Download Archimede`s method for approximating Pi

Calculating Pi
From Antiquity to the Present
by George Beck and Michael Trott
Why is p the same in C = 2 p r and A = p r 2 ?
Approximating p Using Areas of Polygons—the Method of Archimedes
Sandwiching p
A circle of radius r has area p r2 , so the area of a circle of radius 1 is p. Therefore p is sandwiched between the areas of the
regular n-sided polygons inscribed and circumscribed around the circle. PiPolygons[n] draws the three figures and
prints the corresponding inequality for the areas.
PiPolygons[5];
2.37764 § p § 3.63271
CalculatingPi.nb
2
‡ Finding the Area of the Inscribed Polygon
Break the inscribed polygon into n triangles of height Cos@a ê 2D and base 2 Sin@a ê 2D , where a is the angle 2 p ê n.
The area is n µ 1 ê 2 µ base µ height = n ê 2 H2 Sin@a ê 2D Cos@a ê 2DL = n ê 2 Sin@aD .
‡ Finding the Area of the Circumscribed Polygon
For the circumscribed polygon, the n triangles each have height 1 and base 2 Tan@a ê 2D . The area is
n µ 1 ê 2 µ 2 µ Tan@a ê 2D = n Tan@a ê 2D .
This shows a triangle from an inscribed and circumscribed 7-sided polygon.
PiTriangles@7D;
Vieta's Formula for p
We continue the discussion on approximating p by calculating the areas of a circle using inscribed and circumscribed
regular polygons. We illustrate Vieta's formula, developed in 1593, the oldest exact result derived for p.
The Formula
Vieta's formula expresses ÅÅÅÅp2 as an infinite product of nested square roots.
CalculatingPi.nb
3
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j
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z
j ÅÅÅÅ
z
ÅÅÅÅ ã j ÅÅÅÅ 2 z j
2+ 2 z
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2+ 2+ 2 z
…
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j2
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p
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Quality of Approximation
The graph in the middle is the approximation from Vieta's formula, which calculates the area of a regular polygon of 2n+1
sides. In contrast, the upper and lower graphs show the estimates to p based on the areas of the n-sided regular polygons
circumscribed and inscribed around a circle of radius 1.
PiSqueeze@100D;
3.18
3.16
3.14
3.12
20
40
60
80
100
The second column shows the number of sides of the regular polygons, while the third shows the approach to p. The
numbers in the fourth column are the differences between p and the partial products of the infinite product.
CalculatingPi.nb
4
TableFormA
2
TableA9n, 2n+1 , tmp = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ , p - tmp=, 8n, 15<EE
p
NProductACosA ÅÅÅÅÅÅ
E, 8i, 2, n + 1<E
i
2
1
2
3
4
5
6
7
8
4
8
16
32
64
128
256
512
2.82843
3.06147
3.12145
3.13655
3.14033
3.14128
3.14151
3.14157
0.313166
0.0801252
0.0201475
0.00504416
0.0012615
0.000315403
0.0000788524
0.0000197132
9
10
11
12
13
14
15
1024
2048
4096
8192
16384
32768
65536
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
4.92831 µ 10-6
1.23208 µ 10-6
3.0802 µ 10-7
7.70049 µ 10-8
1.92512 µ 10-8
4.81281 µ 10-9
1.2032 µ 10-9
Why is It True?
Computing p with Series
Computing p by Iterative Processes
Implementation