ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.
|
|
|
- ひでか さわい
- 5 years ago
- Views:
Transcription
1 (1 C205) 4 10 (2 C206) 4 11 (2 B202) (2013) , 2007 ( ).,. 2. P. G., J. C., ,,.
2 ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.
3 (1). (2).,. (3),,. 1.2 ( ), Ω., E. E Ω. ( ), E P (E) = E Ω..,.., Ω. 1.1 ( ) ( ). ( )., , 2 (A,K,Q,J) , 1. [1023/1024] , ? [1/221, 1/270725]
4 ( ) , 1, , 3. [2/10] , 10. [2/10] 1.4 ( ) A,B 2. A 2/5, B 3/5. 3, A 2, B 1.? [.] 1.5 ( ) ( ). 3, ( ) 2.,. 1.,, 1 ( ).? 1.5 A,B , A 3, B 2,? 1.3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ).. ( ) ( ) ( ) ( ), ( ), ( ) ( )
5 : Ω: ( ) = (, F: ( ) P : 2.1 ( ), Ω., E P (E) = E Ω,. 2.2 (Ω ( ) ),,., P (X = k) = λk k! e λ, k = 0, 1, 2,...,., λ > 0. λ. 2.3 (Ω ). 2. 1, ( ), , 3.
6 , 30cm 40cm, 5cm. 2.4 ( )?, 3 1 : 2 : Ω E. P (E) = E Ω,..,,,, ,. 2 (, ). 2.2 E P (E), 3, P Ω., P (E) E. (i) 0 P (E) 1. (ii) P (Ω) = 1. (iii) [ ] E 1, E 2, F (, i j E i E j = ), ( ) P E n = P (E n ). n=1, 3 (Ω, F, P ). [ ] Ω,. ω Ω n=1
7 E F (E = F. a < b.) Ω E c E F, E 1 E n E F, E 1 E n E F = 2.3 AB 3. (, 3 3.) B A O [ ], ( )?. 1 2 ( ) ,. (1) ( A,K,Q,J,10) (2) (3) , 00001,..., (1) 9 1.
8 6 2 (2) 9 2. (3) 0, 1,..., (4) 0, 1,..., ( ) , 3. 4 A,B 2. A p, B q = 1 p. 5, A 3, B 2.? 5 1 P, 2 P. 1/3. 6 λ.? [ e λ e λ ] 7 A,B,C,. A,B,C.
9 ,. ( ). (1) 1, 0. (2). (3) 5. (4). (5) 1, ( )., x, y, z, t,...., 0 x 1, x 0 1.,,.,,.., X, Y, Z, T,....,,,, ( ) 3.1 X, X {1, 2, 3, 4, 5, 6}., P (X = 1) = P (X = 2) = = P (X = 6) = 1 6
10 8 3. X,, X ( ). X {a 1, a 2,..., a i,... }, P (X = a i ) = p i, i = 1, 2,...,, X., ( ) X., p i 0, p i = X. (1) 1, 0 X. (2) 2 X. i L, X, X. X L/2 X L., x,, P (X = x) = 0, 3.1.,. X ( ), F (x) = P (X x), x R, X.. ( 3.2 ), 0, x L/2, 2x L F (x) =, L/2 x L, L 1, x L,
11 3.4. ( ) X, X. F (x). (1) [ ] x 1 x 2 F (x 1 ) F (x 2 ). (2) lim F (x) = 0, lim F (x) = 1. x x (3) [ ] lim ϵ +0 F (x + ϵ) = F (x). 3.4 ( ) 3.1 X, P (X x) = F (x) = x f(t)dt F (x) = f(x) f(x) X. (F (x).), P (a X b) = b a f(x)dx. f(x),. 3.4 ( 3.2 ) L, X. X. 3.2, f(x). (1) f(x) 0. (2) + f(x)dx = X , X. X. 3.4 O R 1, O X. X.
12
13 p n, X ( ) n P (X = k) = p k (1 p) n k, k = 0, 1, 2,.... k, B(n, p). 4.1 B(4, 1/2) B(4, 1/4) p, X P (X = k) = p(1 p) k, k = 0, 1, 2,.... p. ( ), p, ( 1 ) Y. P (Y = k) = p(1 p) k 1, k = 1, 2, X λ > 0, P (X = k) = λk k! e λ, k = 0, 1, 2, λ = 0.5, λ = 1, λ = 2.
14 (1) 1. (2) X {a 1, a 2,..., }, p i = P (X = a i ), p i 0, p i = 1 (p i = 0 a i, p i = 0 )., i m = i a i p i, σ 2 = i (a i m) 2 p i = i a 2 i p i m 2. X, X,, E[X], V[X] , 2 100, 1 50, ,. (m) (σ 2 ) (2 ) B(1, p) p p(1 p) B(n, p) np np(1 p) ( p) (1 p)/p (1 p)/p 2 ( λ) λ λ ,.
15 ( ) {0, 1, 2,... } X, G(z) = z k P (X = k) k=0 X ( X )., E(X) = G (1), E(X 2 ) = G (1) + G (1), V(X) = G (1) + G (1) G (1) 2.,. 9 2 ( ) X, ( ) Y. X Y. 4.2 f(x) [a, b] 1. 1 f(x) = b a, a x b 0, λ > 0 f(x) = { λe λx, x 0 0, x < ( ) N(m, σ 2 ): m, σ 2 ( ) { } 1 f(x) = exp (x m)2 2πσ 2 2σ 2 N(0, 1):
16 14 4, χ 2 -( ), t-, F - ( ) f(x) m = xf(x) dx, σ 2 = (x m) 2 f(x) dx = x 2 f(x) dx m 2. X, X,, E[X], V[X].. (m) (σ 2 ) [a, b] (a + b)/2 (b a) 2 /12 ( λ) 1/λ 1/λ 2 N(m, σ 2 ) m σ 2 4.4,.,, ( ). + e x2 dx = π 4.5 L 2, X. X,,,. 10 1, X. X,,,. 11 O R 1, O X. X,,,.
17 A, B 2. P (A) > 0, A B P (B A) = P (A B) P (A) 5.1 ( ) 10, ,,? E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = P (E c ), P (E F c ), P ((E F c ) c ), P (E F ), P (E F c ), P (E F E F ) T, P (T m + n T m) = P (T n), m, n = 0, 1, 2,..., X, P (X a + b X a) = P (X b), a, b 0,.
18 Ω = A 1 A 2, A 1 A 2 =, B, P (A 1 B) = P (A 1 )P (B A 1 ) P (A 1 )P (B A 1 ) + P (A 2 )P (B A 2 ) ( ). 5.2, A B, 95%, 2% , A ? 12, A B, 95%, 100p %... p , 5, 2, 5.? (2,.) 14 2 E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = P (E c ), P (E F c ), P ((E F c ) c ), P (E F ), P (E F c ), P (E F E F ) 15 ( ) (1) 1., 6. (2) 1., 6.
