The story of and related puzzles · Time line I: Ancient period The story starts in ancient Egypt...

The story of

πand related puzzles

Narrator: Niraj KhareCarnegie Mellon University Qatar

Being with math is being with the truth and eternity!

Oct, 30, 2017

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Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000years ago!

• The Rihnd Papyrus of Ahmes from 1650 BC givesapproximation π ≈ 4(64)

81 = 3.16049.

• Late 5th century BCE, Antiphone and Baryson of Heracleainscribe and circumscribe regular polygons to a circle.

• Around 450 BCE, Anaxagoras proposes ‘squaring thecircle’ from a prison! The puzzle was finally ‘settled’ in1882 AD.

• Around 250 BC, Archimedes proves that3.1408 < 310

71 < π < 317 ≈ 3.1428.

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81 = 3.16049.

71 < π < 317 ≈ 3.1428.

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81 = 3.16049.

71 < π < 317 ≈ 3.1428.

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81 = 3.16049.

71 < π < 317 ≈ 3.1428.

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81 = 3.16049.

71 < π < 317 ≈ 3.1428.

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81 = 3.16049.

71 < π < 317 ≈ 3.1428.

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Time to pause and ponder

Are we on solid ground?

Is the ratio of circumference to its diameter for a circle is alwaysa constant?

limx→0

sin(x)

3 / 27

limx→0

sin(x)

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limx→0

sin(x)

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“Cosine rule” but Pythagoras truly rules!

The oldest, shortest words “yes” and “no” are those which require the most thought. -

Pythagoras

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π is a constant.(full stop)

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Time line II

• In 263 AD, Liu Hui of China using regular inscribedpolygons with sides 12 to 192 showed that 3.14159 < π.

• Towrds the end of 5th century AD, Tsu Chung-chih andTsu keng chih use regular polygons with 24, 576 sides toshow 3.1415926 < π < 3.1415927.

• Some mathematician started using inaccurate values suchas√

10 ≈ 3.1622 and for centuries it continued in India andother places!

• Madhava (1340 c.1425) of Sangamagrama (India) found πaccurately to 11 decimal places.

• Jamshid al-Kashi had calculated π to an accuracy of 16decimal digits in 1424 AD.

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Time line II

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Time line II

6 / 27

Time line II

6 / 27

Time line II

6 / 27

Time line II

6 / 27

Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500million sides calculated π calculated π to an accuracy of 20decimal digits by 1596. By the time he died in 1610, heaccurately found 35 digits! The digits were carved into histombstone.

• In 1647, the ratio of circumference of a circle to itsdiameter gets its name and symbol π by William Oughtred.Made popular by Leonhard Euler.

• Time to pause and ponder (II). Ludolph Van Ceulen mustbe fictional!

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Ludolph Van Ceulen:3.14159265358979323846264338327950288...

Ludolph van CeulenDutch-German mathematicianLudolph van Ceulen was a German-Dutch mathematician fromHildesheim. He emigrated to the Netherlands. Wikipedia:

Born:January 28, 1540, Hildesheim, GermanyDied:December 31, 1610, Leiden, NetherlandsKnown for: piInstitution: Leiden UniversityNotable student: Willebrord Snellius

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Archimedes’ approximation of π: Angle bisector

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Archimedes’ approximation of π (I): Upper bound

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Archimedes’ approximation of π (II)Achimedes’ iteration

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Archimedes’ approximation of π (III)Repeated use of Achimedes’ iteration

⇒ OD

⇒ OE

⇒ OF

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⇒ OD

⇒ OE

⇒ OF

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⇒ OD

⇒ OE

⇒ OF

12 / 27

⇒ OD

⇒ OE

⇒ OF

12 / 27

⇒ OD

⇒ OE

⇒ OF

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Archimedes’ approximation of π (III)(b)Pythagoras comes to rescue!

153<OC

√((571)2 + (153)2

(153)2

116218

153<OD

AE11721

116218

233414

153<OE

AF23391

233414

467312

153<OF

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153<OC

√((571)2 + (153)2

(153)2

116218

153<OD

117218

116218

233414

153<OE

AF23391

233414

467312

153<OF

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153<OC

√((571)2 + (153)2

(153)2

116218

153<OD

AE11721

116218

233414

153<OE

233914

233414

467312

153<OF

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153<OC

√((571)2 + (153)2

(153)2

116218

153<OD

AE11721

116218

233414

153<OE

AF23391

233414

467312

153<OF

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153<OC

√((571)2 + (153)2

(153)2

116218

153<OD

AE11721

116218

233414

153<OE

AF23391

233414

467312

153<OF

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153<OC

√((571)2 + (153)2

(153)2

116218

153<OD

AE11721

116218

233414

153<OE

AF23391

233414

467312

153<OF

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Archimedes’ approximation of π: upper bound

