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WorksheetsNumerical Interpolation
Total questions: 30
Worksheet time: 2hrs 27mins
Estimate f(1.5) using Lagrange interpolation.
−0.8673
-0.7214
−0.9773
-0.9113
Find f(2) for the data f(0) = 1, f(1) = 3 and f(3) = 55 using Newton's divided difference formula.
25
21
23
32
Find f(0.25) using Newton's divided difference formula
3.826
3.2567
3.2113
3.912
3.125
3.327
3.542
3.837
Given the two points (a,f(a)),(b,f(b)) , the linear Lagrange polynomial f1(x) that passes through these two points is given by
f1(x)=a−bx−bf(a)+a−bx−af(b)
f1(x)=b−axf(a)+b−axf(b)
f1(x)=f(a)+b−af(b)−f(a)(b−a)
f1(x)=a−bx−bf(a)+b−ax−af(b)
As you were provided with 3 data points, which formula you can use in the next step?
P2(x) =Lo fo+L1 f1
P3(x) =Lo fo+L1 f1
P2(x) =Lo fo+L1 f1+L2 f2
P3(x) =Lo fo+L1 f1+L2 f2
Based on the data given, which is the suitable type of polynomial to do the approximation?
Linear
Quadratic
Cubic
Quartic
What is the value of Lo fo ? Please provide the answer in 2 decimal points.
(a)
What is the value of L1 f1 ? Please provide the answer in 2 decimal points.
(a)
The methods to find the polynomials approximation are ..
Lagrange interpolation
Newton's Divided Difference
Natural Cubic Spline
Clamped Cubic Spline
(1+ Δ ) (1+ ∇ ) =
0
1
2
3
Δ∇
∇−Δ
Δ−∇
Δ+∇
none
∇=
E ∇
E−1Δ
E−1∇
EΔ
f(x)-f(x-h)=
Δf(x)
∇f(x)
E(x)
none
(E-1)/E
Δ
∇
Δ∇
none
E[E(y0)] =
y0
Y1
Y2
Y
If k is constat then Δk is
k
1
0
∇k
Backward difference operator pronounced as
delta
lamda
nabla
alpha
If h=1 then Δ(x2)=
2x
2x-1
2x+1
1
E(f(x))=
f(x-h)
f(x)
f(x+h)
f(x+2h)
What is the value of z?
6
-6
-36
4
What is an appropriate first row command to solve this by Gaussian Elimination?
2R2 + R2 -> R2
-2R1 + R2 -> R2
R3 +R2 --> R3
2R1 + R2 --> R2
Which row command is an appropriate next step?
7R2 + R3 -> R3
-7R1 + R3 -> R3
7R3 -> R3
-7R2 + R3 -> R3
Δ=
E−1
E+1
e−1
None of these
δ=
21(E21−E−21)
21(E21+E−21)
21(E−21−E21)
2h(E−21−E21)
μ=
21(E21+E−21)
21(E21−E−21)
E−2f(x)=
f(x−2h)
f(x−2h)−f(x)
f(x+2h)
f−2(x)
Δ2y0=
Δy1−Δy0
Δy0−Δy1
Δ2y1−Δ2y0
Δy1+Δy0
∆2x=…………….
2(x+h)-2x
2(x+h)
2(x-h)-2x
2(x-h)
If f(x)=a0xn+a1xn−1+...+an−1x+an, a0=0, then Δnf(x)=
a0hn
n!hn
a0n!hn
n!
