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
|
|
|
- しげのぶ さんきち
- 5 years ago
- Views:
Transcription
1 1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) P (A) 1 (1) 2. P 1 P A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B
2 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 (A B) = P (B A)P (A) = P (A B)P (B) (6) P (B A) = P (A B) P (A) = P (A)P (B) P (A) = P (B) (7) A B Ω Ω = A 1 + A A n (8) 2
3 P (B) = P (B A i )P (A i ) (9) i=1 A i 1.4 Ω Ω = A 1 + A A n (10) P (B A i )P (A i ) = P (A i B)P (B) (11) P (A i B) = P (B A i)p (A i ) P (B) = P (B A i )P (A i ) n i=1 P (B A i)p (A i ) (12) P (A i ) B P (A i B) B ( ( ) ) P ( ) = P ( )P ( ) P ( )P ( ) + P ( )P ( ) + P ( )P ( ) (13) 3
4 P ( ) = 2 10, P ( ) = 5 10, P ( ) = 3 10 (14) P ( ) = 6 11, P ( ) = 5 11, P ( ) = 5 11 (15) P ( ) = 3 13 (16) 2 2/10 = 2.6/13 0.4/13 99% (+) 1% 10 (0.001%) P ( +) = P (+ )P ( ) P (+ )P ( ) + P (+ )P ( ) (17) P ( ) = , P ( ) = 1 P ( ) = (18) P ( +) = P (+ ) = 0.99, P ( ) = 0.01 (19) (20) % 4
5 X F (x) F (x) = P (X x) (21) X x X a b P (a X b) = F (b) F (a) (22) F X x i f i = P (X = x i ) F (x) = f i (23) x i x x i x x i X X 0 1 l l ( 0) mm 100g 5
6 f(x) F (x) f(x) = df (x) dx F (x) f(x) x x X x + x (24) P (x X x + x) = F (x + x) F (x) (25) x x x + x F (x) f(x) F (x + x) = F (x) + f(x) x (26) P (x X x + x) = f(x) x (27) f(x) f(x) a X b F (b) F (a) = b a f(x)dx (28) f(x) F (x) = x f(x)dx (29) 0 f(x) 1 (30) f(x)dx = 1 (31) F (x) 2.2 ( ) X ( )E(X) E(X) = xf(x)dx ( ) (32) E(X) = i x i f i dx ( ) (33) 6
7 X h(x) E{h(X)} = h(x)f(x)dx ( ) (34) E{h(X)} = i h(x i )f i dx ( ) (35) h(x) = X k (k ) X k E{X k } = x k f(x)dx (36) X 1 X X V (X) V (X) = E(X 2 ) {E(X)} 2 (37) 2 E{(X E(X)) 2 } = E{X 2 2XE(X)+(E(X)) 2 } = E(X 2 ) 2E{XE(X)}+E{E(X) 2 } (38) E(X) E(X) E(X) E{(X E(X)) 2 } = E(X 2 ) 2E(X)E(X) + {E(X)} 2 = E(X 2 ) {E(X)} 2 (39) 2 S S = V (X) (40) 2.3 (Moment generating function) M(t) = E{e tx } = d k M(t) dt k t=0 = d k dt k etx t=0 f(x)dx = e tx f(x)dx (41) x k f(x)dx = E(X k ) (42) M(t) k k t=0 t t = 0 e x = 1 + x x+ 1 3! x3 + = 7 k=0 x k k! (43)
8 { } (tx) k M(t) = E = k! k=0 k=0 1 k! E(Xk ) = 1 + E(X) E(X2 ) + 1 3! E(X3 ) + (44) (characteristic function) Ψ(t) = E{e itx } = e itx f(x)dx (45) i (i 2 = 1) X d k Ψ(t) dt k t=0 = i k E(X k ) (46) ( ) Ψ(t) = M(it) (47) Ψ(it) = M( t) (48) ( ) X k (cumulant)c k G(t) log M(t) = k=1 1 k! C k (49) k = 1 G(t) M(t) t log(1 + x) = x 1 2 x x3 (50) C 1 = E(X) (51) C 2 = E(X 2 ) {E(X)} 2 = V (X) (52) k k 1 2 8
