Answer
$f(x) = 1 + \frac{x}{3} + \frac{1}{9 \times 2!} x^{2} - \frac{1}{27 \times 3!} x^{3} + \frac{1}{81 \times 4!}x^{4}$
Work Step by Step
Given: $f(x) = e^{\frac{x}{3}}$
$f(x) = f(0) + f'(0)x + \frac{f^{''}(0)}{2!}x^{2} + \frac{f^{'''}(0)}{3!}x^{3} + ... + \frac{f^{n}(0)}{n!}x^{n}$
$f(0) = e^{0} = 1$
$f^{'}(0) = \frac{1}{3}e^{0} = \frac{1}{3}$
$f^{''}(0) = \frac{1}{9}e^{0} = \frac{1}{9}$
$f^{'''}(0) = \frac{1}{27}e^{0} = \frac{1}{27}$
Maclaurin polynomial for $n=4$ is
$f(x) = 1 + \frac{x}{3} + \frac{1}{9 \times 2!} x^{2} - \frac{1}{27 \times 3!} x^{3} + \frac{1}{81 \times 4!}x^{4}$
