# MTH603 Quiz 2 Solution and Discussion

• If y(x) is approximated by a polynomial Pn(x) of degree n then the error is given by

ε(x)=y(x)+Pn(x)
ε(x)=y(x)−Pn(x)
ε(x)=y(x)×Pn(x)
ε(x)=y(x)÷Pn(x)

• Forthegivendividedifferencetablex 1 4 7y2.23.54.11stD.D0.43330.22ndD.D−0.0389theNewton′sdividedifferenceinterpolationformulawillbe

y=f(x)=2.2+(x−1)(−0.0389)+(x−1)((x−4)(0.4333)

y=f(x)=2.2+(x−1)(0.4333)+(x−1)((x−4)(−0.0389)

y=f(x)=−0.0389+(x−1)(0.4333)+(x−1)((x−4)(2.2)

y=f(x)=−0.0389+(x−1)(2.2)+(x−1)((x−4)(0.4333) Solution:  • Forthegivendividedifferencetablex 1 4 7y2.23.54.11stD.D0.43330.22ndD.D−0.0389theNewton′sdividedifferenceinterpolationformulawillbe

y=f(x)=2.2+(x−1)(−0.0389)+(x−1)((x−4)(0.4333)

y=f(x)=2.2+(x−1)(0.4333)+(x−1)((x−4)(−0.0389)

y=f(x)=−0.0389+(x−1)(0.4333)+(x−1)((x−4)(2.2)

y=f(x)=−0.0389+(x−1)(2.2)+(x−1)((x−4)(0.4333) • Whatwillbethevalueof′a′inthegivendividedifferencetable? x2468y0.51.11.72…21stD.D0.30.30.252ndD.Da−0.01253rdD.D−0.0021 • For the given data points (1,0.3),(3,1),and(5,1.2) the divide difference table will be given as  Forthegivendatapoints(1,0.3),(3,1),and(5,1.2)thedividedifferencetablewillbegivenas

• For the given data points (1,0.3),(3,1),and(5,1.2) the divide difference table will be given as  • Δ=----- ?

E−1/2
1-E
E-1
None • Δ=----- ?

• In Lagrange’s interpolation, for the given five points we can represent the function f (x) by a polynomial of degree

3
4
5
6

• In Simpson’s 1/3 rule, the global error is of ………………

O(h2)
O(h3)
O(h4)
None of the given choices

• Integration is a ………………process.

Subtracting
Summing
Dividing
None of the given choices

• Which of the following is the Richardson’s Extrapolation limit: F1(h/2) provided that F(h/2) = F(h) = 1 ?

0
1
3
4

• In the process of Numerical Differentiation, we differentiate an interpolating polynomial in place of ------------.

actual function

extrapolating polynomial

Lagrange’s polynomial

Newton’s Divided Difference Interpolating polynomial

• Richardson extrapolation is method also known as …………

Sequence acceleration method
Series acceleration method

Ref
In numerical analysis, Richardson extrapolation is a sequence acceleration method, used to improve the rate of convergence of a sequence. It is named after Lewis Fry Richardson, who introduced the technique in the early 20th century.

• We prefer ………over the Lagrange’s interpolating method for economy of computation.

Newton’s forward difference method

Newton’s backward difference method

Newton’s divided difference method

none.

• Which of the following is the Richardson’s Extrapolation limit: F3(h/8)
provided that F2(h/8) = F2(h/4) = 1 ?

63
64
1
-1  1

1

127

1

5

9

4
• ## ZOO505 Assignment 2 Solution and Discussion ZOO505 - Cell and Molecular Biology • zoo505 assignment 1 solution discussion fall 2019 • • zareen

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