9 3 3.1 DU(n) X n x 1, x 2,..., x n f i = P (X = x i ) = 1 n DU(2) DU(6) X = 1, 2, 3,..., n E(X) = 1 n E{X 2 } = 1 n k=1 k 2 = k=1 k = n (n + 1)(2n + 1) 6 V (X) = E{X 2 } {E(X)} 2 = n (53) (54) (55) (56) 3.2 B N (n, p) p 1 p ( ) n X X = x( n x) f(x) = n C x p x (1 p) n x (57) n C x n x nc x = x! n!(n x)! (58) f(x) (a + b) n = nc m a m b n m (59) 1 = {p + (1 p)} n ( n ) m=0 1 = {p + (1 p)} n = nc x p x (1 p) n x = f(x) (60) x=0 x=0 9
10 E(X) = x n C x p x (1 p) n x = n x=0 n 1C x 1 p x (1 p) n x (61) x=1 n 1 = np n 1C x p x (1 p) n x 1 = np{p + (1 p)} n 1 = np (62) x=0 E(X 2 ) E{X(X 1)} E{X(X 1)} = x(x 1) n C x p x (1 p) n x = n(n 1) x=0 n 2C x 2 p x (1 p) n x (63) n 2 = n(n 1)p 2 n 2C x p x (1 p) n x 2 = n(n 1)p 2 {p + (1 p)} n 1 = n(n 1)p 2 (64) x=0 V (X) = E(X 2 ) {E(X)} 2 = E{X(X 1)} + E(X) {E(X)} 2 = np(1 p) (65) n x f(x) = n C x ( 1 2 x=2 ) x ( ) n x 1 (66) 2 E(X) = np = n 2 (67) V (X) = np(1 p) = n 2 (68) P 0 (λ) ( ) ( ) 10
11 X λ f(x) = λx x! e λ (69) Ψ(t) = x=0 λ x x! e λ e itx = e λ E(X) = 1 i V (X) 2 x=0 1 ( λe it ) x { ( = exp λ e it 1 )} (70) x! dψ(t) dt = λ (71) t=0 V (X) = d2 log Ψ(t) dt 2 t=0 = λ (72) 3.4 U(α, β) X α X β f(x) = { 1 β α (α x β) 0 ( ) (73) E(X) = β α E(X 2 ) = xf(x)dx = 1 β xdx = α + β β α α 2 β α x 2 f(x)dx == 1 β 3 α 3 β α 3 V (X) = (β α)2 12 (74) (75) (76) N(µ, σ 2 ) µ σ 2 f(x) = 1 2πσ e (x µ)2 2σ 2 (77) 11
12 4.2 N(0, 1) ϕ(x) ϕ(x) = 1 2π e x2 /2 (78) N(µ, σ 2 ) z Z = X µ σ ϕ(z) σ (79) ϕ(x)dx = 1 (80) e x2 /2 dx = 2π (81) I = e x2 /2 dx = e y2 /2 dy (82) ( ) ( ) I 2 = e x2 /2 dx e y2 /2 dy = dx dy e (x2 +y 2 )/2 (83) xy x 2 + y 2 (x, y) 2 r 2 = x 2 + y 2 (84) dx dy (85) r r + r r 2πr r r dx 12 dy = 0 2πrdr (86)
13 I 2 = 0 2πre r2 /2 dr (87) d 2 dr e r /2 = re r2 /2 I 2 = 2π 0 (88) ( ) d 2 dr e r /2 dr (89) I 2 = 2π [e ] r2 /2 = 2π (90) E(X) = 1 2π xe x2 /2 dx = 0 (91) xe x2 /2 0 ( ) dj(λ) dλ = λ=0 J(λ) = e λx2 /2 dx (92) ( ) d ( ) 2 dλ e λx /2 x 2 dx = 2 λ=0 2 e λx /2 dx = frac12 x 2 e x2 /2 dx λ=0 (93) λ 2 2π e λx /2 dx (94) N(0, 1/ λ) 1 J(λ) = 2π λ (95) ( E(X 2 ) = 2 1 dj(λ) = 2 1 ) d 2π 2π dλ λ=0 2π dλ λ = 1 (96) λ=0 13
14 V (X) = E(X 2 ) {E(X)} 2 = 1 (97) 0 1 N(µ, σ 2 ) Z X = µ + σz (98) Z E(X) = E(µ + σz) = µ + σe(z) = µ (99) V (X) = V (µ + σz) = E{(µ + σz) 2 } {E(µ + σz)} 2 = σ 2 (100) µ σ 2 ( σ ) 4.4 M(t) = E{e tx } = 1 e tx (x µ)2 /2σ 2 dx = exp (µt + σ2 t 2 ) 2πσ 2 (101) G(t) = log M(t) = µt + σ2 t 2 (102) 2 t Z 0 Z z Z P ( σ X µ σ) P ( 2σ X µ 2σ)
