Theπ story of and related puzzles

Narrator: Niraj Khare Carnegie Mellon University Qatar

Being with math is being with the truth and eternity!

Oct, 30, 2017

1 / 27 • Late 5th century BCE, Antiphone and Baryson of Heraclea inscribe and circumscribe regular polygons to a circle. • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD. • Around 250 BC, Archimedes proves that 10 1 3.1408 < 3 71 < π < 3 7 ≈ 3.1428.

Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • The Rihnd Papyrus of Ahmes from 1650 BC gives 4(64) approximation π ≈ 81 = 3.16049.

2 / 27 • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD. • Around 250 BC, Archimedes proves that 10 1 3.1408 < 3 71 < π < 3 7 ≈ 3.1428.

Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • The Rihnd Papyrus of Ahmes from 1650 BC gives 4(64) approximation π ≈ 81 = 3.16049. • Late 5th century BCE, Antiphone and Baryson of Heraclea inscribe and circumscribe regular polygons to a circle.

2 / 27 • Around 250 BC, Archimedes proves that 10 1 3.1408 < 3 71 < π < 3 7 ≈ 3.1428.

Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • The Rihnd Papyrus of Ahmes from 1650 BC gives 4(64) approximation π ≈ 81 = 3.16049. • Late 5th century BCE, Antiphone and Baryson of Heraclea inscribe and circumscribe regular polygons to a circle. • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD.

2 / 27 Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • The Rihnd Papyrus of Ahmes from 1650 BC gives 4(64) approximation π ≈ 81 = 3.16049. • Late 5th century BCE, Antiphone and Baryson of Heraclea inscribe and circumscribe regular polygons to a circle. • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD. • Around 250 BC, Archimedes proves that 10 1 3.1408 < 3 71 < π < 3 7 ≈ 3.1428.

2 / 27 Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • The Rihnd Papyrus of Ahmes from 1650 BC gives 4(64) approximation π ≈ 81 = 3.16049. • Late 5th century BCE, Antiphone and Baryson of Heraclea inscribe and circumscribe regular polygons to a circle. • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD. • Around 250 BC, Archimedes proves that 10 1 3.1408 < 3 71 < π < 3 7 ≈ 3.1428.

2 / 27 Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • The Rihnd Papyrus of Ahmes from 1650 BC gives 4(64) approximation π ≈ 81 = 3.16049. • Late 5th century BCE, Antiphone and Baryson of Heraclea inscribe and circumscribe regular polygons to a circle. • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD. • Around 250 BC, Archimedes proves that 10 1 3.1408 < 3 71 < π < 3 7 ≈ 3.1428.

2 / 27 sin(x) lim = 1 x→0 x

Time to pause and ponder

Are we on solid ground?

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

3 / 27 Time to pause and ponder

Are we on solid ground?

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

sin(x) lim = 1 x→0 x

3 / 27 Time to pause and ponder

Are we on solid ground?

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

sin(x) lim = 1 x→0 x

3 / 27 “Cosine rule” but Pythagoras truly rules!

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

Pythagoras

4 / 27 π is a constant.(full stop)

5 / 27 • Some√ mathematician started using inaccurate values such as 10 ≈ 3.1622 and for centuries it continued in India and other places! • Madhava (1340 c.1425) of Sangamagrama (India) found π accurately to 11 decimal places. • Jamshid al-Kashi had calculated π to an accuracy of 16 decimal digits in 1424 AD.

Time line II

• In 263 AD, Liu Hui of China using regular inscribed polygons with sides 12 to 192 showed that 3.14159 < π. • Towrds the end of 5th century AD, Tsu Chung-chih and Tsu keng chih use regular polygons with 24, 576 sides to show 3.1415926 < π < 3.1415927.

6 / 27 • Madhava (1340 c.1425) of Sangamagrama (India) found π accurately to 11 decimal places. • Jamshid al-Kashi had calculated π to an accuracy of 16 decimal digits in 1424 AD.

