3.1415926535897932384626433832795028841971693993751058209749445920899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190HelpI’mTrappedInAUniverseFactory914564856692346034…
In search of π
Aidan Randle‐Conde
22/7 (/2009)
1
Why talk about π?
• In the states March 14th is designated π Day (3‐14) • In the UK July 22nd is wriLen 22/7. So today is π ApproximaRon Day!
– Although today is all about approximaRon, it’s ironically more accurately named than π Day!
• Depending on your taste there are a few other days that can help you celebrate π: – November 10th is the 314th day of the year. – At 1:13am on December 21st the day/Rme is 335/113, an ancient Chinese
approximaRon.
– By April 26th the Earth travels through 1/π of its orbit since the start of the year. – In 2015 we’ll have Super π Day, 3‐14‐15! – 00:37 on July 21st 2069 will be the π epoch (3141592654 seconds since the Unix
epoch) Less than a day away from π ApproximaRon Day!
2
What is π?
• π doesn’t really need an introducRon – It’s the raRo of the circumference of a circle to its diameter. – It’s easy to understand, and yet its exact value has eluded mathemaRcians for
millenia.
– It pops up in the most unusual and unexpected of places. • What we know about π:
– It’s not an integer – It’s not raRonal – It’s not algebraic
• What we don’t know about π: – Is it normal? – For that maLer, is our obsession with it normal?!
3
• One of the first methods used (by Archimedes) was the method of exhausRon. • Sandwich a circle between two regular polygons, measure their perimeters and
place limits on the value of π.
• This method also gives the familiar limits:
θ ½
pS
pL
ExhausRon
• The length of the side of the smaller polygon, pS, is:
• And for the larger polygon, pL, is:
• So for an n‐sided polygon the approximaRon becomes: €
pS = sinθ
€
pL = tanθ
€
n sinθ < π < n tanθ
€
θ = π /n
€
;
€
sinθ →θ
€
lim θ → 0[ ]
€
tanθ →θ
€
tanθ > sinθ4
ExhausRon
• Since θ = π/n, for an arbitrary value of n the argument will become circular (ie we would need to know the value of π to esRmate π.)
• But we know so we can use the half‐angle formulae:
• Square roots weren’t easy to calculate for the ancient Greeks. • Using various tricks Archimedes got as far as a 96‐sided polygon
before moving on to other calculaRons.
• He calculated:
• All that computaRon got π correct to three decimal places. Not bad…
€
cos π /4( ) = sin π /4( ) = 12
€
cos(θ /2) = 1+ cosθ2
€
sin(θ /2) = 1− cosθ2
€
3.1410 < π < 3.1427
5
• The first few iteraRons improve the approximaRons quite rapidly:
ExhausRon
n Lower limit
Upper limit
4 2.828427 4
8 3.061467 3.313708
16 3.121445 3.182597
32 3.136548 3.151724
64 3.140331 3.144118
128 3.141277 3.14222
256 3.141514 3.141750
512 3.141573 3.141632
1024 3.141588 3.14160
2048 3.141591 3.141595
6
RaRonal approximaRons
• Over the coming centuries various mathemaRcians approximated π using the following fracRons: – Archimedes used the areas of his 96‐gons to find the raRonal limits:
€
31071
< π < 31070
– He used the laLer approximaRon, 22/7, which is sRll occasionally used today.
– Ptolemy got slightly closer using the same method with 377/120 = 3.1416…
– Tsu Ch’ung‐Chi used 355/113 = 3.141593… but we don’t know a great deal about him or his methods.
– Nothing much changes unRl the development of infinite series.
