REA Question
REA Question
Let S = {tan(k) : k = 1,2,...}. Find the set of limit points of S on the real line.
The answer to this question is the whole real line could someone explain why this is true?
The answer to this question is the whole real line could someone explain why this is true?

 Posts: 44
 Joined: Tue Aug 09, 2011 6:18 pm
Re: REA Question
Here is my though process for solving the problem:
** The tangent function is periodic with some relation to pi. Like $$\tan(x)=\tan(x+2 \pi)$$ or something.
** To determine the points in S, you can just look at $$S' = \{ \tan( k \pmod{ 2 \pi}) : k \in \mathbb{N} \}$$
** Pi is irrational.
** The range of tan is the entire real line.
** The tangent function is periodic with some relation to pi. Like $$\tan(x)=\tan(x+2 \pi)$$ or something.
** To determine the points in S, you can just look at $$S' = \{ \tan( k \pmod{ 2 \pi}) : k \in \mathbb{N} \}$$
** Pi is irrational.
** The range of tan is the entire real line.

 Posts: 44
 Joined: Tue Aug 09, 2011 6:18 pm
Re: REA Question
Can you show that the set $$\{ n \pmod{2 \pi} : n \in \mathbb{Z} \}$$ is dense in the interval $$[0,2 \pi]$$ ?
Re: REA Question
Actually no? What does n mod(2\pi) mean?

 Posts: 44
 Joined: Tue Aug 09, 2011 6:18 pm
Re: REA Question
For each real numbers $x,y \in \mathbb{R} we define an equiv relation by
x~y if there exists an integer n such that x = 2*\pi n + y
Naturally, for every real number $x$, there is a unique number $y \in [0,2 \pi)$ such that $x~y$. We define $y = x mod 2 \pi$
If the set $$\{ n \pmod{ 2 \pi} : n \in \mathbb{Z} )\}$$ is finite, then there would be integers $n,m$ such that $n = m * (2\pi)$.....
x~y if there exists an integer n such that x = 2*\pi n + y
Naturally, for every real number $x$, there is a unique number $y \in [0,2 \pi)$ such that $x~y$. We define $y = x mod 2 \pi$
If the set $$\{ n \pmod{ 2 \pi} : n \in \mathbb{Z} )\}$$ is finite, then there would be integers $n,m$ such that $n = m * (2\pi)$.....
Re: REA Question
tan(k), k=1,2,3,.... can not be the entire real line since its countable. However, what seems more plausible is the fact that {tan(k)} is dense in R, so that all other points of R can be obtained from a sub seq of {tan(k)}. But proving denseness may be difficult.Topoltergeist wrote:For each real numbers $x,y \in \mathbb{R} we define an equiv relation by
x~y if there exists an integer n such that x = 2*\pi n + y
Naturally, for every real number $x$, there is a unique number $y \in [0,2 \pi)$ such that $x~y$. We define $y = x mod 2 \pi$
If the set $$\{ n \pmod{ 2 \pi} : n \in \mathbb{Z} )\}$$ is finite, then there would be integers $n,m$ such that $n = m * (2\pi)$.....
Re: REA Question
someone proved denseness of tan(k) here http://math.stackexchange.com/questions ... inmathbbr