Given f ( x ) = e − x and x 0 = 0, I need to find the smallest value of n such that | P n ( x ) − f ( x ) | &lt; 10 − 5 for all x ∈ [ 0 , 1 ]

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Given   f  (  x  )  =      e          −      x       and       x    0    =  0, I need to find the smallest value of n such that       |        P    n    (  x  )  −  f  (  x  )      |    &amp;lt;      10          −      5       for all   x  ∈  [  0  ,  1  ]

Kyler Crawford · Accepted Answer

Given   f  (  x  )  =      e          −      x       and       x    0    =  0, find the smallest value of n such that       |        P    n    (  x  )  −  f  (  x  )      |    &amp;lt;      10          −      5       for all   x  ∈  [  0  ,  1  ]The Maclaurin series of this function to be             f      ~        =      ∑          n      =      0              ∞                  (      −      1              )        n                    n      !            x    n  To satisfy       |        P    n    (  x  )  −  f  (  x  )      |    &amp;lt;      10          −      5      , look for an error term so that       |                  x                  n          +          1                            (        n        +        1        )        !                    f              (        n        +        1        )              (    ξ    )    |    &amp;lt;      10          −      5      , where   0  &amp;lt;  ξ  &amp;lt;  x. For all   x  ∈  [  0  ,  1  ]      |        f          (      n      +      1      )        (  ξ  )      |    ≤  1 (this happens since       e          −      x       is strictly decreasing as Gabor pointed out). Now,       |                  x                  n          +          1                            (        n        +        1        )        !                    f              (        n        +        1        )              (    ξ    )    |    &amp;lt;      10          −      5        ⇒      |                  x                  n          +          1                            (        n        +        1        )        !              ⋅    1    |    &amp;lt;      10          −      5       is always nonnegative in the specified interval, this becomes             x              n        +        1                    (      n      +      1      )      !        &amp;lt;      10          −      5      Find a bound for       x          n      +      1      . Since   x  ∈  [  0  ,  1  ]       x          n      +      1        =  0 when   x  =  0      x          n      +      1        =  1, so       1          (      n      +      1      )      !        &amp;lt;      10          −      5      By algebra, this is equivalently   (  n  +  1  )  !  &amp;gt;      10    5  . It is evident   8  !  =  40  ,  320 and   9  !  =  362  ,  880, so   n  +  1  =  9  ⇒  n  =  8 the smallest possible n satisfying       |        P    n    (  x  )  −  f  (  x  )      |    &amp;lt;      10          −      5       for all   x  ∈  [  0  ,  1  ] is   n  =  8

Given f ( x ) = e <mrow class="MJX-TeXAtom-ORD"> −

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