From a proof of Simpson's rule using Taylor polynomial where f ∈ [ x 0 , x 2 ] and, for x 1 = x 0 + hwhere h = x 2 − x 0 2 it got: ∫ x 0 x 2 f ( x ) d x ≅ 2 h f ( x 1 ) + h 3 f ″ ( x 1 ) 3 + h 5 f ( 4 ) ( ξ ) 60 and then, it changed f ″ ( x 1 ) by f ( x 0 ) − 2 f ( x 1 ) + f ( x 2 ) h 2 Where it came?

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From a proof of Simpson&#039;s rule using Taylor polynomial where   f  ∈  [      x          0        ,      x          2        ] and, for      x          1        =      x          0        +  hwhere  h  =                    x                  2                    −              x                  0                      2  it got:      ∫                  x                  0                                    x                  2                      f  (  x  )  d  x  ≅  2  h  f  (      x          1        )  +      h          3                          f        ″            (              x                  1                    )        3    +      h          5                          f                  (          4          )                    (      ξ      )        60  and then, it changed       f    ″    (      x          1        ) by            f      (              x                  0                    )      −      2      f      (              x                  1                    )      +      f      (              x                  2                    )              h              2            Where it came?

britspears523jp · Accepted Answer

f    ″    (      x    1    )  with its second-order central finite difference formula. Given a small   h  &amp;gt;  0 which represents the difference between grid points       x    j   and       x          j      −      1      , the formula can be derived by adding the two Taylor expansions  f  (  x  −  h  )  =  f  (  x  )  −  h      f    ′    (  x  )  +                    h        2                    f        ″            (      x      )              2      !        −                    h        3                    f        ‴            (      x      )              3      !        +      O    (      h    4    )and  f  (  x  +  h  )  =  f  (  x  )  +  h      f    ′    (  x  )  +                    h        2                    f        ″            (      x      )              2      !        +                    h        3                    f        ‴            (      x      )              3      !        +      O    (      h    4    )to form  f  (  x  −  h  )  +  f  (  x  +  h  )  =  2  f  (  x  )  +      h    2        f    ″    (  x  )  +      O    (      h    4    )which after subtracting   2  f  (  x  ) and dividing by       h    2   produces      f    ″    (  x  )  ≈            f      (      x      −      h      )      −      2      f      (      x      )      +      f      (      x      −      h      )              h      2      in your case the grid points are   x  =      x    1  ,   x  −  h  =      x    0  , and   x  +  h  =      x    2   therefore      f    ″    (      x    1    )  ≈            f      (              x        0            )      −      2      f      (              x        1            )      +      f      (              x        2            )              h      2

From a proof of Simpson's rule using Taylor polynomial where f ∈ [ x

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