As OAAG >

4673 12

153 , AGOA < 153

4673 12

. Thus,

Perimeter of the circlediameter < 96 times (2 times the length of AG)

2 times length of OA

<96× 153

467312

=14688

467312

= 3 +6671

467312

< 3 +1

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As OAAG >

4673 12

153 , AGOA < 153

4673 12

. Thus,

<96× 153

467312

=14688

467312

= 3 +6671

467312

< 3 +1

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As OAAG >

4673 12

153 , AGOA < 153

4673 12

. Thus,

<96× 153

467312

=14688

467312

= 3 +6671

467312

< 3 +1

14 / 27

As OAAG >

4673 12

153 , AGOA < 153

4673 12

. Thus,

<96× 153

467312

=14688

467312

= 3 +6671

467312

< 3 +1

14 / 27

As OAAG >

4673 12

153 , AGOA < 153

4673 12

. Thus,

<96× 153

467312

=14688

467312

= 3 +6671

467312

< 3 +1

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Archimedes’ approximation of π: Lower bound

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Archimedes’ approximation of π: Lower bound

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Infinite series era: time line III(a) continues

• French mathematician Viete (1540 1603) and later in 1650John Wallis found infinite products for π.

• By 1682, James Gregory and Leibniz found a famous“useless” series:

arctan(t) = t− t3

5− t7

9+ · · · .

4= 1− 1

5− 1

9+ · · ·

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arctan(t) = t− t3

5− t7

9+ · · · .

4= 1− 1

5− 1

9+ · · ·

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arctan(t) = t− t3

5− t7

9+ · · · .

4= 1− 1

5− 1

9+ · · ·

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arctan(t) = t− t3

5− t7

9+ · · · .

4= 1− 1

5− 1

9+ · · ·

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arctan(t) = t− t3

5− t7

9+ · · · .

4= 1− 1

5− 1

9+ · · ·

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Time line III (b): series expressions for π

• Useless for 10000 terms are required to get four accuratedigits! To compute 100 digits ”you need to add up moreterms than there are particles in the universe” [Blanter,page 42].

• In 1706, an English professor of Astronomy, John Machinusing arctan(x) + arctan(y) = arctan( x+y1−xy ) found:

4= acrtan(

119)−acrtan(

239) = 4 arctan(

5)−arctan(

5− 1

3(5)3+

5(5)5− 1

7(5)7+ · · ·

239− 1

3(239)3+

5(239)5− 1

7(239)7+ · · ·

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4= acrtan(

119)−acrtan(

239) = 4 arctan(

5)−arctan(

5− 1

3(5)3+

5(5)5− 1

7(5)7+ · · ·

239− 1

3(239)3+

5(239)5− 1

7(239)7+ · · ·

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4= acrtan(

119)−acrtan(

239) = 4 arctan(

5)−arctan(

5− 1

3(5)3+

5(5)5− 1

7(5)7+ · · ·

239− 1

3(239)3+

5(239)5− 1

7(239)7+ · · ·

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4= acrtan(

119)−acrtan(

239) = 4 arctan(

5)−arctan(

5− 1

3(5)3+

5(5)5− 1

7(5)7+ · · ·

239− 1

3(239)3+

5(239)5− 1

7(239)7+ · · ·

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Time line IV: squaring the circle put to rest

• Using the above formula Machin could calculate πaccurately till 100 places by hand!

• Using the same above formula many mathematicians fornext 150 years found more and more digits of π.

• In 1873, William Shanks used the formula to calculate 707digits of which only the first 527 were correct. [Berggren,page 627]

• In 1761 to 1776, Lambert and Legendre proved that π isnot a ratio of two integers.[Cajori, page 246]

• In 1882, Ferdinand von Lindemann proved transcendenceof π (i.e., squaring the circle is impossible). [Berggren, page407]

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Time line V(a): π in modern times

• By 1949 using hand calculations and toil of many years, weknew first 1000 digits correctly!

• In 1949 the ENIAC (Electronic, Numerical, Intrigrator andCaculator) was built that calculated the first 2037 digits ofπ in less than 70 hours! [Beckmann, page 180]

• Now using super computers and faster algorithms π isknown over few trillion digits.

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Time line V(b): Do we need to know any more?

• Just 39 decimal places would be enough to compute thecircumference of a circle surrounding the known universe towithin the radius of hydrogen atom. [Berggren, 656]

• At present time the only tangible application of all thosedigits is to test the computers and computer chips for bugs.[The history of pi by Devid Wilson]

• To quench the curiosity and to test if the given digits arefrom π.

