Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
1
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
2
Netherlands Mathematics Congress NMC
Intro
Ludolph’s π
I
22–23 april
I
Very interesting programme for Master students
I
Friday: History, teaching
I
Friday: Mastermath Lunch
I
Student fee: EUR 15,-
I
www.math.uu.nl/nmc2010
Antiquity
Motivation
Conclusion
3
Van Ceulenjaar 2010
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
1610
Ludolph van Ceulen †
2010
www.ludolphvanceulen.nl
4
Intro
Ludolph’s π
Circle Quadrature: Straight or Crooked
Antiquity
Motivation
Conclusion
Steven Wepster
Departement Wiskunde
Universiteit Utrecht
5
Intro
Intro
Ludolph’s π
Ludolph’s π
Antiquity
Motivation
Conclusion
Antiquity
Motivation
Conclusion
6
Ludolph van Ceulen 1540–1610
Intro
Ludolph’s π
1540 Hildesheim – 1610 Leiden
Taught fencing and reckoning
in Delft and Leiden
Antiquity
Motivation
Conclusion
Later: professor at the
Duytsche Mathematique
Friend of Stevin, Snellius,
Maurits of Orange
International fame for his
‘π’-approximation
7
Stevin and Maurits in the “sailing waggon”
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
8
Approximations of π
Intro
Ludolph’s π
Antiquity
Vanden Circkel
Fondamenten
tombstone
Jamshı̄d al-Kāshı̄
Adriaan van Roomen
François Viète
1596
1615
1610
1424
1593
1593
20Motivation
digits
Conclusion
32
35
16
14
10
9
π in Vanden Circkel
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
10
π in Vanden Circkel
Intro
Ludolph’s π
Antiquity
Principle:
bracket π between in- and
circumscribed polygons
(with emphasis on poly)
by repeated doubling of the
number of sides
Motivation
Conclusion
11
Doubling
Given edge CE = in of inscribed n-gon
To find the edge BC = i2n of 2n-gon
Intro
Ludolph’s π
Antiquity
Motivation
C
Conclusion
A
D
B
E
12
Doubling
Given edge CE = in of inscribed n-gon
To find the edge BC = i2n of 2n-gon
Intro
BC 2 = CD 2 + BD 2
Ludolph’s π
= CD 2 + (AB − AD)2
Motivation
p
= CD 2 + (AB − AC 2 − CD 2 )2 Conclusion
Antiquity
C
A
D
B
E
12
Doubling
Given edge CE = in of inscribed n-gon
To find the edge BC = i2n of 2n-gon
Intro
BC 2 = CD 2 + BD 2
Ludolph’s π
= CD 2 + (AB − AD)2
Motivation
p
= CD 2 + (AB − AC 2 − CD 2 )2 Conclusion
Antiquity
C
Substitute:
A
D
B
BC = i2n
CD = 21 in
AB = AC = 1
E
12
Doubling
Given edge CE = in of inscribed n-gon
To find the edge BC = i2n of 2n-gon
Intro
BC 2 = CD 2 + BD 2
Ludolph’s π
= CD 2 + (AB − AD)2
Motivation
p
= CD 2 + (AB − AC 2 − CD 2 )2 Conclusion
Antiquity
C
Substitute:
A
D
B
BC = i2n
CD = 21 in
AB = AC = 1
E
Yields (after a bit of tweaking):
r
q
i2n = 2 − 4 − in2
12
Inscribed 6-12-etc-gon
i2n =
q
2−
p
4 − in2 applied:
Intro
i6 = 1
q
√
i12 = 2 − 3
r
q
2−
i24 =
s
2+
r
Ludolph’s π
Antiquity
Motivation
Conclusion
√
3
√
2− 2+ 2+ 3
v
s
u
r
u
q
√
t
= 2− 2+ 2+ 2+ 3
q
i48 =
i96
so π > 48i96 > 3.14103
13
To get the circumscribed polygon
Given edge CE = in of inscribed n-gon
To find edge FG = un of circumscribed n-gon
Intro
Ludolph’s π
F
C
Antiquity
Motivation
Conclusion
A
D
B
E
G
14
To get the circumscribed polygon
Given edge CE = in of inscribed n-gon
To find edge FG = un of circumscribed n-gon
Similarity:
F
FG : AB = CE : AD
C
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
A
D
B
E
G
14
To get the circumscribed polygon
Given edge CE = in of inscribed n-gon
To find edge FG = un of circumscribed n-gon
Similarity:
F
FG : AB = CE : AD
C
A
D
Intro
Ludolph’s π
Antiquity
Motivation
B
Pythagoras:
p
AD = AC 2 − CD 2
q
= 1 − ( 12 in )2
q
= 12 4 − in2
Conclusion
E
G
14
