Two Ways to Corner π

Begin with a circle of diameter 1. Draw a regular polygon inside it, with all its corners on the circle, and another regular polygon outside it, with all its sides touching the circle.

The inner polygon is shorter round than the circle, because a straight line between two points on the circle is shorter than the arc between them. The outer polygon is longer round than the circle, for a similar reason. So if the circle has diameter 1, its circumference is pi and

(inner perimeter) < pi < (outer perimeter).

That is the whole idea, and it is already a proof of something. With six sides the inner perimeter is exactly 3, so pi is bigger than 3, and no computation was involved in establishing that.

Double the number of sides in each polygon. Both polygons hug the circle more closely, both perimeters move toward pi from their own side, and the two statements above stay true the whole time. Six sides, twelve, twenty-four, forty-eight, ninety-six: at ninety-six the bracket is 3.1410319 to 3.1427146, which is where Archimedes stopped and which is usually quoted as 223/71 < pi < 22/7.

The third view on this page does something entirely different. It drops needles on a floor ruled with parallel lines and counts how many cross a line. The fraction that cross is related to pi, so the count produces a number near it. That number is not a bound and never becomes one, no matter how many needles fall, and holding the two kinds of answer next to each other is what this page is for.

Why the hexagon is exactly 3. Take a regular hexagon inscribed in a circle and join its corners to the centre. You get six triangles, each with two sides equal to the radius and the angle between them equal to a sixth of a full turn, which is sixty degrees. A triangle with two equal sides and sixty degrees between them is equilateral, so the third side, which is one side of the hexagon, is also a radius. Six radii is three diameters. There is no approximation anywhere in that paragraph.

Why doubling is easy. Doubling the sides of a polygon sounds like it should need new geometry each time, and it does not. If $a$ is the circumscribed perimeter and $b$ the inscribed one, both divided by the diameter, then doubling the sides gives

The first is a harmonic mean and the second a geometric mean, and neither knows anything about circles: they are two lines of trigonometry, done once in layer 5 and then applied over and over. So the whole ladder is one square root per rung and nothing else.

Where the difficulty actually is. That square root. Every other operation is exact arithmetic on fractions, so a computer can do it with whole numbers and no error at all. A square root of a fraction is almost never a fraction, so something has to give, and what most programs do is take a floating-point square root and hope. That turns a bracket into an estimate without changing anything you can see on the screen.

What this page does instead is refuse to compute the square root. It computes two fractions with the root between them, rounded so that the lower one is definitely below and the upper one is definitely above, and carries both. The bracket then stays a bracket for the whole ladder, and the only thing the rounding costs is a slightly wider interval than the perfect one.

Why the needles are different. A needle of length $l$ dropped on lines a distance $d$ apart crosses a line with probability $2l/(\pi d)$, when $l\le d$. Turn that around and you can estimate pi from a shower of needles: throw $N$, count the $C$ that crossed, and $2lN/(dC)$ should be near pi. It is near pi. It is never provably near pi, and the difference between those two sentences is the reason this page has three views instead of two.

Put these two lines side by side.

3.1410319 < pi < 3.1427146, from four doublings of a hexagon.

pi is about 3.1405, from two hundred thousand needles.

The second is closer. The first is worth more, and the gap between what they are worth is not a matter of degree.

The first line is a statement. It can be checked, it can be argued with, and it happens to be true, with a proof that fits on this page. If you doubt it, there is something specific to doubt: the geometry, the recurrence, the direction of the rounding. Every one of those is written down.

The second line is a number. There is nothing in it to doubt and nothing to check. It came from a particular shower of needles and a different shower gives a different number. Ask how close it is to pi and the honest answer is that nobody knows without already knowing pi. Ask again after a hundred times as many needles and the answer is the same: still no bound, and now ten times more work for one more likely digit.

A computation that produces a number has not produced a bound. A computation that produces two numbers and a proof that the answer is between them has produced something else entirely, even when the two numbers are further apart than the one number was.

This is the same distinction the rest of this site keeps running into from the other direction. Goldbach's conjecture has been checked past four quintillion and that is evidence rather than proof. Here, four doublings of a hexagon check nothing at all and prove a theorem. The size of the computation was never what settled it.