7
Infinite series
• Using discoveries involving infinite series and calculus various expressions for π were found. Most of them are very preLy:
€
π 2
6=112
+122
+132
+142
+152
+162
+172
…
€
π − 34
=1
2 × 3× 4−
14 × 5 × 6
+1
6 × 7 × 8…
€
π4
=11−13
+15−17
+19−111
+113
−115
…
€
π2
=21×23×43×45×65×67×87×89…
€
8,553,103× 224 ×π 26
3× 26!=1126
+1226
+1326
…
John Wallis
Leonhard Euler
Gokried Leibniz
Leonhard Euler
8
Infinite series
• Although the infinite series are very elegant they can be slow to converge:
n
€
π2
=21×23×43×45×65×67×87×89…
€
π 2
6=112
+122
+132
+142
+152
+162
+172
…
€
π − 34
=1
2 × 3× 4−
14 × 5 × 6
+1
6 × 7 × 8…
€
π4
=11−13
+15−17
+19−111
+113
−115
…
€
8,553,103× 224 ×π 26
3× 26!=1126
+1226
+1326
…
9
Example: π2/6 • Out of these (and many other) series one of the most interesRng to derive is π2/6. • Euler considered the funcRon
• This equaRon has roots at x=±nπ, so it can also be wriLen:
• EquaRng the terms in x2 gives:
• The same method can be used to get higher order terms in π2n. • You can get the same result using Fourier Series, Reinmann Zeta funcRons etc…
€
sin xx
= 1− xnπ
n=−∞
∞,x≠0
∏ = 1− x2
n2π 2
n=1
∞
∏€
sin xx
=−1( )n x 2n
2n +1( )!n= 0
∞
∑
€
−x 2
n2π 2n=1
∞
∑ = −x2
6⇒
π 2
6=
1n2n=1
∞
∑
10
Machin’s method • Archimedes’ method could be extended to polygons of arbitrary numbers of sides
and the known series expansions could be extended to arbitrary number of terms.
• Each step requires more computaRon‐ some people spent their working lives churning out more digits!
• In 1706 Machin discovered a more efficient method using the differences arctan funcRons:
€
π4
= 4arctan 12
− arctan
3117
π4
= 4arctan 13
− arctan
1731
π4
= 4arctan 14
− arctan
79401
€
arctan(x) = x − x3
3+x 5
5−x 7
7+x 9
9…
• Proof is in the backup slides. • The second term converges rapidly. • People oren made mistakes:
• Shanks got the last 180 of 707 digits incorrect.
• This was discovered when the digit 7 didn’t occur “oren enough”
11
Geung irraRonal • For those who didn’t have Rme to sum hundreds of terms, or calculate 26!,
Ramanujan gave a number of approximaRons, using geometrical arguments and series:
€
92 + 192
22
14
= 3.14159265262…
632517 +15 5( ) 7 +15 5( ) = 3.141592652…992
2206 2= 3.14159265…
€
1π
= 8 (1103+ 26390n)(2n −1)!!(4n −1)!!994n+232n (n!)3n= 0
∞
∑
• The last two come from a remarkable collecRon of equaRons published in 1914, such as
12
IrraRonal, transcendental, normal? • In 1761 Lambert showed that π is irraRonal. • All aLempts to find an exact value of π in the form π = a/b are doomed to failure. • Worse sRll, in 1882 Lindemann showed that π is transcendental, so all expressions
of the form:
where there are a finite number of terms and all an and n are raRonal and non‐zero are also doomed to failure.
• One of the remaining problems is to determine whether or not π is normal‐ does every digit appear with equal frequency in all bases? – StaRsRcally, the answer is yes. – As far as mathemaRcians are concerned, the quesRon is unresolved. – Perhaps we can’t use the frequency of the digit 7 to find errors arer all…
• None of these facts stopped people trying to find ever‐more precise approximaRons.
€
0 = anπn∑
13
Buffon’s needle • Towards the end of the 18th century some experiments were done dropping
needles onto lined paper.
• The probability that a needle of length l crossing lines of spacing s is:
• By dropping needles and counRng the how many cross a line you can approximate the value of π. (One poor student of deMorgan got π ≈ 3.317 by dropping 600 needles!)
€
P(l | s) = lcosθ2πs0
2π
∫ dθ = 2lπs
• I cheated and used a computer: • l = 40 • s = 50 • 100,000 needles • 50,794 needles cross a line • π ≈ 3.15
• That’s not bad for a virtual pack of needles and a virtual pad of lined paper.
14
State of play 1946 • By the early 20th century there were a number of different ways to calculate π to
arbitrary degrees of precision.
