009C Sample Final 3, Problem 5 Detailed Solution

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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=3,}   i.e. write  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   in the form

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sum_{n=0}^\infty a_n(x-3)^n.}

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Background Information:
The Taylor polynomial of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)}   at  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a}   is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty}c_n(x-a)^n}   where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_n=\frac{f^{(n)}(a)}{n!}.}

Solution:

(a)

Step 1:
We have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f''(x)=\bigg(-\frac{1}{3}\bigg)^2 e^{-\frac{1}{3}x},}
and
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(3)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x}.}
If we compare these three equations, we notice a pattern.
Thus,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\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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