Derive Simpson's rule with error term by using ∫ x 0 x 2 f ( x ) d x = a 0 f ( x 0 ) + a 1 f ( x 1 ) + a 2 f ( x 2 ) + k f ( 4 ) ( ξ )Find a 0 , a 1 , and a 2 from the fact that Simpson's rule is exact for x n when n = 1, 2, and 3. Then find k by applying the integration formula with f ( x ) = x 4 .We certainly get the correct coefficients, but I don't see how this tells us we can combine ξ 1 and ξ 2

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Derive Simpson&#039;s rule with error term by using      ∫                  x        0                            x        2              f  (  x  )    d  x  =      a    0    f  (      x    0    )  +      a    1    f  (      x    1    )  +      a    2    f  (      x    2    )  +  k      f          (      4      )        (  ξ  )Find       a    0  ,       a    1  , and       a    2   from the fact that Simpson&#039;s rule is exact for       x    n   when   n  =  1,   2, and   3. Then find k by applying the integration formula with   f  (  x  )  =      x    4  .We certainly get the correct coefficients, but I don&#039;t see how this tells us we can combine       ξ    1   and        ξ    2

Cahokiavv · Accepted Answer

Let   F be an anti-derivative of   f,       F    ′    =  f. W.l.o.g.       x    1    =  0, set   x  =  h then we are interested in the error expression  g  (  x  )  =  F  (  x  )  −  F  (  −  x  )  −      x    3    (  f  (  x  )  +  4  f  (  0  )  +  f  (  −  x  )  )  .This has derivatives                              g          ′                (        x        )                            =                  2          3                (        f        (        x        )        −        2        f        (        0        )        +        f        (        −        x        )        )                                  −                  x          3                (                  f          ′                (        x        )        −                  f          ′                (        −        x        )        )                                      g          ″                (        x        )                            =                  1          3                (                  f          ′                (        x        )        −                  f          ′                (        −        x        )        )                                  −                  x          3                (                  f          ″                (        x        )        +                  f          ″                (        −        x        )        )                                      g          ‴                (        x        )                            =                                  −                  x          3                (                  f          ‴                (        x        )        −                  f          ‴                (        −        x        )        )        .            Consequently, by the extended mean value theorem            g      (      x      )              x      m        =                    g        ′            (              x        1            )              m              x        1                  m          −          1                      =                    g        ″            (              x        2            )              m      (      m      −      1      )              x        2                  m          −          2                      =                    g        ‴            (              x        3            )              m      (      m      −      1      )      (      m      −      2      )              x        3                  m          −          3                    with   0  &amp;lt;      x    3    &amp;lt;      x    2    &amp;lt;      x    1    &amp;lt;  x. Using   m  =  5 this gives            g      (      x      )              x      5        =                    g        ‴            (              x        3            )              60              x        3        2              =  −      1    90    ·                    f        ‴            (              x        3            )      −              f        ‴            (      −              x        3            )              2              x        3              =  −      1    90    ·      f          (      4      )        (      x    4    )  .with       |        x    4        |    &amp;lt;      x    3    &amp;lt;  x. This results in the error formula  g  (  x  )  =  −      1    90    ·      f          (      4      )        (      x    4    )  ·      x    5    .or after translating the initial simplifications back,                              ∫                                    x              0                                                          x              2                                      f        (        x        )                d        x                            =                  h          3                [        f        (                  x          0                )        +        4        f        (                  x          1                )        +        f        (                  x          2                )        ]        −                              h            5                    90                          f                      (            4            )                          (        ξ        )            with   ξ  ∈  (      x    0    ,      x    2    )

Derive Simpson's rule with error term by using <msubsup> ∫ <mrow c

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