|
|
|
- きみえ ほがり
- 4 years ago
- Views:
Transcription
1
2 ,. 1 ( )...,.,, F k (a) = σ(a) a + k. b = F k (a), a = F k (b), a b,(a, b) k. (a, b) k. k. k = 0 (a, b) k = 1 (a, b). 1. k = 2 (a, b) k = 1 (a, b). 1. k = 2 (a, b).. 1 Amicable Pairs, a Survey (2003) By 1) Mariano Garcia(11802 SW 37th Terrace, Miami, Florida 33175, USA), 2) Jan Munch Pederse (Vitus Bering CVU, Chr. M. Ostergaardsvej 4, DK-8700 Horsens, Denmark) 3) Herman te Riele ( 1090 GB Amsterdam, The Netherlands) 2
3 maxima g_sigma_k_factor(k,aa,bb):= for c:aa thru bb do(y:divsum(c)-c+k, x:divsum(y)-y+k,if (c=x and x>y ) then print("c=",c,"tab",factor(c),"y=",y,"tab",factor(y)) else 1=1); g_sigma_k_factor(-2,1,200000);. 1: k = A ( 2 e p), D ( : 2 e qr). D D.. A D.. wiki.. 3,
4 a = 2 e r, b = 2 f pq., e = f. 4
5 2 Euler Euler. 1 (p, q, r). n, m(n > m). K = 2 n m + 1 p = 2 m K 1, q = 2 n K 1, r = 2 n+m K 2 1. A = 2 n pq, B = 2 n r, coσ(a) = B. coσ(b) = A. (A, B). Euler a, a = 2 n p, (p = 2 n+1 1: ). 1 D A..,... A = 2 n pq, B = 2 n r, coσ(a) = B Proof. N = 2 n+1 1 coσ(a) = coσ(2 n pq) = N(p + 1)(q + 1) 2 n pq. p + 1 = 2 m K, q + 1 = 2 n K, pq = 2 n+m K 2 K(2 n + 2 m ) + 1. N(p + 1)(q + 1) 2 n pq = N(2 n+m K 2 ) 2 n (2 n+m K 2 K(2 n + 2 m ) + 1) = (2 n+1 1)(2 n+m K 2 ) 2 n (2 n+m K 2 K(2 n + 2 m ) + 1) = 2 2n+m+1 K 2 2 n+m K 2 2 n (2 n+m K 2 K(2 n + 2 m ) + 1) = 2 n (2 n+m K 2 2 m K 2 + K(2 m + 2 n ) 1) = 2 n X. r = K 2 2 n+m 1, X = 2 n+m K m K 2 + K(2 m + 2 n ) = r + L., L = 2 m K 2 + K(2 m + 2 n ) = KL 0 L 0 = 2 m K + (2 m + 2 n ) = (2 m + 2 n ) + (2 m + 2 n ) = 0., coσ(a) = N(p + 1)(q + 1) 2 n pq = 2 n r = B., coσ(a) = B coσ(b) = A.,. 5
6 2.1 coσ(b) = A, A = 2 n pq, B = 2 n r,σ(a) A = coσ(a) = B σ(a) = A+B. σ(b) = A + B. coσ(a) = B. σ(a) = N(p+1)(q+1) = N(r+1) = σ(b), σ(b) = σ(a) = A + B ).B = 284 = , A = 220 = , 2)B = = , A = = ,. A = 2 n pq, B = 2 n r, coσ(a) = B, coσ(b) = A. B = coσ(a) = σ(a) A, A = coσ(b) = σ(b) B, σ(a) = A + B = σ(b). N = 2 n+1 1 σ(a) = N(p + 1)(q + 1), σ(b) = N(r + 1),N(p + 1)(q + 1) = N(r + 1)., (p + 1)(q + 1) = (r + 1).. p = 23, q = 47, r = 1151 p + 1 = 24 = 2 3 3, q + 1 = 48 = 2 4 3, r + 1 = = ( ): p + 1 = 2 n K, q + 1 = 2 m K, (K : ). r + 1 = (p + 1)(q + 1) = 2 n+m K 2, r = 2 n+m K 2 1., p, q n m. 2 n pq = A = coσ(b) = σ(b) B = N(r + 1) 2 n r = 2 n r + N r r = 2 n r + N 2 n pq = 2 n (r pq) + N. r = N + 2 n (r pq). p = 2 n K 1, q = 2 m K 1, r = 2 n+m K 2 1, r pq = 2 + K(2 n + 2 m ). 2 n+m K 2 1 = r = N + 2 n (r pq) = N + 2 n ( 2 + K(2 n + 2 m )) = n K(2 n + 2 m )). 2 n+m K 2 = 2 n K(2 n + 2 m ). 2 m K = 2 n + 2 m,k = 2 n m ( ), ( )
