Introduction
Computations
Circle quadratures of Kāshānı̄ and Van Ceulen
Motivation
Conclusion
Steven Wepster
Departement Wiskunde
Universiteit Utrecht
1
Introduction
Introduction
Computations
Motivation
Computations
Conclusion
Motivation
Conclusion
2
Ludolph van Ceulen 1540–1610
Introduction
1540 Hildesheim (Germany) –
1610 Leiden (Netherlands)
Computations
Motivation
Conclusion
Teacher of fencing and
reckoning
Professor at new engineering
school
Friend of Stevin, Snellius,
Maurice of Orange
Renowned for his
‘π’-approximations
3
π-approximations
Introduction
Computations
Motivation
Vanden Circkel
Fondamenten
tombstone
Jamshı̄d Kāshānı̄
Adriaan v. Roomen
François Viète
1596
1615
1610
1424
1593
1593
Conclusion
20 digits
32
35
16
14
10
4
π in Vanden Circkel
Introduction
Computations
Motivation
Principle:
Bracket π between in- and
circumscribed polygons (with
emphasis on poly)
by repeated doubling of the
number of sides
Conclusion
5
Doubling the number of sides
Given: edge CE = in of inscribed n-gon
To find: side BC = i2n of 2n-gon
Introduction
Computations
Motivation
Conclusion
C
A
D
B
E
6
Doubling the number of sides
Given: edge CE = in of inscribed n-gon
To find: side BC = i2n of 2n-gon
Introduction
BC 2 = CD 2 + BD 2
Computations
= CD 2 + (AB − AD)2
Conclusion
p
= CD 2 + (AB − AC 2 − CD 2 )2
Motivation
C
A
D
B
E
6
Doubling the number of sides
Given: edge CE = in of inscribed n-gon
To find: side BC = i2n of 2n-gon
Introduction
BC 2 = CD 2 + BD 2
Computations
= CD 2 + (AB − AD)2
Conclusion
p
= CD 2 + (AB − AC 2 − CD 2 )2
Motivation
C
Substitute:
A
D
B
BC = i2n
CD = 21 in
AB = AC = 1
E
6
Doubling the number of sides
Given: edge CE = in of inscribed n-gon
To find: side BC = i2n of 2n-gon
Introduction
BC 2 = CD 2 + BD 2
Computations
= CD 2 + (AB − AD)2
Conclusion
p
= CD 2 + (AB − AC 2 − CD 2 )2
Motivation
C
Substitute:
A
D
B
BC = i2n
CD = 21 in
AB = AC = 1
E
Yields (after some tweaking):
r
q
i2n =
2−
4 − in2
6
Inscribed 6-12-etc-gon
i2n
q
p
= 2 − 4 − in2 applied:
Introduction
i6 = 1
q
√
i12 = 2 − 3
r
q
i24 =
Motivation
Conclusion
√
3
q
√
2+ 2+ 3
2−
v
s
u
r
u
q
√
t
= 2− 2+ 2+ 2+ 3
i48 =
i96
2− 2+
s
r
Computations
so π > 48i96 > 3.14103
7
To get the circumscribed polygon
Given: edge CE = in of inscribed n-gon
To find: side FG = un of circumscribed n-gon
Introduction
Computations
F
C
A
D
Motivation
Conclusion
B
E
G
8
To get the circumscribed polygon
Given: edge CE = in of inscribed n-gon
To find: side FG = un of circumscribed n-gon
By similarity:
F
FG : AB = CE : AD
C
A
D
Introduction
Computations
Motivation
Conclusion
B
E
G
8
To get the circumscribed polygon
Given: edge CE = in of inscribed n-gon
To find: side FG = un of circumscribed n-gon
By similarity:
F
FG : AB = CE : AD
C
A
D
Introduction
Computations
Motivation
Conclusion
B
Pythagoras:
p
AD = AC 2 − CD 2
q
= 1 − ( 12 in )2
q
= 12 4 − in2
E
G
8
To get the circumscribed polygon
Given: edge CE = in of inscribed n-gon
To find: side FG = un of circumscribed n-gon
By similarity:
F
FG : AB = CE : AD
C
A
D
Introduction
Computations
Motivation
Conclusion
B
Pythagoras:
p
AD = AC 2 − CD 2
q
= 1 − ( 12 in )2
q
= 12 4 − in2
so:
E
G
2in
un = p
4 − in2
8
Bracket the circle
un = √2in
4−in2
applied to the 96-gon
Introduction
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
Computations
Motivation
Conclusion
which implies: 3.14103 < 48i96 < π < 48u96 < 3.14272
1
Archimedes (250 BC):
3 10
71 < π < 3 7
9
Just a bit more digits. . .
