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This function is continuous, it follows by M-Weierstrass Test. But proving non-differentiability, I think it's too hard. Does someone know how can I prove this? Or at least have a paper with the proof? The function is $$ f(x) = \sum_{k=1}^\infty \frac{\sin((k + 1)!\;x )}{k!}$$ Thanks!

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I wonder if it can be proven using induction and the fundamental theorem of calculus. – Ahmed Masud Oct 31 '11 at 19:44
See also this question -- not an exact duplicate but with much the same properties. – Henning Makholm Oct 31 '11 at 19:57

(Edited: handwaving replaced by rigor)

For conciseness, define the helper functions $\gamma_k(x)=\sin((k+1)!x)$. Then $f(x)=\sum_k \frac{\gamma_k(x)}{k!}$.

Fix an arbitrary $x\in\mathbb R$. We will construct a sequence $(x_n)_n$ such that $$\lim_{n\to\infty} x_n = x \quad\land\quad \lim_{n\to\infty} \left|\frac{f(x_n)-f(x)}{x_n-x}\right| = \infty$$ Such a sequence will directly imply that $f$ is not differentiable at $x$.

Let $x'_n$ be the largest number less than $x$ such that $|\gamma_n(x'_n)-\gamma_n(x)|=1$. Let $x''_n$ be the smallest number larger than $x$ such that $\gamma_n(x''_n)=\gamma_n(x'_n)$. One of these, to be determined later, will become our $x_n$.

No matter which of these two choices of $x_n$ we have $|x_n-x|<\frac{2\pi}{(n+1)!}$ so $\lim x_n=x$.

To estimate the difference quotient, write $$f(x) = \underbrace{\sum_{k=1}^{n-1}\frac{\gamma_k(x)}{k!}}_{p(x)}+ \underbrace{\frac{\gamma_n(x)}{n!}}_{q(x)}+ \underbrace{\sum_{k=n+1}^{\infty} \frac{\gamma_k(x)}{k!}}_{r(x)}$$ and so, $$\underbrace{f(x_n)-f(x)}_{\Delta f} = \underbrace{p(x_n)-p(x)}_{\Delta p} + \underbrace{q(x_n)-q(x)}_{\Delta q} + \underbrace{r(x_n)-r(x)}_{\Delta r}$$ Of these, by construction of $x_n$ we have $|\Delta q| = \frac{1}{n!}$.

Also, $r(x)$ is globally bounded by the remainder term in the series $\sum 1/n! = e$, which by Taylor's theorem is at most $\frac{e}{(n+1)!}$. So $|\Delta r| \le \frac{2e}{(n+1)!}$.

$\Delta p$ is not dealt with as easily. In some cases it may be numerically larger than $\Delta q$, ruining a simple triange-equality based estimate. But it can be tamed by a case analysis:

  • If $p$ is strictly monotonic on $[x'_n, x''_n]$, then $p(x'_n)-p(x)$ and $p(x''_n)-p(x)$ will have opposite signs. Since $q(x'_n)=q(x''_n)$, we can choose $x_n$ such that $\Delta p$ and $\Delta q$ has the same sign. Therefore $|\Delta p+\Delta q|\ge|\Delta q|=\frac{1}{n!}$.

  • Otherwise, $p$ has an extremum between $x'_n$ and $x''_n$; select $x_n$ such that the extremum is between $x$ and $x_n$. Because $p$ is a finite sum of $C^\infty$ functions, we can bound its second derivative separately for each of its terms: $$\forall t: |p''(t)| \le \sum_{k=1}^{n-1}\left|\frac{\gamma''_k(t)}{k!}\right| \le \sum_{k=1}^{n-1}\frac{(k+1)!^2}{k!} \le \sum_{k=1}^{n-1} (k+1)!(k+1) \le 2n!n $$ Therefore the maximal variation of $p$ in an interval of length $\le\frac{2\pi}{(n+1)!}$ that contains a stationary point must be $\left(\frac{2\pi}{(n+1)!}\right)^2 2n!n = \frac{8\pi^2n}{(n+1)^2}\frac{1}{n!}$. The $\frac{8\pi^2n}{(n+1)^2}$ factor is less than $1/2$ for $n>16\pi^2$, so for large enough $n$ we have $|\Delta p+\Delta q|\ge \frac{1}{2n!}$.

Thus, for large $n$ we always have $$|\Delta f| \ge \frac{1}{2n!} - \frac{2e}{(n+1)!} = \frac{1}{n!}\left(\frac{1}{2}-\frac{2e}{n+1}\right)$$ and therefore $$\left|\frac{f(x_n)-f(x)}{x_n-x}\right| \ge \frac{(n+1)!}{2\pi}|\Delta f| \ge \frac{n+1}{2\pi}\left(\frac{1}{2}-\frac{2e}{n+1}\right) = \frac{n+1}{4\pi}-\frac{e}{\pi} \to \infty$$ as promised.

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this site should have a +10 arrow :) Nice argument! Thanks for sharing this. – Leandro Nov 1 '11 at 0:24
@Henning Makholm Sorry for my stupid question Henning, but I only see that $ \Delta p $ it´s bigger than $ \frac{1} {{n!}}\left( {1 - \frac{{n^2 }} {{n + 1}} - \frac{{2e}} {{n + 1}}} \right) $ but the factor $ \left( {1 - \frac{{n^2 }} {{n + 1}} - \frac{{2e}} {{n + 1}}} \right) $ it´s negative. The other step that i don´t understand it´s why also it´s bigger than $$ \frac{1} {{n!}}\left( {1 - \frac{{2e}} {{n + 1}}} \right) $$ since $$ \left( {1 - \frac{{2e}} {{n + 1}}} \right) > \left( {1 - \frac{{2e}} {{n + 1}} - \frac{{n^2 }} {{n + 1}}} \right) $$ – Susuk Nov 1 '11 at 2:58
@Susuk, that's what the handwaving at the end was for. I will edit the proof to make it more explicit how this part works. – Henning Makholm Nov 1 '11 at 12:27
@Susuk, now de-handwaved. That was harder than I had expected ... – Henning Makholm Nov 1 '11 at 13:53
@HenningMakholm Nice answer! Thanks! – Susuk Nov 1 '11 at 16:38

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