Time line II

• In 263 AD, Liu Hui of China using regular inscribed polygons with sides 12 to 192 showed that 3.14159 < π. • Towrds the end of 5th century AD, Tsu Chung-chih and Tsu keng chih use regular polygons with 24, 576 sides to show 3.1415926 < π < 3.1415927. • Some√ mathematician started using inaccurate values such as 10 ≈ 3.1622 and for centuries it continued in India and other places!

6 / 27 • Jamshid al-Kashi had calculated π to an accuracy of 16 decimal digits in 1424 AD.

Time line II

• In 263 AD, Liu Hui of China using regular inscribed polygons with sides 12 to 192 showed that 3.14159 < π. • Towrds the end of 5th century AD, Tsu Chung-chih and Tsu keng chih use regular polygons with 24, 576 sides to show 3.1415926 < π < 3.1415927. • Some√ mathematician started using inaccurate values such as 10 ≈ 3.1622 and for centuries it continued in India and other places! • Madhava (1340 c.1425) of Sangamagrama (India) found π accurately to 11 decimal places.

6 / 27 Time line II

• In 263 AD, Liu Hui of China using regular inscribed polygons with sides 12 to 192 showed that 3.14159 < π. • Towrds the end of 5th century AD, Tsu Chung-chih and Tsu keng chih use regular polygons with 24, 576 sides to show 3.1415926 < π < 3.1415927. • Some√ mathematician started using inaccurate values such as 10 ≈ 3.1622 and for centuries it continued in India and other places! • Madhava (1340 c.1425) of Sangamagrama (India) found π accurately to 11 decimal places. • Jamshid al-Kashi had calculated π to an accuracy of 16 decimal digits in 1424 AD.

6 / 27 Time line II

• In 263 AD, Liu Hui of China using regular inscribed polygons with sides 12 to 192 showed that 3.14159 < π. • Towrds the end of 5th century AD, Tsu Chung-chih and Tsu keng chih use regular polygons with 24, 576 sides to show 3.1415926 < π < 3.1415927. • Some√ mathematician started using inaccurate values such as 10 ≈ 3.1622 and for centuries it continued in India and other places! • Madhava (1340 c.1425) of Sangamagrama (India) found π accurately to 11 decimal places. • Jamshid al-Kashi had calculated π to an accuracy of 16 decimal digits in 1424 AD.

6 / 27 Time line II

• In 263 AD, Liu Hui of China using regular inscribed polygons with sides 12 to 192 showed that 3.14159 < π. • Towrds the end of 5th century AD, Tsu Chung-chih and Tsu keng chih use regular polygons with 24, 576 sides to show 3.1415926 < π < 3.1415927. • Some√ mathematician started using inaccurate values such as 10 ≈ 3.1622 and for centuries it continued in India and other places! • Madhava (1340 c.1425) of Sangamagrama (India) found π accurately to 11 decimal places. • Jamshid al-Kashi had calculated π to an accuracy of 16 decimal digits in 1424 AD.

6 / 27 • In 1647, the ratio of circumference of a circle to its diameter gets its name and symbol π by William Oughtred. Made popular by Leonhard Euler. • Time to pause and ponder (II). Ludolph Van Ceulen must be fictional!

Time line III (a): series expressions for π

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

7 / 27 • Time to pause and ponder (II). Ludolph Van Ceulen must be fictional!

Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500 million sides calculated π calculated π to an accuracy of 20 decimal digits by 1596. By the time he died in 1610, he accurately found 35 digits! The digits were carved into his tombstone. • In 1647, the ratio of circumference of a circle to its diameter gets its name and symbol π by William Oughtred. Made popular by Leonhard Euler.

7 / 27 Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500 million sides calculated π calculated π to an accuracy of 20 decimal digits by 1596. By the time he died in 1610, he accurately found 35 digits! The digits were carved into his tombstone. • In 1647, the ratio of circumference of a circle to its diameter gets its name and symbol π by William Oughtred. Made popular by Leonhard Euler. • Time to pause and ponder (II). Ludolph Van Ceulen must be fictional!

7 / 27 Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500 million sides calculated π calculated π to an accuracy of 20 decimal digits by 1596. By the time he died in 1610, he accurately found 35 digits! The digits were carved into his tombstone. • In 1647, the ratio of circumference of a circle to its diameter gets its name and symbol π by William Oughtred. Made popular by Leonhard Euler. • Time to pause and ponder (II). Ludolph Van Ceulen must be fictional!