And there is a second lesson in the needles, which is quieter. Make the needle longer than the gap between the lines and the formula goes on producing numbers. They are near 4 rather than near 3.14, which is obvious here only because you already know the answer. A formula outside its hypotheses does not fail loudly; it just stops meaning anything while continuing to compute.

Let $a_n$ be the perimeter of the regular circumscribed $n$-gon and $b_n$ that of the regular inscribed $n$-gon, both around a circle of diameter 1. Then:

1. $b_n<\pi <a_n$ for every $n\ge 3$, and $$a_{2n}=\frac{2a_n{b}_n}{a_n+b_n},b_{2n}=\sqrt{a_{2n}{b}_n},$$ with $b_6=3$ and $a_6=2\sqrt{3}$. Proved, in layer 5.
2. $\frac{223}{71}<\pi <\frac{22}{7}$. Proved, as the case $n=96$ of 1, and reproduced by this page in exact rational arithmetic.
3. Every bracket this laboratory prints for the ladder contains $\pi$. Exhaustively checked over the nine rungs it draws, against fifty decimal places of $\pi$ entered by hand, and separately against the enclosure of 5. Not proved: 1 is the theorem, this is a statement about a program.
4. $\frac{\pi }{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$, and truncating each series after $t$ terms leaves an error smaller in size than the first omitted term and of that term's sign. Proved, in layer 5, for the remainder bound.
5. A needle of length $l\le d$ dropped with position and angle independent and uniform on lines a distance $d$ apart crosses a line with probability $2l/(\pi d)$. Proved, in layer 5. The standard deviation of the resulting estimate of $\pi$ from $N$ throws is proportional to $1/\sqrt{N}$. Proved, cited, not reproduced here.
6. For $l>d$ the formula $2lN/(dC)$ estimates nothing. Proved, in layer 5, by identifying the step of the argument that fails.
7. $\pi$ is irrational, and indeed transcendental. Proved, cited, not reproduced, and no part of this page is evidence for either.

Statement 3 is the one that is easiest to overstate and it is deliberately the weakest here. The theorem is 1. What a test suite can establish about an implementation of it is that these particular printed numbers are right, which is a different and much smaller thing.

The polygons bracket the circle. The inscribed polygon's perimeter is a sum of chords, and each chord is shorter than the arc it subtends, so $b_n<\pi$. The circumscribed polygon contains the circle as a convex region contains a convex region, and the perimeter of a convex region containing another is at least as large, so $\pi <a_n$. Both inequalities hold for every $n$, which is what makes every rung of the ladder a statement rather than a stage. ∎

The two recurrences. With the circle of diameter 1, a regular $n$-gon inscribed has perimeter $b_n=n\sin (\pi /n)$ and circumscribed has $a_n=n\tan (\pi /n)$. Write $\theta =\pi /(2n)$, so that $a_{2n}=2n\tan \theta$ and $b_{2n}=2n\sin \theta$. Then

using $\tan 2\theta =2\tan \theta /(1-{\tan }^2\theta )$ and $\sin 2\theta =2\sin \theta \cos \theta$; and

so $b_{2n}=\sqrt{a_{2n}{b}_n}$, both quantities being positive. The starting values are the hexagon: $b_6=3$ from the equilateral triangles of layer 2, and $a_6=2\sqrt{3}=\sqrt{12}$. ∎

The rounding keeps the bracket true. Both recurrences are strictly increasing in each of their arguments: for positive $a$ and $b$, $\partial /\partial a[2ab/(a+b)]=2b^2/(a+b{)}^2>0$, and the square root is increasing. So if ${\alpha }_{\text{lo}}\le a_n\le {\alpha }_{\text{hi}}$ and ${\beta }_{\text{lo}}\le b_n\le {\beta }_{\text{hi}}$, feeding the low pair through gives a value at or below $a_{2n}$ and the high pair a value at or above it. The square root of a rational is bracketed by writing $D=10^p$ and taking $k=\lfloor \sqrt{r{D}^2}\rfloor$ as an exact integer square root; then $k/D\le \sqrt{r}<(k+1)/D$, and squaring gives $(k/D{)}^2\le r\le ((k+1)/D{)}^2$ as an inequality between whole numbers. Finally each low bound is rounded down onto the grid and each high bound up, which can only widen an interval that already contains the truth. By induction every rung brackets $\pi$. ∎

This is the step where a page like this usually goes wrong, so it is worth saying what the failure would look like: a single floating-point square root anywhere in that induction leaves every printed digit unchanged and every claim on this page false. The kernel is checked by a test that reads it as text and fails if the string appears in it.

