# For what $x\in[0,1]$ is $y = \sum\limits_{k = 1}^\infty\frac{\sin( k!^2x )}{k!}$ differentiable?

For what $x\in[0,1]$ will the function $y = \sum\limits_{k = 1}^\infty\frac{\sin( k!^2x )}{k!}$ be differentiable? How do you know?

Here is the equation expressed more clearly on Wolfram Alpha. The only difference is that 10 should be Infinity (Wolfram apparently can't handle that yet).

I'm trying to understand for what $x\in[0,1]$ this function is differentiable. I've used a computer to plot the graph of $y'$ (the derivative of the function) with the upper limit (top number of sigma) as 10, and then with the upper limit as 11, 12... it looks like these "zig-zags" continue to exist as you "go deeper" into the function.

...so I'm thinking the values of $x\in[0,1]$ that make the function differentiable are all of them... But is my line of thinking correct? How can I validate?

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On the contrary, it looks to me as if the sum is nowhere differentiable. –  Henning Makholm Sep 24 '11 at 5:49
Have you tried taking the derivative of the sum? –  JSchlather Sep 24 '11 at 6:05
@Jacob, that is the goal. Now, the sum of the derivatives would be something (but that diverges wildly at every rational multiple of $\pi$, which probably ought to tell us something). –  Henning Makholm Sep 24 '11 at 6:10
Can you answer the question at ONE value of $x$ to start with? Say, the obvious first choice, $x=0$. And, there is no reason to restrict $x$ to $[0,1]$, the function is perfectly well-defined for all $x$, and periodic with period $2\pi$. –  GEdgar Sep 24 '11 at 13:54
@daze, it is nowhere differentiable because for every $x_0$, the difference quotient $\frac{f(x)-f(x_0)}{x-x_0}$ can be made arbitrarily large for $x$ arbitrarily close to $x_0$. As you observe yourself the "zig-zags continue to exist as you go deeper into the function". That means it can't be differentiable, because differentiable means that the function must begin to look linear as you zoom closer in. The sum of the derivatives is $\sum_k k!\cos(k!^2x)$ which fails spectacularly to converge, so I don't see how you could even graph it. –  Henning Makholm Oct 7 '11 at 12:13

Thanks to great comments from people like Henning Makholm, I have an answer, and I decided to summarize it here. (Please edit this if I have any inaccuracies.)

First of all, I was mistaken in my supposed graphing of the derivative. The derivative of y equates to the sum of the derivatives, namely $y' = \sum\limits_{k = 1}^\infty\ k!cos(k!^2x)$

This sum diverges, meaning it should be impossible to graph the derivative. We know the sum diverges because cos oscillates between -1 and 1.

As Henning Makholm says, "[The graph of y] is nowhere differentiable because for every $x_0$, the difference quotient $\frac{\ f(x)−f(x_0)}{x−x_0}$ can be made arbitrarily large for $x$ arbitrarily close to $x_0$." This means that we have points of non-differentiability at every point in the graph. Also, if we examine the graph of y with increasing magnification, we see that the "zig zags" continue to appear, and this further confirms the non-differentiability of all points.

Therefore, for no $x\in[0,1]$ is the function y differentiable.

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A minor addition: I think it is possible to construct special values of $x$ such that $\sum_k k!\cos(k!^2 x)$ converges -- but it diverges for a dense set of $x$'s, namely at least all rational multiples of $\pi$: If $x=p/q$, then $k!^2x$ is a multiple of $2\pi$ for all $k>q+1$, so from that point the cosine is always $1$, so the remaining terms of the series are all just $k!$ and clearly diverge. (Note that $f$ does not have to be differentiable just because the sum of derivatives converges, at least not as long as it doesn't converge in an open interval). –  Henning Makholm Oct 8 '11 at 0:29
Also, thanks for making sure the question will not appear unanswered in the question list. –  Henning Makholm Oct 8 '11 at 0:30