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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2I'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. – J. M. ain't a mathematician Aug 26 '10 at 00:11
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4Have you read http://en.wikipedia.org/wiki/Pi#Use_in_mathematics_and_science ? – Qiaochu Yuan Aug 26 '10 at 00:17
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17Also, let's get one thing straight here: circles are eerily important. You will never stop running into circles in mathematics. – Qiaochu Yuan Aug 26 '10 at 00:26
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5(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: http://mathoverflow.net/questions/18180/what-are-some-fundamental-sources-for-the-appearance-of-pi-in-mathematics) – Qiaochu Yuan Aug 26 '10 at 00:52
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Qiaochu: good that you linked to it; I like gowers's "circles and rotations appear a lot." – J. M. ain't a mathematician Aug 26 '10 at 01:01
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4Fundamental source of $\pi$ is circle nothing else. It may be difficult to find it but it is always there. – Pratik Deoghare Aug 26 '10 at 12:43
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are any of these places simplified by replacing π with τ/2? http://tauday.com/ – endolith Mar 23 '11 at 19:59
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Since you can define $\pi$ to be the ratio of a circle's circumference to its diameter, you can always reverse-engineer some context in which $\pi$ appears to get something to do with circles. Proof: Suppose we have some interesting fact about $\pi$, and let $P$ be a totally self-contained proof of that fact. Then there must be a first statement $s$ in $P$ at which the number $\pi$ appears. Now suppose we have defined $\pi$ to be the ratio of a circle's circumference to its diameter. But then in order to introduce $\pi$ into – John Gowers Apr 23 '12 at 14:51
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(contd.) our proof we have to work straight from the definition of $\pi$ (since $P$ is self contained). So we have to introduce circles into our proof at some point before $p$. – John Gowers Apr 23 '12 at 14:53
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Hmmm.. I seem to have allowed for the possibility that there exists some statement about $\pi$ which is true but which is not provable in any order of logic, and which has nothing to do with circles. Now that is an interesting thought. – John Gowers Apr 23 '12 at 14:55
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On a related note: http://www.smbc-comics.com/?db=comics&id=2420#comic – John Gowers Apr 23 '12 at 18:51
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Often convergence includes PI. For example, Hardy's proof of the Dirichlet eta function – BryanH Oct 18 '12 at 17:15
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Sometimes it seems one will never stop running in circles either. :-) – ShreevatsaR Jun 09 '13 at 16:27
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http://numbers.computation.free.fr/Constants/Pi/piSeries.html – Sawarnik Oct 27 '13 at 14:19
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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 $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$.
damiano
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16You had my upvote at "it is difficult to know if a circle is not lurking somewhere..." – J. M. ain't a mathematician Aug 26 '10 at 00:22
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@asmeurer: Actually it has to do with the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. – ShreevatsaR Jun 09 '13 at 16:29
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@ShreevatsaR you're just restating what $\zeta(2)$ is. That doesn't explain where the $\pi$ comes from, fundamentally. – asmeurer Jun 09 '13 at 19:18
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@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. – ShreevatsaR Jun 10 '13 at 02:01
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@ShreevatsaR Yeah we know that the proof does not use circles to compute $\zeta(2)$. That is the point of damiano's comment "difficult to know if a circle is not lurking somewhere" – Ant Sep 23 '14 at 11:22
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@Ant: My comment was a reply to asmeurer's speculation that it had to do with "angles and the 2D lattice" -- my comment was not a reply to anything in this (damiano's) answer. – ShreevatsaR Sep 23 '14 at 16:40
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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.
Qiaochu Yuan
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7This 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. – Qiaochu Yuan Aug 26 '10 at 00:31
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1I 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 http://terrytao.wordpress.com/2010/01/02/254a-notes-0a-stirlings-formula/ . – Qiaochu Yuan Jan 14 '11 at 00:53
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@QiaochuYuan You might be interested in Kunth's "Why Pi?" Lecture. He shows how this is related to cirlces! – Pedro Feb 27 '12 at 05:19
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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.
John Smith
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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....
KenWSmith
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1How does it explain the sums of reciprocals of squares involving pi? – George Lowther Jan 13 '11 at 23:02
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1@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: http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2 – Qiaochu Yuan Jan 14 '11 at 00:52
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@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 $\pi$. – George Lowther Jan 14 '11 at 01:41
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I'll have a look through the alternative proofs in that link though. – George Lowther Jan 14 '11 at 01:42
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(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. – George Lowther Jan 14 '11 at 01:56
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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.
Tob Ernack
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