How small must the grid spacing, h, be such that using the Trapezoidal Rule to c

Question

How small must the grid spacing, h, be such that using the Trapezoidal Rule to compute integral_1^3 In xdx gives an error no more than 10^-8 using the error bound? So how many grid points are needed? How would the grid spacing, change if the asymptotic error were used instead of the error bound? B. How small must the grid spacing, h, be such that using Simpson's Rule to compute integral_1^3 In xdx gives an error no more than 10^-8 using the error bound? So how many grid points are needed? How would the grid spacing, change if the asymptotic error were used instead of the error bound?

Explanation / Answer

A) Using Trapezoidal rule andd Error Bound Theorem,

    ET(f) = -1/12*h2(b-a)f''(cn) for some cn in interval [a,b]

    f(x) = lnx, so f'(x) = d/dx(lnx) = 1/x and f''(x) = -1/x2

    max |f''(x)| in [1,3] is 1

    Therefore, |E(f) = 1/12*h2*(3-1)*1

                            = h2/6

    We want h2/6<= 10-8 or h<= 2.45*10-4 = .000245

    No of grid points n >= (b-a)/h = 2/h = 8163

    For Trapezoidal rule, Asymptotic Error formula is E = -h2/12 (f'(b) - f'(a))

    f'(b) = f'(3) = 1/3 and f'(1) = 1

    So, E = -h2/12(1/3 - 1) = h2/18

    h2/18 <= 10-8 yields h <= .000424 and n >= 4717

B) For the Simpson's rule E = -1/180 h4*(b-a)f(4)(cn)

   f(x) = ln has 4th derivative f4 = -6/x4

   max|f4| in the interval [1,3] is 6

So, |E| = h4/15 and h4/15 <= 10-8 gives h<= .02, n >= 100

Using Asymptotic error, |E| = -1/180h4(f'''(b) - f'''(a))

f'''(x) for lnx = 2/x3 ; f'''(3) = 2/27 and f'''(1) = 2

E = -h4/180*(2/27 - 2) = .01h4

.01h4 <= 10-8 means h <= .032 and n> = 63

   

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