Erroneously Finding the Lagrange Error BoundConsider f ( x ) = sin ⁡ ( 5 x + π 4 ) and let P(x) be the third-degree Taylor polynomial for f about 0. I am asked to find the Lagrange error bound to show that |(f(1/10)−P(1/10))|&lt;1/100. Because P(x) is a third-degree polynomial, I know the difference is in the fourth degree term.

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Erroneously Finding the Lagrange Error BoundConsider   f  (  x  )  =  sin  ⁡      (    5    x    +                  π        4              )   and let P(x) be the third-degree Taylor polynomial for f about 0. I am asked to find the Lagrange error bound to show that |(f(1/10)−P(1/10))|&amp;lt;1/100. Because P(x) is a third-degree polynomial, I know the difference is in the fourth degree term.

Joanna Vang · Accepted Answer

f  (  x  )  =  sin  ⁡      (    5    x    +                  π        4              )  The third degree Taylor Polynomial is given by:      T    3    (  x  )  =  −            125              x        3                    6              2              −            25              x        2                    2              2              +            5      x              2        +      1          2      The error term is given by:      R          n      +      1        =                              f                      (            n            +            1            )                          (        c        )                    (        n        +        1        )        !              (  x  −  a      )          n      +      1        ≤            M              (        n        +        1        )        !              (  x  −  a      )          n      +      1      We have n=3, thus:                    d        4                    d                  x          4                          (    sin    ⁡          (      5      x      +                        π          4                    )        )    =  625  sin  ⁡      (    5    x    +          π      4        )  We have   a  =  0  ,  n  =  3  ,  x  =            1      10       and the max of sine is 1, so this yields:      R          n      +      1        ≤            M              (        n        +        1        )        !              (  x  −  a      )          n      +      1        =            625              4        !                        (                        1          10                    )        4    =            1      384

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