How do I find Maclaurin series for the function:
$$\sqrt[3]{\sin(x^3)}$$
The answer should be:
$$ x – \frac {x^7}{18} – \frac {{x}^{13}}{3240} + o(x^{13})$$
I tried:
$$\sin x = x – \frac {x^3}{3!} + \frac {x^5}{5!} – \frac {x^7}{7!} + …$$
So, I changed $x$ to $x^3$ to get:
$$\sin(x^3) = x^3 – \frac {x^9}{3!} + \frac {x^{15}}{5!} – \frac {x^{21}}{7!} + …$$
But, I'm stuck when it comes to power of 1/3:
$$\sqrt[3]{x^3 – \frac {x^9}{3!} + \frac {x^{15}}{5!} – \frac {x^{21}}{7!} + …} = a_0+a_1x+a_2x^2+a_3x^3+…$$
Best Answer
If you factor out $x^3$ from cubic root you'll get $$ x\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} $$ Now, use power series expansion for cubic root $$ \sqrt[3]{1-x} = 1-\frac x3-\frac {x^2}9+o(x^2) \\ \sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} = 1-\frac 13\left(\frac{x^6}{3!}-\frac{x^{12}}{5!} \right)-\frac 19\left(\frac {x^{12}}{3!3!}+o(x^{12}) \right ) = \\ \\= 1-\frac {x^6}{3\cdot 3!}+\left ( \frac 1{3\cdot5!}-\frac 1{9\cdot 3! \cdot 3!}\right)x^{12}+o(x^{12}) = 1-\frac {x^6}{18}-\frac {x^{12}}{3240}+o(x^{12}) $$ After multiplying to $x$ you'll get your answer.