Homework Solution: For trapezoid and Simpson methods, the numerical integration I=int_{0}^{1}sqrt{1-x^2}dx h…

    For trapezoid and Simpson methods, the numerical integration I=int_{0}^{1}sqrt{1-x^2}dx have relative error varepsilon _{N}=[I_{exact}-I(N)]/I_{exact}sim1/N^{1.5} where I(N) is the numerical result using N intervals between [0, 1]. (i) Explain the power-law scaling with an exponent 1.5 (instead of 2 and 4 as expected for the trapezoid and Simpson's method, respectively). (ii) How can one circumvent this problem and restore the desired error scaling ?

    Expert Answer

     
    power law scaling:

    Ce trapezoid and Simpson courses, the numerical integration I=int_{0}^{1}sqrt{1-x^2}dx have not-absolute fault varepsilon _{N}=[I_{exact}-I(N)]/I_{exact}sim1/N^{1.5}

    where I(N) is the numerical upshot using N intervals among [0, 1].

    (i) Explain the power-principle scaling with an propounder 1.5 (instead of 2 and 4 as expected ce the trapezoid and Simpson’s course, respectively).

    (ii) How can single outwit this completion and reinstate the desired fault scaling ?

    Expert Counter-argument

     

    power principle scaling:

    power principle scaling is a scaling with a fixed which simpliy
    multiplies the originalpower principle proportion with a
    constant.

    ii)

    ce trapeziod ERROR SCALING

    I=0.2