19 17 第 6 章 正規分布 6.1 標準正規分布 N (0, 1): 標準正規分布 例 題 6.1 確率変数 Z の分布が標準正規分布である (このことを Z N (0, 1) と書く) とする. 標準正規分布表を用いて, (1) 次の確率を求めよ: P (Z 1.15), P (Z 1.23), P ( Z < 2.4) (2) 次の等式が成り立つような a を求めよ. P (Z a) = 0.33, P (Z < a) = 0.75, P ( Z a) = 0.4 問 6.1 Z N (0, 1) に対して, 確率 P (Z 1.82), P (Z 2.13), P ( Z > 1.5) を求めよ. 問 6.2 Z N (0, 1) とするとき, 次の等式が成り立つような a を求めよ. P (Z a) = 0.39, P (Z < a) = 0.91, P ( Z a) = 0.72 定 理 6.1 (標準化) X N (m, σ 2 ) のとき, Z= X m N (0, 1) σ 例 題 6.2 X N (2, 52 ) のとき, P (X 3), P (X 0), P ( X 4) を求めよ. 問 6.3 (1) 確率変数 X が正規分布 N (20, 42 ) に従うとき, P (X > 17.8) を求めよ. (2) 確率変数 Y が正規分布 N ( 2, 52 ) に従うとき, P ( Y 1) を求めよ. 問 6.4 X N (50, 102 ) のとき, P (X > a) = を満たす a を求めよ.
20 18 6 ( )., x = x 1 y = y 1, x = x 2 y = y 2, x 1 < x < x 2 y : y = y 2 y 1 x 2 x 1 (x x 1 ) + y B(100, 0.4) 6.2,. B(n, p) N(np, np(1 p)), 0 < p < 1, n , 225 ( ). 6.5 (1) 1000, 550. (2) 250, , 225.? 6.3 ( ),, ( ).,, n, x 1, x 2,..., x n 1 n n i=1 x i
21 ,.,,? 1 ( ). X. X.,, 1 X 1, 2 X 2,..., n X n., X 1, X 2,..., X n n.,., n, n. 6.3 m, σ 2, : X = 1 n ) X k N (m, σ2. n n k=1 6.4 n,. 16 X N(0, 1), P (X x) = 1 2π x e t2 /2 dt, Y = ax + b., a 0, b. (a ) , 23.5 ( ) ?
22 20 6 I(z) = 1 2π z 0 e x2 /2 dx z
23 ,? ( ) 5 12 ( ) ( ) 10 (%) 13/05/06( ) 21: /05/07( ) 8: /05/12( ) 20: /05/09( ) 22: /05/11( ) 21: /05/11( ) 21: /05/11( ) 21: /05/08( ) 22: /05/06( ) 14: /05/07( ) 22: ,., 27, PM.,, PM 600, 200. ( ) : ( ) 7.2 E 2. ( E p). X 1, X 2,..., X n n ( p ) { 1, i E, X i = 0, i E
24 22 7 ˆp = 1 n, ˆp ( ) (! ˆp )., ˆp,. n k=1, n X 1, X 2,..., X n, X = 1 n.,. n k=1 X k X k 7.3 ( ) m, σ 2, : X = 1 n ) X k N (m, σ2 n n p [ = p, = σ 2 = p(1 p)]., n ) ˆp N (p, σ2 ˆp p n σ/ N(0, 1). n k=1, Z N(0, 1) ( ) P ( z Z z) = 1 α z α α N α z z
25 p 1 α [ ] ˆp(1 ˆp) ˆp(1 ˆp) ˆp z, ˆp + z n n. 90%(α = 0.1, z = 1.64) 95%(α = 0.05, z = 1.96) 99%(α = 0.01, z = 2.58) ,, 1 α p.,! 2 ( ) ˆp p z p(1 p) n ˆp p z ˆp(1 ˆp) n 7.1 ( ) %. 95%, 0.21(1 0.21) 0.21 ± ± , 95% 0.01,? 90%? , 12.. [ ] 90%, 0.12 ± (1 0.12) ± ,,. 20, , % ( ) 95 %.
26
27 ( ),,. n ( ) X n, X n, X 1, X 2,... (iid)., (X k ), X 1, X 2,....,. X 1, X 2,..., n = 1, 2, 3,... x 1, x 2,..., x n P (X 1 x 1,, X n x n ) = P (X 1 x 1 ) P (X n x n )., X i x i X i = x i. 8.1 X 1, X 2,..., X n. (1) [ ] E(X 1 X 2 X n ) = E(X 1 )E(X 2 ) E(X n ) (2) [ ] V(X 1 + X X n ) = V(X 1 ) + V(X 2 ) + + V(X n ) X, Y ( ), α, E(X + Y ) = E(X) + E(Y ), E(αX) = αe(x), V(αX) = α 2 V(X). 8.1 (1) 2 X. X. (2) 2 Y. Y. 8.2 X 1, X 2,... (, ). m, σ 2. X = 1 n X k n. k=1
28 26 8 X m : E[ X] = m ( n, X m : P lim n ) X = m = 1 ( ) 8.2 ( ) X 1, X 2,..., X = 1 n X k n. n k=1 :, X = 1 n ) X k N (m, σ2 n n n., k=1 X m σ/ n = 1 n n k=1 X k m σ N(0, 1) n.,, ( ) lim P a 1 n X k m b = 1 b e x2 /2 dx, a < b. n n σ 2π k=1 a 8.3 ( ) m ( ), σ 2 X 1, X 2,..., X n : n ( (iid) ) 1 n : X = X k n k=1 m 1 α [ X z σ n, X + z σ n ], z N(0, 1) α
29 8.4. ( ) %(α = 0.1, z = 1.64) 95%(α = 0.05, z = 1.96) 99%(α = 0.01, z = 2.58) g., 8g , 95% 1g? 21, 200, 2.2 g., 1.5 g., g?. [1.992, 2.408] 8.4 ( ) m, σ 2 n X 1,..., X n, U 2 = 1 n 1 n (X i X) 2, S 2 = 1 n i=1 n (X i X) 2 i=1,. (,, ) 8.2 U 2 : E(U 2 ) = σ 2.,., n, S 2 U N(m, σ 2 ) n X 1,..., X n. X = 1 n n i=1 X i ( ) U 2 = 1 n 1 n (X i X) 2 ( ) i=1, T = X m U/ n t n 1 (n 1) t-,.
30 28 8 n t- 1 n B ( n 2, 1 2) ( ) n t2 2 n = Γ( n+1 2 ) n Γ( n 2 )Γ( 1 2 ) ( ) n t2 2 n n n n (1) Γ. Γ(x) = 0 t x 1 e t dt, x > 0. (2) B. B(x, y) = 1 0 t x 1 (1 t) y 1 dt = Γ(x)Γ(y), x > 0, y > 0. Γ(x + y) (3) n = t- N(0, 1). (4), n 30 N(0, 1). m 1 α [ X t U n, X + t U n ], t t n 1 α 8.5 8,. 90% ,. 95%
31 8.4. ( ) 29 t P ( T t n (α)) = α n\α , 4% ,. 24 ( ) m, σ, ( ) = x m σ,., 20 80, % %.
32
33 , , 200, 2.2 g., 1.5 g., g?. 22,. 95% , 4% ,. 4 A,B 2. A p, B q = 1 p. 5, A 3, B 2.? 10 1, X. X,,, E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = P (E c ), P (E F c ), P ((E F c ) c ), P (E F ), P (E F c ), P (E F E F ) 12, A B, 95%, 100p %... p. 6.3 (1) X N(20, 4 2 ), P (X > 17.8). (2) Y N( 2, 5 2 ), P ( Y 1). 6.4 X N(50, 10 2 ), P (X > a) = a.
