MATH3242.03/COSC3122.03

FW03


Solutions to Assignment 2



1. Elementary rules

Integral

(a)

(b)

Exact value

0.38866

0.67474

Trapezoidal rule

-0.61228

0.64390

Simpson's Rule

0.41609

0.67440

Midpoint rule

0.93027

0.68966



2. Composite rules using 5 points:

Integral

(a) h = 11/24

(a) h = 0.2

Trapezoidal rule

0.33239

0.67276

Simpson's rule

0.39018

0.67475

Midpoint rule

0.50577

0.67873




3. Integral from tabulated data: n = 6, h = 0.1 (N.B. 3 significant figures is all that can be justified from data but the 4th figure is shown in brackets):


Trapezoidal rule

2.00(8)

Simpson's rule

1.98(8)

Midpoint rule

1.94(8)




4. The magnitude of the error in Simpson's Rule is

where we have taken the largest value for 2x-3 for x Î [1,3] which occurs at x = 1. For this error to be less than 5´10-4 (an accuracy of 3 decimal places) we must have n4 > 2×104/5×180 or n > 5.16. Thus take n = 6. (Remember n has to be even). Thus the composite Simpson's rule yields 2.94381 for the integral. The exact value is 2.94376. Thus the error limit is satisfied.


5. Romberg integration to evaluate the integral


k

h

Rk(1)

Rk(2)

Rk(3)

Rk(4)

1

2.

3.29584




2

1.

3.03421

2.94700



3

0.5

2.96657

2.94402

2.94382


4

0.25

2.94947

2.94377

2.94375

2.94375

which agrees with the exact value given in question 4.



6. Apply Richardson extrapolation to the composite Trapezoidal Rule using the notation of Romberg integration. Thus

using equation (4.32) with mk= 2k-1.. Writing out Rk-1(1) using the form of the composite trapeziodal rule from Theorem 4.5 we get

Noting that 2hk = hk-1 we get

which is equivalent to the form of the composite Simpson's rule given in Theorem 4.4 with n=2k-1.



7. The values from the MAPLE program for using n+1 points are


n

integral

4

3.829179

8

3.820280

16

3.820198

32

3.820198


Since the last two values agree to the number of figures shown, we can conclude that the value of the integral is 3.82020 correct to 5 decimal places. Note that this method does not guarantee the correctness since there may be false convergence. If you want to be sure, evaluate the integral with n = 64 as another check.