7 / 27 Ludolph Van Ceulen: 3.14159265358979323846264338327950288...

Ludolph van Ceulen Dutch-German mathematician Ludolph van Ceulen was a German-Dutch mathematician from Hildesheim. He emigrated to the Netherlands. Wikipedia: Born: January 28, 1540, Hildesheim, Germany Died: December 31, 1610, Leiden, Netherlands Known for: pi Institution: Leiden University Notable student: Willebrord Snellius

8 / 27 Archimedes’ approximation of π: Angle bisector

9 / 27 Archimedes’ approximation of π (I): Upper bound

10 / 27 Archimedes’ approximation of π (II) Achimedes’ iteration

11 / 27 OE OA OA ⇒ + = AE AE AF OF OA OA ⇒ + = AF AF AG

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

OC OA OA + = AC AC AD OD OA OA ⇒ + = AD AD AE

12 / 27 OF OA OA ⇒ + = AF AF AG

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

OC OA OA + = AC AC AD OD OA OA ⇒ + = AD AD AE OE OA OA ⇒ + = AE AE AF

12 / 27 Archimedes’ approximation of π (III) Repeated use of Achimedes’ iteration

OC OA OA + = AC AC AD OD OA OA ⇒ + = AD AD AE OE OA OA ⇒ + = AE AE AF OF OA OA ⇒ + = AF AF AG

12 / 27 Archimedes’ approximation of π (III) Repeated use of Achimedes’ iteration

OC OA OA + = AC AC AD OD OA OA ⇒ + = AD AD AE OE OA OA ⇒ + = AE AE AF OF OA OA ⇒ + = AF AF AG

12 / 27 Archimedes’ approximation of π (III) Repeated use of Achimedes’ iteration

OC OA OA + = AC AC AD OD OA OA ⇒ + = AD AD AE OE OA OA ⇒ + = AE AE AF OF OA OA ⇒ + = AF AF AG

12 / 27 1162 1 OD OA OA 8 < + = 153 AD AD AE 1172 1 1162 1 2334 1 OE OA OA 8 + 8 = 4 < + = 153 153 153 AE AE AF 2339 1 2334 1 4673 1 OF OA OA 4 + 4 = 2 < + = 153 153 153 AF AF AG

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

571 OC OA OA < + = 153 AC AC AD s 591 1 (571)2 + (153)2  OD 8 < < 153 (153)2 AD

13 / 27 1172 1 1162 1 2334 1 OE OA OA 8 + 8 = 4 < + = 153 153 153 AE AE AF 2339 1 2334 1 4673 1 OF OA OA 4 + 4 = 2 < + = 153 153 153 AF AF AG

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

571 OC OA OA < + = 153 AC AC AD s 591 1 (571)2 + (153)2  OD 8 < < 153 (153)2 AD 1162 1 OD OA OA 8 < + = 153 AD AD AE

13 / 27 2339 1 2334 1 4673 1 OF OA OA 4 + 4 = 2 < + = 153 153 153 AF AF AG

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

571 OC OA OA < + = 153 AC AC AD s 591 1 (571)2 + (153)2  OD 8 < < 153 (153)2 AD 1162 1 OD OA OA 8 < + = 153 AD AD AE 1172 1 1162 1 2334 1 OE OA OA 8 + 8 = 4 < + = 153 153 153 AE AE AF

13 / 27 Archimedes’ approximation of π (III)(b) Pythagoras comes to rescue!

571 OC OA OA < + = 153 AC AC AD s 591 1 (571)2 + (153)2  OD 8 < < 153 (153)2 AD 1162 1 OD OA OA 8 < + = 153 AD AD AE 1172 1 1162 1 2334 1 OE OA OA 8 + 8 = 4 < + = 153 153 153 AE AE AF 2339 1 2334 1 4673 1 OF OA OA 4 + 4 = 2 < + = 153 153 153 AF AF AG

13 / 27 Archimedes’ approximation of π (III)(b) Pythagoras comes to rescue!

