Problem 3 :

(i) p = -1/2 

%% PS: In this case (-1<p<0), Simpson's rule can only be 
%% applied on the interval [2h,1].

Exact integral = 2 ( 1 - (2h)^(1/2) )

h = 0.01 => Exact_1 = 1.717157287525381

h = 0.005 => Exact_2 = 1.8

Numerical :

h = 0.01 => S_98 = 1.717224277756064

h = 0.005 => S_198 = 1.800047369313522

Order : Order = log_2((S_98-Exact_1)/(S_198-Exact_2))

Numerical Order = 0.4999979553633133

Order of convergence : O(h^(1/2))

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(ii) p = 1/2

Exact integral = 2/3

Numerical :

h = 0.01 => S_100 = 0.6665854820667237

h = 0.005 => S_200 = 0.6666379635700298

h = 0.0025 => S_400 = 0.6666565185891514

Order : Order = log_2((S_100-S_200)/(S_200-S_400))

Numerical Order = 1.499999558029851

Order of convergence : O(h^(3/2))

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(iii) p = 5/2

Exact integral = 2/7

Numerical :

h = 0.01 => S_100 = 2.857142837421070

h = 0.005 => S_200 = 2.857142855372715

h = 0.0025 => S_400 = 2.857142856984712

Order : Order = log_2((S_100-S_200)/(S_200-S_400))

Numerical Order = 3.477195416857635

Order of convergence : O(h^(7/2))