• It wasn’t uncommon for people to get a whole series of digits wrong!
Year Source Method Decimal places
2000 BC Rhind papyrus (16/9)2 1
250 BC Archimedes 96‐gon exhausRon 3
263 Liu Hui 192‐gon exhausRon 5
480 Tsu Ch’ung‐Chi 355/113 7
1400 Madhava Gregory’s series 11
1430 Al’Kashi 3x228‐gon exhausRon 14
1695 Van Ceulen 262‐gon exhausRon 35
1706 Machin arctan 100
1946 Fergurson arctan 620
15
Precision wars • Al’Kashi wanted to calculate the circumference of
the universe precise to the width of a hair. – He thought that the universe was about 600,000 Rmes
wider than the Earth. – Using his 805,306,368‐gon he calculated 14 digits of π.
That’s about right, if a hair is 1mm wide.
• Changing our sense of scale to something more meaningful: – The width of the universe if about 30 billion light years
(≈3x1026m.)
– The width of a Hydrogen atom is ≈1x10‐10m. – It turns out that 40 digits of π is more than enough for
nearly all calculaRons. (Add another 25 digits if you want to use the Planck scale.)
• Anything else is just showing off. And yet the calculaRons conRnue to this day.
16
• What use are the remaining digits of π? – It’s a great way to benchmark a new supercomputer. The digits of π give an endless supply of
calculaRons for the mulR‐petaflop machines. – The digits of π provide a good source of random numbers (True randomness is a constant
annoyance in the world of compuRng science and staRsRcs.) – Some staRsRcians run tests on the digits to see if the digits really are random and whether π really
is a normal number. So far π has passed all tests. – People are sRll amazed by this number for some reason.
• Current record is 1,241,100,000,000 (1.24 trillion) digits.
The age of the computers
• The numbers of decimal places obtained since 1947 span 9 orders of magnitude!
• (I needed two graphs to fit all the points in.)
• You can see Moore’s law at work.
17
The age of the computers • There are quite a few commonly used algorithms for calculaRng π to very large
numbers of decimal places.
• One of the fastest to converge is the Gauss‐Legendre algorithm:
€
an+1 = an + bn( ) /2
€
b0 =1/ 2
€
c0 =1/4
€
d0 =1
€
a0 =1
€
bn+1 = anbn
€
cn+1 = cn − dn an − bn( )2 /4
€
dn+1 = 2dn
Itera9on π ≈ Decimals
0 2.9142135623730949 0
1 3.1405792505221686 2
2 3.1415926462135428 7
3 3.141592653589794 11 €
π ≈an + bn( )
2
4cn
Ler: Yasumasa Kanada with the Hitachi SR‐8000 supercomputer, that calculated 15,000,000,000 digits of π. 18
The age of the computers • In 1995, Bailey, Borwein and Plouffe discovered a fascinaRng formula:
• This can be used to determine any given hexidecimal digit (base 16 digit) if π, without needing to calculate any of the rest.
€
π =4
8n +1−
28n + 4
−1
8n + 5−
18n + 6
116
n
n= 0
∞
∑
Ler: π in binary. Right: 22/7 in binary
• ConverRng a hexidecimal number to a binary number is trivial. This method was used to approximate π to 40,000,000,000,000 bits.
• There is now a search for an algorithm for finding digits in base 10.
19
Fun facts about π • Since we have all these digits we may as well do
something fun with them: – Take combinaRons of 3 digits, divide by 37 and take the
remainder. Assign 1=A, 2=B, 3=C and so on. – This gives the π code. – Using the first 1,000,000 digits, “PI” appears 299 Rmes.
“SASS” never appears and neither does “AIDAN”. For some reason “E” appears 9,208 Rmes. Why should e appear more Rmes than π? Hmm…
• Plenty of people have searched for paLerns: – The three digits around digit 315 are 3‐1‐5, and the three
digits around digit 360 are 3‐6‐0. – The famous “Feynman point” consists of six nines from digit
762, and Feynman used to joke that π is equal to 3.14159…999999… etc
• But that’s just the start…
20
Indiana #246 • In 1897 Edwin Goodwin proposed a Bill to the Indiana Legislature
to declare the value of π to be 3.2 (or 4, the mathemaRcs behind it was ambiguous).
• He put forward some rather incredible claims: – He could trisect the angle with just a pair of compasses and straight edge – He doubled the cube (made 3√2 raRonal) – He squared the circle (made π algebraic)
• All these claims had already been shown to be impossible. • Even so, the bill was sent (via the CommiLee on Swamp Lands) to
the CommiLee on EducaRon.
• The bill passed the first reading with unanimous consent. • Prof Waldo, who was passing through Indianapolis at the Rme,
read the bill.
• He convinced the Legislature to postpone the second reading indefinitely.
• This seems to be the only aLempt in history to decrease our knowledge of π. 21
Mnemonics • If you’re having trouble remembering the first few digits of π there are plenty of
mnemonic phrases to help you. The number of leLers in each word represents a single digit:
How I wish I could recollect pi easily today!How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics!
• Or if you’re a fan of sonnets iambic pentameter: Now I defy a tenet gallantly
Of circle canon law: these integersImporting circles' quotients are, we see,
Unwieldy long series of cockle bursPut all together, get no clarity;
Mnemonics shan't describeth so reformedCreating, with a grammercy plainly,A sonnet liberated yet conformed.
Strangely, the queer'st rules I manipulateBeing followéd, do facilitate
Whimsical musings from geometric bard.This poesy, unabashed as it's distressed,
Evolvéd coherent - a simple test,Discov'ring poetry no numerals jarred. 22
π everywhere • π seems to appear in all over the place. The most obvious example is Euler’s
relaRon. (Which is preLy weird when you think about it.)
• We’ve already seen it in some infinite series:
• And staRsRcs:
• And error funcRons:
• And integral calculus:
• And if you divide the length of a river by the distance between its source and mouth, and the river is sufficiently old and long, you get π. And nobody knows why.
• And…
€
e−x2
−∞
∞
∫ dx = π
€
Γ(1/2) = π
€
π2
=21×23×43×45×65…
€
1− x 2−∞
∞
∫ dx = π /2
23
π and primes • If you take two integers at random, what is the probability that they share no
common divisor? (Sound impossible?)
• The probability that an integer is not divisible by a prime q is 1‐1/q. • So the probability that two integers share no divisors is:
• Consider mulRplying the following series:
€
P = 1− 1q2
primes∏
€
112
+122
+132
+142
+152
+162
+172
+182
+192
…
× 1−
122
=112
+132
+152
+172
+192
…
112
+132
+152
+172
+192
…
× 1−
132
=112
+152
+172
…24
π and primes • This process conRnues unRl every term is removed, except 1. • So we get the idenRty:
• Using the idenRty we derived earlier:
• gives the probability that no two randomly selected integers share a divisor as:
• (And if you don’t find that interesRng then the chances are I just wasted 40 minutes of your Rme…)
€
1n2n=1
∞
∑
1−
1q2
primes∏
=1
€
1n2n=1
∞
∑ = π2
6
€
P = 6π 2
= 0.61…
25
Conclusion • We take π for granted. Today geung its value to 9 or 10 decimal places is
extremely easy.
• The true story behind the search for π reveals that geung just a few decimal places is laborious task.
• MathemaRcians have spent centuries geung to know π beLer – the more we know the more we realize π finds its way into every part of mathemaRcs, oren in
strange places. – Despite know more unique decimal places of π than any other number there are sRll lots of
unanswered quesRon.
• Learning about π can be fun, fascinaRng and informaRve.