15 P ( 3σ X µ 3σ) X Y X Y F (x, y) = P (X x, Y y) (103) X Y 1 F 1 (x) = P (X x) = F (x, ) (104) F 2 (y) = P (Y y) = F (, y) (105) f(x, y) = 2 F (x, y) (106) x y XY x y dx f(x, y) x y (107) f(x, y) 0 (108) X Y f 1 (x) = df 1(x) dx f 2 (x) = df 2(y) dy X Y dyf(x, y) = 1 (109) = = f(x, y)dy (110) f(x, y)dx (111) f(x, y) = f 1 (x)f 2 (y) (112) 15
16 5.2 2 h(x, Y ) E{h(X, Y )} = E(X) = µ 1, E(Y ) = µ 2 (113) V (X) = σ 2 1, V (Y ) = σ 2 2 (114) dx dyh(x, y)f(x, y) (115) h(x, Y ) X Y X Y h(x, Y ) = h 1 (X)h 2 (Y ) (116) E{h(X, Y )} = E{h 1 (X)}E{h 2 (Y )} (117) X Y 2 2 Cov(X, Y ) = E{(X µ 1 )(Y µ 2 )} = E(XY ) E(X)E(Y ) = E(XY ) µ 1 µ 2 (118) X Y Z Z 1 = X µ 1 σ 1, Z 2 = Y µ 2 σ 2 (119) ρ ρ Corr(X, Y ) Cov(Z 1, Z 2 ) = Cov(X, Y ) V (X)V (Y ) (120) ρ = ±1 1 ρ 1 (121) Z 2 = ±Z 1 (122) Y = µ 2 ± σ 2 σ 1 (X µ 1 ) (123) Y X 16
17 X Y E(XY ) = E(X)E(Y ) ρ > 0 X Y ρ < 0 X Y X Y U = ax + by (124) E(U) = E(aX + by ) = ae(x) + be(y ) = aµ 1 + bµ 2 (125) V (U) = E{(aX + by ) 2 } {E(aX + by )} 2 = a 2 V (X) + b 2 V (Y ) + 2ab Cov(X, Y ) (126) U X Y N 2 (µ, Σ) µ = ( µ1 µ 2 ) (127) Σ = ( V (X) Cov(X, Y ) Cov(X, Y ) V (Y ) ) ( σ 2 = 1 ρσ 1 σ 2 ρσ 1 σ 2 σ2 2 ) (128) [ { 1 f(x, y) = 2πσ 1 σ 2 (1 ρ2 ) exp 1 (x µ1 ) 2 2(1 ρ 2 ) σ ρ(x µ 1)(y µ 2 ) σ 1 σ 2 + (y µ 2) 2 X Y z [ ] 1 f(z 1, z 2 ) = 2π (1 ρ 2 ) exp 1 2(1 ρ 2 ) (z2 1 2ρz 1 z 2 + z2) 2 σ 1 σ 2 2 }] (129) (130) ρ = 0 [ 1 f(x, y) = exp 1 ( ) ] [ 2 x µ1 1 exp 1 ( ) ] 2 y µ2 = f 1 (x)f 2 (y) (131) 2πσ1 2 2πσ σ 2
18 2 0 2 x + y f(x, y) = [ 1 exp 1 2πσ 1 σ 2 2 ( ) ] 2 x µ1 σ 1 [ 1 exp 1 2πσ2 2 f(z 1, z 2 ) ( ) ] 2 y µ2 = f 1 (x)f 2 (y) (132) u = z 1 + z 2, v = z 1 z 2 (133) σ 2 1 2(1 ρ 2 ) (z2 1 2ρz 1 z 2 + z2) 2 = 1 ( ) u ρ + v2 1 ρ (134) U = Z 1 + Z 2 V = Z 1 Z 2 2 Z 1 Z 2 (ρ = 0) U f(u) = 1 2π e u2 /2 (135) X 1, X 2,..., X n µ σ 2 ( ) X n = 1 n (X 1 + X X n ) (136) n E( X n ) = µ (137) V ( X n ) = 1 n 2 [ E{(X1 + + X n ) 2 } {E(X X n )} 2] = σ2 n (138) 1/ n 18
19 6.2 (law of large numbers) n X n µ X n µ E( X n ) = µ X n n X n µ 1 ( )µ 6.3 (central limit theorem) n n X n N(µ, σ 2 ) ( ) ( n ) n N(µ, σ2 n ) : n i 1 0 X i (i = 1, 2,..., n) M M = X 1 + X X n (139) M B N (n, 1/2) n n M N(n/2, n/4) n µ σ 2 X 1, X 2,..., X n ( 19
