Slices of PI
Richmond U3A Maths Group
5 November 2021
Donald Bell
donald@marchland.org
Lots of slices of π to sample!
 Definition of π
 Some nice books on π
 Archimedes Approximation (250 BC)
 Vieti's exact formula for π based on √2
 The arctan technique
 Machin's formula and other arctan formulas
 Finding our own formula
 a trillion digits of π
 Extending the accuracy of π
 Buffon's Needle and other statistical curiosities
(part 2 later)
 proof that π is irrational
 Euler's Harmonic Series
 Using the Zeta Function with even arguments
 Gauss and the Arithmetic Geometric Mean
 The (wrong) proofs that PI is rational
Definition of π
π is the perimeter of the
circle divided by its diameter
(d), or twice its radius (2r).
So the perimeter = 2 π r
A circle can be approximated
by a polygon and the
segments of the polygon can
be rearranged to make an
approximation to a
parallelogram of height r
and length πr
In the limit the area is πr2
How many digits do we need for π?
In precision mechanical engineering, like boring the cylinder of a car, the smallest
unit of measurement is 0.0001 inch, so 4 or 5 significant digits will do.
The definition of the metre is the distance travelled by light in 1/ 299 792 458 of a
second, so 9 or 10 significant digits will do.
In numerical analysis, there is always the problem of "rounding errors" so modern
computers are usually "64-bit". The IEEE 754 double-precision binary floatingpoint format uses 11 bits for the exponent and 53 for the mantissa (including the
sign bit) so the precision is about 16 significant decimal digits
the largest 64-bit integer is 9,223,372,036,854,775,807 ie 19 significant digits
The Scientific Calculator, built in to Windows 10, has 32 significant digits
But π has been calculated to 31,415,926,535,897 digits
(Amanda Kooser of Google on March 14, 2019)
why did Amanda choose that day and that number of digits?
PI Unleashed
by Jorg Arndt and
Christoph Haenel
Translated from the German
by Catriona and David Lischka
published by Springer (2001)
Title of the original German
edition:
Pi. Algorithmen, Computer,
Arithmetik
A delightful collection of short
articles about π
"The Joy of π" by David Blatner
A collection of
miscellaneous stories and
equations all about π,
from ancient times to the
present day.
And in the margins one
million digits of π.
The millionth digit, by the
way, is a 1.
PI: a source book
by Lennart Berggren,
Jonathan Borwein and Peter
Borwein
Published by Springer
(2004)
Many interesting facts and
figures, including many
original papers.
Alas, this book is now an
(expensive) collector's item.
An important "tool of the trade"
the 32-digit Windows Scientific Calculator
Another important "tool of the trade"
the "Ink Equation Editor" in Microsoft Word
Another important "tool of the trade"
the Python programming language
•
•
•
•
•
•
•
•
•
•
•
Widely available
Versions for Windows, Linux etc
Free
"Object Oriented" Language
Suitable for both small and large programs
Well supported
Easy to learn and use
Good documentation
Lots of specialist libraries
Integers can be of any size (in Python 3.0 onwards)
Floating point libraries permit high precision
Example
Using Python to
search for things,
like:
a3 + b3 + c3 = d3
#!/usr/bin/python
# cube_triplet.py - find integers a,b,c,d
# so that:
a³ + b³ + c³ = d³
def is_cube(n):
root = round(n**(1/3))
if n == root*root*root: return True
return False
Just try every
def main():
combination of
limit = 20
numbers a, b, c up to
print ("cube triplets up to", limit)
a given limit such
for a in range(3, limit+1):
that:
for b in range (2, a):
limit ≥ a > b > c
for c in range(1, b):
sum = a*a*a + b*b*b + c*c*c
if is_cube(sum):
print (a,b,c,round(sum**(1/3)))
the function called:
is_cube
tests a number to see if __name__ == "__main__":
if it is a perfect cube
main()
Console output from the cube_triplet Python program
#!/usr/bin/python
# cube_triplet.py - find integers a,b,c,d
# so that:
a³ + b³ + c³ = d³
def is_cube(n):
root = round(n**(1/3))
if n == root*root*root: return True
return False
def main():
limit = 20
print ("cube triplets up to", limit)
for a in range(3, limit+1):
for b in range (2, a):
for c in range(1, b):
sum = a*a*a + b*b*b + c*c*c
if is_cube(sum):
print (a,b,c,round(sum**(1/3)))
if __name__ == "__main__":
main()
cube triplets up
to 20
5 4 3 6
8 6 1 9
10 8 6 12
15 12 9 18
16 12 2 18
17 14 7 20
18 10 3 19
20 16 12 24
750 Digits of π
Early estimates of π
The true value of π is about
3.141592653
√10 is
3.1622777
π/4 was thought to be (8/9)2 or
0.7901234
giving a value of
3.1604938
Archimedes inequalities (about 250 BC)
Archimedes showed that :
or
3.142857 > π > 3.140845
He considered a polygon of 96 sides
inside a circle and another of 96
sides outside the circle.
The proof is quite long, but it is
completely described in the
"Archimedes" chapter of:
"God Created the Integers" by
Stephen Hawking
π and the Great Pyramid of Giza
The Great Pyramid stands
on a square base.
When it was built, the base
was 440 royal cubits on
one side, so the perimeter
was 1760 cubits.
The height was 280 cubits,
so the ratio of perimeter to
height was:
1760/280 = 6.2
But 2π is 6.28
(and many books have been written on this coincidence!)
Fractional Approximation of π
The value of π to 15 decimal digits is :
3.141592653589793
A reasonable approximation is given by the fraction
22/7 (= 3.14285)
and a much better one by
355/113 (= 3.14159292)
How can these fractions be arrived at? And is there a still better
approximation?
Write π as 3.0 + 0.141592653589793
then 0.141592653589793 can be re-written as
1/7.06251330593105766
so π is approximately
3 + 1/7 or 22/7
Fractional Approximation of π (2)
π is approximately
3.0 + 1.0 / 7.06251330593105766
but 7.06251330593105766 can be written as
7.0 + 0.06251330593105766
and 0.06251330593105766 can be written as
1/15.996594406682676
so π is approximately
3 + 1/(7 + 1/16) or 3 + 1/(113/16) or 355/113
π = 3.0 + 1/(7.0 + 1/15.996594406682676)
0.996594406682676 = 1/1.003417231016437
so that the next approximations are:
103993 / 33102, 104348/33215, 208341 / 66317, etc
Improvements on Archimedes' Methods
Archimedes began with the hexagon, and
considered the inscribed circle and the
hexagon inside that again. He then repeatedly
doubled the number of sides until he got to 96.
He showed that
3.142857 > π > 3.140845
and the average of these numbers is 3.141851
In about 1600, Ludolf van Ceulen repeatedly doubled the number of
sides until he had about 32,000,000,000 of them. Eventually he
determined π to 35 decimal places.
In 2003, Gopal Chakrabarti and Richard Hudson re-examined
Archimedes technique and showed that the true vale of π was 2/3 of
the way between Archimedes' limits, so that (to 6 decimal places):
π = (3.142857 + 2 * 3.140845) / 3 = 3.141516
Viète's Method of calculating π (around 1596)
Starting with a square, François Viète repeatedly doubled the
number of sides of the polygon.