571 OC OA OA < + = 153 AC AC AD s 591 1 (571)2 + (153)2  OD 8 < < 153 (153)2 AD 1162 1 OD OA OA 8 < + = 153 AD AD AE 1172 1 1162 1 2334 1 OE OA OA 8 + 8 = 4 < + = 153 153 153 AE AE AF 2339 1 2334 1 4673 1 OF OA OA 4 + 4 = 2 < + = 153 153 153 AF AF AG

13 / 27 Archimedes’ approximation of π (III)(b) Pythagoras comes to rescue!

571 OC OA OA < + = 153 AC AC AD s 591 1 (571)2 + (153)2  OD 8 < < 153 (153)2 AD 1162 1 OD OA OA 8 < + = 153 AD AD AE 1172 1 1162 1 2334 1 OE OA OA 8 + 8 = 4 < + = 153 153 153 AE AE AF 2339 1 2334 1 4673 1 OF OA OA 4 + 4 = 2 < + = 153 153 153 AF AF AG

13 / 27 96 × 153 < 1 4673 2 14688 = 1 4673 2 1 667 2 = 3 + 1 4673 2 1 < 3 + 7

Archimedes’ approximation of π: upper bound

1 OA 4673 2 AG 153 As AG > 153 , OA < 1 . Thus, 4673 2

P erimeter of the circle 96 times (2 times the length of AG) diameter < 2 times length of OA

14 / 27 14688 = 1 4673 2 1 667 2 = 3 + 1 4673 2 1 < 3 + 7

Archimedes’ approximation of π: upper bound

1 OA 4673 2 AG 153 As AG > 153 , OA < 1 . Thus, 4673 2

P erimeter of the circle 96 times (2 times the length of AG) diameter < 2 times length of OA 96 × 153 < 1 4673 2

14 / 27 1 < 3 + 7

Archimedes’ approximation of π: upper bound

1 OA 4673 2 AG 153 As AG > 153 , OA < 1 . Thus, 4673 2

P erimeter of the circle 96 times (2 times the length of AG) diameter < 2 times length of OA 96 × 153 < 1 4673 2 14688 = 1 4673 2 1 667 2 = 3 + 1 4673 2

14 / 27 Archimedes’ approximation of π: upper bound

1 OA 4673 2 AG 153 As AG > 153 , OA < 1 . Thus, 4673 2

P erimeter of the circle 96 times (2 times the length of AG) diameter < 2 times length of OA 96 × 153 < 1 4673 2 14688 = 1 4673 2 1 667 2 = 3 + 1 4673 2 1 < 3 + 7

14 / 27 Archimedes’ approximation of π: upper bound

1 OA 4673 2 AG 153 As AG > 153 , OA < 1 . Thus, 4673 2

P erimeter of the circle 96 times (2 times the length of AG) diameter < 2 times length of OA 96 × 153 < 1 4673 2 14688 = 1 4673 2 1 667 2 = 3 + 1 4673 2 1 < 3 + 7

14 / 27 Archimedes’ approximation of π: Lower bound

15 / 27 Archimedes’ approximation of π: Lower bound

16 / 27 t3 t5 t7 t9 arctan(t) = t − + − + + ··· . 3 5 7 9 π 1 1 1 1 = 1 − + − + + ··· 4 3 5 7 9 .

Infinite series era: time line III(a) continues

• French mathematician Viete (1540 1603) and later in 1650 John Wallis found infinite products for π. • By 1682, James Gregory and Leibniz found a famous “useless” series:

17 / 27 π 1 1 1 1 = 1 − + − + + ··· 4 3 5 7 9 .

Infinite series era: time line III(a) continues

• French mathematician Viete (1540 1603) and later in 1650 John Wallis found infinite products for π. • By 1682, James Gregory and Leibniz found a famous “useless” series:

t3 t5 t7 t9 arctan(t) = t − + − + + ··· . 3 5 7 9

17 / 27 Infinite series era: time line III(a) continues

• French mathematician Viete (1540 1603) and later in 1650 John Wallis found infinite products for π. • By 1682, James Gregory and Leibniz found a famous “useless” series:

t3 t5 t7 t9 arctan(t) = t − + − + + ··· . 3 5 7 9 π 1 1 1 1 = 1 − + − + + ··· 4 3 5 7 9 .

