D 1 l θ lsinθ y L D 2 2: D 1 y l sin θ (1.2) θ y (1.1) D 1 (1.2) (θ, y) π 0 π l sin θdθ π [0, π] 3 sin cos π l sin θdθ = l π 0 0 π Ldθ = L Ldθ sin θdθ

1 1 (Buffon) 1 l L l < L 1: 2 D 1 D 2 D 1 P 1 2 θ D 1 y θ y π θ π; 0 y L (1.1) 1 Georges Louis Leclerc Comte de Buffon Born: 7 Sept 1707 in Montbard, Cōte d Or, France Died: 16 April 1788 in Paris, France At the age of 20 Georges Buffon discovered the binomial theorem. He corresponded with Cramer on mechanics, geometry, probability, number theory and the differential and integral calculus. His first work Sur le jeu de franc-carreau introduced differential and integral calculus into probability theory. He next wrote Th?orie de la terre and became the most important natural historian of his day having great influence across a wide scientific field. He is remembered most in mathematics for a probability experiment which he carried out calculating π by throwing sticks over his shoulder onto a tiled floor and counting the number of times the sticks fell across the lines between the tiles. This experiment caused much discussion among mathematicians which helped towards an understanding of probability. 1

D 1 l θ lsinθ y L D 2 2: D 1 y l sin θ (1.2) θ y (1.1) D 1 (1.2) (θ, y) π 0 π l sin θdθ π [0, π] 3 sin cos π l sin θdθ = l π 0 0 π Ldθ = L Ldθ sin θdθ = l[ cos θ] π 0 = l[cos θ] π 0 = l(( 1) (1)) = 2l π π π 1dθ = 2πL P P 1 = l πl P = 2P 1 = 2l πl 2 (1.3) (1.4)

L lsinθ π 0 π 3: (1.4) N n n/n P n N 2l πl N n π π 2l N L n π ) ([1], [2]) (1.5) 1: π l N n 2lN/Ln 1850 0.8 5000 2532 3.1596 1855 0.6 3204 1218.5 3.1553 1860 1 600 382.5 3.137 1864 0.75 1030 489 3.1595 1901 0.83 3408 1808 3.1415929 1925 0.5419 2520 859 3.1795 L = 1.0 0.5 L = 10cm l = 7.85398cm π 2l N L n = 2 0.785398 2 = 3.141592 3

2 n 1, 2, 3,... x y n n x y n n 2n (2n+1) (2n+1) = (2n+1) 2 (p, q) (0, 0) n p 2 + q 2 < n 2 (2.1) (2n + 1) 2 n -n n -n 4: (2n) 2 (0, 0) n πn 2 s n π s n (2n + 1) 2 πn2 (2n) 2 = π 4 π 4s n (2n + 1) 2 n (2n + 1) 2 (2.1) π π 4

3 π π π π π arctan 4 ( π = lim n2 n n 2 1 2 + n 2 2 2 + + ) n 2 n 2 π = 4 (1 13 + 15 17 ) +... π = lim n π = 3 + 2 4n (n!) 4 n((2n)!) 2 7 + 15 + 1 + 1 1 1 1 292 + 1 1 + 5 7 + 1 9 11 + ( 1 π = 8 1 3 + 1 ) ( 4( 1) n π = 4 (2n + 1) 5 2n+1 ( 1) n ) (2n + 1) 239 2n+1 n=0 π = 6 1 + 1 4 + 1 9 + 1 16 + ( π = 2 1 + 1 3 + 1 2 3 5 + 1 2 3 3 5 7 + 1 2 3 4 ) 3 5 7 9 + π = 3 + 1 ( 1 + 1 ( 4 + 1 ( 1 + 1 ( )))) 5 + 10 10 10 10 π = 2 + 1 ( 2 + 2 ( 2 + 3 ( 2 + 4 ( )))) 2 + 3 5 7 9 ( π = 9801 ) 1 (4n)!(1103 + 26390n) 8 (n!) 4 396 4n π = ( 12 n=0 n=0 ( 1) n (6n)!(13591409 + 545140134n) (3n!)(n!) 3 640320 3n+3/2 ) 1 π = 16 arctan 1 5 4 arctan 1 239 ( 1 4 π = 16 n 8n + 1 2 8n + 4 1 8n + 5 1 ) 8n + 6 n=0 π Yes, I have a number. π May I have a large container of coffee? π Wie? O! Dies π macht ernstlich so vielen Müh! π 5

4 π 1000 4 = +4.000000........................... 000000 4/3 = 1.333333........................... 333333 4/5 = +0.800000........................... 000000 4/7 = 0.571428........................... 571428 1000 2400 C int a=10000,b,c=8400,d,e,f[8401],g;main(){for(;b-c;) f[b++]=a/5;for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a), e=d%a)for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b);} 5 π π π 2 1200 ( ) 2002 12 06 3 2000. 1 2411 6 1999 9 2061 1 2 1 2411 5 9 600 11 24 6 1 500 ( N ) l L L > l L 2 L 3 N 2 3 3 http://satellite.nikkei.co.jp/news/main/ 6