Van Ceulen starts with a 15-gon and doubles 31 times.
This givs him the edges of in- and circumscribed
32212254720-gon and he concludes:
Introduction
Computations
Motivation
Conclusion
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. . .
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Motivation
Introduction
Computations
Motivation
Conclusion
11
Van Ceulen’s motivation I
Simon van der Eycke 1584:
‘π’ =
39
22
2
Introduction
Computations
Motivation
Conclusion
12
Van Ceulen’s motivation I
Simon van der Eycke 1584:
‘π’ =
39
22
2
Adriaan Anthoniszoon: nonsense!
‘π’ ≈
Introduction
Computations
Motivation
Conclusion
355
113
which he derived from
15
17
3 106
< π < 3 120
12
Van Ceulen’s motivation I
Simon van der Eycke 1584:
‘π’ =
39
22
2
Adriaan Anthoniszoon: nonsense!
‘π’ ≈
Introduction
Computations
Motivation
Conclusion
355
113
which he derived from
15
17
3 106
< π < 3 120
He asks Van Ceulen’s opinion.
Van Ceulen shows that the
circumscribed 192-gon is already
smaller than Vander Eycke’s circle!
12
Van Ceulen’s motivation II
Vander Eycke makes new attempt
(1586) and proposes a geometrical
construction which amounts to:
q
√
π=
320 − 8
Introduction
Computations
Motivation
Conclusion
≈ 3.14461
13
Van Ceulen’s motivation II
Vander Eycke makes new attempt
(1586) and proposes a geometrical
construction which amounts to:
q
√
π=
320 − 8
Introduction
Computations
Motivation
Conclusion
≈ 3.14461
That is even worse!
misdone rather than wisdom
says Van Ceulen, and he commences
his study of Archimedes.
13
Van Ceulen’s motivation II
Vander Eycke makes new attempt
(1586) and proposes a geometrical
construction which amounts to:
q
√
π=
320 − 8
Introduction
Computations
Motivation
Conclusion
≈ 3.14461
That is even worse!
misdone rather than wisdom
says Van Ceulen, and he commences
his study of Archimedes.
p√
Interestingly, Antoniszoon and others relate
320 − 8 to
Nikolaus Cusanus. . .
13
Nikolaus Cusanus
Introduction
Computations
Nicholas of Cusa 1401–1464
Filosopher and theologician;
Roman-Catholic cardinal
Fascinated by infinity
Motivation
Conclusion
14
Nikolaus Cusanus
Introduction
Computations
Nicholas of Cusa 1401–1464
Filosopher and theologician;
Roman-Catholic cardinal
Fascinated by infinity
Produces a diluge of circle
quadratures (most had been disproved
Motivation
Conclusion
before Vander Eycke)
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A quadrature of Cusanus
A
Introduction
E
Computations
B
C
F
D
G
Motivation
Conclusion
Choose G such that AG = 2 DF
15
A quadrature of Cusanus
A
Introduction
E
Computations
B
C
F
D
G
Motivation
Conclusion
Choose G such that AG = 2 DF
Prolong CB so that DE = 4 DF
15
A quadrature of Cusanus
A
Introduction
E
Computations
B
C
F
D
G
Motivation
Conclusion
Choose G such that AG = 2 DF
Prolong CB so that DE = 4 DF
Assertion:
ED = arc BAC
4EAD = area BACD
15
A quadrature of Cusanus
A
Introduction
E
Computations
B
C
F
Motivation
D
G
Conclusion
Choose G such 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
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Further observations
Protection K: Ulugh Beg, vC: Maurice of Orange
Culture K: well-established scholarly tradition;
vC: Europe is awakening
Introduction
Computations
Motivation
Conclusion
Motivation K: fix the ratio of circumference and diameter as
accurately as ever needed; application to
trigonometrical tables
vC: demonstrate faults of circle-squarers;
trigonometry as by-product
Method both based on Archimedes; K. is first significant
improvement ever
Possible practical applications K: astronomy, vC: surveying
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Conclusion
Van Ceulen disproves many other “exact” quadratures:
Cusanus, Bouvelle, Barbel, Reimers, Scaliger, . . . .
Introduction
Computations
Motivation
Conclusion
His argument is always that the false π-values lie outside his
carefully calculated bounds.
Kāshānı̄ sets himself the goal to compute π as accurately as will
ever be needed. He fixes the accuracy before he sets out.
Kāshānı̄’s result could have satisfied Van Ceulen’s purpose, but
he did not know it.
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