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Find $A$ such that: $f(x)=\lfloor{x}\rfloor \cos(Ax)$ and $ x \in \mathbb{R}$ is continuous on $\mathbb{R.}$

We have to ensure continuity for integers. I do not think that $A$ exists.

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We need "$f(n)\approx f(n-\epsilon)$", i.e. $n\cos(nA)= (n-1)\cos(nA)$ for all $n\in\mathbb Z$, hence $\cos(nA)=0$ for all $n$, thus $nA$ is an odd multiple of $\frac\pi2$ for all $n$. Especially, both $A$ and $2A$ must be such odd multiples, $\frac{2A}\pi$ and $2\cdot \frac{2A}\pi$ are odd integers, contradiction.

D'oh, even more clearly $$\lim_{x\to0^-} f(x)=-1\ne 0=f(0).$$

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Thanks for the heads up and lol @ how easy it is. – Git Gud Jan 16 '13 at 20:22
Yeah, all that complicated thinking better suits for $\lfloor x\rfloor \sin A x$ (and there you do find such $A$). – Hagen von Eitzen Jan 16 '13 at 20:27
@Hagen von Eitzen I do not understand your method – aiki93 Jan 16 '13 at 21:18
@aiki93 For $-1\le x<0$, $\lfloor x\rfloor =-1$ and of course $\cos(Ax)\to 1$ as $x\to 0$. – Hagen von Eitzen Jan 16 '13 at 21:26
ok, thanks you. Could you tell me what happens for sin instead cos? – aiki93 Jan 16 '13 at 22:27

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