In this article on Non-analytic smooth function seems to offer the following argument:

$|F^{(n)}(x_o)| \geq e^{-\sqrt{n}}n^n + O(q^n) \ $ implies $ \ \displaystyle \limsup_{n \rightarrow \infty} \left( \frac{|F^{(n)}(x_o)|}{n!} \right)^{1/n} = \infty$

where $q = 2^m$ for some positive integer $m$. My initial thought was the implication followed by comparison since $$ \limsup_{n \rightarrow \infty} \left(\frac{e^{-\sqrt{n}}n^n + O(q^n)}{n!}\right)^{1/n} = \infty \qquad (\text{false})$$ But, I believe the limit above actually converges. In particular, this example calculation in wolfram alpha and similar calculations show the $O(q^n)$-type term is irrelevant and the limit of $\left(e^{-\sqrt{n}}n^n/n! \right)^{1/n} \rightarrow e$ as $n \rightarrow \infty$. Since the limsup matches the limit when it exists we cannot conclude the limsup of $\left(\frac{e^{-\sqrt{n}}n^n + O(q^n)}{n!}\right)^{1/n}$ diverges. Therefore, there is no comparison to be made and I don't see how we are able to say anything about $\limsup_{n \rightarrow \infty} \left( \frac{|F^(n)(x_o)|}{n!} \right)^{1/n}$.

Question: is Wikipedia wrong here? Show me the error of my ways please.


1 Answer 1


That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows $$\limsup_{n \to \infty} \:\biggl(\frac{\lvert F^{(n)}(x_0)\rvert}{n!}\biggr)^{\frac{1}{n}} \geqslant e\,.$$

To fix it, replace $\cos (kx)$ with $\cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.

However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain

$$F^{(n)}(x_0) \geqslant e^{-n}n^{2n} - C\cdot q^n$$

and that yields

$$\biggl(\frac{\lvert F^{(n)}(x_0)\rvert}{n!}\biggr)^{\frac{1}{n}} \geqslant n\cdot \bigl(1 + o(1)\bigr)\,.$$

  • $\begingroup$ Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^{3n}$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks. $\endgroup$ Commented Jul 20, 2018 at 20:50
  • $\begingroup$ Every $r > 1$ works, the dominant term becomes $e^{-\sqrt{n}}n^{rn}$, and $$\biggl(\frac{n^{rn}}{e^{\sqrt{n}}n!}\biggr)^{1/n} \sim en^{r-1}\,.$$ $\endgroup$ Commented Jul 20, 2018 at 20:57
  • $\begingroup$ Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat. $\endgroup$ Commented Jul 20, 2018 at 20:58
  • $\begingroup$ I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer. $\endgroup$ Commented Jul 20, 2018 at 22:01
  • $\begingroup$ but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m \approxeq n^2$ for the argument? $\endgroup$ Commented Jul 21, 2018 at 1:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .