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

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1 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

2 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)

3 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 L = L = 10cm l = cm π 2l N L n = =

4 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

5 3 π π π π π arctan 4 ( π = lim n2 n n n ) n 2 n 2 π = 4 ( ) +... π = lim n π = n (n!) 4 n((2n)!) ( 1 π = ) ( 4( 1) n π = 4 (2n + 1) 5 2n+1 ( 1) n ) (2n + 1) 239 2n+1 n=0 π = ( π = ) π = ( ( ( ( )))) π = ( ( ( ( )))) ( π = 9801 ) 1 (4n)!( n) 8 (n!) n π = ( 12 n=0 n=0 ( 1) n (6n)!( n) (3n!)(n!) n+3/2 ) 1 π = 16 arctan arctan ( 1 4 π = 16 n 8n n n ) 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

6 4 π = /3 = /5 = /7 = 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 ( ) ( N ) l L L > l L 2 L 3 N

7 2: π BC ( 8 ) = BC2000 = BC BC = n 5 π ˆπ 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 x m ) 7

8 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) = π 95% m = 10 [ x s, x s ] (7.2) : 9 t- 2.5% 8 1. L L π 8

9 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, February 23, 1855) b Guinness William S. Gosset 1908 Student Student s t distribution t 9

10 m 1 t g m 1 (x) t g m 1 (x)dx = t g m 1 (x)dx = 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 ( ) = 0.95 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, [2] π 2001 [3] Java Applet 10

11 3: t

t χ 2 F Q t χ 2 F 1 2 µ, σ 2 N(µ, σ 2 ) f(x µ, σ 2 ) = 1 ( exp (x ) µ)2 2πσ 2 2σ 2 0, N(0, 1) (100 α) z(α) t χ 2 *1 2.1 t (i)x N(µ, σ 2 ) x µ σ N(0, 1

t χ 2 F Q t χ 2 F 1 2 µ, σ 2 N(µ, σ 2 ) f(x µ, σ 2 ) = 1 ( exp (x ) µ)2 2πσ 2 2σ 2 0, N(0, 1) (100 α) z(α) t χ 2 *1 2.1 t (i)x N(µ, σ 2 ) x µ σ N(0, 1 t χ F Q t χ F µ, σ N(µ, σ ) f(x µ, σ ) = ( exp (x ) µ) πσ σ 0, N(0, ) (00 α) z(α) t χ *. t (i)x N(µ, σ ) x µ σ N(0, ) (ii)x,, x N(µ, σ ) x = x+ +x N(µ, σ ) (iii) (i),(ii) z = x µ N(0, ) σ N(0, ) ( 9 97.