MATH105 April 2017
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q3 • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q5 (e) • Q6 (a) • Q6 (b) •
Question 01 (l)

Find the first three nonzero terms of the Maclaurin series for the function $f(x)=\tan ^{1}\left({\frac {x}{3}}\right)$.
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Hint

Recall that a Maclaurin series is a Taylor series expansion of a function about 0,
$f(x)=f(0)+f^{'}(0)x+{\frac {f^{''}(0)}{2!}}x^{2}+{\frac {f^{'''}(0)}{3!}}x^{3}+...+{\frac {f^{n}(0)}{n!}}x^{n}+....$

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Solution

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First note that $f(0)=\tan ^{1}(0)=0.$
$f^{'}(x)={\frac {1/3}{x^{2}/9+1}}={\frac {3}{x^{2}+9}};f^{'}(0)={\frac {1}{3}}.$
$f^{''}(x)={\frac {3}{(x^{2}+9)^{2}}}^{\cdot }2x={\frac {6x}{(x^{2}+9)^{2}}};f^{''}(0)=0.$
$f^{'''}(x)={\frac {6}{(x^{2}+9)^{2}}}+{\frac {6x}{(x^{2}+9)^{3}}}\cdot (2)\cdot (2x)={\frac {6}{(x^{2}+9)^{2}}}+{\frac {24x^{2}}{(x^{2}+9)^{3}}};f^{'''}(0)={\frac {2}{27}}.$
$f^{''''}(x)={\frac {6}{(x^{2}+9)^{3}}}\cdot (2)\cdot (2x)+{\frac {48x}{(x^{2}+9)^{4}}}+{\frac {24x^{2}}{(x^{2}+9)^{4}}}\cdot (3)\cdot (2x)={\frac {72x}{(x^{2}+9)^{3}}}+{\frac {144x^{3}}{(x^{2}+9)^{4}}};f^{''''}(0)=0.$
$f^{(5)}(x)={\frac {72}{(x^{2}+9)^{3}}}+....;f^{(5)}(0)={\frac {8}{81}}.$
Thus, the first three nonzero terms are ${\frac {x}{3}}{\frac {2}{27\times 3!}}x^{3}+{\frac {8}{81\times 5!}}x^{5}=\color {blue}{\frac {x}{3}}{\frac {1}{81}}x^{3}+{\frac {1}{1215}}x^{5}.$
