Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I admit that I'm probably out of my depth with this question, but I can't help but feel curious.

I wanted to show that, in the sequence $\{\sin(n)\}$, there is never a largest term (the sequence never attains its limit superior). My reasoning was that, given $$ \left| \sin(x) - \sin(y) \right| \leq \left| x - y \right|,$$ if we can furnish an even $M$ with $M = N\pi \pm \epsilon$, then either $\frac{3}{2}M$ or $\frac{1}{2}M$ will nearly equal an odd multiple $K$ of $\frac{\pi}{2}$ such that $\sin(K\frac{\pi}{2}) = 1$. By the inequality, the difference between $\sin(\frac{3}{2}M)$ and $1$ -- or $\sin(\frac{1}{2}M)$ and $1$, whichever -- would be at most $\frac{3}{2} \epsilon$. Thus, the problem reduces to showing that we can get an arbitrarily small $\epsilon$.

(I recognize that the inequality above is pretty watered down: the mean value theorem shows that the inequality is as stark as $\cos(\xi) \leq 1$ for $\xi \in (x,y)$, which, if both points $x$ and $y$ are close to $(2k+ \frac{1}{2}) \pi$, is really much stronger than what I've got. This seems like a hard way to prove the claim, so if anybody has a better one, I'd also like to hear about that.)

But my main question, which I came to because of the above, is about approximating multiples of $\pi$ by integers. If $\pi = \frac{p}{q} + \epsilon$, then $q\pi - q\epsilon = p$; thus, the size of $q$ becomes important to the accuracy, since the $\epsilon$ we were considering above is $q\epsilon$ in these terms.

Spivak's Calculus has a little discussion about this when he proves $e$ is transcendental. He notes that the proof of $e$'s irrationality shows that $\sum_{k=1}^n \frac{n!}{k!} = n!e - R_n,$ with $|R_n| < \frac{3}{n+1}$. The sum on the left can be controlled for parity, since choosing $n$ odd leaves $(\cdots + n + 1)$ at the tail of this sum, and all other terms multiplied by $(n-1)$. So there must exist, given $\epsilon > 0$, an $N$ and an even $M$ such that $M = Ne \pm \epsilon$.

(In other words, if $e$ were $\pi$, I'd be home already!)

Spivak mentioned that this property - good approximations existing with small denominators - is somehow characteristic of transcendental numbers.

"The number $e$ is by no means unique in this respect: generally speaking, the better a number can be approximated by rational numbers, the worse it is."

So I wonder:

  • Can we furnish an approximation $\frac{p}{q}$ to $\pi$ with $q\epsilon$ arbitrarily small? (For my purposes, can we do better and furnish one with an even $p$?)
  • More generally (and I am out of my depth here, but would enjoy references), what can we prove about the "goodness" of rational approximations to transcendental numbers, in the sense of small denominators?
share|improve this question
4  
    
As a fellow enthusiastic amateur, I highly recommend finding a copy of Exploring the Number Jungle: A Journey into Diophantine Analysis by Edward B. Burger. It is entertaining reading, and a gentle introduction to such problems. –  Byron Schmuland Feb 19 '13 at 22:49

3 Answers 3

In the spirit of experimental mathematics, I did a calculation of the quantity $|q\epsilon|$ for the first $100$ rational approximations for $\pi$ up to $10^{-n}$ accuracy, and it looks like it in fact decreases exponentially:

enter image description here

So if this trend continues, you can indeed approximate $\pi$ by large enough integers. Of course, this isn't a proof or full solution to your question, just a suggestion in the right direction.

share|improve this answer
    
How are you ordering them? By accuracy? –  Chris Feb 19 '13 at 22:35
    
Ordered by tolerance. –  nbubis Feb 19 '13 at 22:39

The following general result of Dirichlet in particular answers your question about approximations to $\pi$. Let $\alpha$ be irrational. Then there exist infinitely many (reduced) fractions $\frac{p}{q}$ such that $$\left|\alpha-\frac{p}{q}\right|\lt \frac{1}{q^2}.$$ For discussion of the general topic, and further links, please see this Wikipedia entry.

share|improve this answer

Here's an alternative approach to your original problem that doesn't use Diophantine approximation.

A topological result (cf. Kelley's Topology, page 58) says that an additive subgroup $G$ of the real line is either dense, or of the form $r\mathbb{Z}$ for some $r\in\mathbb{R}$. If we consider the group $G$ generated by $1$ and $2\pi$, we see that the purported $r$ would be both rational and irrational. Such $r$ doesn't exist so $G$ must be dense.

Letting $(m_n)$ and $(\ell_n)$ be integer sequences with $m_n+\ell_n\,2\pi\to\pi/2$, we see that $\sin(m_n)=\sin(m_n+\ell_n\,2\pi)\to\sin(\pi/2)=1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.