Show that one extrapolation of the trapezoid rule leads to Simpson's rule.

Question

Show that one extrapolation of the trapezoid rule leads to Simpson&#039;s rule.

Braeden Shannon · Accepted Answer

Indeed, Simpson&#039;s method can be obtained by applying Richardson&#039;s extrapolation to the trapezoidal method. We begin with the trapezoidal method:                    (1)                              ∫          a          b                f        (        x        )                d        x        ≈                  h          2                          {          f          (                      x            0                    )          +          2          f          (                      x            1                    )          +          2          f          (                      x            2                    )          +          2          f          (                      x            3                    )          +          …                    ⋯          +          2          f          (                      x                          n              −              1                                )          +          f          (                      x            n                    )          }                    where   h  =  (  b  −  a  )      /    n and       x    i    =  a  +  i  h. The error of trapezoidal method is known to be of order       h          2      . Therefore, if we double   n, thus halving   h, the error should go down by about the factor of   4. We can then subtract   1      /    4 of the original formula (1) from the new one, thus canceling out the error term with       h    2  . The result is Simpson&#039;s formula.But I find it notationally easier to proceed in the opposite direction: using   h and   2  h, not   h and   h      /    2. To this end: let   n in the formula (1) be even. Next, write down the trapezoidal rule with step   2  h. It will involve only       x    0    ,      x    2    ,      x    4    ,  …  ,      x    n  :                    (2)                              ∫          a          b                f        (        x        )                d        x        ≈        h                  {          f          (                      x            0                    )          +          2          f          (                      x            2                    )          +          2          f          (                      x            4                    )          +          2          f          (                      x            6                    )          +          …                    ⋯          +          2          f          (                      x                          n              −              2                                )          +          f          (                      x            n                    )          }                    The error of formula (2) is about   4 times the error in formula (1), since we doubled the step size. So we divide both sides of (2) by   4 and subtract the result from (1), hoping to cancel out the error.      3    4        ∫    a    b    f  (  x  )    d  x  ≈      h    2        {    f    (          x      0        )    +    2    f    (          x      1        )    +    2    f    (          x      2        )    +    2    f    (          x      3        )    +    …        ⋯    +    2    f    (          x              n        −        1              )    +    f    (          x      n        )    }      −      h    4        {    f    (          x      0        )    +    2    f    (          x      2        )    +    2    f    (          x      4        )    +    2    f    (          x      6        )    +    …        ⋯    +    2    f    (          x              n        −        2              )    +    f    (          x      n        )    }  Collect like terms:      3    4        ∫    a    b    f  (  x  )    d  x  ≈      h    4        {    f    (          x      0        )    +    4    f    (          x      1        )    +    2    f    (          x      2        )    +    4    f    (          x      3        )    +    …        ⋯    +    4    f    (          x              n        −        1              )    +    f    (          x      n        )    }  Finally, multiply both sides by       4    3   and you&#039;ll get Simpson&#039;s formula.

Show that one extrapolation of the trapezoid rule leads to

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