \(\frac{3x^2-22x+36}{8} \)
b) 3x2-22x+36
c) \(\frac{3x^2+22x+36}{2} \)
d) \(\frac{3x^2-19x+36}{8} \)
View Answer
Explanation: Here,
| x | y | Δy | Δ2y | Δ3y |
| 4 | 1 | 2 | 3 | 0 |
| 6 | 3 | 5 | 3 | |
| 8 | 8 | 8 | ||
| 10 | 16 |
y0 is 1 since it is forward interpolation formula.
Δy0 = 2
Δ2y0 = 3
Δ3y0 = 0
x = x0 + nh,
x = 4 + n(2)
Hence
n = (x-4)/2
Substituting these values in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3!,
\(f(x)=1+\frac{(x-4)}{2} 2+\frac{(x-4)}{2} (\frac{x-4}{2}-1) \frac{3}{2}+0 \)
\(=\frac{3x^2-22x+36}{8}.\)
5. Using Newton’s Forward formula, find sin(0.1604) from the following table.
| x | 0.160 | 0.161 | 0.162 |
| f(x) | 0.1593182066 | 0.1603053541 | 0.1612923412 |
a) 0.169713084
b) 0.159713084
c) 0.158713084
d) 0.168713084
View Answer
Explanation: Here,
x0 = 0.160
x = 0.1604
h = 0.001
x = x0 + nh,
0.1604 = 0.160 + n(0.001)
n = 0.4
| x | y | Δy | Δ2y |
| 0.160 | 0.1593182066 | 9.871475*10-4 | -1.604*10-7 |
| 0.161 | 0.1603053541 | 9.869871*10-4 | |
| 0.162 | 0.1612923412 |
y0 is 0.1593182066 since it is forward interpolation formula.
Δy0 = 9.871475*10-4
Δ2y0 = -1.604*10-7
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! ,
f(0.1604) = 0.1593182066+(0.4)(9.871475*10-4)+(0.4)(-0.6) \(\frac{(-1.604*10^{-7})}{2} \)
= 0.159713084.
6. Find f(5) using Newton’s Forward interpolation formula from the following table.
| x | 0 | 2 | 4 | 6 | 8 |
| f(x) | 4 | 26 | 58 | 112 | 466 |
a) 71.109375
b) 61.103975
c) 70.103957
d) 71.103957
View Answer
Explanation: Here,
x0 = 0
x = 5
h = 2
x = x0 + nh,
5 = 0 + n(2)
n = 2.5
| x | y | Δy | Δ2y | Δ3y | Δ4y |
| 0 | 4 | 22 | 10 | 12 | 266 |
| 2 | 26 | 32 | 22 | 278 | |
| 4 | 58 | 54 | 300 | ||
| 6 | 112 | 354 | |||
| 8 | 466 |
y0 is 4 since it is forward interpolation formula.
Δy0 = 22
Δ2y0 = 10
Δ3y0 = 12
Δ4y0 = 266
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
f(5)=4+(2.5)(22)+\(\frac{(2.5)(1.5)(10)}{2}+\frac{(2.5)(1.5)(0.5)(12)}{6}+\frac{(2.5)(1.25)(0.5)(-0.5)(266)}{24} \)
f(5) = 71.109375.
7. Find f(0.18) from the following table using Newton’s Forward interpolation formula.
| x | 0 | 0.1 | 0.2 | 0.3 | 0.4 |
| f(x) | 1 | 1.052 | 1.2214 | 1.3499 | 1.4918 |
a) 1.18878784
b) 1.8878784
c) 1.9878785
d) 0.8878784
View Answer
Explanation: Here,
x0 = 0
x = 0.18
h = 0.1
x = x0 + nh,
0.18 = 0 + n(0.1)
n = 1.8
| x | y | Δy | Δ2y | Δ3y | Δ4y |
| 0 | 1 | 0.052 | 0.1174 | -0.1583 | 0.2126 |
| 0.1 | 1.052 | 0.1694 | -0.0409 | 0.0543 | |
| 0.2 | 1.2214 | 0.1285 | 0.0134 | ||
| 0.3 | 1.3499 | 0.1419 | |||
| 0.4 | 1.4918 |
y0 is 1 since it is forward interpolation formula.
Δy0 = 0.052
Δ2y0 = 0.1174
Δ3y0 = -0.1583
Δ4y0 = 0.2126
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
\(f(0.18)=1+(1.8)(0.052)+\frac{(1.8)(0.8)(0.1174)}{2}+\frac{(1.8)(0.8)(-0.2)(-0.1583)}{6}\)
\(+\frac{(1.8)(0.8)(-0.2)(-1.2)(0.2126)}{24} \)
f(0.18) = 1.18878784.
8. Find f(2.75) using Newton’s Forward interpolation formula from the following table.
| x | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| y | 3.375 | 7 | 13.625 | 24 | 38.875 | 59 |
a) 1.8296875
b) 18.296875
c) 22.296875
d) 24.296875
View Answer
Explanation: Here,
x0 = 1.5
x = 2.75
h = 0.5
x = x0 + nh,
2.75 = 1.5 + n(0.5)
n = 2.5
| x | y | Δy | Δ2y | Δ3y | Δ4y | Δ5y |
| 1.5 | 3.375 | 3.625 | 3 | 0.75 | 0 | 0 |
| 2 | 7 | 6.625 | 3.75 | 0.75 | 0 | |
| 2.5 | 13.625 | 10.375 | 4.5 | 0.75 | ||
| 3 | 24 | 14.875 | 5.25 | |||
| 3.5 | 38.875 | 20.125 | ||||
| 4 | 59 |
y0 is 3.375 since it is forward interpolation formula.
Δy0 = 3.625
Δ2y0 = 3
Δ3y0 = 0.75
Δ4y0 = 0
Δ5y0 = 0
Substituting in the formula,
f(x) = y0 + nΔy0 + n(n-1)Δ2y0/2! + n(n-1)(n-2) Δ3y0 /3! +….
f(2.75)=3.375+(2.5)(3.625)+\(\frac{(2.5)(1.5)(3)}{2}+\frac{(2.5)(1.5)(0.5)(0.75)}{6}\)+0+0
f(2.75) = 18.296875.
9. Find n if x0 = 0.75825, x = 0.759 and h = 0.00005.
a) 1.5
b) 15
c) 2.5
d) 25
View Answer
Explanation: Given
x0 = 0.75825
x = 0.759
h = 0.00005
Substituting in the formula,
x = x0 + nh,
0.759 = 0.75825 + n(0.00005)
Therefore, n = 15.
10. Find x if x0 = 0.6, n = 2.6 and h = 0.2.
a) 12
b) 1.2
c) 1.12
d) 1.22
View Answer
Explanation: Given
x0 = 0.6
n = 2.6
h = 0.2
Substituting in the formula,
x = x0 + nh
x = 0.6+(0.2)(2.6)
x = 1.12.
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