26
3.14159265358979323846264338327950288419716939937510582097494459208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460104543266482133936072602491412737
Backup
27
DerivaRon of Machin’s formula • The equaRon for the tangent of a sum of angles is:
• Applying this recursively gives:
• And then:
• Make some subsRtuRons:
€
tan(A + B) = tanA + tanB1− tanA tanB
€
tan(4A) = 4 tanA 1− tan2 A
1− 6tan2 A + tan4 A
€
tan(4A − B) = 4 tanA(1− tan2 A) − tanB(1− 6tan2 A + tan4 A)
1− 6tan2 A + tan4 A + 4 tanA tanB(1− tan2 A)
€
4A − B = arctan(1/ x)
€
A = arctan(1/(x + n))
€
B = arctan(y)
28
DerivaRon of Machin’s formula • From the previous slide:
• Rearranging gives:
• Then let x=1, n=4, which gives y=1/239:
€
tan(4A − B) = 4 tanA(1− tan2 A) − tanB(1− 6tan2 A + tan4 A)
1− 6tan2 A + tan4 A + 4 tanA tanB(1− tan2 A)
€
4A − B = arctan(1/ x)
€
A = arctan(1/(x + n))
€
B = arctan(y)
€
y =4x (x + n)3 − (x + n)( ) − (x + n)2 − 6 x + n( )2 +1( )4 (x + n)3 − x + n( )( ) + x (x + n)2 − 6 x + n( )2 +1( )
€
arctan(1/ x) = 4arctan(1/(x + n)) − arctan(y)
€
π4
= 4arctan 15
− arctan
1239
29
π as an integer?
• Most socieRes realize quite quickly that π is not an integer. • However, if you take a look at the bible:
– ‘And he made a molten sea, ten cubits from the one brim to the other: it was round all about, and his height was five cubits: and a line of thirty cubits did compass it about.’ (I Kings 7, 23)
• So π = 3? Oops! • This approximaRon actually comes from the Great Temple of
Solomon and dates back to around 1000BC.
30
π 0
0
1
∫
x 4 (1− x)4
1+ x 2= x 6 − 4x 5 + 5x 4 − 4x 2 + 4( ) − 41+ x 2
I = x 6 − 4x 5 + 5x 4 − 4x 2 + 4( )0
1
∫ dx − 41+ x 2
0
1
∫
=227− 4arctan x[ ]0
1
=227−π
31
Bibliography • There are loads of books out there that look into π and its properRes. Here are a
few that are currently siung on my shelf: – I’d recommend the Penguin DicRonary of Curious and InteresRng Numbers (David Wells) for a quick overview of π and lots of easy
recreaRonal mathemaRcs.
– Finding the probability of a shared divisor comes from Excursions in Number Theory, another easy book that leads onto harder things.
– For the gory details of Archimedes’ method of exhausRon (as well as loads on planetary moRons) check out 100 Great Problems of Elementary MathemaRcs. (They use the original derivaRons‐ it’s difficult stuff!)
• There’s also a great deal online, in the obvious places. The usual suspects link to lots of other sites: – Wiki:
• hLp://en.wikipedia.org/wiki/%CE%A0 • hLp://en.wikipedia.org/wiki/Numerical_approximaRons_of_%CF%80 • hLp://en.wikipedia.org/wiki/Category:Pi_algorithms
– Mathworld Wolfram: • hLp://mathworld.wolfram.com/PiFormulas.html
– Machin derivaRon: • hLp://milan.milanovic.org/math/english/pi/machin.html
– Chronology of π approximaRons: • hLp://www‐groups.dcs.st‐and.ac.uk/~history/HistTopics/Pi_chronology.html
32
Images • If I don’t credit an image/plot I’ve made it myself.
– Slide 3: hLp://plus.maths.org – Slide 4: hLp://archimedes.galilei.com/archimedes_body.htm – Slide 7: hLp://jeff560.tripod.com/stamps.html – Slide 8: hLp://en.wikipedia.org/wiki/John_Wallis , hLp://en.wikipedia.org/wiki/Leonhard_Euler ,
hLp://en.wikipedia.org/wiki/Leibniz
– Slide 12: hLp://www.alcorngallery.com/Portraits/Portraits_display.php?i=24 – Slide 16: hLp://www.danstopicals.com/laRtude1.htm , hLp://www.nasa.gov/ – Slide 18: hLp://www.educ.fc.ul.pt/icm/icm2001/icm34/kanada.htm – Slide 19: hLp://mathworld.wolfram.com/PiDigits.html – Slide 21: hLp://www.civicheraldry.com/page/5735 , hLp://www.agecon.purdue.edu/ – Slide 26: hLp://areason2write.wordpress.com/
33
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