To get the circumscribed polygon
Given edge CE = in of inscribed n-gon
To find edge FG = un of circumscribed n-gon
Similarity:
F
FG : AB = CE : AD
C
A
D
Intro
Ludolph’s π
Antiquity
Motivation
B
Pythagoras:
p
AD = AC 2 − CD 2
q
= 1 − ( 12 in )2
q
= 12 4 − in2
Conclusion
so:
E
G
2in
un = p
4 − in2
14
Bracket the circle
un = √2in
4−in2
applied to the inscribed 96-gon
Intro
i96
u96
v
s
u
r
u
q
√
t
= 2− 2+ 2+ 2+ 3
vs
u
r
q
u
p
√
u
u 2− 2+ 2+ 2+ 3
u
= 2u s
r
u
q
p
t
√
2+ 2+ 2+ 2+ 3
Ludolph’s π
Antiquity
Motivation
Conclusion
which implies: 3.14103 < 48i96 < π < 48u96 < 3.14272
1
Archimedes (250 BC):
3 10
71 < π < 3 7
15
Just a bit more digits. . .
Van Ceulen starts with a 15-gon and doubles 31 times.
This gives him the edges of in- and circumscribed
32212254720-gon and he concludes:
Intro
Ludolph’s π
Antiquity
Motivation
When the Diameter of a circle is 1 / then its circumference is
more than
14159265358979323846
3
100000000000000000000
and less than
14159265358979323847
3
.
100000000000000000000
You may get closer if you mind.
(Wie lust heeft can naerder comen)
Conclusion
16
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
17
Excursion: Greek antiquity
Intro
Ludolph’s π
Antiquity
3 classical problems:
Given an angle: divide it in 3 equal parts
Motivation
Conclusion
Given a cube: make a cube of double the volume
Given a circle: make a square of the same area
18
Quadrature ∼ rectification
It was well known that
area of circle =
1
2 circumference × radius.
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
19
Quadrature ∼ rectification
It was well known that
area of circle =
1
2 circumference × radius.
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
Also known:
Given a triangle: make a square of
the same area.
19
Quadrature ∼ rectification
It was well known that
area of circle =
1
2 circumference × radius.
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
Also known:
Given a triangle: make a square of
the same area.
Therefore: an area construction is
equivalent to a circumference
construction.
19
Quadratrix
Intro
Ludolph’s π
D
Antiquity
Deinostratos (ca. 335 vChr) had
already solved the problem
A
P
Motivation
Conclusion
B
20
Quadratrix
Intro
Ludolph’s π
D
Antiquity
Deinostratos (ca. 335 vChr) had
already solved the problem
Motivation
Conclusion
arcDB : AB = AB : AP
therefore AP =
A
P
2
π
B
20
Motivation
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
21
Van Ceulen’s motivation I
Simon van der Eycke 1584:
‘π’ =
39
22
2
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
22
Van Ceulen’s motivation I
Simon van der Eycke 1584:
‘π’ =
39
22
2
Adriaan Anthoniszoon: nonsense!
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
355
‘π’ ≈
113
which he derived from
15
17
3 106
< π < 3 120
22
Van Ceulen’s motivation I
Simon van der Eycke 1584:
‘π’ =
39
22
2
Adriaan Anthoniszoon: nonsense!
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
355
‘π’ ≈
113
which he derived from
15
17
3 106
< π < 3 120
He asks for Van Ceulen’s opinion.
Van Ceulen shows: the
circumscribed 192-gon is already
smaller than Vander Eycke’s circle!
22
Van Ceulen’s motivation II
Vander Eycke makes a new attempt
(1586) and proposes a geometrical
construction which amounts to:
q
√
320 − 8
π=
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
≈ 3.14461
23
Van Ceulen’s motivation II
Vander Eycke makes a new attempt
(1586) and proposes a geometrical
construction which amounts to:
q
√
320 − 8
π=
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
≈ 3.14461
That is even worse!