2: π BC2000 3 + 1 ( 8 ) = 3.125 1 2 16 BC2000 = 3.16049 1 9 BC1200 3 0 BC250 3.14185 3 130 10 = 3.1622 1 1000 1220 3.141818 3 1665 16 1700 10 1706 100 1719 112 1874 527 20 1949 2037 1958 10000 1961 100265 1973 1001250 1982 16777206 1987 134217700 1989 1011196691 1997 51539600000 1999 206158430000 2002 1241100000000 4 n 5 π ˆπ 6 3 5 10 π 10 ˆπ 1, ˆπ 2,, ˆπ 10 7 ˆπ 1, ˆπ 2,, ˆπ 10 95% 7 m x 1, x 2,, x m (ˆπ 1, ˆπ 2,, ˆπ 10 ) 1. x x = 1 m (x 1 + x 2 + + x m ) 7

2. m m (x i x) 2 = x 2 i ( m i=1 x i) 2 = m i=1 i=1 m x 2 i m x 2 i=1 3. s 2 m s 2 i=1 = (x i x) 2 m 1 4. s s = s 2 5. µ 95% x s m µ 95% t 2.5% t(m 1, 0.025) [ ] P x t(m 1, 0.025) s < µ < x + t(m 1, 0.025) s = 0.95 (7.1) m m P [ ] [ ] µ x t(m 1, 0.025)s/ m x + t(m 1, 0.025)s/ m 0.95 t(9, 0.025) = 2.262 π 95% m = 10 [ x 2.262 s, x + 2.262 s ] 10 10 (7.2) 0.3 0.025 0.2 0.1 0.025-4 -2 2 4 5: 9 t- 2.5% 8 1. L L π 8

2. l L N = 100, 500, 1000 ˆπ π n n = n ± 1 π π 3. Student t µ σ 2 ( N(µ, σ 2 ) ) f(x) f(x) = 1 ) (x µ)2 exp ( 2πσ 2σ 2 a n g n (x) = 1 Γ((n + 1)/2) 1 nπ Γ(n/2) (1 + x 2 /n) (n+1)/2 n t b Γ( ) Γ(x) = 0 t x 1 exp( t)dt X 1,..., X m N(µ, σ 2 ) X = m i=1 X i m, S = m i=1 (X i X) 2 m 1 m( X µ) m 1 t S a Johann Carl Friedrich Gauss (April 30, 1777 - February 23, 1855) 10 10 b Guinness William S. Gosset 1908 Student Student s t distribution t 9

m 1 t g m 1 (x) t g m 1 (x)dx = 0.975 + t g m 1 (x)dx = 0.025 t ( t(m 1, 0.025) 2.5% ) t y a t(m 1,0.025) t(m 1,0.025) g m 1 (x)dx = 1 (0.025 + 0.025) = 0.95 m( X µ) t(m 1, 0.025) < < t(m 1, 0.025) S 0.95 X t(m 1, 0.025) S m < µ < X + t(m 1, 0.025) S m 95% µ X t(m 1, 0.025)S/ m X + t(m 1, 0.025)S/ m t(m 1, 0.025) 2 a [1] N.T. Gridgeman, Geometric Probability and the number π. Scripta Mathematica, 25 (1960), 3, 183 195. [2] π 2001 [3] Java Applet http://www.mste.uiuc.edu/reese/buffon/buffon.html 10

3: t 0.10 0.05 0.025 0.01 0.005 0.001 1. 3.078 6.314 12.706 31.821 63.657 318.313 2. 1.886 2.920 4.303 6.965 9.925 22.327 3. 1.638 2.353 3.182 4.541 5.841 10.215 4. 1.533 2.132 2.776 3.747 4.604 7.173 5. 1.476 2.015 2.571 3.365 4.032 5.893 6. 1.440 1.943 2.447 3.143 3.707 5.208 7. 1.415 1.895 2.365 2.998 3.499 4.782 8. 1.397 1.860 2.306 2.896 3.355 4.499 9. 1.383 1.833 2.262 2.821 3.250 4.296 10. 1.372 1.812 2.228 2.764 3.169 4.143 11. 1.363 1.796 2.201 2.718 3.106 4.024 12. 1.356 1.782 2.179 2.681 3.055 3.929 13. 1.350 1.771 2.160 2.650 3.012 3.852 14. 1.345 1.761 2.145 2.624 2.977 3.787 15. 1.341 1.753 2.131 2.602 2.947 3.733 16. 1.337 1.746 2.120 2.583 2.921 3.686 17. 1.333 1.740 2.110 2.567 2.898 3.646 18. 1.330 1.734 2.101 2.552 2.878 3.610 19. 1.328 1.729 2.093 2.539 2.861 3.579 20. 1.325 1.725 2.086 2.528 2.845 3.552 11