Pi appears a LOT in trigonometry, but only because of its 'circle-significance'. Does pi ever matter in things not concerned with circles? Is its only claim to fame the fact that its irrational and an important ratio?
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asked Aug 26, 2010 at 0:05
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2$\begingroup$ I'd rather this be a comment instead, so: $\pi$ turns up in the expression for the so-called "probability integral" (a.k.a. the "error function") among other things. How circles relate to this is a bit of a long-winded explanation though. $\endgroup$– J. M. ain't a mathematicianAug 26, 2010 at 0:11
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4$\begingroup$ Have you read en.wikipedia.org/wiki/Pi#Use_in_mathematics_and_science ? $\endgroup$– Qiaochu YuanAug 26, 2010 at 0:17
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17$\begingroup$ Also, let's get one thing straight here: circles are eerily important. You will never stop running into circles in mathematics. $\endgroup$– Qiaochu YuanAug 26, 2010 at 0:26
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5$\begingroup$ (For example, although the Fourier transform is "concerned with circles" (functions on the circle being the same thing as periodic functions) it penetrates into the deepest parts of modern mathematics. Many appearances of pi are because of a Fourier transform lurking somewhere in the background. You might also want to read this MO thread where I asked a similar question: mathoverflow.net/questions/18180/…) $\endgroup$– Qiaochu YuanAug 26, 2010 at 0:52
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4$\begingroup$ Fundamental source of $\pi$ is circle nothing else. It may be difficult to find it but it is always there. $\endgroup$– Pratik DeoghareAug 26, 2010 at 12:43
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5 Answers
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It is difficult to know if a circle is not lurking somewhere, whenever there is $\pi$, but the values of the Riemann zeta function at the positive even integers have a lot to do with powers of $\pi$: see here for the values.
For instance, you can prove that the probability that two "randomly chosen" positive integers are coprime is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.
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16$\begingroup$ You had my upvote at "it is difficult to know if a circle is not lurking somewhere..." $\endgroup$ Aug 26, 2010 at 0:22
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$\begingroup$ Maybe that has something to do with angles and the 2D lattice. $\endgroup$– asmeurerDec 20, 2012 at 20:03
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$\begingroup$ @asmeurer: Actually it has to do with the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. $\endgroup$ Jun 9, 2013 at 16:29
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$\begingroup$ @ShreevatsaR you're just restating what $\zeta(2)$ is. That doesn't explain where the $\pi$ comes from, fundamentally. $\endgroup$– asmeurerJun 9, 2013 at 19:18
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$\begingroup$ @asmeurer: True, but I'm saying it has nothing to do with "angles and the 2D lattice". Nothing I know of, anyway. The proof that $\zeta(2)$ has that value (see the linked article) doesn't use those, and the proof that the probability of two "randomly chosen" prime numbers is $\frac{1}{\zeta(2)}$ is straightforward. $\endgroup$ Jun 10, 2013 at 2:01
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$\pi$ appears in Stirling's approximation, which is not obviously related to circles. This means that $\pi$ appears in asymptotics related to binomial coefficients, such as
$$\displaystyle {2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}.$$
In other words, the probability of flipping exactly $n$ heads and $n$ tails after flipping a coin $2n$ times is about $\frac{1}{\sqrt{\pi n}}$. This asymptotic also suggests that on average you should flip between $n + \sqrt{\pi n}$ and $n - \sqrt{\pi n}$ heads.
answered Aug 26, 2010 at 0:22
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7$\begingroup$ This is closely related to J. Mangaldan's comment about the probability integral. Somehow I think it all ties back to the fact that e^{-x^2} is its own Fourier transform. $\endgroup$ Aug 26, 2010 at 0:31
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1$\begingroup$ I guess that comment is worth explaining: the relationship is that the constant in Stirling's approximation can be computed from the central limit theorem. This is explained at terrytao.wordpress.com/2010/01/02/… . $\endgroup$ Jan 14, 2011 at 0:53
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$\begingroup$ @QiaochuYuan You might be interested in Kunth's "Why Pi?" Lecture. He shows how this is related to cirlces! $\endgroup$– Pedro ♦Feb 27, 2012 at 5:19
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Because of the formula $e^{i\pi}+1=0$ you will find $\pi$ appearing in lots of places where it's not clear there is a circle, e.g in the normal distribution formulae.
answered Jan 13, 2011 at 15:56
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Yes, the ratio $\pi$ of a circle's circumference to its diameter shows up in many, many places where one might not expect it!
One partial explanation (similar in spirit to "circles lurk everywhere") is that the equation for a circle is a $quadratic$ (eg. $x^2+y^2 = r^2$.) After nice linear functions, the next most commonly used functions are quadratic functions and everywhere one runs into a quadratic function, a trig substitution (e.g. $x = r \cos \theta; y=r\sin \theta$) may be useful, turning the quadratic function into something involving $\pi.$ This explains the antiderivative $\int \frac{1}{1+x^2} dx$ involving $\pi$, the sum of reciprocals of squares $\sum^\infty\frac{1}{k^2}$ involving $\pi$ and the area under the Gaussian distribution involving $\pi$. And so on....
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1$\begingroup$ How does it explain the sums of reciprocals of squares involving pi? $\endgroup$ Jan 13, 2011 at 23:02
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1$\begingroup$ @George: there are a few elementary proofs of sum 1/k^2 = pi^2/6 where pi creeps in for reasons at least analogous to a trig substitution: math.stackexchange.com/questions/8337/… $\endgroup$ Jan 14, 2011 at 0:52
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$\begingroup$ @Qiaochu: Most of the proofs I know apply equally well to evaluating $\sum 1/n^d$ (for d even) and even $\sum (-1)^d/n^d$ (for d odd), which also involve $\pi$. So, the fact that the terms are squares doesn't seem particularly significant to the appearance of $\pi$. $\endgroup$ Jan 14, 2011 at 1:41
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$\begingroup$ I'll have a look through the alternative proofs in that link though. $\endgroup$ Jan 14, 2011 at 1:42
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$\begingroup$ (I meant $\sum(-1)^d/(2n+1)^d$ above). I always thought of these sums involving $\pi$ for similar reasons, and not just the $d=2$ case in isolation. $\endgroup$ Jan 14, 2011 at 1:56
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I am late to the party, but I found a curious appearance of the decimal expansion of $\pi$ in the lecture notes for the Dynamics and Relativity course by David Tong, in section 5.2.2.
We have a one-dimensional problem with two bouncing balls of mass $M$ and $m$ which can collide elastically. The leftmost ball has mass $M = 16 \times 100^Nm$, the smaller ball is on the right of the large ball. There is a perfect wall on the right of both balls (the wall does not move upon collision). $N$ is an arbitrary integer.
If a push to the right is given to the ball with mass $M$, it will move toward the smaller ball and upon collision, the smaller ball will start moving towards the wall, until it bounces back toward the larger ball. They will again collide, taking away some of the momentum of the large ball and allowing the smaller ball to move toward the wall again.
After a finite number of collisions, the large ball will start moving in the left direction. The number $p(N)$ of collisions of the balls $M$ and $m$ required to reach this situation is the first $N+1$ digits of $\pi$ (so $p(0) = 3$, $p(1) = 31$, $p(2) = 314$, $p(3) = 3141$, etc). (More precisely, it is given by $p(N) = \lfloor 10^N \pi \rfloor$).
For further details, I invite you to read the lecture notes by David Tong linked above. Basically there is a matrix recurrence relation for the velocities of the balls after each collision, and the solution to this recurrence equation involves the $\sin$ and $\cos$ functions.
