Question

Suppose you know that

\[{{\rm{f}}^{{\rm{(n)}}}}{\rm{(4) = }}\frac{{{{{\rm{( - 1)}}}^{\rm{n}}}{\rm{n!}}}}{{{{\rm{3}}^{\rm{n}}}{\rm{(n + 1)}}}}\]

and the Taylor series of f centered at 4 converges to f(x) for all x in the interval of convergence. Show that the fifth-degree Taylor polynomial approximates f(5) with error less than 0.0002. Formula

Short answer

Using n^th degree polynomial, in given Taylor series of f the fifth-degree Taylor polynomial approximates f(5)with error less than 0.0002.

Step-by-step solution

nth degree of Taylor polynomial

n^thdegree Taylor polynomial centered at a is\[{{\rm{T}}_{\rm{n}}}{\rm{(x) = }}\sum\limits_{{\rm{n = 0}}}^{\rm{n}} {\frac{{{{\rm{f}}^{{\rm{(n)}}}}{\rm{(a)}}}}{{{\rm{n!}}}}} {{\rm{(x - a)}}^{\rm{n}}}\].

Given parameters

Taylor series of f centered at 4converges to f(x) for all x in the interval of convergence.

\[{f^{(n)}}(4) = \frac{{{{( - 1)}^n}n!}}{{{3^n}(n + 1)}},\;\;\;a = 4\]

Calculating 5th and 6th degree of Taylor polynomial

Suppose 5^th degree Taylor polynomial be T₅(x)

Write T₆(x) as-

\[\begin{array}{c}{f^{(6)}}(4) = \frac{{{{( - 1)}^6}6!}}{{{3^6}(6 + 1)}}\\ = \frac{{80}}{{567}}\end{array}\]

We can write next Taylor series term as-

\[\frac{{80}}{{567}} \cdot \frac{{{{(x - 4)}^6}}}{{6!}} = \frac{{{{(x - 4)}^6}}}{{5103}}\]

To check the error in \[f(5)\], put \[x = 5\]

\[\begin{array}{c}\left| {\frac{{{{(5 - 4)}^6}}}{{5103}}} \right| = \frac{1}{{5103}}\\f(5) \approx 0.000196 < 0.0002\end{array}\]

Therefore, the fifth-degree Taylor polynomial approximates f(5) with error less than 0.0002.