The perimeter of a polygon with n sides is written as P(n)
Then if n is a power of 2,
P(n) / P(2n) = cos (π/2n)
and P(n) gets closer to π
So P(4) = 2√2 = 2.828427 (rather a long way from π)
Then P(8) = P(4) / cos(22.5°) = 2.828427 / 0.9238795 = 3.061467
And P(16) = P(8) / cos(11.25°) = 3.061467 / 0.980785 = 3.121445
And P(32) = p(16) / cos(5.625°) = 3.121445 / 0.99518 = 3.136548
Calculating cos(π/2N) - Viète's Method
cos 2θ = cos2 θ – sin2 θ,
cos2 θ = (cos 2θ + 1)/2,
cos 2θ = 2 cos2 θ – 1
cos θ = √((cos 2θ + 1)/2)
so if θ = π/8, then cos 2θ = cos π/4 = √2/2
cos(π/8) = √((√2/2 + 1)/2)
cos(π/16) = √((√((√2/2 + 1)/2)+ 1)/2)
and so on
Viète actually calculated 2/π, about 0.63662, since it was
simpler:
2/π = √2/2 * cos(π/8) * cos(π/16) * cos(π/32) * . . .
He had discovered the first exact method of calculating π.
But it was a lot of work for just a few digits.
Viète's Method of calculating PI
Suppose we have reached the
stage where AB is one of the edges
of our 2N sided polygon.
We now replace AB (= AC + CB)
with AD and DB.
Thus increasing the perimeter by a
ratio of (DB/CB) or 1/cos(θ/2)
Viète actually calculated 2 / π, which is: 0.63661977
It can be written as:
2 / π = √2/2 * cos(π/8) * cos(π/16) * cos(π/32) . . .
= 0.707107 * 0.923879 * 0.980785 * 0.995185
= 0.6376434
Further applications of cos(π/64), cos(π/128), etc would get
nearer to the true value
Viète's Method of calculating PI (summary)
The square is the first approximation to π
So π = (AB + BF + FG + GA)/AF = 2√2 or about 2.828
Replace AB with AD+DB and this will multiply by 1/cos(DBA)
Angle DBA is ¼ AOB or π/8, so a better approximation is:
π = (2√2)/ cos(π/8) = 2.828 / 0.9239 = 3.061467
or
2/π = √2/2 * cos(π/8)
then
2/π = √2/2 * cos(π/8) * cos(π/16) * cos(π/32) * . . .
Archimedes' Method (in modern notation)
Archimedes started with a hexagon, of side 1,
so its perimeter was 6 units long. He
repeatedly doubled the number of sides.
What is the formula for the (increased)
length of the perimeter?
In the diagram, AB is one of the sides
of the polygon and we want to replace
it with the two sides AC and CB.
The radius of the circumscribing circle, OB, is
r, the length of the original side, AB, is x, and
the length of the new sides, AC and CB is z.
We need a formula for z in terms of r and x
Archimedes' Method
In the right angled (green) triangle ODB
w2 = r2 – (x/2)2 = (4r2 – x2)/4
w = ½ √(4r2 – x2)
Then since OC = r, y = (r-w)
In the right angled (pink) triangle BCD,
z2 = y2 + (x/2)2 = (r-w)2 + (x/2)2
z2 = r2 – 2rw + w2 + (x2/4)
expanding, and substituting for w and w2
z2 = r2 – 2r(½ √(4r2 – x2)) + (4r2 – x2)/4 + (x2/4)
z2 = 2r2 – r√(4r2 – x2)
finally:
z = √( 2r2 – r√(4r2 – x2) )
and if r = 1
z = √( 2 – √(4 – x2) )
Calculating π by repeated
doubling of sides
Begin with a regular hexagon, of side, s0 = 1. Its
half-perimeter is therefore 3, an approximation, p0,
to a semicircle, whose length is π.
When the number of sides is doubled to 12, the
length of each side, s1, becomes 0.5176, and so the
approximation, p1 becomes 6 * 0.5176 or 3.1058
in general,
pn = 3*2n * sn
We derived the equation:
z = √( 2 – √(4 – x2) )
and this can be re-written as a recurrence relation
s(n+1) = √( 2 – √(4 – sn2) )
for calculating the lengths of the sides
#!/usr/bin/python
# pi_polygon.py - calculate pi by
# repeatedly dividing polygon sides
import math
def main():
sides = 6
x = 1.0
cycles = 10
for n in range (cycles):
pi = (sides//2) * x
print (n, sides, x, pi)
sides = sides * 2
x = math.sqrt(2 - math.sqrt(4-x*x))
if __name__ == "__main__":
main()
Results of running the Python program, repeatedly
doubling the number of sides of a polygon
each cycle, on average, adds one more useful digit to the value of π
Calculate π by doubling edges of a polygon
Instead of measuring the perimeter of a polygon when we double
its sides, we will look at what happens to the area
Suppose we have a polygon with N sides.
Call its area AN . What is A2N
AB is one side of the N-polygon. The angle
θ is π/N. O is the centre of the circle.
The area of the N-polygon is N times the area of OAB or 2N times
the area of OAC.
The area of OAC is ½ AC * OC = ½ sin θ cos θ = ¼ sin 2θ
So the area of the N-polygon is (N/2) sin (2π/N)
And the area of a 2N-polygon, A2N , will then be N sin (π/N)
But sin (2π/N) = 2 sin (π/N) * cos (π/N), so
𝐴2𝑁
1
=
𝜋
𝐴𝑁
cos
𝑁
Calculate π by doubling edges of a polygon
We derived the formula
𝐴2𝑁
1
=
𝜋
𝐴𝑁
cos
𝑁
Suppose we start with a hexagon, N = 6
The area A6 will be (3/2)√3 = 2.5981
If we double the sides to make a 12-polygon,
then A12 will be A6 / cos(π/6) or
2.5981 / 0.8660 = 3.0000
Doubling again gives A24 = A12 / cos(π/12) or
3.0000 / 0.9659 = 3.1058
Doubling again gives A48 = A24 / cos(π/24) or
3.1058 / 0.9914 = 3.1326
getting closer to π
Calculate π by doubling the edges of a polygon
(putting all the bits together)
The area of a circle is πr2 or, if the radius is unity, then just π.
Start with a regular polygon, inscribed in this unit circle: hexagon,
square, pentagon or even a triangle whose area, AN , we can calculate
exactly.
For a square A4 is 2, for a hexagon A6 is (3/2)√3 = 2.5981
Choose a number N, which is of the form s*2n so that we can
determine AN
For example, if N = 48, then s = 6 and n = 3, since 48 = 6 * 23 . Then:
𝐴12 𝐴24 𝐴48
𝐴48 = 𝐴6
∗
∗
𝐴6 𝐴12 𝐴24
𝐴2𝑁
1
We derived the formula
=
𝜋
𝐴𝑁
cos
𝑁
We just need to be able to calculate all those cosines
Calculate π by doubling the edges of a polygon
(putting all the bits together)
We know that A6 is (3/2)√3 = 2.5981, and
𝐴12 𝐴24 𝐴48
𝐴48 = 𝐴6
∗
∗
𝐴6 𝐴12 𝐴24
𝐴2𝑁
1
Also
=
𝜋
𝐴𝑁
cos
𝑁
so
𝐴12
1
=
𝜋
𝐴6
cos
etc
6
And the half-cosine rule is
cos θ/2 = √((cos θ + 1)/2)
Since cos(π/6) is ½√3, we can calculate cos(π/12), cos(π/24) etc
So A48 = 2.598076 / (0.8660254 * 0.9659258 * 0.9914449)
= 2.598076 / 0.8293597 = 3.1326285
Similarly, we can calculate A40 (starting with a pentagon)
Calculating π from Pythagorean Triangles
The quadrant BCDE is drawn with a
radius of 5 inside a 5x5 grid of unit
squares as shown.
The grey triangles ADF and ACJ are both
Pythagorean triangles with sides 3, 4, 5.
The length of the quadrant BCDE is
(5/2)π which is about 7.8540
Note that the triangles EFD, DKC and CJB (shown in colour) are all
right-angled with integer sides.
We can calculate the lengths of the three edges BC, CD and DE
They are √10 + √2 + √10 = 3.1623 + 1.4142 + 3.1623 = 7.7388
giving a value of π of (2/5)*7.7388 = 3.0955, about 1.5% too low.
