# 009C Sample Final 3, Problem 5

Consider the function

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

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

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

${\displaystyle f(x)=\sum _{n=0}^{\infty }a_{n}(x-3)^{n}.}$
Foundations:
The Taylor polynomial of ${\displaystyle f(x)}$  at ${\displaystyle a}$ is

${\displaystyle \sum _{n=0}^{\infty }c_{n}(x-a)^{n}}$ where ${\displaystyle c_{n}={\frac {f^{(n)}(a)}{n!}}.}$

Solution:

(a)

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

(b)

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

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