MTH603 Quiz 2 Solution and Discussion
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Please share Quiz and Please discuss the each question!
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1st ordered divided difference formula is defined as…
y[x0,x1]=(y1-y0)/(x1-x0)
y[x0,x1]=(y1+y0)/(x1-x0)
y[x0,x1]=(y1-y0)/(x1+x0) -
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
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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.
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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. -
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
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Which of the following is the Richardson’s Extrapolation limit: F1(h/2) provided that F(h/2) = F(h) = 1 ?
0
1
3
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Integration is a ………………process.
Subtracting
Summing
Dividing
None of the given choices -
In Simpson’s 1/3 rule, the global error is of ………………
O(h2)
O(h3)
O(h4)
None of the given choices -
In Lagrange’s interpolation, for the given five points we can represent the function f (x) by a polynomial of degree
3
4
5
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Δ=----- ?
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For the given data points (1,0.3),(3,1),and(5,1.2) the divide difference table will be given as
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@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
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Whatwillbethevalueof′a′inthegivendividedifferencetable? x2468y0.51.11.72…21stD.D0.30.30.252ndD.Da−0.01253rdD.D−0.0021
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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)