How can the value of e be found to five-place accuracy using Higher Order Mean Value Theorem. f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) ( x − a ) 2 2 ! + . . . + f n ( a ) ( x − a ) n n ! + f ( n + 1 ) ( a ) ( x − a ) ( n + 1 ) ( n + 1 ) !

Question

How can the value of e be found to five-place accuracy using Higher Order Mean Value Theorem.  f  (  x  )  =  f  (  a  )  +      f    ′    (  a  )  (  x  −  a  )  +                    f        ″            (      a      )      (      x      −      a              )        2                    2      !        +  .  .  .  +                    f        n            (      a      )      (      x      −      a              )        n                    n      !        +                    f                  (          n          +          1          )                    (      a      )      (      x      −      a              )                  (          n          +          1          )                            (      n      +      1      )      !

timmeraared · Accepted Answer

Instead, we shall use   e  &amp;lt;  3.  e  =      ∑          n      =      0              ∞            1          n      !        =      ∑          n      =      0        m        1          n      !        +      ∑          n      =      m      +      1        ∞        1          n      !            ∑          n      =      m      +      1        ∞        1          n      !        &amp;lt;      1          (      m      +      1      )      !        (      ∑          n      =      0        ∞        1          n      !        )  =      1          (      m      +      1      )      !        e  &amp;lt;      3          (      m      +      1      )      !      Therefore, we have   0  &amp;lt;  e  −      ∑          n      =      1        m        1          n      !        &amp;lt;      3          (      m      +      1      )      !       for all mWhen   m  =  8,       3          9      !        &amp;lt;      10          −      5      , hence we know that e agrees with       ∑          n      =      1        8        1          n      !        =  2.71827... at least up to 2.7182.

How can the value of e be found to five-place accuracy using Higher Order Mean Value Theorem. f

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