009C Sample Final 3, Problem 5

Consider the function

$f(x)=e^{-{\frac {1}{3}}x}$

(a) Find a formula for the  $n$th derivative  $f^{(n)}(x)$  of  $f$  and then find  $f'(3).$

(b) Find the Taylor series for  $f(x)$  at  $x_{0}=3,$  i.e. write  $f(x)$  in the form

$f(x)=\sum _{n=0}^{\infty }a_{n}(x-3)^{n}.$

Foundations:
The Taylor polynomial of $f(x)$  at $a$ is
$\sum _{n=0}^{\infty }c_{n}(x-a)^{n}$ where $c_{n}={\frac {f^{(n)}(a)}{n!}}.$

Solution:

(a)

Step 1:
We have
$f'(x)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-{\frac {1}{3}}x},$
$f''(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{2}e^{-{\frac {1}{3}}x},$
and
$f^{(3)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{3}e^{-{\frac {1}{3}}x}.$
If we compare these three equations, we notice a pattern.
Thus,
$f^{(n)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-{\frac {1}{3}}x}.$

Step 2:
Since
$f'(x)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-{\frac {1}{3}}x},$
we have
$f'(3)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-1}.$

(b)

Step 1:
Since
$f^{(n)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{3}e^{-{\frac {1}{3}}x},$
we have
$f^{(n)}(3)={\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-1}.$
Therefore, the coefficients of the Taylor series are
$c_{n}={\frac {{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-1}}{n!}}.$

Step 2:
Therefore, the Taylor series for  $f(x)$  at  $x_{0}=3$  is
$\sum _{n=0}^{\infty }{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}{\frac {1}{e(n!)}}(x-3)^{n}.$

Final Answer:
(a)    $f^{(n)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-{\frac {1}{3}}x},~f'(3)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-1}$
(b)    $\sum _{n=0}^{\infty }{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}{\frac {1}{e(n!)}}(x-3)^{n}$

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