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) 
@zaasmi said in MTH603 Quiz 2 Solution and Discussion:
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

@zaasmi said in MTH603 Quiz 2 Solution and Discussion:
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= ?

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 methodRef
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