34
35 Sir Ronald Aylmer Fisher ( ) , 220.? (1), (2), (3) H 0 H T ( ), H 0, < α < 1 P (T W ) = α W R ( ) H T t, W. t W. T, α., H 0 H 1. t W. T, α., H 0. (1),,. (2), ( ). (3),, 5%, 1%. (4) 2,. (5), H 0, (2 ). H 0,.
36 34 9 α α α W W W W %., ? 9.3 N 35%., 37 %., Z N(0, 1), α = P ( Z z) = 1 1 2π z z e x2 /2 dx, z 0 z α α α -z z H 0, 4. \ H 0 H 0 H 0 2 H 0 1, 1, 2. α: 1 = β: 2
37 , 215.? 2. θ θ β α c c 26 A ,. 1000, 545, , , % ( ),,. 29 ( ) ( ) 10,. p. H 0 : p = 1 2, H 1: p T. {T = 0, 1, 9, 10}., H 0 2 β p. H 0 P (2 T 8).
38
39 William Sealy Gosset ( ) 1. H 0 H T ( ), H 0, < α < 1 P (T W ) = α W R ( ) H 1 ( α-, α ). 4. T t, W. t W. T, α., H 0 H 1. t W. T, α., H 0.., (,, ), (, t-, χ 2 -, F - ), ( ) m, σ 2 n, X = 1 n n ) X k N (m, σ2 n k=1 X m σ/ n N(0, 1),, n (. N(m, σ 2 ) ) ( ), m = 60 (g).,, m 50 70, σ = 3 ( )., 25,, m = 60?
40 ( ). 120,., , m. : H 0 : m = 120 H 1 : m > ( ) B(n, p) N(np, np(1 p)) np:, np(1 p):, ˆp : ( ) p(1 p) ˆp N p, n p:, n: 10.3 ( ) 400, 175., ( ) m, σ 2 n X 1,..., X n, U 2 = 1 n (X i n 1 X) 2,. X, i=1 T = X m U/ n t n 1 (n 1) t (kg), kg,
41 10.4. ( ) cm cm, 4.63 cm. [ 1% ] 32, 100 g 2g., 200, 2.2 g., 1.5 g. [ 5% ] 33 8%., 175, , 38, 62. [ 5% ] g g, 10 2 g.,? 5%. 1% A , A. A. [ 5% ]
42
43 (x, y): x =, y = 11.2 m x = 1 n σx 2 = 1 n n x i, m y = 1 n i=1 n (x i m x ) 2, σy 2 = 1 n i=1 n y i, i=1 n (y i m y ) 2, (X, Y ), m X = E[X], σx 2 = V[X] = E[(X m X ) 2 ] = E[X 2 ] E[X] (1) n = 205 m x = m y = 50.9 σx 2 = = σy 2 = = (2) n = 917 m x = m y = 63.8 σx 2 = = σy 2 = = i=1
44 x, y., BMI( ),,,,. ( ) x ( ) y y = f(x)., 1,.,.,, (x i, y i ) (i = 1, 2,..., n) x, y 11.1 σ xy = 1 n n (x i m x )(y i m y ) = 1 n i=1 r xy = σ xy σ x σ y n x i y i m x m y i=1 (X, Y ), σ XY = E[(X m X )(Y m Y )] = E[XY ] E[X]E[Y ], r XY = σ XY σ X σ Y r XY = σ XY σ X σ Y [ X mx = E σ X Y m Y σ Y ]. X, Y., X = X m X σ X, Ỹ = Y m Y σ Y r XY = σ XỸ = r XỸ
45 r xy 1. r xy > 0, r xy < 0., r xy > 0.8, r xy < (1) : σ xy = 19.96, r xy = 0.64 (2) : σ xy = 19.97, r xy = (x 1, y 1 ),..., (x n, y n ), y = ax + b. y i = ax i + b + ϵ i, n n Q = ϵ 2 i = (y i ax i b) 2 i=1 a, b ( )., n Q = (yi 2 + a 2 x 2 i + b 2 2ax i y i 2by i + 2abx i ) i=1 Q, i=1 = y 2 i + a 2 x 2 i + b 2 n 2a x i y i 2b y i + 2ab x i. Q a = 2a x 2 i 2 x i y i + 2b x i = 0, Q b = 2bn 2 y i + 2a x i = 0
46 44 11 a, b : a = σ xy σ 2 x, b = ȳ a x 11.3 (1) : y = 0.72x (2) : y = 0.69x X, Y, a > 0, r XY = r ax,y ( ) X, ( ) Y. (1) X. (2) Y. (3) X, Y. (4) X, Y. 1. = 7 24 = 7 25 = ,. 4. ( ),. 5..,.
47 Karl Pearson ( ) ( n ) x n 2 1 e x 2, x > 0, f n (x) = 2 n/2 Γ 2 0, x 0, n 2 (χ 2 - ). (χ 2.), χ 2 n., Γ(t). n = n = n = n = n = χ 2 - (1) X 1, X 2,..., X n,, N(0, 1), n χ 2 n = i=1 n χ 2 -. (2) X 1, X 2,..., X n,, N(m, σ 2 )., χ 2 n 1 = 1 σ 2 n i=1 n 1 2. X 2 i (X i X) 2, X = 1 n n i=1 X i ( )
48 A 1, A 2,..., A k k. n, X 1, X 2,..., X k. A 1 A 2 A k p 1 p 2 p k 1 X 1 X 2 X k n, p 1, p 2,..., p k m i = np i, χ 2 k 1 = k (X i m i ) 2 m i=1 i, m 1,..., m k (m i = np i 5), k , 120.? ,. 4 : 3 : 2 : 1.,? A O B AB , , 5.,, 5 1:1? [ ] : 0:5 1:4 2:3 3:2 4:1 5: , 45, 55.? (1) (2), 2.
49 A = {A 1,..., A r }, B = {B 1,..., B s }, χ 2 = n r i=1 s j=1 ( Xij n X i n X i n X j n ) 2 X j n, n (X ij 5), (r 1)(s 1) 2. B 1 B 2 B s A 1 X 11 X 12 X 1s X 1 A 2 X 21 X 22 X 2s X 2. A r X r1 X r2 X rs X r X 1 X 2 X s n 12.2.? ?
50 : P (χ 2 n χ 2 n(α)) = α α χ n α n\α (n = 1 ).
51 William Feller ( ) B, A. A B 1, B 1. n X n, P (X n = +1) = P (X n = 1) = 1 2, X 1, X 2,.... n A : n S n = k=1 X k n = (1). (2) S n.
52 n, (+1) A n, ( 1) B n, n = A n + B n, S n = A n B n,, ( P lim n A n = S n + n 2, B n = S n n 2 A n n = 1 ) ( B n = P lim 2 n n = 1 ) = 1. 2, S n N(0, n), X 1, X 2,..., X n n., m = E(X 1 ) = 0, σ 2 = V[X 1 ] = 1., 1 ) ( n S n N (m, σ2 = N 0, 1 ) n n., S n N(0, n). 42 ( ) P ( S ).
