• Cyberian's Gold

    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)

  • Cyberian's Gold

    @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)

    1d767974-39ce-409c-818f-505d317238cc-image.png

    Solution:

    c58d97cf-c1fa-46aa-add5-3cb89de6f935-image.png 74e3b9b6-be5c-46dc-8ecc-0c4f775cfe60-image.png

  • Cyberian's Gold

    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)

    1d767974-39ce-409c-818f-505d317238cc-image.png

  • Cyberian's Gold

    Whatwillbethevalueof′a′inthegivendividedifferencetable? x2468y0.51.11.72…21stD.D0.30.30.252ndD.Da−0.01253rdD.D−0.0021

    e0d49474-f272-41dc-bb69-4fd9f6e1af61-image.png

  • Cyberian's Gold

    @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

    219e6e10-947a-4ef6-8072-8df332b2257e-image.png

    0e268c65-269c-4352-b66a-ee4fd412055e-image.png

    Forthegivendatapoints(1,0.3),(3,1),and(5,1.2)thedividedifferencetablewillbegivenas

  • Cyberian's Gold

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

    219e6e10-947a-4ef6-8072-8df332b2257e-image.png

    0e268c65-269c-4352-b66a-ee4fd412055e-image.png

  • Cyberian's Gold

    @zaasmi said in MTH603 Quiz 2 Solution and Discussion:

    Δ=----- ?

    E−1/2
    1-E
    E-1
    None

    b050ee38-f1b6-4728-919c-2ea1eb3b4dd4-image.png

  • Cyberian's Gold

    Δ=----- ?

  • Cyberian's Gold

    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

  • Cyberian's Gold

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

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

  • Cyberian's Gold

    Integration is a ………………process.

    Subtracting
    Summing
    Dividing
    None of the given choices

  • Cyberian's Gold

    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

  • Cyberian's Gold

    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

  • Cyberian's Gold

    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.

  • Cyberian's Gold

    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.

  • Cyberian's Gold

    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

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