Prove that the simple Simpson’s rule <msubsup> &#x222B;<!-- ∫ --> a b </msubs

migongoniwt

migongoniwt

Answered question

2022-06-18

Prove that the simple Simpson’s rule
a b f ( x ) d x ( b a ) 6 [ f ( a ) + 4 f ( a + b 2 ) + f ( b ) ]
is exact for all cubic polynomials.

Answer & Explanation

Dustin Durham

Dustin Durham

Beginner2022-06-19Added 31 answers

The change of variable allows to replace the range [ a , b ] by [ 0 , 1 ], to simplify the notation. Then setting
g ( t ) := f ( a + t ( b a ) ) ,
we have
a b f ( x ) d x = ( b a ) 0 1 g ( t ) d t ( b a ) 1 6 [ g ( 0 ) + 4 g ( 1 2 ) + g ( 1 ) ] .
Now we can show that the relation holds exactly for all powers of x up to third:
0 1 d t = 1 = 1 6 [ 1 + 4 1 + 1 ] ,
0 1 t d t = 1 2 = 1 6 [ 0 + 4 1 2 + 1 ] ,
0 1 t 2 d t = 1 3 = 1 6 [ 0 + 4 1 4 + 1 ] ,
0 1 t 3 d t = 1 4 = 1 6 [ 0 + 4 1 8 + 1 ] .
By linearity, it holds for all cubic polynomials.

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