53 {S n } 2n ( 0) ( ) L 2n = max{0 m 2n ; S m = 0} P (L 2n = 2k) = ( 2k k )( 2n 2k n k ) ( 1 4 ) n, k = 0, 1, 2,..., n, (13.1). (.) 13.1: : L 100, : 13.3 ( ) L 2n, ( ) 1 lim P n 2n L 2n a = 1 a dx = 2 π x(1 x) π arcsin a, 0 a 1, (13.2). 0 0 a < b 1. P (2an L 2n 2bn). [2an, 2bn] 2k 1, 2k 2, P (2an L 2n 2bn) = k 2 k=k 1 P (L 2n = 2k) = k 2 k=k 1 np (L 2n = 2k) 1 n (13.3). 13.2, ( )( ) ( ) n 2k 2n 2k 1 np (L 2n = 2k) = n k n k 4 n π k(n k)
54 52 13, (13.3), P (2an L 2n 2bn) k 2 k=k 1 n π k(n k) k2 1 n =, n,., b a dx π x(1 x) b lim P (2an L 2n 2bn) = n a 1 ( k=k 1 k π dx n π x(1 x) 1 k n ) 1 n (13.4). (..) n! ( n ) n 2πn e b n, a n b n lim = 1., n n a n, n. P ( ) ( ) 1 1 2n L 2n 0.1 = P 2n L 2n k 2n, (k 1, S k 1 ) (k.s k ) x k H 2n, H 2n L 2n ( 13.2).
55 ( 1, ) ( ) [1] [6]. [7] [8] 1. (100 ).. (,, ).,. [1] ( ) 10 2 (1 ). 3, 1, 2, 3. (10 ) [2] ( ) O R 1, O X. X F (x) = P (X x).,, X,,. (20 ) [3] ( ). (10 ) (1) X N(4, 5 2 ) P (X 5.6). (2), 10%., 65, 8.,. [4] ( ) 100 A %,, 95% 22.1 ± 3.3%. (20 ) (1)?. (2) 95%, 22.1% ± 0.33%,?. [5] ( ), A B, 90%, 5%.. (%, 1 ). (10 )
56 [6] ( ) ? 5%. 1%? 2. (10 ) [7] ( ) 2 X, Y.,, X = Y. (20 ) (1) P (X = 4). (2) P (Y 2 X 5). (3) X E[X]. (4) X, Y σ XY = E[(X E[X])(Y E[Y ])]. [8] ( ) 400, % 2., 2 5%. (20 ) P = 1 z e x2 /2 dx 2π z
57 ( ) [1] 1 A, 2 B, 3 C.,,.. 1, P (A) = 2 10 P (A c ) = P (B A) = 1 9, P (B Ac ) = 2 9, P (B) = P (B A)P (A) + P (B A c )P (A c ) = 1 9, 2, = = 2 10 = P (A)., P (C A B) = 0, P (C A B c ) = P (C A c B) = 1 8, P (C Ac B c ) = 2 8 P (C) = P (C A B)P (A B) + P (C A B c )P (A B c ) + P (C A c B)P (A c B) + P (C A c B c )P (A c B c ) = = = 2 10 = P (A),. [2] x X 0 X R, x < 0 F (x) = 0, x R F (x) = 1., 0 x R. X x 1 O x, O x., F (x) = P (X x) = πx2 πr 2 = x2 R 2., 0, x 0, x F (x) = 2 R 2. 0 x R, 1, x R.
58 : 2x f(x) = R 2. 0 x R, 0,., σ 2 = R 0 m = x 2 f(x)dx m 2 = R 0 R 0 xf(x)dx = 2 3 R 2x 3 ( ) R 2 dx m2 = R R = 1 18 R2 [3] (1) X N(4, 5 2 ) Z = X 4 N(0, 1), 5 ( X 4 P (X 5.6) = P ) = P (Z 0.32) = = (2) X N(65, 8 2 ), P (X a) = 0.1 a., Z N(0, 1), P (Z 1.28) = 0.1., Z = X X 75.2, 76. [4] (1). (2) 95% X ± 1.96 σ n, 1/ [5] A P (A) = 4 100, P (Ac ) = B,,, P (B A) = 0.9 P (B A c ) = 0.05 P (A)P (B A) P (A B) = P (A)P (B A) + P (A c )P (B A c ) = = 3.6 = %
59 [6] H 0 : p = 1 2, H 1 : p 1, α = X, X B(400, 1/2) N(200, 10 2 ). H 1 W : x x = 222 W., 5% H 0,. α = 0.01, W : x , x = 222 W., 1% H 0,. [7] 2 X, Y. 1 i, 2 j,, Ω = {ω = (i, j) ; i, j {1, 2, 3, 4, 5, 6}}, P ({ω}) = P ({(i, j)}) = X, Y Ω, X : Ω ω = (i, j) max{i, j}, Y : Ω ω = (i, j) min{i, j},. X, Y,. i\j X(i, j) Y (i, j) (1) {X = 4} 7, i\j P (X = 4) = 7 36 (2), {X 5} = {X 5, Y 2} = (1, 5) (1, 6) (2, 5) (2, 6) (3, 5) (3, 6) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) (5, 1) (5, 2) (6, 1) (6, 2) (1, 5) (1, 6) (2, 5) (2, 6),
60 , (3) X. P (Y 2 X 5) = P (Y 2, X 5) P (X 5) = 8/36 20/36 = 2 5. E(X) = k P (X = k) 1/36 3/36 5/36 7/36 9/36 11/36 6 kp (X = k) k=1 = = (4) σ XY = E[XY ] E[X]E[Y ]., XY = ij i, j, E[XY ] = E[ij] = E[i]E[j] = = 49 4 (3),, E(Y ) = σ XY = = = 36 2 = 36 2 = [8] H 0 : p = 1 2, H 1 : p 1, α = X, X B(400, 1/2) N(200, 10 2 ). H 1 W : x x = 215 W., 5% H 0,.,, 2., p, 2., p = ( ) p = 0.6 ( ). p = 0.525, p = 0.5,, β 0.5.,,, ( 2 ). p = 0.6, 2, β , β = 0.05 p 0.6., 10 2, = 236 p = 0.59., (p = 0.5), p 0.59,,
61 p p p p β β
62 ( 2, ) ( ) [1] [7]. [8] [9] 1. (100 ).. (,, ).,. [1] ( ) L 1 2., 1.5. (10 ) [2] ( ) 2 X. (, X.) X. (10 ) [3] ( ) 2 E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = 2 3., E c E. (10 ). (1) P (E c F ). (2) P (E F c ). [4] ( ). (10 ) (1) X N(2, 3 2 ) P (X a) = a. (2) (6 ) 720, [5] ( ) 100 A %,, 95% 22.1 ± 3.3%.. (20 ) (1),. (2) 99%,,. (3) 95% 1/ % ± 0.33% 10. (4) , 95% 10. [6] ( ) ? 5%. 1%? 2. (10 )
63 [7] ( ), A B, 90%, 5%.. (%, 1 ). (10 ) [8] ( ) 400, % 2., 2 5%. (20 ) [9] ( ) O 1 1, O X. X F (x) = P (X x).,,,,. (20 ) P = 1 z e x2 /2 dx 2π z
64 ( ) [1] 2/5 2/5, 4/5. [2] 1 i, 2 j,, Ω = {ω = (i, j) ; i, j {1, 2, 3, 4, 5, 6}}, P ({ω}) = P ({(i, j)}) = X = min{i, j},... i\j k P (X = k) 11/36 9/36 7/36 5/36 3/36 1/36, E[X] = 6 kp (X = k) k=1 = = V[X] = E[X 2 ] E[X] 2 = ( ) 91 2 = [3] E F E F (1) P (E F ) = P (E) + P (F ) P (E F ),,, 2 3 = P (E F ). 2 P (E F ) = 1 6.