But that is just the start – we can do better.
Calculating π from Pythagorean Triangles
We observe that it is always possible to make a Pythagorean
triangle, with sides a, b, c where the longest side, c, is a power of 5.
And if the length of c is 5N then
the number of different
Pythagorean triangles is also N.
35² + 120² = 15,625 = 125²
44² + 117² = 15,625 = 125²
75² + 100² = 15,625 = 125²
3² + 4² = 5²
175² + 600² = 390,625 = 625²
220² + 585² = 390,625 = 625²
336² + 527² = 390,625 = 625²
375² + 500² = 390,625 = 625²
7² + 24² = 625 = 25²
15² + 20² = 625 = 25²
So we can get better and better values for π by using grids of size:
25x25, 125x125, 625x625 and so on
Calculating π from Pythagorean Triangles
A quarter circle in a 25x25 grid.
The points on the circle with
integer coordinates are:
(25,0), (24,7), (20, 15), (15, 20),
(7, 24) and (0, 25)
The shorter sides of the red,
green and yellow triangles are:
(1, 7), (4, 8) and (5, 5)
So all five of the long edges are:
√50, √80, √50, √80, √50
The total length of the five edges is
3 * 7.071068 + 2 * 8.944272 = 39.101747
Giving a value for π of (2/25) * 39.101747 or 3.12814
About 99.6% of the accepted value
Pythagorean Triangles with side 5N
A recurrence relation can be used to calculate the various triangle
sides:
Begin with x0 = 0 and y0 = 1
The odd and even members of the sequences are calculated slightly
differently. So, for k = 0, 1, 2, . . .
x2k+1 = 3y2k – 4x2k
x2k+2 = 4y2k+1 – 3x2k+1
y2k+1 = 4y2k + 3x2k
y2k+2 = 3y2k+1 + 4x2k+1
This gives:
x1 = 3*1 – 4*0 = 3, y1 = 4*1 + 3*0 = 4
x2 = 4*4 – 3*3 = 7, y2 = 3*4 + 4*3 = 24
x3 = 3*24 – 4*7 = 44, y3 = 4*24 + 3*7 = 117
x4 = 4*117 – 3*44 = 336, y4 = 3*117 + 4*44 = 527
and we can easily verify that
3362 + 5272 = 11896 + 277,729 = 390,625 = 6252
Pythagorean Triangles with side 5N
The Pythagorean triangles so far discovered are:
[ 3, 4, 5 ], [ 7, 24, 25 ], [ 44, 17, 125 ], [ 336, 527, 625 ]
Multiply them by 125, 25 and 5 to get:
[ 375, 500, 625 ], [ 175, 600, 625 ], [ 220, 585, 625 ], [ 336, 527, 625 ]
If we have a quarter circle with radius 625, then the points on it with
integer coefficients are (in descending x order)
( 625, 0 ), ( 600, 175 ), ( 585, 220 ), ( 527, 336 ), ( 500, 375 ),
( 375, 500 ), ( 336, 527 ), ( 220, 585 ), ( 175, 600 ), ( 0, 625 )
The x-difference and y-difference numbers are:
( 25, 175 ), ( 15, 45 ), ( 58, 116 ), ( 27, 39 ) – all twice, and (125, 125)
So the lengths of the sides are
2 * (√31250 + √2250 + √13456 + √2250 ) + √31250
= 2 * (176.777 + 47.434 + 129.692 + 47.434) + 176.777 = 979.451
So the approximation to π is (2/625) * 979.451 = 3.1342432
Shanks' Method for Improving Accuracy of π
Suppose that we know π to n significant digits. Then Daniel
Shanks (1961) showed how to calculate π to 3n significant digits.
Suppose p0 is our best estimate of π and the error is ε so that
p0 = π + ε
Recall that sin (π + ε) = - sin ε
Calculate p1 = p0 + sin p0
Then p1 = π + ε + sin (π + ε)
= π + ε – sin ε
= π + ε – (ε – ε3/3! + ε5/5! – ε7/7! + . . .)
= π + ε3/6 - . . .
So if ε, the error in p0, is less than 10-n , then ε3/6, the error in p1
will be less than 10-3n.
sin x = x – x3/3! + x5/5! – x7/7! + . . .
Where x = 3, the first few terms will be quite large and the value
will fluctuate wildly, but quite soon it will settle down close to its
final value of 0.1411200 (bottom right corner)
N
xN
N!
xN/N!
sum
1
3
1
3.0
3.00
3
27
6 -4.5
-1.5
5
243
120
7
2187
5040 -0.4339286 0.0910714
9
19683
11
177147
362880
2.025
0.525
0.0542411 0.1453125
39916800 -0.0044379 0.1408745
13
1594323
6227020800 0.0002560 0.1411306
15 14348907
1307674368000 -0.0000110 0.1411196
17 129140163 355687428096000 0.0000004 0.1411200
Shanks' Method for Improving Accuracy of π
If p0 , our best estimate for π, is 3.0, then we can calculate p1
p1 = p0 + sin p0 = 3.0 + sin 3.0
= 3.0 + 0.14112000805986722210074480280811
= 3.14112000805986722210074480280811
now calculate p2 = p1 + sin p1
sin p1 = 0.00047264551232833663414376533276947
p2 = 3.1415926535721955587348885681409
now calculate p3 = p2 + sin p2
sin p2 = 0.000000000017597679727754815138622229
p3 = 3.1415926535897932384626433832795
compare with the accepted value
π = 3.1415926535897932384626433832795
All about arctan (x) and calculating π
Defined as the inverse of the trigonometric function tan( )
So, if x = tan (θ), then θ = arctan(x)
And π/4 = 45° = arctan (1)
Angles can be combined like this:
arctan(x) + arctan(y) = arctan( (x + y)/(1 – xy) )
and so:
𝑎
𝑐
𝑎𝑑 + 𝑏𝑐
𝑎𝑟𝑐𝑡𝑎𝑛
+ 𝑎𝑟𝑐𝑡𝑎𝑛
= 𝑎𝑟𝑐𝑡𝑎𝑛
𝑏
𝑑
𝑏𝑑 − 𝑎𝑐
Series expansion for arctan (x)
arctan (x) = x – x3/3 + x5/5 – x7/7 + . . .
The series converges rapidly if x is small, and a lot less than 1.
There are many formulas for π/4 in terms of arctan, eg
π/4 = arctan(1/2) + arctan(1/3)
𝒙𝟑 𝒙𝟓 𝒙𝟕
𝐏𝐫𝐨𝐨𝐟 𝐨𝐟 𝒂𝒓𝒄𝒕𝒂𝒏 𝒙 = 𝒙 −
+
− …
𝟑
𝟓
𝟕
Start with the derivative of a quotient:
𝑑
𝑢
𝑣
𝑣
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥
𝑑𝑥
𝑣2
𝑢
𝑑𝑦
if 𝑦 = then
= =
𝑣
𝑑𝑥
𝑑𝑥
set u = sin x and v = cos x
𝑠𝑖𝑛 𝑥
2 𝑥 + 𝑠𝑖𝑛2 𝑥
𝑑
𝑑(𝑡𝑎𝑛 𝑥)
𝑐𝑜𝑠
1
𝑐𝑜𝑠
𝑥
2 𝑥 = 1 + 𝑡𝑎𝑛2 𝑥
=
=
=
=
𝑠𝑒𝑐
𝑑𝑥
𝑑𝑥
𝑐𝑜𝑠 2 𝑥
𝑐𝑜𝑠 2 𝑥
----------------------------------------------------------------------------
now if y = arctan(x), then x = tan y and
or
integrating,
𝑑𝑥
= 1 + 𝑡𝑎𝑛2 𝑦 = 1 + 𝑥 2
𝑑𝑦
𝑑𝑦
1
2 + 𝑥4 − 𝑥6 + ⋯
=
=
1
−
𝑥
𝑑𝑥
1+𝑥 2
𝑦 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥
𝑥3
𝑥5
𝑥7
=𝑥− + − +⋯
3
5
7
Showing how π/4 = arctan(1/2) + arctan(1/3)
A 3 x 2 grid of unit squares.
Triangle ABC has angles of
45° and 90°
and 45° = π/4
So angle CAB is π/4
and CAD = φ and DAE = θ
and so π/4= θ + φ
tan θ = DE/AE = 1/3 and tan φ = CD/AC = 1/2
so arctan(1/3) = θ and arctan(1/2) = φ
hence π/4 = arctan(1/2) + arctan(1/3)
π/4 = arctan(1/2) + arctan(1/5) + arctan(1/8)
Using this 16x10 grid, Carl Frederick Gauss (1777-1855) used a
similar technique to show that
π/4 = arctan(1/2) + arctan(1/5) + arctan(1/8)
With the help of Zacharias Dahse, a lightning fast calculator, he
used this formula to compute π to 200 decimal places.
π/4 = 2 arctan(1/3) + arctan(1/7)
Grid is 21 units wide by 15 high. Pink and green triangles
have sides in the ratio 1:3. Cyan one is 1:7
π/4 = 2 arctan(1/2) - arctan(1/7)
Grid is 30 units wide by 24 high. Pink and green triangles have
sides in the ratio 1:2. Cyan one is 1:7
Finding a good arctan formula for 𝝅
The two best-known arctan formulas for π are:
π
1
1
= arctan
+ arctan
4
2
3
π
1
1
= 4 arctan
− arctan
4
5
239
For quicker convergence, we would like to have the arctan
arguments as small as possible. Can we find a better formula?
Assume that the formula will be something like this:
π
𝑏
𝑒
= 𝑎 arctan
± 𝑑 arctan
4
𝑐
𝑓
where a, b, c, d, e, f are integers and b/c and e/f are both small
numbers. This implies that a, c, d and f are large integers and b
and e are small integers.
What is the best strategy for finding a suitable set of numbers?
Finding a good arctan formula for 𝝅
We are looking for suitable integer values of a, b, c, d, e, f in:
π
𝑏
𝑒
= 𝑎 𝑎𝑟𝑐𝑡𝑎𝑛
± 𝑑 𝑎𝑟𝑐𝑡𝑎𝑛
4
𝑐
𝑓
Assuming that a large, fast computer is available, here is one way
of finding a set of integers.
if θ = a arctan(b/c) generate a large number of sets of numbers of
the form [a, b, c, θ] and sort these sets into ascending values of θ.
Take one of the sets [a1, b1, c1, θ1] and look for a suitable matching
set [a2, b2, c2, θ2] so that θ2 = π/4 - θ1
If a suitable value exists, it can be quickly found because all the θ
values are in ascending order.
It turns out that there are very few suitable sets of a, b, c, d, e, f
The best one for numbers under 1000 is 5, 29, 278, 7, 3, 79
Finding a good arctan formula for 𝝅
π
𝑏
𝑒
= 𝑎 arctan
± 𝑑 arctan
4
𝑐
𝑓
The best set of values of a, b, c, d, e, f is 5, 29, 278, 7, 3, 79 so:
π
29
3
= 5 arctan
+ 7 arctan
4
278
79
The arguments 29/278 and 3/79 are small, just 0.1043 and 0.0380
So we can expect rapid convergence.
A quick check (using the Windows 32 digit Scientific Calculator):
5 arctan(29/278) = 0.51970304707924787518575118474499
7 arctan(3/79)
= 0.26569511631820043442990966107489
making π/4
= 0.78539816339744830961566084581988
multiply by 4, so π = 3.1415926535897932384626433832795
correct to 32 digits.
some arctan formulas for calculating π
π/4 = arctan(1)
(very slow to converge)
π/4 = arctan(1/2) + arctan(1/3)
(easy to explain but also quite slow to converge)
π/4 = 4 * arctan(1/5) - arctan(1/239)
(Machin's formula, the first of the "practical" ones)
π/4 = 12 * arctan(1/38) + 20 * arctan(1/57)
+ 7 * arctan(1/239) + 24 * arctan(1/268)
(a formula with small values for the arctan arguments,
giving fast convergence)
Devising new arctan formulas for π/4
The formulas that we already know about are these:
𝜋
1
1
= arctan
+ arctan
4
2
3
𝜋
1
1
= 2 arctan
+ arctan
4
3
7
𝜋
1
1
= 4 arctan
− arctan
4
5
239
They all seem to have the general form:
𝜋
1
𝑐
= 𝑛 − 1 arctan
+ arctan
4
𝑛
𝑑
What would they look like for other values of n?
Devising new arctan formulas for π/4
Investigating the formula
𝜋
1
𝑐
= 𝑛 − 1 𝑎𝑟𝑐𝑡𝑎𝑛
+ 𝑎𝑟𝑐𝑡𝑎𝑛
4
𝑛
𝑑
An obvious question - How did John Machin (1686 – 1751)
derive his famous formula?
𝜋
1
1
= 4 𝑎𝑟𝑐𝑡𝑎𝑛
− 𝑎𝑟𝑐𝑡𝑎𝑛
4
5
239
Probably by repeatedly subtracting arctan(1/5) from arctan(1/1)
Recall that
𝑎
𝑐
𝑎𝑑 + 𝑏𝑐
arctan
+ arctan
= arctan
𝑏
𝑑
𝑏𝑑 − 𝑎𝑐
Then:
arctan( 1 / 1 ) - arctan( 1 / 5 ) = arctan( 2 / 3 )
arctan( 2 / 3 ) - arctan( 1 / 5 ) = arctan( 7 / 17 )
arctan( 7 / 17 ) - arctan( 1 / 5 ) = arctan( 9 / 46 )
arctan( 9 / 46 ) - arctan( 1 / 5 ) = arctan( -1 / 239 )
𝜋
1
𝑐
= 𝑛 − 1 𝑎𝑟𝑐𝑡𝑎𝑛
+ 𝑎𝑟𝑐𝑡𝑎𝑛
4
𝑛
𝑑
What about other values of n, such as 4, 6, 7 etc
arctan( 1 / 1 ) - arctan( 1 / 4 ) = arctan( 3 / 5 )
arctan( 3 / 5 ) - arctan( 1 / 4 ) = arctan( 7 / 23 )
arctan( 7 / 23 ) - arctan( 1 / 4 ) = arctan( 5 / 99 )
arctan( 1 / 1 ) - arctan( 1 / 6 ) = arctan( 5 / 7 )
arctan( 5 / 7 ) - arctan( 1 / 6 ) = arctan( 23 / 47 )
arctan( 23 / 47 ) - arctan( 1 / 6 ) = arctan( 91 / 305 )
arctan( 91 / 305 ) - arctan( 1 / 6 ) = arctan( 241 / 1921 )
arctan( 241 / 1921 ) - arctan( 1 / 6 ) = arctan( -475 / 11767 )
So
𝜋
1
475
= 5 arctan
− arctan
4
6
11767
Or, using the last but one line:
𝜋
1
241
= 4 arctan
+ arctan
4
6
1921
𝜋
1
𝑐
= 𝑚 𝑎𝑟𝑐𝑡𝑎𝑛
+ 𝑎𝑟𝑐𝑡𝑎𝑛
4
𝑛
𝑑
The original concept of m = (n - 1) works for some values of n
𝜋
1
5
= 3 𝑎𝑟𝑐𝑡𝑎𝑛
+ 𝑎𝑟𝑐𝑡𝑎𝑛
4
4
99
But it is best modified to m = (n-2) or (n-3) or even more to get the
smallest value for c/d
(a computer was used to derive these big numbers)
π/4 = 5 arctan ( 1 / 7 ) + arctan ( 237 / 3116 )
π/4 = 6 arctan ( 1 / 8 ) + arctan ( 15247 / 388079 )
π/4 = 7 arctan ( 1 / 9 ) + arctan ( 4765 / 441284 )
π/4 = 8 arctan ( 1 / 10 ) – arctan ( 1758719 / 147153121 )
or even the monster:
π/4 = 16 arctan( 1 / 20 )
- arctan (13176476709447727679 / 945426570789006031681)
Evaluating the monster formula
π/4 = 16 arctan ( 1 / 20 )
- arctan (13176476709447727679/ 945426570789006031681)
derived by repeatedly subtracting arctan (1/20) from arctan (1/1)
13176476709447727679/ 945426570789006031681
= 0.01393707043631246268103774635514
whose arctan is
0.01393616815363587294443974673764
arctan(1/20) is
0.04995839572194276141000628703484
times 16
0.79933433155108418256010059255752
minus 0.01393616815363587294443974673764
0.78539816339744830961566084581988
times 4 to get π
3.1415926535897932384626433832795
Summary of the arctan formulas that are
commonly used for π/4
(1) the ones with 2 terms
π/4 = arctan(1/2) + arctan (1/3)
2 arctan(1/3) + arctan (1/7)
2 arctan(1/2) - arctan (1/7)
4 arctan(1/5) - arctan (1/239)
5 arctan(1/7) + 2 arctan (3/79)
5 arctan(29/278) + 7 arctan (3/79)