17 / 27 Infinite series era: time line III(a) continues

• French mathematician Viete (1540 1603) and later in 1650 John Wallis found infinite products for π. • By 1682, James Gregory and Leibniz found a famous “useless” series:

t3 t5 t7 t9 arctan(t) = t − + − + + ··· . 3 5 7 9 π 1 1 1 1 = 1 − + − + + ··· 4 3 5 7 9 .

17 / 27 Infinite series era: time line III(a) continues

• French mathematician Viete (1540 1603) and later in 1650 John Wallis found infinite products for π. • By 1682, James Gregory and Leibniz found a famous “useless” series:

t3 t5 t7 t9 arctan(t) = t − + − + + ··· . 3 5 7 9 π 1 1 1 1 = 1 − + − + + ··· 4 3 5 7 9 .

17 / 27 • π 120 1 1 1 = acrtan( )−acrtan( ) = 4 arctan( )−arctan( ). 4 119 239 5 239

• π 1 1 1 1  = 4 − + − + ··· 4 5 3(5)3 5(5)5 7(5)7  1 1 1 1  − − + − + ··· . 239 3(239)3 5(239)5 7(239)7

Time line III (b): series expressions for π

• Useless for 10000 terms are required to get four accurate digits! To compute 100 digits ”you need to add up more terms than there are particles in the universe” [Blanter, page 42]. • In 1706, an English professor of Astronomy, John Machin x+y using arctan(x) + arctan(y) = arctan( 1−xy ) found:

18 / 27 • π 1 1 1 1  = 4 − + − + ··· 4 5 3(5)3 5(5)5 7(5)7  1 1 1 1  − − + − + ··· . 239 3(239)3 5(239)5 7(239)7

Time line III (b): series expressions for π

• Useless for 10000 terms are required to get four accurate digits! To compute 100 digits ”you need to add up more terms than there are particles in the universe” [Blanter, page 42]. • In 1706, an English professor of Astronomy, John Machin x+y using arctan(x) + arctan(y) = arctan( 1−xy ) found: • π 120 1 1 1 = acrtan( )−acrtan( ) = 4 arctan( )−arctan( ). 4 119 239 5 239

18 / 27 Time line III (b): series expressions for π

• Useless for 10000 terms are required to get four accurate digits! To compute 100 digits ”you need to add up more terms than there are particles in the universe” [Blanter, page 42]. • In 1706, an English professor of Astronomy, John Machin x+y using arctan(x) + arctan(y) = arctan( 1−xy ) found: • π 120 1 1 1 = acrtan( )−acrtan( ) = 4 arctan( )−arctan( ). 4 119 239 5 239

• π 1 1 1 1  = 4 − + − + ··· 4 5 3(5)3 5(5)5 7(5)7  1 1 1 1  − − + − + ··· . 239 3(239)3 5(239)5 7(239)7

18 / 27 Time line III (b): series expressions for π

• Useless for 10000 terms are required to get four accurate digits! To compute 100 digits ”you need to add up more terms than there are particles in the universe” [Blanter, page 42]. • In 1706, an English professor of Astronomy, John Machin x+y using arctan(x) + arctan(y) = arctan( 1−xy ) found: • π 120 1 1 1 = acrtan( )−acrtan( ) = 4 arctan( )−arctan( ). 4 119 239 5 239

• π 1 1 1 1  = 4 − + − + ··· 4 5 3(5)3 5(5)5 7(5)7  1 1 1 1  − − + − + ··· . 239 3(239)3 5(239)5 7(239)7

18 / 27 • In 1873, William Shanks used the formula to calculate 707 digits of which only the first 527 were correct. [Berggren, page 627] • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

Time line IV: squaring the circle put to rest

• Using the above formula Machin could calculate π accurately till 100 places by hand! • Using the same above formula many mathematicians for next 150 years found more and more digits of π.

