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
$\sum _{n=0}^{\infty }{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}{\frac {1}{e(n!)}}(x-3)^{n}.$ (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)    $\sum _{n=0}^{\infty }{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}{\frac {1}{e(n!)}}(x-3)^{n}$ 