20 ) X = 1 n (X 1 + X X n ) (140) n N(µ, σ2 n ) SX 2 = 1 (X i n X) 2 = 1 Xi 2 n X 2 (141) i 1 V ( X) 2 2 i 1 E( X) = µ, V ( X) = E{ X 2 } {E( X)} 2 = σ2 n (142) E( X 2 ) = σ2 n + µ2 (143) X i µ σ 2 E(X 2 i ) = σ 2 + µ 2 (144) E(SX) 2 = σ 2 + µ 2 σ2 n µ2 = σ 2 σ2 n (145) σ2 n (bias) (unbiased) ˆσ 2 x = 1 n 1 (X i X) 2 (146) i 1 x 1, x 2,..., x n ( ) x = 1 n (x 1 + x x n ) (147) n N(µ, σ2 n ) s 2 x = 1 n (x i x) 2 = 1 n i 1 20 x 2 i x 2 (148) i 1
21 2 ˆσ 2 x = 1 n 1 (x i x) 2 (149) i 1 x ˆσ 2 x 7.2 x x 100 X B N (100, 1/2) N(50, 25) z Z = X = X 5 10 (150) Z N(0, 1) 3 Z 3 99% (99.7%) Z z 3 X (151) 35 X 65 (152) % 2 Z % p X B N (100, p) 21
22 N(100p, 100p(1 p)) Z Z = X 100p 100p(1 p) (153) Z N(0, 1) Z 99.7% 3 Z 3 Z 100p 3 10 p(1 p) X 100p p(1 p) (154) X 100p X p(1 p) p X p(1 p) (155) p p p p X 100 p X X 100 ( 1 X ) p X X 100 ( 1 X ) 100 (156) p X p 99.7% 99.7% 99% 95.4% % 99% Z % 1.96 Z % p 0.25 p % 0.3 p 0.5 p = 0.5 ( ) 99% 99% 99% ( ) 99% p p N(100p, 100p(1 p)) 99% ( 40 ) 7.3 ( ) σ 2 n ( n ) X Z n( X µ) Z = (157) σ 22
23 N(0, 1) 95.4% µ 2 Z 2 (158) X 2 σ n µ X + 2 σ n (159) N(0, 1) X zα Z zα 1 α zα α z 0.05 = 1.96 µ 1 α X zα σ µ X + z σ n α (160) n 7.4 ( ) σ 2 ˆσ n 2 σ 2 ˆσ n 2 Z T T = n( X µ) ˆσ n (161) N(0, 1) T n 1 t n t t n 1 t α t n 1,α µ 1 α X t ˆσ n n 1,α µ X + t ˆσ n n n 1,α (162) n 7.5 χ 2 Z 1, Z 2,, Z n N(0, 1) n 2 S 2 = Z1 2 + Z Zn 2 (163) n χ 2 χ 2 n n = 1 N(0, 1) Z 2 S = Z 2 P (0 S s) = P ( Z s) = 2 s e x2 /2 dx (164) 2π 23 0
24 2 s 2π 0 y = x 2 (165) dy dx = 2x = 2 y (166) e x2 /2 dx = 1 2π s 0 1 y e y/2 dy (167) S f(s) = dp (0 S s) ds = 1 2π 1 y e y/2 (168) (null hypothesis) :40 24
25 1/2 B(100, 1/2) 100 n P (n) = C n (169) 60 1% p(n 60) = C n (170) n=60 B N (100, 1/2) N(50, 25) Z Z = X = X 5 10 (171) N(0, 1) X = 60 Z = 2 P (Z 2) P (Z 2) = (172) 2 Z Z 2 P (Z 2) = 1 P (Z 0) P (0 Z 2) = P ( 2 Z 2)/2 (173) % p p 2.3%
26 ( ) ( ) () p p 26
27 p
..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 }.
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
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!
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...........
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
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
untitled
2 : n =1, 2,, 10000 0.5125 0.51 0.5075 0.505 0.5025 0.5 0.4975 0.495 0 2000 4000 6000 8000 10000 2 weak law of large numbers 1. X 1,X 2,,X n 2. µ = E(X i ),i=1, 2,,n 3. σi 2 = V (X i ) σ 2,i=1, 2,,n ɛ>0
ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.
