Solutions for homework assignment #2

Problem1

The purpose of this assignment was to investigate the scaling of the error as a function of the number of intervals used. Below is a link to a program that integrates the function using both integration formulas in the same run. It uses the trick of using a previously calculated integral when doubling the number of intervals.

Solution program [openint.f90]

A) Non-singular integrand

These are results along with lines with slope 2 and 3. The expected error scaling for small h is confirmed. Note that there are small deviations from the linear behavior for the largest h, showing the presence of higher-order corrections to the scaling.

B) Singular integrand

The exponent can be extracted by plottng the logarithm of the error versus log(h) [or log(N) or n].

Using the line fitting program from the previous assignment (with all errors set to 1 since they are irrelevant but should be the same for every point to give them equal weight in the fit) gave the following exponents (slopes): 
   alpha=1/2, 1st order: slope = 0.49983
              2nd order: slope = 0.50003
   alpha=3/4, 1st order: slope = 0.24996
              2nd order: slope = 0.25001
This is in very good agreement with the expectation; slope=1-alpha.

C) Almost singular integrand

The figure below shows results using alpha=1/2 and epsilon=0.00001.

The lines show the expected slopes 1/2 (for h much larger than epsilon) 2 (for h much less than epsilon with 1st-order formula) and 3 (for h much less than epsilon with second-order formula).

Problem 2

Solution program [inertia.f90]

A run for the case of the 5 cm sphere with a 1 cm inner cylinder, made of copper and gold, respectively, gave the following results (average and errors of Ix and Iz after each 10^5 step bin, stopping when the relative error of both quantities is less than eps=0.001): 
    1          0.0047045734             NaN        0.0049770784    0.0000000000
    2          0.0047070661    0.0000017626        0.0049626889    0.0000101749
    3          0.0047234163    0.0000134015        0.0049655108    0.0000071639
    4          0.0047133494    0.0000133053        0.0049550252    0.0000105513
    5          0.0047077014    0.0000117822        0.0049545036    0.0000084539
    6          0.0047021198    0.0000110619        0.0049462451    0.0000103182
    7          0.0047012592    0.0000095150        0.0049481774    0.0000090233
    8          0.0046955375    0.0000098975        0.0049449019    0.0000084690
    9          0.0046920539    0.0000093909        0.0049416094    0.0000081429
   10          0.0046938560    0.0000086230        0.0049441229    0.0000077068
   11          0.0046944982    0.0000078630        0.0049461666    0.0000072721
   12          0.0046958480    0.0000073227        0.0049493473    0.0000073288
   13          0.0046938449    0.0000070280        0.0049500340    0.0000067971
   14          0.0046933723    0.0000065419        0.0049494446    0.0000063371
   15          0.0046948479    0.0000062699        0.0049514169    0.0000062140
   16          0.0046935844    0.0000060040        0.0049492709    0.0000061851
   17          0.0046934146    0.0000056532        0.0049491679    0.0000058221
   18          0.0046938952    0.0000053596        0.0049498738    0.0000055413
   19          0.0046936391    0.0000050836        0.0049495672    0.0000052581
   20          0.0046936416    0.0000048294        0.0049494278    0.0000049970
   21          0.0046939780    0.0000046112        0.0049500226    0.0000047944
Since there are fluctuations also in the statistical errors, the number of bins needed to achieve this accuracy will vary somewhat.