65 , (2), P (E c F ) = P (F ) P (E F ) = = 1 3. P (E F c ) = P (E) P (E F ) = = 1 6., P (E F c ) = P (E F c ) P (F c ) = 1/6 1/2 = 1 3 [4] (1), Z N(0, 1), P (Z 1.16) = = N(0,1),, a = X 2 3 = Z 1.16 X 1.48 (2) B(720, 1/6) N(120, 10 2 ).,, ( ) ( ) X X 120 P (X 135) = P (X 134.5) = P = P ,, = [5] X ± z σ n,., 95% z = 1.96, 99% z = (1) 95%,. (2) 99%, 99%,,. (3) 2z σ n., 1/ (4),.
66 [6] H 0 : p = 1 2, H 1 : p 1, α = X, X B(400, 1/2) N(200, 10 2 ). H 1 W : x x = 224 W., 5% H 0,. α = 0.01, W : x , x = 224 W., 1% H 0,. [7] A P (A) = 3 100, P (Ac ) = B,,,, 36%. P (B A) = 0.9 P (B A c ) = 0.05 P (A)P (B A) P (A B) = P (A)P (B A) + P (A c )P (B A c ) = = = = [8] H 0 : p = 1 2, H 1 : p 1, α = X, X B(400, 1/2) N(200, 10 2 ). H 1 W : x x = 215 W., 5% H 0,.,, 2., p, 2., p = ( ) p = 0.6 ( ). p = 0.525, p = 0.5,, β 0.5.,,, ( 2 ). p = 0.6, 2, β , β = 0.05 p 0.6., 10 2, = 236 p = 0.59., (p = 0.5), p 0.59,,
67 p p p p β β [9] x X 0 X 1, x < 0 F (x) = 0, x 1 F (x) = 1., 0 x 1. X x 1 O x, O x., F (x) = P (X x) = πx2 π = x2., : 0, x 0, F (x) = x 2, 0 x 1, 1, x 1. f(x) = { 2x, 0 x 1, 0,., σ 2 = 1 0 m = 1 x 2 f(x)dx m 2 = 0 xf(x)dx = x 3 dx m 2 = 1 2 ( 2 3 ) 2 = 1 18
68 ( 2, ) ( ) [1] [6]. [7] [8] 1. (100 ).. (,, ).,. [1] ( ) A,B 2. A 2/5, B 3/5. 5, A 4, B 2.? (10 ) [2] ( ) 2 X. (, X.) X. (10 ) [3] ( ). (20 ) (1) (6 ) 720, (2), 5%., 68, 8.,. [4] ( ), A B, 90%, 5%... (10 ) [5] ( ) 100 A %,, 95% 22.1 ± 3.3%. (20 ) (1)?. (2) 95%, 22.1% ± 0.33%,?.
69 [6] ( ) ? 5%. 1%? (10 ) [7] ( ) 400, % 2., 2 5%. (20 ) [8] ( ) O 1 1, O X. X F (x) = P (X x).,,,,. (20 ) P = 1 z e x2 /2 dx 2π z
70 ( ) [1] A 1, B 3,, 98 : 27. P (A) = = = P (B) = = A : = 7840, B : = 2160 [2] 1 i, 2 j,, Ω = {ω = (i, j) ; i, j {1, 2, 3, 4, 5, 6}}, P ({ω}) = P ({(i, j)}) = X = min{i, j},... i\j k P (X = k) 11/36 9/36 7/36 5/36 3/36 1/36,, E[X] = E[X 2 ] = 6 kp (X = k) k=1 = = k 2 P (X = k) k=1 = = V[X] = E[X 2 ] E[X] 2 = ( ) 91 2 =
71 [3] (1) X, X B(720, 1/6) N(120, 10 2 )., Z = X 120 N(0, 1)., 10 ( ) X P (X 105) = P (X 105.5) = P = P (Z 1.45) ,, = (2) X N(68, 8 2 ), P (X a) = 0.05 a., Z N(0, 1), P (Z 1.64) = 0.05., Z = X X 81.12, 82. [4] A P (A) = 2 100, P (Ac ) = B,,, P (B A) = 0.9 P (B A c ) = 0.05 P (A)P (B A) P (A B) = P (A)P (B A) + P (A c )P (B A c ) = = = = = 40% [5] (1). (2) 95% X ± 1.96 σ n, 1/ [6] H 0 : p = 1 2, H 1 : p 1, α = X, X B(400, 1/2) N(200, 10 2 ). H 1 W : x x = 224 W., 5% H 0,. α = 0.01, W : x
72 , x = 224 W., 1% H 0,. [7] H 0 : p = 1 2, H 1 : p 1, α = X, X B(400, 1/2) N(200, 10 2 ). H 1 W : x x = 215 W., 5% H 0,.,, 2., p, 2., p = ( ) p = 0.6 ( ). p = 0.525, p = 0.5,, β 0.5.,,, ( 2 ). p p p p β β p = 0.6, 2, β , β = 0.05 p 0.6., 10 2, = 236 p = 0.59., (p = 0.5), p 0.59,, [8] x X 0 X 1, x < 0 F (x) = 0, x 1 F (x) = 1., 0 x 1. X x 1 O x, O x., F (x) = P (X x) = πx2 π = x2.
73 , 0, x 0, F (x) = x 2, 0 x 1, 1, x 1. : f(x) = { 2x, 0 x 1, 0,., σ 2 = 1 0 m = 1 x 2 f(x)dx m 2 = 0 xf(x)dx = x 3 dx m 2 = 1 2 ( 2 3 ) 2 = 1 18
ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.
24(2012) (1 C106) 4 11 (2 C206) 4 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 (). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5... 6.. 7.,,. 8.,. 1. (75%)
ii 3.,. 4. F. (), ,,. 8.,. 1. (75% ) (25% ) =9 7, =9 8 (. ). 1.,, (). 3.,. 1. ( ).,.,.,.,.,. ( ) (1 2 )., ( ), 0. 2., 1., 0,.
23(2011) (1 C104) 5 11 (2 C206) 5 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 ( ). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5.. 6.. 7.,,. 8.,. 1. (75%
ii 2. F. ( ), ,,. 5. G., L., D. ( ) ( ), 2005.,. 6.,,. 7.,. 8. ( ), , (20 ). 1. (75% ) (25% ). 60.,. 2. =8 5, =8 4 (. 1.) 1.,,
(1 C205) 4 8 27(2015) http://www.math.is.tohoku.ac.jp/~obata,.,,,..,,. 1. 2. 3. 4. 5. 6. 7.... 1., 2014... 2. P. G., 1995.,. 3.,. 4.. 5., 1996... 1., 2007,. ii 2. F. ( ),.. 3... 4.,,. 5. G., L., D. ( )
..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A
![..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A](/thumbs/93/112175219.jpg "..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A")
.. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.