(1 C205) 4 10 (2 C206) 4 11 (2 B202) 4 12 25(2013) http://www.math.is.tohoku.ac.jp/~obata,.,,,..,,. 1. 2. 3. 4. 5. 6. 7. 8. 1., 2007 ( ).,. 2. P. G., 1995. 3. J. C., 1988. 1... 2.,,. ii 3.,. 4. F. ( ),..
( )/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
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,
n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
.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,
( 30 ) 30 4 5 1 4 1.1............................................... 4 1.............................................. 4 1..1.................................. 4 1.......................................
() 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 ()
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%
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
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%)
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 ),
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.
Acrobat Distiller, Job 128
(2 ) 2 < > ( ) f x (x, y) 2x 3+y f y (x, y) x 2y +2 f(3, 2) f x (3, 2) 5 f y (3, 2) L y 2 z 5x 5 ` x 3 z y 2 2 2 < > (2 ) f(, 2) 7 f x (x, y) 2x y f x (, 2),f y (x, y) x +4y,f y (, 2) 7 z (x ) + 7(y 2)
) ] [ 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
,. 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,
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
II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
![II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2](/thumbs/101/149159646.jpg "II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2")
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
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
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.................................
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
Untitled
II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
(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 ]
st.dvi
9 3 5................................... 5............................. 5....................................... 5.................................. 7.........................................................................
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
M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -
M3............................................................................................ 3.3................................................... 3 6........................................... 6..........................................
微分積分 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. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 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. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
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]
( ) 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
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
5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (
![5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (](/thumbs/92/107866626.jpg "5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (")
5 5.1 [ ] ) d f(t) + a d f(t) + bf(t) : f(t) 1 dt dt ) u(x, t) c u(x, t) : u(x, t) t x : ( ) ) 1 : y + ay, : y + ay + by : ( ) 1 ) : y + ay, : yy + ay 3 ( ): ( ) ) : y + ay, : y + ay b [],,, [ ] au xx
No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
DVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
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,
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
i
i 1 1 1.1..................................... 1 1.2........................................ 3 1.3.................................. 4 1.4..................................... 4 1.5......................................
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
30
3 ............................................2 2...........................................2....................................2.2...................................2.3..............................
.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,
.1.1 Y K L Y = K 1 3 L 3 L K K (K + ) 1 1 3 L 3 K 3 L 3 K 0 (K + K) 1 3 L 3 K 1 3 L 3 lim K 0 K = L (K + K) 1 3 K 1 3 3 lim K 0 K = 1 3 K 3 L 3 z = f(x, y) x y z x-y-z.1 z = e x +xy y 3 x-y ( ) z 0 f(x,
2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37
( 12 ( ( ( ( Levi-Civita grad div rot ( ( = 4 : 6 3 1 1.1 f(x n f (n (x, d n f(x (1.1 dxn f (2 (x f (x 1.1 f(x = e x f (n (x = e x d dx (fg = f g + fg (1.2 d dx d 2 dx (fg = f g + 2f g + fg 2... d n n
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)
2009 IA I 22, 23, 24, 25, 26, a h f(x) x x a h
009 IA I, 3, 4, 5, 6, 7 7 7 4 5 h fx) x x h 4 5 4 5 1 3 1.1........................... 3 1........................... 4 1.3..................................... 6 1.4.............................. 8 1.4.1..............................
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..........
x E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx
x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I
= π2 6, ( ) = π 4, ( ). 1 ( ( 5) ) ( 9 1 ( ( ) ) (
+ + 3 + 4 +... π 6, ( ) 3 + 5 7 +... π 4, ( ). ( 3 + ( 5) + 7 + ) ( 9 ( ( + 3) 5 + ) ( 7 + 9 + + 3 ) +... log( + ), ) +... π. ) ( 3 + 5 e x dx π.......................................................................
