Difference between revisions of "Pi"

Revision as of 12:21, 9 March 2022

Pi is an irrational number (in fact, transcendental number, as proved by Lindemann in 1882) denoted by the Greek letter $\pi$ .

Definition

Pi is the ratio of the circumference (perimeter) of a given circle to its diameter. It is approximately equal to 3.141592653. The number pi is one of the most important constants in all of mathematics and appears in some of the most surprising places, such as in the sum $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ . Some common fractional approximations for pi are $\frac{22}{7} \approx 3.14285$ and $\frac{355}{113} \approx 3.1415929$ .

The number pi often shows up in problems in number theory, particularly algebraic number theory. For example, many class number formulae involve pi.

Approximating pi

$\pi$ can be calculated in several ways, and it can also be approximated. One way to approximate $\pi$ is to inscribe a unit circle in a square of side length 2. Using a computer, random points are placed inside the square. Because the area of the circle is $\pi$ , and the area of the square is 4, the ratio of the amount of points inside the circle to the total number of points approximates $\frac{\pi}{4}$ . This can simply be multiplied by 4 to approximate $\pi$ .

One formula for pi is $4\left( \sum_{i = 0}^\infty (-1)^i \left(\frac{1}{2n+1}\right)\right)$ . This can be computed to the desired degree of accuracy.

Other interesting properties

Letting $\theta = \pi$ in Euler's identity gives $e^{\pi i} + 1 = 0$ , which is considered to be one of the most beautiful results in mathematics since it involves five of the greatest constants: e, pi, i, 1, and 0.

Proof that $\pi$ is not Liouvillian

In this section we'll outline the Beukers proof that $\pi$ is not Liouvillian, i.e., that $\left|\pi-\frac pq\right|\ge \frac 1{q^M}$ for large $q$ with some fixed positive $M$ . The best known value of $M$ is currently (June, 1993) $M=8.02$ and is due to Hata. This article can be read right after its parent article rational approximation of famous numbers, but an unexperienced reader may prefer to read the proof that ln 2 is not Liouvillian first.

Beukers' Proof

Preliminaries

Note that if $x$ is Liouvillian, then so is any rational multiple of any positive integer power of $x$ (this is a simple exercise we leave to the reader), so, it will suffice to show that $\frac{\pi^2}6$ is not Liouvillian.

Euler's identity

$\frac{\pi^2}6=\sum_{m=1}^{\infty}\frac 1{m^2}$

The Beukers integral

Consider the integrals $I(k,\ell):=\iint_{(0,1)\times(0,1)}\frac {x^ky^\ell}{1-xy}\,dx\,dy$ . Expanding $\frac 1{1-xy}=\sum_{j=0}^\infty x^jy^j$ and integrating term by term, we get $I(k,\ell)=\sum_{j=0}^{\infty}\frac1{(k+j+1)(\ell+j+1)}.$

If $k=\ell$ , we get the sum of all inverse squares starting with $\frac 1{(k+1)^2}$ , so in this case,

$I(k,\ell)=\frac {\pi^2}{6}-\frac 1{1^2}-\frac 1{2^2}-\dots-\frac 1{k^2}.$

On the other hand, if $k>\ell$ , we get

$I(k,\ell)=\frac 1{k-\ell}\sum_{j=0}^\infty\left[\frac 1{\ell+j+1}-\frac {1}{k+j+1}\right]=\frac 1{k-\ell}\left[\frac 1{\ell+1}+\frac 1{\ell+2}+\dots+\frac 1k\right]$

Similarly, if $\ell>k$ , we get

$I(k,\ell)=\frac 1{\ell-k}\left[\frac 1{k+1}+\frac 1{k+2}+\dots+\frac 1\ell\right]$

Let now $D_n$ be the least common multiple of $1,2,\dots,n$ . It follows immediately from the above formulae for $I(k,\ell)$ that for every polynomial $R(x,y)$ with integer coefficients whose degree in both $x$ and $y$ does not exceed $n$ , we have

$D_n^2\iint_{(0,1)\times(0,1)}\frac {R(x,y)}{1-xy}\,dx\,dy=P_n-Q_n\frac{\pi^2}{6}$ where $P_n,Q_n\in \mathbb Z$ .

The trick is to choose a polynomial $R(x,y)$ that makes the integral very small. We put

$R(x,y)=\frac 1{n!}(1-y)^n\left(\frac d{dx}\right)^n [x^n(1-x)^n].$

Note that $R(x,y)$ has integer coefficients as a product of two polynomials with integer coefficients.