(2) arctan formulas for π/4 with 3 terms
π/4 = arctan(1/2) + arctan (1/5) + arctan (1/8)
= 5 arctan(1/6) - arctan (1/43) - 2 arctan (1/117)
= 5 arctan(1/7) + 4 arctan(1/53) + 2 arctan(1/4443)
= 6 arctan(1/8) + 2 arctan (1/57) + arctan (1/239)
= 8 arctan(1/10) - arctan(1/239) - 4 arctan(1/515)
16 arctan(1/21) + 3 arctan(1/239) + 4 arctan(3/1042)
for some reason, arctan(1/57) and arctan(1/239) appear
frequently in the arctan formulas
(3) arctan formulas for π/4 with four terms
π/4 = 22 arctan(1/28) + 2 arctan (1/443)
- 5 arctan (1/1393) - 10 arctan(1/11018)
π/4 = 24 arctan(1/36) + 8 arctan (1/57)
- 5 arctan (1/239) - 12 arctan(1/23382)
π/4 = 12 arctan(1/38) + 20 arctan (1/57)
+ 7 arctan (1/239) + 24 arctan(1/268)
π/4 = 24 arctan(1/43) + 20 arctan (1/57)
- 12 arctan (1/117) - 5 arctan(1/239)
π/4 = 12 arctan(1/49) + 32 arctan (1/57)
- 5 arctan (1/239) + 12 arctan(1/110443)
π/4 = 44 arctan(1/57) + 7 arctan (1/239)
- 12 arctan (1/682) + 24 arctan(1/12943)
(4) an arctan formula for π/4 with five terms
π/4 = 88 arctan (1/172)
+ 51 arctan (1/239)
+ 32 arctan (1/682)
+ 44 arctan (1/5357)
+ 68 arctan (1/12943)
Note that the arguments in brackets are all very small numbers so
they should converge rapidly.
However it is necessary to do five calculations
The trillion digit calculation of π
In 2002, Yasumasa Kanada calculated π to 1.2 trillion digits.
There was only one way to check the accuracy of the calculation.
Do it twice, using different formulas, and compare the two results.
He used the two arctan identities:
The repeated use of 1/57 and 1/239 means that their values can be
used twice, thereby saving a lot of computing time.
Using the (32 digit) Windows calculator, we can check the sums.
Checking the (32 digit) calculation of π
= 0.24486396823845703731274008015512
+ 0.56134592183687960900789657403783
- 0.02092038001037361932269107479643
+ 0.00010865333248528261771526642336
= 0.78539816339744830961566084581988
= 0.77185064252570946238585778930201
+ 0.02928853201452306705176750471500
- 0.01759529530814347689864424032988
+ 0.00185428416535925707667979213274
= 0.78539816339744830961566084581987
Summary of the arctan formulas
• they are usually used for calculating π/4 since π/4 = arctan(1)
• there are many formulas that can be used, with 2, 3, or more
arctan terms.
• The first practical one was Machin's formula
π/4 = 4 arctan(1/5) – arctan(1/239)
• Gregory's sequence is used to calculate each of the angles
arctan (x) = x – x3/3 + x5/5 – x7/7 + . . .
• The formulas are suitable for calculating π to enormous numbers
of digits
• calculations can be checked by using two different formulas
Buffon's Needle
Suppose some needles of unit length are scattered
at random on a grid of lines spaced one unit apart.
What proportion of the needles will cross a grid
line? Theoretically it is 2/π or 0.637
The problem can be analyzed by considering one
needle and two lines, at x-coordinates 0 and 1.
One end of the needle is at the point (x1, y1) and it
is at an angle θ. The numbers x1, y1 and θ are
chosen at random, with:
0 < x1 < 1, 0 < y1 < 1, 0 < θ < 2π
Then the other end is at (x2, y2), where:
x2 = x1 + cos θ
y2 = y2 + sin θ
Then the needle crosses a grid line if
x2 < 0 or x2 > 1
The Buffon Needle
experimental calculation of
π using a parallel line grid
or a square grid
The spacing between the lines is the
same as the length of the needle.
For the line grid, the probability of a
needle crossing a line is:
2/π, or about 0.6366
For the square grid, it is
3/π, or about 0.9549
A simple Python program can show this
#!/usr/bin/python
# buffon_demo.py calculate pi using Buffon
import math
import random
def main():
needles = 1000000
count = 0
two_pi = 2 * math.pi
for j in range (needles):
x1 = random.random()
theta = two_pi * random.random()
x2 = x1 + math.cos(theta)
if x2 < 0 or x2 > 1:
count = count + 1
ratio = count/needles
print (count, "of", needles, "needles crossing lines")
print ("estimate of pi, (2/ratio)", 2/ratio)
if __name__ == "__main__":
main()
Result of running the Buffon Needle program
with 1000000 needles (six times)
Console output:
637049 of 1000000 needles crossing lines
estimate of pi, (2/ratio) 3.1394759272834585
636887 of 1000000 needles crossing lines
estimate of pi, (2/ratio) 3.140274491393293
636459 of 1000000 needles crossing lines
estimate of pi, (2/ratio) 3.1423862338343866
636760 of 1000000 needles crossing lines
estimate of pi, (2/ratio) 3.140900810352409
636022 of 1000000 needles crossing lines
estimate of pi, (2/ratio) 3.144545314470254
637219 of 1000000 needles crossing lines
estimate of pi, (2/ratio) 3.138638364518321
Python program for the Buffon experiment using a square grid
As well as calculating x2 = x1 + cos θ, we also need y2 = y1 + sin θ
The needle crosses a grid line if x2 < 0 or x2 > 1 or y2 < 0 or y2 > 1
Probability is 3/π, so π = 3/(ratio of crossings to total needles)
needles = 1000000
count = 0
two_pi = 2 * math.pi
for j in range (needles):
x1 = random.random()
y1 = random.random()
theta = two_pi * random.random()
x2 = x1 + math.cos(theta)
y2 = y1 + math.sin(theta)
if x2 < 0 or x2 > 1 or y2 < 0 or y2 > 1:
count = count + 1
ratio = count/needles
print ("count, needles, ratio =", count, needles,
count/needles)
print ("estimate of pi, (3/ratio)", 3/ratio)
Result of running the Buffon Grid program with
1000000 needles (six times)
Console output:
count, needles, ratio = 955091 1000000 0.955091
estimate of pi, (3/ratio) 3.1410619511648625
count, needles, ratio = 954705 1000000 0.954705
estimate of pi, (3/ratio) 3.142331924521187
count, needles, ratio = 954879 1000000 0.954879
estimate of pi, (3/ratio) 3.141759322385349
count, needles, ratio = 954812 1000000 0.954812
estimate of pi, (3/ratio) 3.141979782407427
count, needles, ratio = 954760 1000000 0.95476
estimate of pi, (3/ratio) 3.142150907034228
count, needles, ratio = 955198 1000000 0.955198
estimate of pi, (3/ratio) 3.140710093614099
Calculate π from random points in a square
Scatter a large number of points (x, y) at
random in a unit square.