6.1 (P (P (P (P (P (P (, P (, P.
(011 30 7 0 ( ( 3 ( 010 1 (P.3 1 1.1 (P.4.................. 1 1. (P.4............... 1 (P.15.1 (P.16................. (P.0............3 (P.18 3.4 (P.3............... 4 3 (P.9 4 3.1 (P.30........... 4 3.
6.1 (P (P (P (P (P (P (, P (, P.101
(008 0 3 7 ( ( ( 00 1 (P.3 1 1.1 (P.3.................. 1 1. (P.4............... 1 (P.15.1 (P.15................. (P.18............3 (P.17......... 3.4 (P................ 4 3 (P.7 4 3.1 ( P.7...........
統計学のポイント整理
.. September 17, 2012 1 / 55 n! = n (n 1) (n 2) 1 0! = 1 10! = 10 9 8 1 = 3628800 n k np k np k = n! (n k)! (1) 5 3 5 P 3 = 5! = 5 4 3 = 60 (5 3)! n k n C k nc k = npk k! = n! k!(n k)! (2) 5 3 5C 3 = 5!
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
2 1,, x = 1 a i f i = i i a i f i. media ( ): x 1, x 2,..., x,. mode ( ): x 1, x 2,..., x,., ( ). 2., : box plot ( ): x variace ( ): σ 2 = 1 (x k x) 2
1 1 Lambert Adolphe Jacques Quetelet (1796 1874) 1.1 1 1 (1 ) x 1, x 2,..., x ( ) x a 1 a i a m f f 1 f i f m 1.1 ( ( )) 155 160 160 165 165 170 170 175 175 180 180 185 x 157.5 162.5 167.5 172.5 177.5
( )/2 hara/lectures/lectures-j.html 2, {H} {T } S = {H, T } {(H, H), (H, T )} {(H, T ), (T, T )} {(H, H), (T, T )} {1
( )/2 http://www2.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html 1 2011 ( )/2 2 2011 4 1 2 1.1 1 2 1 2 3 4 5 1.1.1 sample space S S = {H, T } H T T H S = {(H, H), (H, T ), (T, H), (T, T )} (T, H) S
renshumondai-kaito.dvi
3 1 13 14 1.1 1 44.5 39.5 49.5 2 0.10 2 0.10 54.5 49.5 59.5 5 0.25 7 0.35 64.5 59.5 69.5 8 0.40 15 0.75 74.5 69.5 79.5 3 0.15 18 0.90 84.5 79.5 89.5 2 0.10 20 1.00 20 1.00 2 1.2 1 16.5 20.5 12.5 2 0.10
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P
1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A
18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C
8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,
³ÎΨÏÀ
2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p
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.
1 (1) () (3) I 0 3 I I d θ = L () dt θ L L θ I d θ = L = κθ (3) dt κ T I T = π κ (4) T I κ κ κ L l a θ L r δr δl L θ ϕ ϕ = rθ (5) l
1 1 ϕ ϕ ϕ S F F = ϕ (1) S 1: F 1 1 (1) () (3) I 0 3 I I d θ = L () dt θ L L θ I d θ = L = κθ (3) dt κ T I T = π κ (4) T I κ κ κ L l a θ L r δr δl L θ ϕ ϕ = rθ (5) l : l r δr θ πrδr δf (1) (5) δf = ϕ πrδr
.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
2011 ( ) ( ) ( ),,.,,.,, ,.. (. ), 1. ( ). ( ) ( ). : obata/,.,. ( )
2011 () () (),,.,,.,,. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.,.. (. ), 1. ( ). ()(). : www.math.is.tohoku.ac.jp/ obata/,.,. () obata@math.is.tohoku.ac.jp http://www.dais.is.tohoku.ac.jp/ amf/, (! 22 10.6; 23 10.20;
> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3
13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >
68 A mm 1/10 A. (a) (b) A.: (a) A.3 A.4 1 1
67 A Section A.1 0 1 0 1 Balmer 7 9 1 0.1 0.01 1 9 3 10:09 6 A.1: A.1 1 10 9 68 A 10 9 10 9 1 10 9 10 1 mm 1/10 A. (a) (b) A.: (a) A.3 A.4 1 1 A.1. 69 5 1 10 15 3 40 0 0 ¾ ¾ É f Á ½ j 30 A.3: A.4: 1/10
201711grade1ouyou.pdf
2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2
Microsoft Word - 表紙.docx
黒住英司 [ 著 ] サピエンティア 計量経済学 訂正および練習問題解答 (206/2/2 版 ) 訂正 練習問題解答 3 .69, 3.8 4 (X i X)U i i i (X i μ x )U i ( X μx ) U i. i E [ ] (X i μ x )U i i E[(X i μ x )]E[U i ]0. i V [ ] (X i μ x )U i i 2 i j E [(X i
() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
分散分析・2次元正規分布
2 II L10(2016-06-30 Thu) : Time-stamp: 2016-06-30 Thu 13:55 JST hig F 2.. http://hig3.net ( ) L10 2 II(2016) 1 / 24 F 2 F L09-Q1 Quiz :F 1 α = 0.05, 2 F 3 H 0, : σ 2 1 /σ2 2 = 1., H 1, σ 2 1 /σ2 2 1. 4
( 30 ) 30 4 5 1 4 1.1............................................... 4 1.............................................. 4 1..1.................................. 4 1.......................................
6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P
6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )
II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11
(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
untitled
yoshi@image.med.osaka-u.ac.jp http://www.image.med.osaka-u.ac.jp/member/yoshi/ II Excel, Mathematica Mathematica Osaka Electro-Communication University (2007 Apr) 09849-31503-64015-30704-18799-390 http://www.image.med.osaka-u.ac.jp/member/yoshi/
II (No.2) 2 4,.. (1) (cm) (2) (cm) , (
II (No.1) 1 x 1, x 2,..., x µ = 1 V = 1 k=1 x k (x k µ) 2 k=1 σ = V. V = σ 2 = 1 x 2 k µ 2 k=1 1 µ, V σ. (1) 4, 7, 3, 1, 9, 6 (2) 14, 17, 13, 11, 19, 16 (3) 12, 21, 9, 3, 27, 18 (4) 27.2, 29.3, 29.1, 26.0,
2 1,2, , 2 ( ) (1) (2) (3) (4) Cameron and Trivedi(1998) , (1987) (1982) Agresti(2003)
3 1 1 1 2 1 2 1,2,3 1 0 50 3000, 2 ( ) 1 3 1 0 4 3 (1) (2) (3) (4) 1 1 1 2 3 Cameron and Trivedi(1998) 4 1974, (1987) (1982) Agresti(2003) 3 (1)-(4) AAA, AA+,A (1) (2) (3) (4) (5) (1)-(5) 1 2 5 3 5 (DI)
populatio sample II, B II? [1] I. [2] 1 [3] David J. Had [4] 2 [5] 3 2
![populatio sample II, B II? [1] I. [2] 1 [3] David J. Had [4] 2 [5] 3 2](/thumbs/49/25550404.jpg "populatio sample II, B II? [1] I. [2] 1 [3] David J. Had [4] 2 [5] 3 2")
(2015 ) 1 NHK 2012 5 28 2013 7 3 2014 9 17 2015 4 8!? New York Times 2009 8 5 For Today s Graduate, Just Oe Word: Statistics Google Hal Varia I keep sayig that the sexy job i the ext 10 years will be statisticias.