1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
4 4. A p X A 1 X X A 1 A 4.3 X p X p X S(X) = E ((X p) ) X = X E(X) = E(X) p p 4.3p < p < 1 X X p f(i) = P (X = i) = p(1 p) i 1, i = 1,,... 1 + r + r
4 1 4 4.1 X P (X = 1) =.4, P (X = ) =.3, P (X = 1) =., P (X = ) =.1 E(X) = 1.4 +.3 + 1. +.1 = 4. X Y = X P (X = ) = P (X = 1) = P (X = ) = P (X = 1) = P (X = ) =. Y P (Y = ) = P (X = ) =., P (Y = 1) =
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
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
(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)
simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =
II 6 ishimori@phys.titech.ac.jp 6.. 5.4.. f Rx = f Lx = fx fx + lim = lim x x + x x f c = f x + x < c < x x x + lim x x fx fx x x = lim x x f c = f x x < c < x cosmx cosxdx = {cosm x + cosm + x} dx = [
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(,,
r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B
1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n
II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (
II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
( ) s n (n = 0, 1,...) n n = δ nn n n = I n=0 ψ = n C n n (1) C n = n ψ α = e 1 2 α 2 n=0 α, β α n n! n (2) β α = e 1 2 α 2 1
(3.5 3.8) 03032s 2006.7.0 n (n = 0,,...) n n = δ nn n n = I n=0 ψ = n C n n () C n = n ψ α = e 2 α 2 n=0 α, β α n n (2) β α = e 2 α 2 2 β 2 n=0 =0 = e 2 α 2 β n α 2 β 2 n=0 = e 2 α 2 2 β 2 +β α β n α!
mugensho.dvi
1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim
i 18 2H 2 + O 2 2H 2 + ( ) 3K
i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................
X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
( 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
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)
4................................. 4................................. 4 6................................. 6................................. 9.................................................... 3..3..........................
20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................
l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r
2 1 (7a)(7b) λ i( w w ) + [ w + w ] 1 + w w l 2 0 Re(γ) α (7a)(7b) 2 γ 0, ( w) 2 1, w 1 γ (1) l µ, λ j γ l 2 0 Re(γ) α, λ w + w i( w w ) 1 + w w γ γ 1 w 1 r [x2 + y 2 + z 2 ] 1/2 ( w) 2 x2 + y 2 + z 2
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
I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT
I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345
5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E
5 5.1 E 1, E 2 N 1, N 2 E tot N tot E tot = E 1 + E 2, N tot = N 1 + N 2 S 1 (E 1, N 1 ), S 2 (E 2, N 2 ) E 1, E 2 S tot = S 1 + S 2 2 S 1 E 1 = S 2 E 2, S 1 N 1 = S 2 N 2 2 (chemical potential) µ S N
http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................
(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
I 1
I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
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
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
n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1)
D d dx 1 1.1 n d n y a 0 dx n + a d n 1 y 1 dx n 1 +... + a dy n 1 dx + a ny = f(x)...(1) dk y dx k = y (k) a 0 y (n) + a 1 y (n 1) +... + a n 1 y + a n y = f(x)...(2) (2) (2) f(x) 0 a 0 y (n) + a 1 y
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
W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................
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;
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.
n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x
2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin
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
II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
![II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re](/thumbs/94/118770263.jpg "II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re")
II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
(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
Part. 4. () 4.. () 4.. 3 5. 5 5.. 5 5.. 6 5.3. 7 Part 3. 8 6. 8 6.. 8 6.. 8 7. 8 7.. 8 7.. 3 8. 3 9., 34 9.. 34 9.. 37 9.3. 39. 4.. 4.. 43. 46.. 46..
Cotets 6 6 : 6 6 6 6 6 6 7 7 7 Part. 8. 8.. 8.. 9..... 3. 3 3.. 3 3.. 7 3.3. 8 Part. 4. () 4.. () 4.. 3 5. 5 5.. 5 5.. 6 5.3. 7 Part 3. 8 6. 8 6.. 8 6.. 8 7. 8 7.. 8 7.. 3 8. 3 9., 34 9.. 34 9.. 37 9.3.
2 1 Introduction
1 24 11 26 1 E-mail: toyoizumi@waseda.jp 2 1 Introduction 5 1.1...................... 7 2 8 2.1................ 8 2.2....................... 8 2.3............................ 9 3 10 3.1.........................
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. ( )
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
1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1
sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V
ohpmain.dvi
fujisawa@ism.ac.jp 1 Contents 1. 2. 3. 4. γ- 2 1. 3 10 5.6, 5.7, 5.4, 5.5, 5.8, 5.5, 5.3, 5.6, 5.4, 5.2. 5.5 5.6 +5.7 +5.4 +5.5 +5.8 +5.5 +5.3 +5.6 +5.4 +5.2 =5.5. 10 outlier 5 5.6, 5.7, 5.4, 5.5, 5.8,
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z