Meer versuft dan vernuft, (misdone
rather than wisdom) says Van
Ceulen.
23
Van Ceulen’s motivation II
Vander Eycke makes a new attempt
(1586) and proposes a geometrical
construction which amounts to:
q
√
320 − 8
π=
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
≈ 3.14461
That is even worse!
Meer versuft dan vernuft, (misdone
rather than wisdom) says Van
Ceulen.
Next, he commences his study of
Archimedes.
23
What drives Van der Eycke?
Simon vander Eycke = Du Chesne
Born in Dôle, France
Reckoning-master in Delft
1584-86 Circle quadrature
1595 works on Longitude problem
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
(to find geographical longitude at sea)
1603 Improves windmills
24
What drives Van der Eycke?
Simon vander Eycke = Du Chesne
Born in Dôle, France
Reckoning-master in Delft
1584-86 Circle quadrature
1595 works on Longitude problem
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
(to find geographical longitude at sea)
1603 Improves windmills
. . . Apparently eager to solve “major problems”
24
What drives Van der Eycke?
Simon vander Eycke = Du Chesne
Born in Dôle, France
Reckoning-master in Delft
1584-86 Circle quadrature
1595 works on Longitude problem
Intro
Ludolph’s π
Antiquity
Motivation
Conclusion
(to find geographical longitude at sea)
1603 Improves windmills
. . . Apparently eager to solve “major problems”
p√
By the way: Antoniszoon and others link
320 − 8 to
Nikolaus Cusanus. . .
24
Nikolaus Cusanus
Intro
Nicholas of Cusa (a place near
Luxembourg) 1401–1464
Philosopher and Theologian;
Cardinal
Fascinated by infinity
Ludolph’s π
Antiquity
Motivation
Conclusion
25
Nikolaus Cusanus
Intro
Nicholas of Cusa (a place near
Luxembourg) 1401–1464
Philosopher and Theologian;
Cardinal
Fascinated by infinity
Produces an abundance of circle
quadratures
Ludolph’s π
Antiquity
Motivation
Conclusion
(of which most had already been
disproved)
25
A quadrature of Cusanus
A
Intro
Ludolph’s π
E
B
C
F
D
G
Antiquity
Motivation
Conclusion
Choose G so that AG = 2 DF
26
A quadrature of Cusanus
A
Intro
Ludolph’s π
E
B
C
F
D
G
Antiquity
Motivation
Conclusion
Choose G so that AG = 2 DF
Prolong CB so that DE = 4 DF
26
A quadrature of Cusanus
A
Intro
Ludolph’s π
E
B
C
F
D
G
Antiquity
Motivation
Conclusion
Choose G so that AG = 2 DF
Prolong CB so that DE = 4 DF
Assertion:
ED = arc BAC
4EAD = area BACD
26
A quadrature of Cusanus
A
Intro
Ludolph’s π
E
B
C
F
Antiquity
D
G
Motivation
Conclusion
Choose G so that AG = 2 DF
Prolong CB so that DE = 4 DF
Assertion:
ED = arc BAC
4EAD = area BACD
After some work it appears that ED =
p√
320 − 8
26
Etcetera
Intro
Ludolph’s π
Antiquity
Van Ceulen disproves a number of other “exact” quadratures:
Cusanus, Bouvelle, Barbel, Reimers, Scaliger, . . . .
Motivation
Conclusion
His argument is always that the false π-values lie outside his
carefully calculated bounds.
27
Conclusion
Intro
Ludolph’s π
I
Circle Quadrature was a popular hobby, but the difference
between exact and approximate was often neglected.
Antiquity
Motivation
Conclusion
28
Conclusion
Intro
Ludolph’s π
I
I
Circle Quadrature was a popular hobby, but the difference
between exact and approximate was often neglected.
Van Ceulen takes action against faulty reasoning, as a
watchdog of mathematics (waakhond der wiskunde).
Antiquity
Motivation
Conclusion
28
Conclusion
Intro
Ludolph’s π
I
Circle Quadrature was a popular hobby, but the difference
between exact and approximate was often neglected.
I
Van Ceulen takes action against faulty reasoning, as a
watchdog of mathematics (waakhond der wiskunde).
I
Bracketing π between narrow bounds is Ludolph’s weapon
against “onwisse wiskonstenaars”
(mistaken mathematicians).
Antiquity
Motivation
Conclusion
28