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This problem comes from John M Lee's Introduction to smooth manifold 2ed, problem 2-5(b).Let $\tilde{f}=f(x^{\frac{1}{3}})$. For necessity, observe that $f'(x)=\tilde{f}'(x^3)\cdot 3x^2$, $f''(x)=\tilde{f}''(x^3)\cdot9x^4+\tilde{f}'(x^3)\cdot6x$, $f'''(x)=\tilde{f}'''(x^3)\cdot9x^6+\tilde{f}''(x^3)\cdot54x^3+\tilde{f}'(x^3)\cdot6$. We can prove by induction that $f^{(3n+1)}(x)=x^2({a_{3n+1}}^{(0)}\tilde{f}^{(n+1)}(x^3)+\cdots {a_{3n+1}}^{(6n)}f^{(3n+1)}(x^3)x^{6n})$, and so on. And obtain that $f^{(3n+1)}=0, f^{(3n+2)}=0$ $\forall n$. The computation is tedious, is there a easier proof without using explicit formula of $f^{(n)}$?

For another direction, we can use L'Hospital's Rule to compute and prove that $\tilde{f}^{(n)}$ exists. For example, $\lim\limits_{x\to 0}\frac{\tilde{f}(x)-\tilde{f}(0)}{x-0}=\lim\limits_{x\to 0}\frac{f(x)-f(0)}{x^3}=\lim\limits_{x\to 0}\frac{f'(x)}{3x^2}=\lim\limits_{x\to 0}\frac{f''(x)}{6x}=\frac{f'''(0)}{6}$. When it comes to $\tilde{f}^{(n)}$, the computation is very hard. Although Taylor expansion suggests that we may have $\tilde{f}^{(n)}=f^{(3n)}(0)\frac{1}{(3n)!}$, we cannot use Taylor expansion to prove this. Since there exists a smooth function that has no convergent Taylor expansion in arbitrary small neighborhood around 0.

More generally, I believe that $f(x^{\frac{1}{k}})$ is smooth iff $f(x)$ is smooth and $f^{(n)}(0)=0$ when $k\nmid n$. But if we still use the method above, the computation will be too complicated. So I want to ask if there is a easier way to prove this.

Nzzz
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    Duplicate of https://math.stackexchange.com/questions/1957525/when-is-the-composite-with-cube-root-smooth . The generalization is straightforward ( for $k$ even the new smooth function is defined only on $[0, \infty)$, the proof is essentially the same) – orangeskid Oct 07 '17 at 15:34
  • The easy implication uses the following (well known?) fact: the Taylor series of the composition is the composition of the Taylor series. – orangeskid Oct 07 '17 at 15:59
  • I worry that Taylor expansion may fail for such functions like $f(x)=e^{(-1/x)}$ if $x\neq 0$, $f(0)=0$. This funciton is not equivalent to its Taylor series in arbitrary small neighborhood of 0, although it is smooth. If we want to compare two power series, we need to differentiate the series and let x=0 to compare the coefficients, but differentiation should be performed on a neighborhood of 0. – Nzzz Oct 07 '17 at 16:07
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    The Taylor series is always an asymptotic expansion, sometimes convergent. Asymptotic expansion is enough, you don't need convergence. – orangeskid Oct 07 '17 at 16:12
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    For example, if $\tilde{f}$ is smooth. It can be expanded as $\tilde{f}=\sum_{n=0}^{\infty}\frac{\tilde{f}^{(0)}}{n!}x^n$. Then $f(x)=\tilde{f}(x^3)=\sum_{n=0}^{\infty}\frac{\tilde{f}^{(0)}}{n!}x^{3n}$, but this equality may only hold for x=0. If we know $f(x)=\sum_{n=0}a_nx^n$ when $x=0$. Let x=0, we know $f(0)= a_0$. But if we want $f'(0)= a_1$ and so on, we need to differentiate both sides of the above equation. But this needs the equation holds for a neighborhood of 0. Since differentiation is a local property, not the pointwise property. – Nzzz Oct 07 '17 at 16:26
  • ref:https://en.wikipedia.org/wiki/Non-analytic_smooth_function – Nzzz Oct 07 '17 at 16:27
  • Have you heard of this ? https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula –  Oct 15 '17 at 17:30

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