# Numerical Analysis Questions and Answers – Newton-Gregory Forward Interpolation Formula

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

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

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 0.242 0.1295 0.054

a) 2.4
b) 3.4
c) 2.6
d) 3.6

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) $$\frac{3x^2-19x+36}{8}$$

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 0.16 0.161 0.162 f(x) 0.159318 0.160305 0.161292

a) 0.169713084
b) 0.159713084
c) 0.158713084
d) 0.168713084

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

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

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

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

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

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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