Estimate of $D_n$

Since the largest possible power of a given prime $p\le n$ that can divide one of the numbers $1,2,\dots,n$ is $\left\lfloor\frac{\log n}{\log p}\right\rfloor$ where $\lfloor\cdot\rfloor$ is the floor function, we have $D_n=\prod_{p\mathrm{\ prime,}\,p\le n}p^{\left\lfloor\frac{\log n}{\log p}\right\rfloor}\approx n^{\pi(n)}$ where $\pi(n)$ is the number of primes not exceeding $n$ . According to the prime number theorem, we have $\pi(n)\approx \frac n{\log n}$ , so $D_n\approx e^n.$ (The exact meaning of the $\approx$ sign here is that, as $n\to\infty$ , the ratio of the $n$ -th roots of the left and the right sides tends to $1$ . Note also that we need the full strength of the prime number theorem here. The elementary Chebyshev's estimate $D_n\le n^{\sqrt n}\cdot 4^n$ is no longer sufficient).

Estimates of the integral $\iint_{(0,1)\times(0,1)}\frac {R(x,y)}{1-xy}\,dx\,dy$

Integrating $n$ times by parts with respect to $x$ , we see that our integral equals $\iint_{(0,1)\times(0,1)}\left[\frac {x(1-x)y(1-y)}{1-xy}\right]^n\,\frac {dx\,dy}{1-xy}$ . We want to estimate $\max_{0<x,y<1}\frac {x(1-x)y(1-y)}{1-xy}$ . The first remark is that it is attained on the diagonal $x=y$ . It becomes clear if we rewrite the fraction as $\frac {xy(1-x-y+xy)}{1-xy}$ and recall that for a fixed product $xy$ the minimum of the sum $x+y$ is achieved when $x=y$ . So, we need to find the maximum of $\frac {x^2(1-x)^2}{1-x^2}=\frac {x^2(1-x)}{1+x}$ over the interval $(0,1)$ . Taking the logarithmic derivative and solving the resulting quadratic equation $\frac 2x-\frac 1{1-x}-\frac 1{1+x}=0$ , we see that the maximum is attained at the point $x=\frac{\sqrt 5-1}{2}$ and equals $\frac{5\sqrt 5-11}2$ . It means that the integral does not exceed $\left(\frac{5\sqrt 5-11}2\right)^n I(0,0)\approx \left(\frac{5\sqrt 5-11}2\right)^n$ .

On the other hand, to estimate the integral from below, it suffices to notice that the integrand is at least $\frac 4{81}$ for $\frac 1 3\le x,y\le\frac 2 3$ . Thus, the integral is not much less than $\left(\frac 4{81}\right)^n$ .

A small miracle

The miracle that makes the proof work is that $e^2\cdot\frac{5\sqrt 5-11}2$ is strictly less than $1$ (just check it on your calculator or prove it by hand yourself). Thus, for large $n$ , the whole product $D_n^2\iint_{(0,1)\times(0,1)}\frac {R(x,y)}{1-xy}\,dx\,dy$ can be estimated from above and from below by the $n$ -th powers of some constants strictly between $0$ and $1$ . Using the main theorem from the parent article, we see that it remains only to show that $Q_n$ grow not faster than some geometric progression.

Estimate of $Q_n$ and the end of the proof

Note that $Q_n$ is just minus the product of $D_n^2$ , which grows like $e^{2n}$ and the sum of the coefficients of the polynomial $R(x,y)$ at the products $x^ky^k$ with $k=0,\dots,n$ . It follows that $|Q_n|$ does not exceed the sum of coefficients of $(1-y)^n$ , which is $2^n$ , times the maximal coefficient of $\frac 1{n!}\left(\frac d{dx}\right)^n[x^n(1-x)^n]$ . But even the sum of the absolute values of the coefficients of this polynomial does not exceed $8^n$ . So $|Q_n|$ grows not much faster than $(16e^2)^n$ and we are done.

@@ Line 71: / Line 71: @@
 ===Estimate of <math>Q_n</math> and the end of the proof===
 Note that <math>Q_n</math> is just minus the product of <math>D_n^2</math>, which grows like <math>e^{2n}</math> and the sum of the coefficients of the polynomial <math>R(x,y)</math> at the products <math>x^ky^k</math> with <math>k=0,\dots,n</math>. It follows that <math>|Q_n|</math> does not exceed the sum of coefficients of <math>(1-y)^n</math>, which is <math>2^n</math>, times the maximal coefficient of <math>\frac 1{n!}\left(\frac d{dx}\right)^n[x^n(1-x)^n]</math>. But even the sum of the absolute values of the coefficients of this polynomial does not exceed   <math>8^n</math>. So <math>|Q_n|</math> grows not much faster than <math>(16e^2)^n</math> and we are done.
+[[Category:Constants]]

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