19 / 27 • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

Time line IV: squaring the circle put to rest

• Using the above formula Machin could calculate π accurately till 100 places by hand! • Using the same above formula many mathematicians for next 150 years found more and more digits of π. • In 1873, William Shanks used the formula to calculate 707 digits of which only the first 527 were correct. [Berggren, page 627]

19 / 27 • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

Time line IV: squaring the circle put to rest

• Using the above formula Machin could calculate π accurately till 100 places by hand! • Using the same above formula many mathematicians for next 150 years found more and more digits of π. • In 1873, William Shanks used the formula to calculate 707 digits of which only the first 527 were correct. [Berggren, page 627] • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246]

19 / 27 Time line IV: squaring the circle put to rest

• Using the above formula Machin could calculate π accurately till 100 places by hand! • Using the same above formula many mathematicians for next 150 years found more and more digits of π. • In 1873, William Shanks used the formula to calculate 707 digits of which only the first 527 were correct. [Berggren, page 627] • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

19 / 27 Time line IV: squaring the circle put to rest

• Using the above formula Machin could calculate π accurately till 100 places by hand! • Using the same above formula many mathematicians for next 150 years found more and more digits of π. • In 1873, William Shanks used the formula to calculate 707 digits of which only the first 527 were correct. [Berggren, page 627] • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

19 / 27 • Now using super computers and faster algorithms π is known over few trillion digits.

Time line V(a): π in modern times

• By 1949 using hand calculations and toil of many years, we knew first 1000 digits correctly! • In 1949 the ENIAC (Electronic, Numerical, Intrigrator and Caculator) was built that calculated the first 2037 digits of π in less than 70 hours! [Beckmann, page 180]

20 / 27 Time line V(a): π in modern times

• By 1949 using hand calculations and toil of many years, we knew first 1000 digits correctly! • In 1949 the ENIAC (Electronic, Numerical, Intrigrator and Caculator) was built that calculated the first 2037 digits of π in less than 70 hours! [Beckmann, page 180] • Now using super computers and faster algorithms π is known over few trillion digits.

20 / 27 Time line V(a): π in modern times

• By 1949 using hand calculations and toil of many years, we knew first 1000 digits correctly! • In 1949 the ENIAC (Electronic, Numerical, Intrigrator and Caculator) was built that calculated the first 2037 digits of π in less than 70 hours! [Beckmann, page 180] • Now using super computers and faster algorithms π is known over few trillion digits.

20 / 27 • To quench the curiosity and to test if the given digits are from π.

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

• Just 39 decimal places would be enough to compute the circumference of a circle surrounding the known universe to within the radius of hydrogen atom. [Berggren, 656] • At present time the only tangible application of all those digits is to test the computers and computer chips for bugs. [The history of pi by Devid Wilson]

21 / 27 Time line V(b): Do we need to know any more?

• Just 39 decimal places would be enough to compute the circumference of a circle surrounding the known universe to within the radius of hydrogen atom. [Berggren, 656] • At present time the only tangible application of all those digits 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 are from π.

21 / 27 Time line V(b): Do we need to know any more?

• Just 39 decimal places would be enough to compute the circumference of a circle surrounding the known universe to within the radius of hydrogen atom. [Berggren, 656] • At present time the only tangible application of all those digits 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 are from π.

21 / 27 [n-times application yields] . . x x x x = 2nsin( )cos( )cos( ) ··· cos( ) 2n 2 22 2n x x x x sin(x) sin( n )cos( )cos( 2 ) ··· cos( n ) ⇒ lim = lim 2 2 2 2 n→∞ n→∞ x x 2n

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

x x sin(x) = 2sin( )cos( ) 2 2 x x x = 22sin( )cos( )cos( ) 22 2 22

22 / 27 . . x x x x = 2nsin( )cos( )cos( ) ··· cos( ) 2n 2 22 2n x x x x sin(x) sin( n )cos( )cos( 2 ) ··· cos( n ) ⇒ lim = lim 2 2 2 2 n→∞ n→∞ x x 2n

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

x x sin(x) = 2sin( )cos( ) 2 2 x x x = 22sin( )cos( )cos( ) 22 2 22 [n-times application yields]

22 / 27 x x x x sin(x) sin( n )cos( )cos( 2 ) ··· cos( n ) ⇒ lim = lim 2 2 2 2 n→∞ n→∞ x x 2n

