Test: Newton-Gregory Forward Interpolation Formula - Civil Engineering (CE) MCQ

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.

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.

Here,

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!,

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.

Here,
x₀ = 0
x = 5
h = 2
x = x₀ + nh,
5 = 0 + n(2)
n = 2.5

y₀ is 4 since it is forward interpolation formula.
Δy₀ = 22
Δ²y₀ = 10
Δ³y₀ = 12
Δ⁴y₀ = 266
Substituting in the formula,

Here,
x₀ = 0.160
x = 0.1604
h = 0.001
x = x₀ + nh,
0.1604 = 0.160 + n(0.001)
n = 0.4

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) 
= 0.159713084.

Here,
x₀ = 1.5
x = 2.75
h = 0.5
x = x₀ + nh,
2.75 = 1.5 + n(0.5)
n = 2.5

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,

Here,
x₀ = 0
x = 0.18
h = 0.1
x = x₀ + nh,
0.18 = 0 + n(0.1)
n = 1.8

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,

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.

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.