I have the following question to answer:
Let S_n be the nth Simpson approximation for a function f.
Let M_n be the nth Midpoint Rule approximation.
Let T_n be the nth Trapezoidal Rule approximation.
Let B = sup{abs(f''(x)}
Show that one has the estimate,
absolute value( (S_n) - integral f(x)dx ) <= [B((b-a)^2)]/18n^2 where B >= abs(f''(x)) for all x in [a,b].
The hint in my book said to use the identity,
(S_2n) = (2/3)(M_n) + (1/3)(T_n).
Therefore, (S_n) = (2/3)(M_n/2) + (1/3)(T_n/2).
I take the error bounds for M_n and T_n as follows:
absolute value ( (2/3)(M_n/2) - (2/3)integral f(x)dx ) <= [B((b-a)^3)]/9n^2
absolute value ( (1/3)(T_n/2) - (1/3)integral f(x)dx ) <= [B((b-a)^3)]/9n^2
If I add them together I get,
absolute value ( (2/3)(M_n/2) + (1/3)(T_n/2) - integral f(x)dx ) <= [2B((b-a)^3)]/9n^2, but this isn't what I want. Why won't this work?
Thanks for any of your help.
Let S_n be the nth Simpson approximation for a function f.
Let M_n be the nth Midpoint Rule approximation.
Let T_n be the nth Trapezoidal Rule approximation.
Let B = sup{abs(f''(x)}
Show that one has the estimate,
absolute value( (S_n) - integral f(x)dx ) <= [B((b-a)^2)]/18n^2 where B >= abs(f''(x)) for all x in [a,b].
The hint in my book said to use the identity,
(S_2n) = (2/3)(M_n) + (1/3)(T_n).
Therefore, (S_n) = (2/3)(M_n/2) + (1/3)(T_n/2).
I take the error bounds for M_n and T_n as follows:
absolute value ( (2/3)(M_n/2) - (2/3)integral f(x)dx ) <= [B((b-a)^3)]/9n^2
absolute value ( (1/3)(T_n/2) - (1/3)integral f(x)dx ) <= [B((b-a)^3)]/9n^2
If I add them together I get,
absolute value ( (2/3)(M_n/2) + (1/3)(T_n/2) - integral f(x)dx ) <= [2B((b-a)^3)]/9n^2, but this isn't what I want. Why won't this work?
Thanks for any of your help.