I need to find Maclaurin series for $(\sin(x^3))^{1/3} $
My attempt:
I found first derivative. $f '(x) = \frac{(x^2 \cdot \cos x^3)}{(\sin x^3)^{2/3}}$. Hence, $f'(0)$ does not exists and there is no Maclaurin series for the function above. Nevertheless, this is incorrect answer.
By the way, I saw the same question on this forum, but the answer doesn't solve my problem
Best Answer
In order to find the Maclaurin series of $(\sin(x^3))^{1/3}$ at $x=0$ up to to $x^{13}$, it is better not to compute it by using the definition. Instead, I suggest to start with the series of $\sin(x^3)$,
$$\sin(x^3)=x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}+O(x^{21}).$$ It follows that $$(\sin(x^3))^{1/3}=\left(x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}+O(x^{21})\right)^{1/3}= x\left(1-\frac{x^6}{3!}+\frac{x^{12}}{5!}+O(x^{18})\right)^{1/3}.$$ Now use the expansion $(1+t)^{1/3}=1+\frac{t}{3}-\frac{t^2}{9}+O(t^3)$ and finish the job.
P.S. It turns out that the first derivative at $0$ exists and $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{(\sin(x^3))^{1/3}}{x}=1.$$