Count the proportion of them where
x² + y² < 1
This will equal π/4.
A suitable Python program looks like this:
p = 10000000
print ("Calculate PI from", p, "random points")
count = 0
for jj in range(p):
x = random.random()
y = random.random()
if x*x + y*y < 1:
count = count + 1
print ("count =", count, "\nPI =", 4*count/p)
Calculate π from random points in a square or cube
The computer program finds that the
ratio of points in the quadrant to those in
the unit square is: 0.7854
We know (from school geometry) that
this is π/4, so π = 0.7854 * 4 = 3.1416
Likewise, in three dimensions, the
computer can calculate the ratio of points
in an eighth of a cube to those in a unit
cube. It is 0.5236.
We know (also from school geometry)
that the volume of a sphere is (4/3)πr3
So (1/8)*(4/3)π = 0.5236
and therefore π = 0.5236 * 6 = 3.1416
The Volumes of Hyperspheres
We know (from school geometry), that the area of a circle is πr2
And the volume of a sphere is (4/3)πr3
By plotting 100,000,000 random points in a unit square or unit
cube and using these formulas, we estimated π to be about 3.1416.
Knowing the value of π can we use the same techniques to derive
plausible FORMULAS for hyperspheres?
If Vn is the volume of a hypersphere in n dimensions, with unit
radius (and writing the "area" in two dimensions as V2), we have
V2 = π and V3 = (4/3)π
So it is reasonable to assume that Vn = (a/b)π and we just have to
find (a/b) for each dimension n.
The Volumes of Hyperspheres
If Rn is the ratio of an n-dimensional slice to an n-dimensional
cube (as calculated by the computer), we had
R2 = 0.7854 and R3 = 0.5236. Each of these needs to be multiplied
by 2n where n is the dimension:
So 22 * R2 = 4 * 0.7854 = 3.1416 = π
and 23 * R3 = 8 * 0.5236 = 4.1888 = (4/3)π
The computer calculates R4 to be 0.30845 so we expect
24 * R4 = (a/b)π or a/b = 16 * 0.30845 / 3.1416 = 1.5709
So what is the nearest rational fraction to 1.5709?
(11/7) is 1.5714, but we believe our fraction to be 1.5709 ± 0.0001
Not good enough.
Our expectation is wrong, maybe we need a π2 factor
The Volumes of Hyperspheres
The expectation that V4 = (a/b) π has turned out to be wrong.
Lets try V4 = (a/b) π2 , so
24 * R4 = (a/b)π2 or
a/b = 16 * 0.30845 / 3.14162 = 0.50000
Which shows immediately that:
V4 = (1/2)π2
The computer calculates R5 to be 0. 16449
So let's try V5 = (a/b) π2 , so
25 * R5 = (a/b)π2 or
a/b = 32 * 0.16449 / 3.14162 = 0.5333
very close to 8/15, so
V5 = (8/15)π2
Calculating volumes of n-dimensional hyperspheres
By scattering random points in unit hypercubes of up to 10
dimensions, we get the following ratios:
R2 = 0.7853970874
R3 = 0.5235924897
R4 = 0.3084200631
R5 = 0.1644924921
R6 = 0.0807471966
R7 = 0.0369103398
R8 = 0.0158542389
R9 = 0.0064431922
R10 = 0.0024908857
(a laptop computer can find these numbers in a few minutes)
So it should be possible to derive formulas for the volumes of the
hyperspheres
Calculating volumes of n-dimensional hyperspheres
So far, we have:
V2 = π, V3 = (4/3π, V4 = (1/2)π2, V5 = (8/15)π2
What about V6 – will it have a factor based on π, or π2 or even π3 ?
R6 = 0.0807435
Try using just π. Then
26 * R6 = (a/b)π or
a/b = 64 * 0.0807435 / 3.1416 = 1.64489
closest fraction is 51/31 = 1.6452 (not quite close enough)
Try using just π2. Then
26 * R6 = (a/b)π2 or
a/b = 64 * 0.0807435 / 3.14162 = 0.52359
Closest fraction is 11/21 = 0.52381 (not quite close enough)
Try using just π3. Then
26 * R6 = (a/b)π3 or
a/b = 64 * 0.0807435 / 3.14163 = 0.16666 = 1/6
V6 = (1/6)π3
Calculating volumes of n-dimensional hyperspheres
So far, we have:
V2 = π, V3 = (4/3)π, V4 = (1/2)π2,
V5 = (8/15)π2, V6 = (1/6)π3
The computer has calculated R7 as 0.036912774
It looks like V7 will be based on π3. Then
27 * R7 = (a/b)π3 or
a/b = 128 * 0.036912774 / 3.14163 = 0.152383
= 16/105
Giving
V7 = (16/105)π3
Having got to this point, it ought to be fairly easy to derive the
formulas for V8, V9 and V10
formulas for V8, V9 and V10
Recall that:
R8 = 0.015858087, R9 = 0.006442008
R10 = 0.002490457
Assume that V8 will be based on π4. Then
28 * R8 = (a/b)π4 or
a/b = 256 * 0.0158581 / 3.14164 = 0.0416765 = 1/24
V8 = (1/24)π4
Assume that V10 will be based on π5. Then
210 * R8 = (a/b)π5 or
a/b = 1024 *0.002490457/ 3.14165 = 0.0083335 = 1/120
V10 = (1/120)π5
V9 is more difficult. Assume that it will be based on π4. Then
29 * R9 = (a/b)π4 or
a/b = 512 * 0.0064420 / 3.14164 = 0.0338498
Best fraction (in this example) is 11/235, but true value is 32/945 so
V9 = (32/945)π4
Summarizing the results for V2 to V10
A pattern begins to emerge for the volumes of the hyper spheres:
V2 = π r2
V4 = (1/2) π2 r4
V6 = (1/6) π3 r6
V8 = (1/24) π4 r8
V10 = (1/120) π5 r10
= ( 1/6! ) π5 r10
V3 = (4/3) π r3
V5 = (8/15) π2 r5
V7 = (16/105) π3 r7
V9 = (32/945) π4 r9
= (25/3 * 5 * 7 * 9) π4 r9
= ( 29 * 4! / 9! ) π4 r9
For even numbered dimensions, 2n:
V(2n) = (1/!n) πn r2n
For odd numbered dimensions 2n+1
V(2n+1) = (2(2n+1) * n!/(2n+1)!) πn r(2n+1)
Proofs that π is irrational
The approximation 22/7 is a useful one for most practical purposes
And 355/113 is even better (being 3.14159292 while the true value
is 3.14159265).
But π cannot be accurately expressed as a rational fraction A/B.
• The first proof came from Lambert in 1761 using the properties
of continued fractions
• In 1873, Hermite proved that π2 is irrational.
• And in about 1960, Ivan Niven published a one-page article with
the title "A simple proof that π is irrational"
• Subsequently Presh Talwalkar published a (rather longer)
explanation of the Ivan Niven paper (recognizing that "simple"
may not be the most appropriate description!)
