# Discuss the differentiability of $y=\sin(\pi(x-[x]))$

Discuss the differentiability of $y=\sin(\pi(x-[x]))$ in $(-\frac{\pi}{2},\frac{\pi}{2})$, where $[x]$ is the largest integer in x which is $\leq x$.

$$y = \left\{ \begin{array}{ll} \sin(\pi(x+2)), & \quad -\frac{\pi}{2}<x<-1 \\ \sin(\pi(x+1)), & \quad -1\leq x<0 \\ \sin(\pi x), &\;\;\; \quad 0\leq x<1 \\ \sin(\pi(x-1)), & \:\:\:\:\quad 1\leq x<\frac{\pi}{2} \end{array} \right.$$

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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Discuss..."), not a request for help, so please consider rewriting it. –  Zev Chonoles Dec 21 '12 at 6:38
@ZevChonoles : I copy this question from a question paper.There was the word "Discuss".You can see I tag it as homework. –  Argha Dec 21 '12 at 6:58
I can see that. But directly copying the question and waiting for people to answer it for you does not really display an appropriate amount of effort or politeness on your part. Even if all you can say is "I don't understand this question, can someone please help?", it will go a long way towards people wanting to help, and will let them know how to formulate their answers. –  Zev Chonoles Dec 21 '12 at 7:04
At any rate, I am glad that you have added to the question now. Just remember to add your own thoughts and work on the question in the future :) –  Zev Chonoles Dec 21 '12 at 7:16

Hint: Because $\sin (x+2k\pi)=\sin x$ and $\sin(x+2k\pi+\pi)=-\sin x$ ($k\in \mathbb{Z}$), your function is in fact: $$f(x)=\begin{cases}\sin(\pi x)&\mbox{if, }[x]\text{ is even}\\ -\sin (\pi x)&\mbox{if, }[x]\text{ is odd} \end{cases}$$ or since we are working in $(-\frac{\pi}{2},\frac{\pi}{2})$, $$f(x)=\begin{cases}\sin(\pi x)&\mbox{if, }0\le x<\frac{\pi}{2}\\ -\sin (\pi x)&\mbox{if, }-\frac{\pi}{2}< x<0 \end{cases}$$ I think you can finish the rest if you remember that $$\lim_{h\to 0}\frac{\sin ah}{h}=a$$

EDIT: Here is the rest:

Obviously $f$ is differentiable in $(-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2})$. We need to check differentiability at $0$ separately with the definition:

$$\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+}\frac{\sin(\pi h)}{h}=\pi$$ while $$\lim_{h\to 0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^-}\frac{-\sin(\pi h)}{h}=-\pi$$ and therefore $f$ is not differentiable at $0$ even though it is continuous there.

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So, here at $x=0$ , $f(x)$ is not differentiable. Am I right? –  Argha Dec 21 '12 at 7:10
Right! I will add this to my answer. –  Nameless Dec 21 '12 at 7:12
Thanks a lot. Now I got this. –  Argha Dec 21 '12 at 7:17
Nice, Nice answer +1 –  B. S. Dec 21 '12 at 7:28