I noticed that I get the exact error, using midpoint rule error bound formula, but with f

Misael Matthews

Misael Matthews

Answered question

2022-06-10

I noticed that I get the exact error, using midpoint rule error bound formula, but with f ( b a 2 ) for K, i.e. :
K ( b a ) 3 24 n 2
f ( b a 2 ) ( b a ) 3 24 n 2

Answer & Explanation

Zayden Andrade

Zayden Andrade

Beginner2022-06-11Added 22 answers

Step 1
What you observe happens because your test example is cubic, so that its second derivative is linear. This gives a certain symmetry about the midpoint.
To highlight this, consider that the integral value over [ 0 , a ] is the same as the one for the symmetrized function
f e ( x ) = 1 2 ( f ( x ) + f ( a x ) ) . .
For this modified function the second derivative is a constant, the linear contributions of both terms in it are complementary. This constant has the value
f e ( x ) = f e ( a / 2 ) = f ( a / 2 ) , ,
so that indeed all the error formulas are correct when inserting this value.

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