Outline of Niven's Proof that π is irrational
(as explained by Presh Talwalkar)
First, let's assume that π is a rational fraction a/b
Then, using the numbers a and b, create a function f(x), of degree 2n
(where n is a number yet to be defined)
The function is defined so that f(0) = 0
and also f(a/b) = 0, so that f(π) = 0, since we are assuming π = a/b
Now create all the derivatives of f(x), which we can call f'(x), f''(x)
etc, up to f(2n) (x). Anything beyond f(2n)(x) will be zero.
From this set of derivatives, form another function g(x)
We can then demonstrate that g is an integer greater that 0 and also
less than 1. This is obviously a contradiction.
So π is not a rational fraction.
Willebrord Snell and Christiaan Huygens
Willebrord Snell is best known for his
contribution to optics in
Snell's Law of Refraction
But he also developed a very accurate
method of calculating π
He drew two very similar diagrams.
In each one there is a circle, radius r,
a line BO, extended to X or Y, a point F
and a tangent at B going to G or H.
Snell suspected, and Huygens proved, that the length of the arc BF
lies between BH and BG. This gave them a quick way to calculate
π to a large number of digits.
Euler's equation for π2 (slide 1 of 3)
Euler solved the "Basel problem" which had defeated many and
showed that
𝟏
𝟏
𝟏
𝟏
𝝅𝟐
+ 𝟐+ 𝟐+ 𝟐+⋯=
𝟐
𝟏
𝟐
𝟑
𝟒
𝟔
Here is a simplified (ie not very rigorous) version of his proof
If f(x) is a polynomial in x, suppose roots of the equation f(x) = 0 are:
r0, r1, r2, . . . etc
suppose also that r0 = 0 and all the other roots are non-zero. Then:
f(x) = Cx(x – r1)(x – r2) . . .
where C is some constant
𝒙
we can re-write each factor (x – r) as −𝒓(𝟏 − ) and so
𝒓
𝒙
𝒙
𝒙
𝒇 𝒙 = 𝑲𝒙 𝟏 −
𝟏−
𝟏−
…
𝒓𝟏
𝒓𝟐
𝒓𝟑
where K is a suitable constant, incorporating all the (-r) constants
Euler's equation for π2 (slide 2 of 3)
recall that f(x) can be written as:
𝒙
𝒇 𝒙 = 𝑲𝒙 𝟏 −
𝒓𝟏
𝒙
𝟏−
𝒓𝟐
𝒙
𝟏−
…
𝒓𝟑
the function sin(x) is zero where x = 0, π, -π, 2π, -2π, 3π, -3π, etc.
So if it is expressed as a polynomial, these will be its roots and it can
be written as:
𝑥
x
𝑥
x
𝑥
x
sin 𝑥 = 𝐾𝑥 1 −
1+
1−
1+
1−
1+
…
π
π
2π
2π
3π
3π
combining pairs of factors, we get:
𝑥2
sin 𝑥 = 𝐾𝑥 1 − 2 2
1 π
𝑥2
1− 2 2
2 π
𝑥2
1− 2 2 …
3 π
Euler's equation for π2 (slide 3 of 3)
We have shown that:
𝒙𝟐
𝒔𝒊𝒏 𝒙 = 𝑲𝒙 𝟏 − 𝟐 𝟐
𝟏 𝝅
𝒙𝟐
𝟏− 𝟐 𝟐
𝟐 𝝅
𝒙𝟐
𝟏− 𝟐 𝟐 …
𝟑 𝝅
but we know that
𝒙𝟑
𝒙𝟓
𝒙𝟕
𝒔𝒊𝒏 𝒙 = 𝒙 −
+
−
+⋯
𝟑!
𝟓!
𝟕!
equating the coefficients of x shows that K = 1
then, equating the coefficients of x3 shows that
𝟏
𝟏
𝟏
𝟏
𝟏
− = − 𝟐 𝟐− 𝟐 𝟐− 𝟐 𝟐− 𝟐 𝟐+⋯
𝟔
𝟏 𝝅
𝟐 𝝅
𝟑 𝝅
𝟒 𝝅
or
𝝅𝟐
𝟏
𝟏
𝟏
𝟏
= 𝟐+ 𝟐+ 𝟐+ 𝟐+⋯
𝟔
𝟏
𝟐
𝟑
𝟒
Euler's Calculation of π2 (on one slide)
suppose f(x) is an infinite polynomial representing sin(x)
Then when f(x) = 0, x = 0, ±π, ±2π, ±3π, etc
So f(x) can be written as:
f(x) = x (x – π)(x + π) (x – 2π)(x + 2π) (x – 3π)(x + 3π) . . .
or f(x) = x (x2 – π2) (x2 – 22π2) (x2 – 32π2) . . .
Introducing a constant K, this can be re-written as:
f(x) = Kx (1 – x2/12π2) (1 – x2/22π2) (1 – x2/32π2) . . .
Now f(x) / x = K + terms in x2 and above
but f(x) / x = sin(x) / x and this tends to 1 as x tends to 0, so K = 1
Extract the coefficient of x3 from f(x), it is
- 1/12π2 - 1/22π2 - 1/32π2 - 1/42π2 - 1/52π2 - . . .
but the expansion of sin(x) is x – x3/3! + x5/5! – x7/7! + . . .
So the coefficient of x3 from sin(x) is -1/6, so:
π2 /6 = 1/12 + 1/22 + 1/32 + 1/42 + 1/52 + 1/62 + . . .
Demonstration of Euler's Calculation of π2
The formula is: π2 /6 = 1/12 + 1/22 + 1/32 + 1/42 + . . .
Unfortunately, this series converges rather slowly.
It is easy to program a computer to calculate these terms.
Using just the first 10 terms, here is what we get:
π2/6 = 1.00000 + 0.25000 + 0.111111 + 0.062500 +
0.04000 + 0.027778 + 0.020408 + 0.015625 +
0.012346 + 0.010000 = 1.549768
so π2 = 9.298606 and π = 3.049362
The values of π for different numbers of terms:
100 terms gives 3.1320765318
1000 terms gives 3.1406380562
10000 terms gives 3.1414971639
100000 terms gives 3.1415831043
1000000 terms gives 3.1415916987
10000000 terms gives 3.1415925581
actual value of π is 3.1415926536
Euler's calculations of π using the Zeta function
The general definition of the Zeta function is
We are interested in the Zeta function for positive integer values,
usually even ones, and especially powers of 2:
Zeta(1) = 1/1 + 1/2 + 1/3 + . . . (it does not converge)
Zeta(2) = 1/12 + 1/22 + 1/32 + 1/42 + . . . ( = π2 / 6 )
Zeta(3) = 1/13 + 1/23 + 1/33 + 1/43 + . . . ( no easy formula )
Zeta(4) = 1/14 + 1/24 + 1/34 + 1/44 + . . . ( = π4 / 90 )
Zeta(6) = 1/16 + 1/26 + 1/36 + 1/46 + . . . ( = π6 / 945 )
Zeta(8) = 1/18 + 1/18 + 1/18 + 1/18 + . . . ( = π8 / 9450 )
Zeta(16) = 1/116 + 1/216 + 1/316 + 1/416 + . . .
= 3617 π16/ 325 641 566 250
Zeta(32) = 1/132 + 1/232 + 1/332 + 1/432 + . . .
= 7709321041217 π32 / 62490220571022341207266406250
Recall that Zeta(32) = 1/132 + 1/232 + 1/332 + 1/432 + . . .