1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +
( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n
( 28 ) ( ) ( ) 0 This note is c 2016, 2017 by Setsuo Taniguchi. It may be used for personal or classroom purposes, but not for commercial purp
( 28) ( ) ( 28 9 22 ) 0 This ote is c 2016, 2017 by Setsuo Taiguchi. It may be used for persoal or classroom purposes, but ot for commercial purposes. i (http://www.stat.go.jp/teacher/c2epi1.htm ) = statistics
I L01( Wed) : Time-stamp: Wed 07:38 JST hig e, ( ) L01 I(2017) 1 / 19
I L01(2017-09-20 Wed) : Time-stamp: 2017-09-20 Wed 07:38 JST hig e, http://hig3.net ( ) L01 I(2017) 1 / 19 ? 1? 2? ( ) L01 I(2017) 2 / 19 ?,,.,., 1..,. 1,2,.,.,. ( ) L01 I(2017) 3 / 19 ? I. M (3 ) II,
( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552
3 3.0 a n a n ( ) () a m a n = a m+n () (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 55 3. (n ) a n n a n a n 3 4 = 8 8 3 ( 3) 4 = 8 3 8 ( ) ( ) 3 = 8 8 ( ) 3 n n 4 n n
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
tokei01.dvi
2. :,,,. :.... Apr. - Jul., 26FY Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 3. (probability),, 1. : : n, α A, A a/n. :, p, p Apr. - Jul., 26FY Dept. of Mechanical Engineering, Saga Univ., JAPAN
2000年度『数学展望 I』講義録
2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53
(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (
![(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (](/thumbs/92/107866524.jpg "(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (")
B 4 4 4 52 4/ 9/ 3/3 6 9.. y = x 2 x x = (, ) (, ) S = 2 = 2 4 4 [, ] 4 4 4 ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, 4 4 4 4 4 k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) 2 2 + ( ) 3 2 + ( 4 4 4 4 4 4 4 4 4 ( (
1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =
1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i
L P y P y + ɛ, ɛ y P y I P y,, y P y + I P y, 3 ŷ β 0 β y β 0 β y β β 0, β y x x, x,, x, y y, y,, y x x y y x x, y y, x x y y {}}{,,, / / L P / / y, P
005 5 6 y β + ɛ {x, x,, x p } y, {x, x,, x p }, β, ɛ E ɛ 0 V ɛ σ I 3 rak p 4 ɛ i N 0, σ ɛ ɛ y β y β y y β y + β β, ɛ β y + β 0, β y β y ɛ ɛ β ɛ y β mi L y y ŷ β y β y β β L P y P y + ɛ, ɛ y P y I P y,,
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
newmain.dvi
数論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/008142 このサンプルページの内容は, 第 2 版 1 刷発行当時のものです. Daniel DUVERNEY: THÉORIE DES NOMBRES c Dunod, Paris, 1998, This book is published
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
II 2 II
II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................
4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.
A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c
x y 1 x 1 y 1 2 x 2 y 2 3 x 3 y 3... x ( ) 2
1 1 1.1 1.1.1 1 168 75 2 170 65 3 156 50... x y 1 x 1 y 1 2 x 2 y 2 3 x 3 y 3... x ( ) 2 1 1 0 1 0 0 2 1 0 0 1 0 3 0 1 0 0 1...... 1.1.2 x = 1 n x (average, mean) x i s 2 x = 1 n (x i x) 2 3 x (variance)
(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n
. 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n
) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
![) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4](/thumbs/92/107866707.jpg ") ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4")
1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
5 Armitage x 1,, x n y i = 10x i + 3 y i = log x i {x i } {y i } 1.2 n i i x ij i j y ij, z ij i j 2 1 y = a x + b ( cm) x ij (i j )
5 Armitage. x,, x n y i = 0x i + 3 y i = log x i x i y i.2 n i i x ij i j y ij, z ij i j 2 y = a x + b 2 2. ( cm) x ij (i j ) (i) x, x 2 σ 2 x,, σ 2 x,2 σ x,, σ x,2 t t x * (ii) (i) m y ij = x ij /00 y
Z: Q: R: C: sin 6 5 ζ a, b
Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,
waseda2010a-jukaiki1-main.dvi
November, 2 Contents 6 2 8 3 3 3 32 32 33 5 34 34 6 35 35 7 4 R 2 7 4 4 9 42 42 2 43 44 2 5 : 2 5 5 23 52 52 23 53 53 23 54 24 6 24 6 6 26 62 62 26 63 t 27 7 27 7 7 28 72 72 28 73 36) 29 8 29 8 29 82 3
(iii) x, x N(µ, ) z = x µ () N(0, ) () 0 (y,, y 0 ) (σ = 6) *3 0 y y 2 y 3 y 4 y 5 y 6 y 7 y 8 y 9 y ( ) *4 H 0 : µ
t 2 Armitage t t t χ 2 F χ 2 F 2 µ, N(µ, ) f(x µ, ) = ( ) exp (x µ)2 2πσ 2 2 0, N(0, ) (00 α) z(α) t * 2. t (i)x N(µ, ) x µ σ N(0, ) 2 (ii)x,, x N(µ, ) x = x + +x ( N µ, σ2 ) (iii) (i),(ii) x,, x N(µ,
A
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4
20 20.0 ( ) 8 y = ax 2 + bx + c 443 ax 2 + bx + c = 0 20.1 20.1.1 n 8 (n ) a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 444 ( a, b, c, d
A A = a 41 a 42 a 43 a 44 A (7) 1 (3) A = M 12 = = a 41 (8) a 41 a 43 a 44 (3) n n A, B a i AB = A B ii aa
1 2 21 2 2 [ ] a 11 a 12 A = a 21 a 22 (1) A = a 11 a 22 a 12 a 21 (2) 3 3 n n A A = n ( 1) i+j a ij M ij i =1 n (3) j=1 M ij A i j (n 1) (n 1) 2-1 3 3 A A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33
応用数学III-4.ppt
III f x ( ) = 1 f x ( ) = P( X = x) = f ( x) = P( X = x) =! x ( ) b! a, X! U a,b f ( x) =! " e #!x, X! Ex (!) n! ( n! x)!x! " x 1! " x! e"!, X! Po! ( ) n! x, X! B( n;" ) ( ) ! xf ( x) = = n n!! ( n
OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P
4 ( ) ( ) ( ) ( ) 4 5 5 II III A B (0 ) 4, 6, 7 II III A B (0 ) ( ),, 6, 8, 9 II III A B (0 ) ( [ ] ) 5, 0, II A B (90 ) log x x () (a) y x + x (b) y sin (x + ) () (a) (b) (c) (d) 0 e π 0 x x x + dx e
2 G(k) e ikx = (ik) n x n n! n=0 (k ) ( ) X n = ( i) n n k n G(k) k=0 F (k) ln G(k) = ln e ikx n κ n F (k) = F (k) (ik) n n= n! κ n κ n = ( i) n n k n
. X {x, x 2, x 3,... x n } X X {, 2, 3, 4, 5, 6} X x i P i. 0 P i 2. n P i = 3. P (i ω) = i ω P i P 3 {x, x 2, x 3,... x n } ω P i = 6 X f(x) f(x) X n n f(x i )P i n x n i P i X n 2 G(k) e ikx = (ik) n
( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (
6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b
24 6 I., X, x X. Radom Samplig with Replacemet ( ) 1,.,, 1 X 1, 2 X 2,..., X., X 1, X 2,..., X ( ).,.,,. Estimate of Populatio Parameters ( ),..,,.. 6
23 第 6 章 母数の推定 I 二項母集団の母比率 6.1 Audiece Ratig Survey (視聴率調査) テレビ局では視聴率の獲得にしのぎを削っているようである. 果たして, コンマ以下の数字に 意味はあるのだろうか? 2016 年 4 月 25 日 (月) 5 月 1 日 (日) ドラマ (関東地区) 視聴率ベスト 10 番組名 放送局 連続テレビ小説 とと姉ちゃん 真田丸 日曜劇場
LLG-R8.Nisus.pdf
d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =
(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t
![(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t](/thumbs/67/56612994.jpg "(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t")
6 6.1 6.1 (1 Z ( X = e Z, Y = Im Z ( Z = X + iy, i = 1 (2 Z E[ e Z ] < E[ Im Z ] < Z E[Z] = E[e Z] + ie[im Z] 6.2 Z E[Z] E[ Z ] : E[ Z ] < e Z Z, Im Z Z E[Z] α = E[Z], Z = Z Z 1 {Z } E[Z] = α = α [ α ]
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.
![() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.](/thumbs/89/98384840.jpg "() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.")
() 6 f(x) [, b] 6. Riemnn [, b] f(x) S f(x) [, b] (Riemnn) = x 0 < x < x < < x n = b. I = [, b] = {x,, x n } mx(x i x i ) =. i [x i, x i ] ξ i n (f) = f(ξ i )(x i x i ) i=. (ξ i ) (f) 0( ), ξ i, S, ε >
, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p,
20 9 19 1 3 11 1 3 111 3 112 1 4 12 6 121 6 122 7 13 7 131 8 132 10 133 10 134 12 14 13 141 13 142 13 143 15 144 16 145 17 15 19 151 1 19 152 20 2 21 21 21 211 21 212 1 23 213 1 23 214 25 215 31 22 33
y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
2009 I 2 II III 14, 15, α β α β l 0 l l l l γ (1) γ = αβ (2) α β n n cos 2k n n π sin 2k n π k=1 k=1 3. a 0, a 1,..., a n α a
009 I II III 4, 5, 6 4 30. 0 α β α β l 0 l l l l γ ) γ αβ ) α β. n n cos k n n π sin k n π k k 3. a 0, a,..., a n α a 0 + a x + a x + + a n x n 0 ᾱ 4. [a, b] f y fx) y x 5. ) Arcsin 4) Arccos ) ) Arcsin
, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x
![, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x](/thumbs/93/113428934.jpg ", 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x")
1 1.1 4n 2 x, x 1 2n f n (x) = 4n 2 ( 1 x), 1 x 1 n 2n n, 1 x n n 1 1 f n (x)dx = 1, n = 1, 2,.. 1 lim 1 lim 1 f n (x)dx = 1 lim f n(x) = ( lim f n (x))dx = f n (x)dx 1 ( lim f n (x))dx d dx ( lim f d
ad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign(
I n n A AX = I, YA = I () n XY A () X = IX = (YA)X = Y(AX) = YI = Y X Y () XY A A AB AB BA (AB)(B A ) = A(BB )A = AA = I (BA)(A B ) = B(AA )B = BB = I (AB) = B A (BA) = A B A B A = B = 5 5 A B AB BA A
21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........
II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K
II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F
2 1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a 2 + b 2 α (norm) N(α) = a 2 + b 2 = αα = α 2 α (spure) (trace) 1 1. a R aα =
1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a + b α (norm) N(α) = a + b = αα = α α (spure) (trace) 1 1. a R aα = aα. α = α 3. α + β = α + β 4. αβ = αβ 5. β 0 6. α = α ( ) α = α
1: *2 W, L 2 1 (WWL) 4 5 (WWL) W (WWL) L W (WWL) L L 1 2, 1 4, , 1 4 (cf. [4]) 2: 2 3 * , , = , 1
![1: *2 W, L 2 1 (WWL) 4 5 (WWL) W (WWL) L W (WWL) L L 1 2, 1 4, , 1 4 (cf. [4]) 2: 2 3 * , , = , 1](/thumbs/49/25412084.jpg "1: *2 W, L 2 1 (WWL) 4 5 (WWL) W (WWL) L W (WWL) L L 1 2, 1 4, , 1 4 (cf. [4]) 2: 2 3 * , , = , 1")
I, A 25 8 24 1 1.1 ( 3 ) 3 9 10 3 9 : (1,2,6), (1,3,5), (1,4,4), (2,2,5), (2,3,4), (3,3,3) 10 : (1,3,6), (1,4,5), (2,2,6), (2,3,5), (2,4,4), (3,3,4) 6 3 9 10 3 9 : 6 3 + 3 2 + 1 = 25 25 10 : 6 3 + 3 3
211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq
49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r
I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1
ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD
18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α
18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t
2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a
1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =
1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
I
I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R
![II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R](/thumbs/95/125540821.jpg "II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R")
II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
keisoku01.dvi
2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP
直交座標系の回転
b T.Koama x l x, Lx i ij j j xi i i i, x L T L L, L ± x L T xax axx, ( a a ) i, j ij i j ij ji λ λ + λ + + λ i i i x L T T T x ( L) L T xax T ( T L T ) A( L) T ( LAL T ) T ( L AL) λ ii L AL Λ λi i axx
数理統計学Iノート
I ver. 0/Apr/208 * (inferential statistics) *2 A, B *3 5.9 *4 *5 [6] [],.., 7 2004. [2].., 973. [3]. R (Wonderful R )., 9 206. [4]. ( )., 7 99. [5]. ( )., 8 992. [6],.., 989. [7]. - 30., 0 996. [4] [5]
1 (1) ( i ) 60 (ii) 75 (iii) 315 (2) π ( i ) (ii) π (iii) 7 12 π ( (3) r, AOB = θ 0 < θ < π ) OAB A 2 OB P ( AB ) < ( AP ) (4) 0 < θ < π 2 sin θ
1 (1) ( i ) 60 (ii) 75 (iii) 15 () ( i ) (ii) 4 (iii) 7 1 ( () r, AOB = θ 0 < θ < ) OAB A OB P ( AB ) < ( AP ) (4) 0 < θ < sin θ < θ < tan θ 0 x, 0 y (1) sin x = sin y (x, y) () cos x cos y (x, y) 1 c
IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト
https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,
SFGÇÃÉXÉyÉNÉgÉãå`.pdf
SFG 1 SFG SFG I SFG (ω) χ SFG (ω). SFG χ χ SFG (ω) = χ NR e iϕ +. ω ω + iγ SFG φ = ±π/, χ φ = ±π 3 χ SFG χ SFG = χ NR + χ (ω ω ) + Γ + χ NR χ (ω ω ) (ω ω ) + Γ cosϕ χ NR χ Γ (ω ω ) + Γ sinϕ. 3 (θ) 180