Does there exist a monic x n , n &gt; 4, for which Simpson’s rule is exact? If not, why? S ( f ) = b − a 6 f ( a ) + 2 ( b − a ) 3 f ( a + b 2 ) + b − a 6 f ( b )

Question

Does there exist a monic       x    n  ,   n  &amp;gt;  4, for which Simpson’s rule is exact? If not, why?  S  (  f  )  =            b      −      a        6    f  (  a  )  +            2      (      b      −      a      )        3    f  (            a      +      b        2    )  +            b      −      a        6    f  (  b  )

Aaron Everett · Accepted Answer

If you take, for sake of simplicity,  a  =  0  ,  b  =  1 (and   f  (  x  )  =      x    n  ) you would require that      1    6    f  (  0  )  +      2    3    f  (  1      /    2  )  +      1    6    f  (  1  )  =      1          n      +      1      i.e.      2    3    (  1      /    2      )    n    +      1    6    =      1          n      +      1        ⇔      1          2              n        −        2              =                    5        −        n                    n        +        1            However, if   n  &amp;gt;  4 this last equality is impossible.

Does there exist a monic x n </msup> , n > 4 , for which Simpson’s rule is e

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