REA Question

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marco
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Joined: Mon Jun 11, 2012 2:27 am

REA Question

Post by marco » Wed Oct 10, 2012 12:34 pm

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?

Topoltergeist
Posts: 44
Joined: Tue Aug 09, 2011 6:18 pm

Re: REA Question

Post by Topoltergeist » Wed Oct 10, 2012 10:13 pm

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.

marco
Posts: 263
Joined: Mon Jun 11, 2012 2:27 am

Re: REA Question

Post by marco » Thu Nov 01, 2012 7:49 pm

Sorry but I don't follow :oops: .

Topoltergeist
Posts: 44
Joined: Tue Aug 09, 2011 6:18 pm

Re: REA Question

Post by Topoltergeist » Fri Nov 09, 2012 11:26 am

Can you show that the set $$\{ n \pmod{2 \pi} : n \in \mathbb{Z} \}$$ is dense in the interval $$[0,2 \pi]$$ ?

marco
Posts: 263
Joined: Mon Jun 11, 2012 2:27 am

Re: REA Question

Post by marco » Fri Nov 09, 2012 1:14 pm

Actually no? What does n mod(2\pi) mean?

Topoltergeist
Posts: 44
Joined: Tue Aug 09, 2011 6:18 pm

Re: REA Question

Post by Topoltergeist » Fri Nov 09, 2012 6:22 pm

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 $n = m * (2\pi)$.....

inferno
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Joined: Sat Nov 10, 2012 3:10 pm

Re: REA Question

Post by inferno » Sat Nov 10, 2012 3:48 pm

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 $n = m * (2\pi)$.....
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.

inferno
Posts: 7
Joined: Sat Nov 10, 2012 3:10 pm

Re: REA Question

Post by inferno » Sat Nov 10, 2012 4:26 pm

someone proved denseness of tan(k) here http://math.stackexchange.com/questions ... in-mathbbr



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