The alternating remainder. For $|x|<1$, $\arctan x=x-x^3/3+x^5/5-\cdots$, whose terms decrease strictly in size and alternate in sign. For such a series the partial sums straddle the limit: $S_t\le S\le S_{t+1}$ when $t$ is even and the reverse when $t$ is odd, since $S-S_t=(-1{)}^t(|u_t|-|u_{t+1}|+\cdots )$ and the bracketed quantity lies strictly between $0$ and $|u_t|$. So the remainder after $t$ terms is smaller in size than the first omitted term and has its sign, which is a bound and not an estimate. Applying that to $\arctan (1/5)$ and $\arctan (1/239)$ separately and combining with $\pi =16\arctan (1/5)-4\arctan (1/239)$ gives an interval, provided the subtraction takes the upper bound of the term being removed to make the lower bound of the difference. ∎

The needle. Let the needle have length $l\le d$ and the lines be $d$ apart. Let $X$ be the distance from the needle's centre to the nearest line, uniform on $[0,d/2]$, and $\Theta$ its acute angle to the lines, uniform on $[0,\pi /2]$, independent. The needle crosses exactly when $X\le (l/2)\sin \Theta$. So

Where that proof stops being true. The integral above computes the expected number of crossings, and the last line silently identifies it with the probability of at least one. Those agree only when the needle cannot cross twice, that is when $l\le d$. For a longer needle the expected count is strictly larger than the probability, so the fraction of needles observed to cross is smaller than $2l/(\pi d)$, and $2lN/(dC)$ therefore comes out too large. Nothing in the arithmetic notices: it returns a number near 4 with exactly the same confidence it returns one near 3.14159. ∎

What this laboratory computes. Nine rungs of the polygon ladder, from six sides to 1536, in exact rational arithmetic on arbitrary-precision integers, with every bound rounded outward onto a grid of $10^{-25}$ by default. Twenty terms of the Machin series, also exact. Up to two million needles, in ordinary double precision, which is entirely adequate because nothing about the needles is a bound.

Why the ladder stops at 1536. Not for want of arithmetic. Each doubling buys about a factor of four in the width of the bracket, so roughly one decimal digit every two rungs, and eight rungs have bought five digits. Getting twenty digits this way means about thirty more doublings and a polygon with more sides than there are atoms in a teaspoon of water. This is why the arctangent formulae replaced polygons in 1706 and never gave the ground back, and why the modern computations use neither.

Why the needles will not get better. The error of an average of $N$ independent trials shrinks like $1/\sqrt{N}$. To gain one decimal digit you need a hundred times as many needles. The two million this page will throw are worth about three digits on a good day, and there is no seed and no patience that turns that into a bound: every one of those digits is a digit that happens to be right, not a digit that is known to be.

What is not settled anywhere. Nothing about the bracket, which has been closed for two thousand years, and nothing about pi's irrationality or transcendence, which are theorems. What is genuinely open is smaller and stranger than either: whether pi is normal, meaning that every string of digits appears in it with the frequency chance would predict. Billions of digits have been examined and they look normal. Nobody has proved that any digit appears infinitely often. That is the same shape of gap this whole site keeps arriving at, and it is worth noticing that even pi, the most computed number in mathematics, has one.

Read this next to 1, 2, 4, 8, 16, and Then Not 32. The other page here that proves its own formula in front of the reader rather than citing it, on the same circle with straight lines drawn across it. One is about a count that a pattern gets wrong at the sixth term; this one is about a number no pattern reaches at any term.