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

x x sin(x) = 2sin( )cos( ) 2 2 x x x = 22sin( )cos( )cos( ) 22 2 22 [n-times application yields] . . x x x x = 2nsin( )cos( )cos( ) ··· cos( ) 2n 2 22 2n

22 / 27 Viete-Wallis series (I): found infinite products for π

x x sin(x) = 2sin( )cos( ) 2 2 x x x = 22sin( )cos( )cos( ) 22 2 22 [n-times application yields] . . x x x x = 2nsin( )cos( )cos( ) ··· cos( ) 2n 2 22 2n x x x x sin(x) sin( n )cos( )cos( 2 ) ··· cos( n ) ⇒ lim = lim 2 2 2 2 n→∞ n→∞ x x 2n

22 / 27 Viete-Wallis series (I): found infinite products for π

x x sin(x) = 2sin( )cos( ) 2 2 x x x = 22sin( )cos( )cos( ) 22 2 22 [n-times application yields] . . x x x x = 2nsin( )cos( )cos( ) ··· cos( ) 2n 2 22 2n x x x x sin(x) sin( n )cos( )cos( 2 ) ··· cos( n ) ⇒ lim = lim 2 2 2 2 n→∞ n→∞ x x 2n

22 / 27 √ p √ q p √ 2 2 2 + 2 2 + 2 + 2 = ··· π 2 2 2

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

sin(t) As limt→0 t = 1, we get sin(x) x x x = cos( )cos( )cos( ) ··· x 2 22 23

q 1+cos(2θ) Using cos(θ) = 2 and the above infinite product at π x = 2 ,

23 / 27 Viete-Wallis series (II): found infinite products for π

sin(t) As limt→0 t = 1, we get sin(x) x x x = cos( )cos( )cos( ) ··· x 2 22 23

q 1+cos(2θ) Using cos(θ) = 2 and the above infinite product at π x = 2 ,

√ p √ q p √ 2 2 2 + 2 2 + 2 + 2 = ··· π 2 2 2

23 / 27 Viete-Wallis series (II): found infinite products for π

sin(t) As limt→0 t = 1, we get sin(x) x x x = cos( )cos( )cos( ) ··· x 2 22 23

q 1+cos(2θ) Using cos(θ) = 2 and the above infinite product at π x = 2 ,

√ p √ q p √ 2 2 2 + 2 2 + 2 + 2 = ··· π 2 2 2

23 / 27 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 Wilson http://sites.math.rutgers.edu/~cherlin/History/ Papers2000/wilson.html • Archimedes’ Approximation of Pi http://itech.fgcu.edu/faculty/clindsey/mhf4404/ archimedes/archimedes.html • Euclid’s Elements https://mathcs.clarku.edu/~djoyce/java/elements/ • Wekipedia: Ludolph Van Ceulen’s biography https://en.wikipedia.org/wiki/Ludolph_van_Ceulen

24 / 27 Formal and informal references: Formal References

George E. Andrews, Peter Paule. Some questions concerning computer-generated proofs of a binomial double-sum identity, J. Symbolic Comput. 16(1993), 147–151. P. Backman, The history of Pi, The Golem Press. Boulder Colorado, 1971. J.L.Berggren, J. , Borwein, P. Borwein, Pi: A Source Book, Springer, 2004. D. Blatner, The joy of Pi, Walker Publishing Company, Inc Newyork, 1997. F. Cajori, A history of Mathematics, MacMillan and Co. London, 1926. Sir T. Heath,A History of Greek Mathematics: From Thales to Euclid, Volume 1, Dover Publications, inc. Newyork,

1981. 25 / 27 J. J. O’Connor, E.F. Robertson, The MacTutor History of Mathematics Archive,World Wide Web., 1996. Acknowledgments

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

26 / 27 A: A half of an egg! Because nothing is better than eternal bliss, and a half of an egg is better than nothing.

Something to carry home!

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

27 / 27 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 eternal bliss, and a half of an egg is better than nothing.

27 / 27 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 eternal bliss, and a half of an egg is better than nothing.

27 / 27