π (pi) is the ratio of the circumference of a circle to its diameter.
There are many ways to derive an approximate value for it, some
highly arcane and others less so, but even the simplest approaches
typically require moderately advanced trigonometry and some calculus.
For example, two solutions I learned in school were to do it by
finding the area of a halfcircle of radius 1 by integrating √1x
^{2},
and by finding the arctangent of 1 by integrating 1/(1+x
^{2}) from
0 to 1. Yet the value of π was known with some accuracy by the
ancient Greeks, who had only rather simple mathematical tools available;
clearly it's possible to do it more simply than the approaches I just
mentioned.
On this page, we'll obtain a reasonably good value for π using nothing
more advanced than
Pythagoras's theorem.
The accuracy of the value we obtain will actually be limited by
accumulated floating point errors  the technique we'll use doesn't
behave particularly well with regard to error control, and so won't easily
produce a highly precise result using ordinary double precision
arithmetic. However, it's sufficiently simple that one could do the
calculations by hand and obtain a moderately accurate result (though on
this page we'll actually use a Perl script for the "heavy lifting").
Overview of the Method
We define the
perimeter of a polygon as the sum of the lengths of
all of its sides. If we imagine an ant walking around the perimeter
of a circle, and then walking around the perimeter of a polygon of the
same radius, it's clear that, as the number of sides of the polygon
increases, the total length walked by the ant in going around the polygon
will approach the total length walked in going around the circle.
So, we'll just find the perimeter of a polygon with a very large
number of sides.
If we start with a circle, we can either inscribe a polygon within the
circle, or inscribe the circle within a polygon. We're actually
going to do it both ways; that way we obtain bounds on the perimeter of
the circle: It's at least as long as the interior polygon, and no
longer than the exterior polygon. And when the accumulated errors
finally blow up and the method diverges, we'll know it, because the new
values we obtain will fall outside the range we already established at
previous steps of the process. Finally, we'll use the average of the
"best" two values we obtain as our estimate for π.
In each approach, we're going to start with a square, then split each side
in half to make an octagon, and then split each side again to make a
16sided figure, and so forth, finding better and better approximations to
π (until we start to lose accuracy as a result of accumulating floating
point errors).
To do this, we'll need to find a recurrence relation which, given an
ngon
of a particular radius with a particular side length, tells us how long a
side will be for a
2ngon of the same radius. Once
we've found that, we can apply it repeatedly to obtain the perimeter of a
figure with as many sides as we like.
Circle Inscribed in a Polygon
In this section we'll find a formula for the perimeter of a polygon in
which a circle is inscribed. (The polygon's perimeter will be, in
each instance,
longer than the perimeter of the inscribed circle.)
We're going to work exclusively with polygons with power of 2 sides
 4, 8, 16, etc. The recurrence formula is easy to find for this
case, and by doubling the number of sides at each step, we get to polygons
with large numbers of sides quickly, which is what we want.
Figure
1: Circle inscribed
in a square:

Circle
in an octagon:

Circle
in a 16gon:

The perimeter of a square which inscribes a circle is easy enough to find
(
figure 1): it's just 8 times the radius.
Figure
2: Recurrence relation,
inscribed circle:

We'll now assume we have a polygon with
n sides, where the
length of one side is
2a, and with radius of the inscribed
circle equal to
r. (We make the side length
2a
rather than
a to make the diagrams simpler.) The
radius,
r, is also the distance from the center of the
polygon to the midpoint of each of its sides.
Given those assumptions, we'll now find the length of one side of a
polygon which has
2n sides.
Let:
(1)
We're going to find the values for the new polygon by "cutting off the
points" on the old polygon in order to double the number of sides.
In
figure 2, we show a piece of the
polygon, with sides of length
2a drawn in red, and the new
segment we've added when we "cut off the point" is shown in green.
We can identify two right triangles in the figure, named here by
their side lengths:
Triangle
r a (r+c)
Triangle
b c (ab)
From Pythagoras's theorem, we then have:
(2a)
(2b)
Multiplying out the square in (2b) and canceling
b^{2}, we
have:
(3)
And from (2a) we have:
(4)
Plugging (4) into (3), multiplying out, canceling terms, and multiplying
through by 1, we obtain:
(5)
Multiplying through by 1/
(2a) and factoring out
r^{2}
we obtain,
(6)
We don't really care about the absolute radius; all we're interested in is
the ratio of the length of one side to the radius. Dividing through
by
r, we obtain
(7)
For clarity we define
(
8)
and then (7) becomes
(9)

This polygon has
2n sides, and
βr is
half
the length of one side. If we let
N=
2n=number
of sides, then the perimeter will be
2Nrβ, and the ratio of
the perimeter to the diameter, which is our approximate value for π, will
be
(10)

Incidentally, since this doesn't depend on the size of the polygons we're
using, we've just shown that π is the same for all circles regardless of
their size.
For a square, we have the following:
(11)
Using the square as a starting point, we can use (
9)
to find the value for β for each succeeding figure, where we double the
number of sides at each step. Along with (
10)
we can use that to find successively better approximations for π. We
will do that a little later, after we find the equivalent to (9) for a
polygon inscribed in a circle. But first, as a simple example, we'll
work out the values for an octagon by hand.
For the octagon, we start with a square and double the number of sides.
So, in (
9) we plug in α=1, and since we have
N=8
for an octagon, we find
(
12)
Polygon Inscribed in a Circle
In this section, we'll find a formula for the perimeter of a polygon which
is inscribed in a circle. (In each case the perimeter of the polygon
will be
shorter than the perimeter of the circle.) Again, we
will be working exclusively with polygons with a power of 2 number of
sides.
Figure
3: Square inscribed
in a circle:

Octagon
inscribed
in a circle:

16gon
in a circle:

The perimeter of a square inscribed in a circle, as we see from
figure
3 (and Pythagoras's theorem), is
r·4√2.
Figure
4: Recurrence relation,
inscribed polygon:

As in the previous section, we'll now proceed to find a recurrence
relation which, given the radius and length of one side of a polygon, will
tell us the length of one side of a polygon with twice as many sides.
We will again define the length of one side of the "old" polygon to
be
2a, and the length of one side of the "new" polygon to
be
2b. The old polygon is assumed to have
n
sides, and the new one will have
2n sides. This time,
since the polygon is inscribed in the circle, the radius,
r,
is the distance from the center to one point. To double the number
of sides, we'll
split each side, and insert an extra vertex in the
middle of the split side. The splitting operation is shown in
figure
4.
We can identify two distinct right triangles in figure 4. Calling
them by the lengths of their sides, they are:
Triangle
c a r
Triangle
a (rc) (2b)
From triangle
c,a,r, we have:
(13)
From triangle
a,(rc),(2b) we have:
(14)
Substituting (13) into (14) we obtain:
(15)
Multiplying out the square and bringing
b^{2} over onto
the left by itself, we have:
(16)
As above, we're interested in the ratio of the side lengths to the radius
rather than the absolute lengths of the sides, so we will again define α
and β as in equations (
8). Dividing (16)
through by
r^{2}, expressing it in terms of α and β,
and taking square roots of both sides, we obtain:
(17)

As in the previous section, the perimeter of the inscribed polygon with
N
sides is
2Nrβ, and our approximate value for π is the
perimeter divided by twice the radius, which leads us again back to
equation (
10).
For the inscribed square, with
N=4, we have:
(18)
Using the square as a starting point, and using the recurrence relation in
(
17), doubling the number of sides at each step, we
can again find successively better values for π. We will, again,
work out the case of the octagon "by hand" (with a small assist from a
calculator for the square roots). Plugging the value of β for the
square into (17), and then multiplying by
N=8 for the
octagon, we have
(19)
Averaging this with the value we obtained above, in equations (
12),
for a circle inscribed in an octagon, we obtain the value
3.188 
not awful for a value which we can obtain with hand calculation, but we
can do much better just iterating a few times, as we'll show in the next
section.
Finding a Better Value Using a Script
We built a small Perl script, which is
here,
which iterates, applying equations (
9) and (
17),
along with (
10), to find the ratio of the perimeter
to twice the radius for polygons with increasing numbers of sides.
The script runs 20 steps, doubling the number of sides each time,
and then goes back over the data to find the first step which went "out of
bounds". We use the "interior" polygon approximation for π as a
lower bound and the "exterior" polygon approximation as an upper bound; at
each step the new values for those should lie between the old values,
unless accumulated errors have become so large that the result is not
valid.
We've summarized the results in table 1, below. The algorithm went
"out of bounds" at step 14, when finding the values for polygons with
65,536 sides. It's interesting to note that, since there were only
14 steps (because we're
doubling the number of sides at each
step), it would have been be very tedious but none the less possible to
compute the values in the table entirely by hand.
Table
1: Results of the Perl script:
Sides 
Interior
Polygon
Value for Pi 
Exterior
Polygon
Value for Pi 
Average 
4 
2.82842712474619 
4 
3.41421356237309 
8 
3.06146745892072 
3.31370849898476 
3.18758797895274 
16 
3.12144515225805 
3.18259787807453 
3.15202151516629 
32 
3.13654849054594 
3.15172490742924 
3.14413669898759 
64 
3.14033115695474 
3.14411838524586 
3.1422247711003 
128 
3.14127725093276 
3.14222362994269 
3.14175044043772 
256 
3.14151380114415 
3.14175036917004 
3.14163208515709 
512 
3.14157294036788 
3.14163208070386 
3.14160251053587 
1024 
3.14158772527996 
3.14160251023713 
3.14159511775855 
2048 
3.14159142150464 
3.14159511763137 
3.141593269568 
4096 
3.14159234561108 
3.14159326953813 
3.14159280757461 
8192 
3.141592576545 
3.1415928080678 
3.1415926923064 
16384 
3.14159263346325 
3.14159268885851 
3.14159266116088 
32768 
3.14159265480759 
3.14159265160561 
3.1415926532066 
65536 
3.14159264532122 
3.14159262180328 
3.14159263356225 

The last row, which finally went "out of bounds" due to accumulation of
errors, is colored red. Our "best" value for π, which is the average
of the interior and exterior polygon values on the step before the
algorithm "went out of bounds", is
3.1415926532066,
where we have again colored the erroneous digits red. The correct
value, rounded to 11 places, is 3.14159265359. The difference is
3.832e10. So, our approximation is good to 10 digits (or 9 decimal
places), which is about the best we can expect from the double precision
floating point package we were using.
Page created on 2/20/2008. Table of values corrected, later on
2/20/2008. Clarifications to last paragraph, 8/21/2009