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Viete-Wallis series (I): found infinite products for π

sin(x) = 2sin(x

2)cos(

= 22sin(x

22)cos(

2)cos(

[n-times application yields]

= 2nsin(x

2n)cos(

2)cos(

22) · · · cos( x

⇒ limn→∞

sin(x)

x= lim

n→∞

sin( x2n )cos(x2 )cos( x

22) · · · cos( x

2n )x2n

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sin(x) = 2sin(x

2)cos(

= 22sin(x

22)cos(

2)cos(

= 2nsin(x

2n)cos(

2)cos(

22) · · · cos( x

⇒ limn→∞

sin(x)

x= lim

n→∞

22) · · · cos( x

2n )x2n

22 / 27

sin(x) = 2sin(x

2)cos(

= 22sin(x

22)cos(

2)cos(

= 2nsin(x

2n)cos(

2)cos(

22) · · · cos( x

⇒ limn→∞

sin(x)

x= lim

n→∞

22) · · · cos( x

2n )x2n

22 / 27

sin(x) = 2sin(x

2)cos(

= 22sin(x

22)cos(

2)cos(

= 2nsin(x

2n)cos(

2)cos(

22) · · · cos( x

⇒ limn→∞

sin(x)

x= lim

n→∞

22) · · · cos( x

2n )x2n

22 / 27

sin(x) = 2sin(x

2)cos(

= 22sin(x

22)cos(

2)cos(

= 2nsin(x

2n)cos(

2)cos(

22) · · · cos( x

⇒ limn→∞

sin(x)

x= lim

n→∞

22) · · · cos( x

2n )x2n

22 / 27

Viete-Wallis series (II): found infinite products for π

As limt→0sin(t)t = 1, we get

sin(x)

x= cos(

2)cos(

22)cos(

23) · · ·

Using cos(θ) =

√1+cos(2θ)

2 and the above infinite product atx = π

√2 +√

√2 +

√2 +√

2· · ·

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sin(x)

x= cos(

2)cos(

22)cos(

23) · · ·

Using cos(θ) =

√1+cos(2θ)

√2 +√

√2 +

√2 +√

2· · ·

23 / 27

sin(x)

x= cos(

2)cos(

22)cos(

23) · · ·

Using cos(θ) =

√1+cos(2θ)

√2 +√

√2 +

√2 +√

2· · ·

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Formal and informal references:Informal References

• Documentaries:• Math and rise of civilizations:http://motionpic.com/catalogue/

math-rise-of-civilization-science-docs-documentaries-2/• BBC: Story of Mathematics: http://www.bbc.co.uk/programmes/b00dxjls/episodes/guide

• Websites:• The history of pi by David Wilsonhttp://sites.math.rutgers.edu/~cherlin/History/

Papers2000/wilson.html• Archimedes’ Approximation of Pihttp://itech.fgcu.edu/faculty/clindsey/mhf4404/

archimedes/archimedes.html• Euclid’s Elementshttps://mathcs.clarku.edu/~djoyce/java/elements/

• Wekipedia: Ludolph Van Ceulen’s biographyhttps://en.wikipedia.org/wiki/Ludolph_van_Ceulen

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Formal and informal references:Formal References

George E. Andrews, Peter Paule. Some questions concerningcomputer-generated proofs of a binomial double-sumidentity, J. Symbolic Comput. 16(1993), 147–151.

P. Backman, The history of Pi, The Golem Press. BoulderColorado, 1971.

J.L.Berggren, J. , Borwein, P. Borwein, Pi: A Source Book,Springer, 2004.

D. Blatner, The joy of Pi, Walker Publishing Company, IncNewyork, 1997.

F. Cajori, A history of Mathematics, MacMillan and Co.London, 1926.

Sir T. Heath,A History of Greek Mathematics: From Thalesto Euclid, Volume 1, Dover Publications, inc. Newyork,1981.

J. J. O’Connor, E.F. Robertson, The MacTutor History ofMathematics Archive,World Wide Web., 1996.

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Acknowledgments

I would specially like to thank:i) Prof. Marion Oliver for his support, interest in history ofmathematics and encouragement for the concept of ‘ExploreMath’.ii) Prof. H. Demirkoparan and Prof. Z. Yilma for their help andsuggestions.iii) Kara, Angela, Catalina and Geetha for promoting the eventand taking care of logistics.iv) Ghost of Ludolph Van Ceulen for haunting me and pushingme to explore more about Pi :).

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Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

A: A half of an egg! Because nothing is better than eternalbliss, and a half of an egg is better than nothing.

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The story of and related puzzles · Time line I: Ancient period The story starts in ancient Egypt...

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