Let $f$ be a smooth function. Suppose that $F(x)=f(x^{\frac{1}{3}})$. Show that $F$ is smooth if and only if $f^{n}(0)=0$ for all $n \ne 3k$, where $n,k \in \mathbb{N}$.
Solution: ($\implies$)
Since $F$ is smooth, differentiating once gives $$F'(x)=f'(x^{\frac{1}{3}})\frac{1}{3}x^{-\frac{2}{3}}$$which is same as $$3x^{\frac{2}{3}}F'(x)=f'(x^{\frac{1}{3}})$$
Taking limit as $x \to 0$ on both sides we get $f'(0)=0$. Now differentiating the equation again and multiplying through out by $3x^{\frac{2}{3}}$ we get $$6x^{\frac{1}{3}}F'(x)+9x^{\frac{4}{3}}F''(x)=f''(x^{\frac{1}{3}})$$
Now taking the limit as $x \to 0$ on both sides we get $f''(0)=0$. Differentiating again we get $$ 54xF''(x)+27x^2F'''(x)+6F'(x)=f'''(x^{\frac{1}{3}})$$ Taking limit as $x \to 0$ on both sides we get $F'(0)=\frac{f'''(0)}{6}$, which is as expected. Now the question is How do I get from here?? I can always differentiate and go further but for how many times?? The usual method in this cases is to use induction and conclude. Now I am unable to find any particular relation which can be exploited in the induction hypothesis.
$(\impliedby)$
Now suppose that $f^{n}(0)=0$ for all $n \ne 3k$. The first thing that comes to my mind is L'Hospital's rule. So I start with it. The only point which needs to be checked at is $0$. Then for $x \ne 0$, we have $$F'(x)=f'(x^{\frac{1}{3}})\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3}\frac{f'(x^{\frac{1}{3}})}{x^{\frac{2}{3}}}$$Thus, $$\lim_{x \to 0}F'(x)=\lim_{x \to 0}\frac{f''(x^{\frac{1}{3}})x^{-\frac{2}{3}}}{6x^{-\frac{1}{3}}}=\lim_{x \to 0}\frac{f''(x^{\frac{1}{3}})}{6x^{\frac{1}{3}}}=\frac{f'''(0)}{6}$$ Hence $F'(x)$ is continuous at $x=0$. I can always apply L'Hospital rule again to establish the continuity of $F''(x)$ and so on. I don't see an obvious way to extend or generalize this thing to higher order derivatives. The other thing that comes to my mind is the Taylor's series expansion which works very well for $F'(x)$ but no further. How do I do that??
The other thing I can do is look at the Taylor Series Expansion of $f$ near $0$. Then $$f(x)=\sum_{k=0}^n\frac{x^{3k}}{(3k)!}f^{3k}(0)+R_{n+1}(x)$$ So $$F(x)=\sum_{k=0}^n\frac{x^{k}}{(3k)!}f^{3k}(0)+R_{n+1}(x)$$
Can I do term by term differentiation to conclude?? Thanks for the help!!