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I was working on a an old worksheet problem here. It asks

Let $S=\{\tan(k):k=1,2,\dots\}$. Find the set of limit points of $S$ on the real line.

The answer is $(-\infty,\infty)$. Intuitively I feel that if we keep evaluating tangent at positive integer points, they will be so scattered over the real line that we could always construct some subsequence converging to any real number. How can this be made rigorous to get this purported conclusion? Thanks.

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I don't have the energy to type out a full answer, but I'd start by thinking "what would it mean if the conclusion was false?" and then trying to show why that doesn't happen (roughly speaking, trying to show that there aren't any points that are "missed") –  Ben Millwood Sep 17 '12 at 0:29
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3 Answers

up vote 6 down vote accepted

For each real number $x$ there is a unique integer $n_x$ such that $n_x\pi\le x<(n_x+1)\pi$; let $\hat x=x-n_x\pi\in[0,\pi)$, and observe that $\hat x$ is the unique element of $[0,\pi)$ such that $\tan\hat x=\tan x$. Thus, $\{\tan k:k\in\Bbb N\}=\{\tan\hat k:k\in\Bbb N\}$. Let $D=\{\hat k:k\in\Bbb N\}$. It suffices to show that $D$ is dense in $[0,\pi)$: the tangent function is continuous and maps $[0,\pi)$ onto $\Bbb R$, so $\tan[D]=\{\tan\hat k:k\in\Bbb N\}$ must then be dense in $\Bbb R$.

Note that for any $x,y\in\Bbb R$, $\hat x=\hat y$ iff $\frac{x}{\pi}-\frac{y}{\pi}\in\Bbb Z$. Thus, instead of showing that $D$ is dense in $[0,\pi)$, we can scale everything by a factor of $1/\pi$ and show that $D_0=\{\hat k/\pi:k\in\Bbb N\}$ is dense in $[0,1)$.

This is a nice application of the pigeonhole principle. Let $n$ be a positive integer, and divide $[0,1)$ into the $n$ subintervals $\left[\frac{k}n,\frac{k+1}n\right)$ for $k=0,\dots,n-1$. Two of the $n+1$ numbers $\frac{\hat k}{\pi}$ for $k=0,\dots,n$ must belong to the same one of these subintervals; say $$\frac{\hat k}{\pi},\frac{\hat\ell}{\pi}\in\left[\frac{i}n,\frac{i+1}n\right)\;,$$ where $0\le k<\ell\le n$ and $0\le i<n$. Then $$0<\left|\frac{\hat\ell}{\pi}-\frac{\hat k}{\pi}\right|<\frac1n\;.$$ Let $m=\ell-k$; then $\dfrac{\hat m}{\pi}\in\left[0,\dfrac1n\right)$ if $\hat\ell-\hat k>0$, and $\dfrac{\hat m}{\pi}\in\left[1-\dfrac1n,1\right)$ if $\hat\ell-\hat k<0$.

In the first case let $N$ be the smallest positive integer such that $\dfrac{N\hat m}{\pi}>1$, and in the second let $N$ be the smallest positive integer such that $N\left(1-\dfrac{\hat m}{\pi}\right)>1$. Then every point of $[0,1)$ is within $1/n$ of one of the multiples $\dfrac{\widehat{jm}}{\pi}$ for $j=1,\dots,N-1$. Thus, every $x\in[0,1)$ is within $1/n$ of some element of $D_0$, and since $n$ was arbitrary, $D_0$ is dense in $[0,1)$.

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Thanks for such a clear answer! –  Nastassja Sep 17 '12 at 1:54
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Very clear. Thanks. –  Bombyx mori Sep 17 '12 at 3:50
    
Why does each of the $n$ subintervals contain exactly one of the multiples $\dfrac{\widehat{jm}}{\pi}$ for $j=1,\dots,n$? I think this would only be the case if $\dfrac{\hat m}{\pi}\in\left[\dfrac{n-1}n\dfrac1n,\dfrac1n\right)$. Generally, you do get multiples in each of the subintervals (since $m$ is too small to skip one of the subintervals), but you need to take higher multiples than $j=n$, no? –  joriki Sep 19 '12 at 9:22
    
@joriki: You’re right: I tried to make it simpler than it actually is. –  Brian M. Scott Sep 19 '12 at 9:33
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For convenience's sake we use $2\pi$ instead of $\pi$. You need to prove the images of the $a\rightarrow a\pmod{2\pi}, a\in \mathbb{N}$ is equidistributed. This can be done by Weyl's criterion that modified with $\pmod{2\pi}$ instead of $\pmod{1}$. We need to prove that $$ \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n-1}_{0}e^{2ilx_{j}}=0, \forall l\in \mathbb{Z} $$ with $x_{j}$ be $j$'s image by the quotient map. The summation is a geometric series with $e^{2il}, e^{2i2l},e^{2i3l}$, etc. So we have $$\sum^{n-1}_{0}e^{2ilx_{j}}=\frac{1-e^{2iln}}{1-e^{2il}}$$ which can be bounded by $$\left|\frac{1-e^{2iln}}{1-e^{2il}}\right|\le \frac{2}{|1-e^{2il}|}$$ which is finite since $l$ is fixed. Therefore the limit $$ \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n-1}_{0}e^{2ilx_{j}}=0, \forall l\in \mathbb{Z} $$must be 0, and the original sequence be equidisributed.

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+1. Note that You need to prove the images... are a dense set (equidistribution is a bonus here). –  Did Sep 17 '12 at 6:09
    
It's not clear to me what you're referring to as "the limit". It seems that the first "the limit" is meant to refer to the limit of the part without $1/n$? In that case, it's wrong, since that part doesn't converge by itself; what you wrote is the analytic continuation of the limit to the unit circle (and the numerator should be $1$ because the sum starts at $j=0$). You need to use the formula for the partial sums of a gemoetric series and bound its absolute value to show that the limit of the full expression including $1/n$ is $0$. –  joriki Sep 17 '12 at 8:22
    
Updated. Thanks for pointing out. –  Bombyx mori Sep 17 '12 at 17:37
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The function $\tan(x)$ is continuous and $2\pi$-periodic. Suppose I want to find a sequence that converges to the real number $y$. Let $y = \tan(x)$ for some real number $x$. Well, we know that the multiples of an irrational number are equidistributed modulo $1$. Thus, we pick some natural number $N_1$ so that $x + 2N_1\pi$ is really close to $0$ modulo $1$. This means that $x + 2N_1\pi$ is really close to some integer, say $M_1$. Then $\tan(x+2N_1\pi) = \tan(x) = y$ is very close to $\tan(M_1)$ by continuity, so $\tan(M_1)$ is really close to $y$. To get the next integer $M_2$, pick $N_1$ so that $x + 2N_2\pi$ is even closer to $0$ modulo $1$, which is to say that $M_2$ is even closer to $x + 2N_2\pi$ then $M_1$ was close to $x + 2N_1\pi$. By continuity, this means the approximation $\tan(M_2)$ should be even close to $y$ than $\tan(M_1)$. Repeat.

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Incidentally, $\tan(x)$ is $\pi$-periodic. –  Sasha Sep 17 '12 at 1:36
    
@Sasha: yeah, I guess my proof cause a misleading effect. –  Bombyx mori Sep 17 '12 at 3:19
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