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:

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

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

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

    (ii) How can undivided diplomatize this bearing and refresh the desired fallacy scaling ?

    Expert Counterpart

     

    power principle scaling:

    power principle scaling is a scaling with a continuous which simpliy
    multiplies the originalpower principle kindred with a
    constant.

    ii)

    control trapeziod ERROR SCALING

    I=0.2