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My 7-year-old son was staring at the graph of tan() and its endlessly-repeating serpentine strokes on the number line between multiples of $\pi$ and he asked me the question in the title. More precisely, is the following true or false?

For any $\epsilon > 0$, there exists some $N \in \mathbb{Z}^+$ such that $|\tan(N)-\lfloor \tan(N) \rceil| < \epsilon$.

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Take $N = 0$. :-) –  Eric O. Korman Feb 27 '11 at 16:31
    
I'd bet for a much stronger property: that we can make it arbitrarily close to ANY integer. –  leonbloy Feb 27 '11 at 16:41
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Your 7-year-old son asked you that precise question?? Wow! –  lhf Feb 27 '11 at 18:10
    
Edited to disqualify $N=0$ :). Also, yes, my son asked that question although he didn't formalize it (I did that) and it was part of a discussion, not something out of the blue. –  Fixee Feb 27 '11 at 18:41
    
Let me know when your son is ready to apply for college! :) –  aaron Feb 28 '11 at 1:59
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2 Answers 2

up vote 27 down vote accepted

Yes, you can. In fact, you can make $\tan(N)$ arbitrarily close to your favorite integer.

Pick your favorite integer, which is probably 17. Consider $x=\arctan 17$. Then $\tan(x)=17$ and also $\tan(x+n\pi)=17$ for all integers $n$. If you could make $x+n\pi$ very close to an integer (say $K$), then since the tangent function is continuous, $\tan(K)$ will be very close to 17.

So the problem is now to make $x+n\pi$ very close to an integer. This can be done because the multiples of $\pi$ (or any irrational number) are dense mod 1, so you can make $n\pi$ arbitrarily close to $-x$ mod 1. This takes proof but it is elementary: think about a circle of circumference 1, and start walking around the circle, around and around, marking a point each time your total distance travelled is a multiple of $\pi$. You'll see the marks start to fill in the circle, and as you go on and on, the largest interval without a mark in it keeps getting smaller. Eventually you will mark a point that's within $\epsilon$ of $-x$.

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(irony on) Usually, when I think about a circle it has circumference $2\pi$ and I always keep hitting the same two spots (irony off) ;-) –  Fabian Feb 27 '11 at 17:23
    
Do you mean the "largest interval without a mark"? –  Fixee Feb 27 '11 at 17:54
    
Oh, yes. Thanks! I edited it. (Although the smallest keeps getting smaller too ... :) –  aaron Feb 27 '11 at 18:09
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So for example $\tan(212) \approx 17.37\ldots$ but $\tan(89650 ) \approx 16.999\ldots$. –  Henry Feb 27 '11 at 23:05
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Fairly random (and definitely unrelated) web surfing turned up this nice short paper of Cheng and Zheng which proves in a very constructive way that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is any continuous function which is periodic with irrational period, then $f(\mathbb{N})$ is dense in $f(\mathbb{R})$.

The ideas are first exhibited with respect to the function $f(x) = \sin x$.

Your question is not literally a special case of this, since $\tan x$ is not continuous on all of $\mathbb{R}$. Nevertheless I think it is close enough for the same methods to be carried over. (For instance, $\tan x$ can be viewed as a continuous function with values in $\mathbb{R} \mathbb{P}^1 \cong S^1$.) In any case the relation here is close enough so that I thought the paper would be of interest to readers of this question.

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