= 7709321041217 π32 / 62490220571022341207266406250
1/132 = 1.0
1/232 = 0.00000000023283064365386962890625
1/332 = 0.00000000000000053965952773542901
1/432 = 0.00000000000000000005421010862427
1/532 = 0.00000000000000000000004294967296
1/632 = 0.00000000000000000000000012564927
sum = 1.00000000023283118336765054828182
62490220571022341207266406250 / 7709321041217
= 8,105,800,788,023,426.4635374292628855
multiply by 1.000000000232831183367650548282
8,105,800,789,910,709.6531553579820859
sqrt is 90,032,220.842933279567130768186585
sqrt is 9,488.5310160705740071285755017288
sqrt is 97.409091034002437236440332677526
sqrt is 9.8696044010893586188344909993098
sqrt is 3.1415926535897932384626433831894
Euler's Prime Number Series for π
𝝅𝟐
𝟏
𝟏
𝟏
𝟏
we recall that
= 𝟐+ 𝟐+ 𝟐+ 𝟐+⋯
𝟔
𝟏
𝟐
𝟑
𝟒
the equations are going to get complicated, so we write 1/N2 simply
as gN and note that g1 = 1 and that gX * gY = gXY
so π2/6 = g1 + g2 + g3 + g4 + g5 + g6 + g7 + g8 + g9 + g10 + g11 +
multiply by the second term, g2, and spread out
g2* π2/6 = g2*g1 + g2*g2 + g2*g3 + g2*g4 + g2*g5 + or
g2* π2/6 =
g2
+ g4 +
g6 + g8 + g10 +
subtract this second equation from the first equation
(1-g2) * π2/6 = g1 + g3 + g5 + g7 + g9 + g11 + g13 + g15 + g17 +
again, multiply by the second term, this time g3, and spread out
g3*(1-g2) * π2/6 = g3*g1
+
g3*g3
+
g3* g5 +
g3*(1-g2) * π2/6 = g3
+
g9
+
g15 +
subtract this equation from the previous one:
(1-g3)(1-g2) π2/6 = g1 + g5 + g7 +
g11 + g13 +
g17 +
Euler's Prime Number Series for π
𝝅𝟐
𝟏
𝟏
𝟏
𝟏
we recall that starting with
= 𝟐+ 𝟐+ 𝟐+ 𝟐+⋯
𝟔
𝟏
𝟐
𝟑
𝟒
or π2/6 = g1 + g2 + g3 + g4 + g5 + g6 + g7 + g8 + g9 + g10 + g11 +
we first removed multiples of g2, then g3 to get:
(1-g3)(1-g2) π2/6 = g1+g5+g7+ g11+ g13+ g17+ g19 + g23 + g25
each operation multiplies the left side by (1 – gN) and removes all
the multiples of gN from the right.
And N is always a prime number (otherwise it would already have
been removed)
so (1-g2)(1-g3)(1-g5)(1-g7)(1-g11)(1-g13)(1-g17) … π2/6 = g1 = 1
giving (for p = prime)
∞
𝒑=𝟐
𝟏
𝟔
𝟏− 𝟐 = 𝟐
𝒑
𝝅
Euler's Prime Number Series for π
∞
𝒑=𝟐
𝟏
𝟔
𝟏− 𝟐 = 𝟐
𝒑
𝝅
where p = prime
Not a rapidly converging product, so 6 significant figures is enough:
For the first 11 primes (ie 2 to 31), (1 – 1/p2) is:
[2, 0.75], [3, 0.888889], [5, 0.96 ], [ 7, 0.979592],
[11, 0.991736 ], [ 13, 0.994082], [ 17, 0.996540], [ 19, 0.997230],
[23, 0.998110], [ 29, 0.998811], [ 31, 0.998959]
product is 0.611706, then 6/product is 9.808634, so π= 3.131874
Using all the primes up to 997 (168 of them) we get a value for π of:
3.141393
only 4 or 5 significant digits
Not very satisfactory for a lot of work!
Euler's Prime Number Series for π
Recall that Zeta(2) =
π2/6 =1/12 + 1/22 + 1/32 + 1/42 + 1/52 + 1/62 + 1/72 + 1/82 + 1/92 +
make a copy, multiply it by ( 1/22 ) and subtract from the original.
This removes all the terms divisible by ( 1/22 ) so
(1 – 1/22) π2/6 = 1/12 + 1/32 + 1/52 + 1/72 + 1/92 +1/112 + 1/132 +
Same again, but with ( 1/32 )
This removes all the terms divisible by ( 1/32 ) so
(1 – 1/22) (1 – 1/32) π2/6 = 1/12 + 1/52 + 1/72 +1/112 + 1/132 +
Then ( 1/52 )
(1 – 1/22) (1 – 1/32) (1 – 1/52) π2/6 =1/12 + 1/72 +1/112 + 1/132 +
Then ( 1/72 ), then (1/112 ), then ( 1/132 ) all through the primes
So, in the end, all the terms on the right, except 1/12 , are gone
(1 – 1/22) (1 – 1/32) (1 – 1/52) . . . (1 – 1/p2) . . . π2/6 = 1/12
or
(1 – 1/22) (1 – 1/32) (1 – 1/52) . . . (1 – 1/p2) . . . = 6/π2
an infinite product of terms of the form (1 – 1/p2) where p is prime.
Defining π without a circle
It is possible to define π without a circle.
if y = cos x,
then π/2 is the first root where y = 0
and cos x can be defined by an infinite series
and cos can also be defined as the solution of a differential
equation:
𝑑2𝑦
2 = -y
𝑑𝑥
Clawson's Limit Equation for π
In his book of "Mathematical Mysteries",
Clawson gives, without proof, the following
amazing equation for π
Although the numbers involved may grow to be quite large, it
should be possible to investigate the limit with a relatively
simple computer program.
The value of √(2π) is about:
2.506628274631000502415765284811
Clawson's Limit
Equation for π
The computer program to calculate these numbers is quite
straightforward, but for n > 700, the factorial and power terms
become very large and the computer is not able to handle them
without special "large number" programming.
The results are these:
n = 50, π = 3.1520819617264695
n = 100, π = 3.1468329896120797
n = 200, π = 3.144211736418174
n = 400, π = 3.142901923001544
n = 700, π = 3.1423407408442308
(very slow convergence – and a lot of calculation!)
Deriving π using Euler's Gamma Constant
(not to be confused with Euler's number e, the logarithm base)
Gamma is defined like this:
the difference between a
logarithm and a harmonic
series.
There is a long and complex Wikipedia entry:
https://en.wikipedia.org/wiki/Euler%27s_constant
and near the end of it, there is this curious equation:
where N1(n) is the number of ones in the binary expression of the
number n. And N0(n) is the number of zeroes. Since the number
87 in binary is 1010111, then N0(87) = 2 and N1(87) = 5.
Not a difficult Python program to write!
#!/usr/bin/python
import math
def bit_diff(number):
n, n0, n1 = number, 0, 0
while n > 0:
digit = n & 1 # rightmost bit
n1 = n1 + digit
n0 = n0 + (1-digit)
n = n>>1
return n1-n0
def main():
sum, limit = 0, 2000
for n in range(1, limit+1):
sum = sum + bit_diff(n)/(2*n*(2*n+1))
pi = 4/math.exp(sum)
print ("value of pi with", limit, "terms =", pi)
if __name__ == "__main__":
main()
Results of the Gamma calculation of π
The convergence of the function is not particularly fast, even for
very large values of n.
Calculating π from the frequency
of relatively prime numbers
If two integers are chosen at random, the probability that they are
relatively prime is 6/π².
A Python program generated 10 000 000 pairs of numbers of size up
to 100 000 000 and calculated their Greatest Common Divisor.
This shows the result of running the program five times and
counting the relatively prime pairs.
#!/usr/bin/python
# pi_relative_prime.py
import math
import random
def main():
scope
= 100000000
attempts = 10000000
count
= 0
for jj in range(attempts):
a = int(scope*random.random())
b = int(scope*random.random())
if math.gcd(a, b) == 1:
count = count + 1
ratio = count/attempts
pi_squared = 6/ratio
pi = math.sqrt(pi_squared)
print ("count, π², π", count, pi_squared, pi)
if __name__ == "__main__":
main()