7 . (p + 1)(q + 1) = (r + 1), p + 1, q + 1, K 1, K 2 p + 1 = 2 n K 1, q + 1 = 2 m K 2 r + 1 = (p + 1)(q + 1) = 2 n+m K 1 K 2. 2 n pq = A = coσ(b) = σ(b) B = N(r + 1) 2 n r = 2 n r + N r r = 2 n r + N 2 n pq = 2 n (r pq) + N. r = N + 2 n (r pq).( ) p = 2 n K 1 1, q = 2 m K 2 1, r = 2 n+m K 1 K 2 1, r pq = 2 + (2 n K m K 2 ). 2 n 2 n (r pq) = 2 n n (2 n K m K 2 ). r = N + 2 n (r pq), 2 n+1 = N 1 r N = 2 n (r pq) = 2 n n (2 n K m K 2 ) = 1 N + 2 n (2 n K m K 2 ). r + 1 = 2 n (2 n K m K 2 ). r + 1 = (p + 1)(q + 1) = 2 n+m K 1 K 2, 2 n+m K 1 K 2 = 2 n (2 n K m K 2 )., 2 m K 1 K 2 = 2 n K m K 2. 2 m K 2 (K 1 1) = 2 n K 1. n m K 2 (K 1 1) = 2 n m K 1. K 1 1, K 1 α (K 1 1)α = 2 n m. α > 1 α. K 2 (K 1 1) = 2 n m K 1, K 2 = αk 1. α α = 1. (K 1 1)α = 2 n m K 1 1 = 2 n m. K 1 = 2 n m + 1. K 2 (K 1 1) = 2 n m K 1 K 2 = K 1. End. 7
8 m < n < 40, (n, m) = (2, 1), (4, 3), (7, 6), (8, 1) (p, q, r).( ) 2: k = 0,Euler (p, q, r) m = 1, n = m = 3, n = (F ermat) m = 6, n = (Descartes) m = 1, n =
9 3: k = : k =
10 5: k = a = 38, b = k = 2 a = 2p, b = 2 e q, (p, q) :,, a = 2p = 38, b = 2 e q = 24. Proof F 2 (a) = F 2 (2p) = 3(p + 1) 2p + 2 = p + 5, F 2 (a) = b = 2 e q, p + 5 = 2 e q. 2p = a = F 2 (b) = F 2 (2 e q) = (2 e+1 1)(q + 1) 2 e q = 2 e q (q + 1) + 2 e p = 2 e q 5 2(2 e q 5) = 2 e q (q + 1) + 2 e e q = 2 e q (q + 1) + 2 e = 2 e (q 2) q
11 2 e q = 2 2 e q + 11 (2 e + 1)q = 2 2 e q 3, 2 e e e., 3 e. 1. e = 3 2 e q = 2 2 e q q = 16 q + 11., 9q = 27, q = 3. p = 2 e q 5 = 24 5 = 19. a = 2 19 = 38, b = 8q = e = 2 2 e q = 2 2 e q q = 8 q e = 1 2 e q = 2 2 e q +11 2q = 4 q +11., q = 5. p = 2 e q 5, p = 2 e q 5 = 10 5 = 5. a = 10, b = 10. a = b. End 11
12 2 a = 2 e p, b = 6q, (p, q) :, q 5, p = e+1, X = 2 e, 6q = 2X X 14. coσ(a) = coσ(2 e p) = 2 e p p + N = b 2 = 6q 2, coσ(b) = coσ(6q) = qq + 12 = a 2. 6: p, q: e X p Y=2X X 14 q=y/ p p p p * * *
13 7: k =
14 m. m = 18, 58, 14.. k k. k = k = 16 8: k = 16 k = a = 4p, b = 12q, (p, q):,. F (a) = F 16 (a) F (a) = σ(a) a 16. a = 4p, b = 6q, (p, q):,. 14
15 a = 4p F (a) = σ(a) a 16 = 7(p + 1) 16 = 3p 9 = b = 12q p 3 = 4q. p 3 = 4q b = 12q F (b) = 28(q + 1) 12q 16 = 16q + 12 = 4(4q + 3) = 4p = a. 2 (a, b) 16. a = 2 e p, b = 2 e qr, (p, q, r < q):,, e = 2, r = 4, p = 4q + 3: (q, p = 4q + 3). Proof. a = 2 e p, N = 2 e+1 1 F (a) = σ(a) a 16 = N(p + 1) 2 e p 16 = Np 2 e p + N 16 = 2 e p p + N 16. F (a) = b = 2 e qr, 2 e p p + N 16 = 2 e qr. 2 e p = p N e qr. (2 e 1)p = N e qr. b = 2 e qr, = q + r F (b) = σ(b) b 16 = N(q + 1)(r + 1) 2 e qr 16 = Nqr 2 e qr 16 + N( + 1) = 2 e qr qr + N( + 1) 16 = a = 2 e p. 2 e qr qr 16 + N( + 1) = 2 e p. 2 e p = p N e qr 2 e qr qr 16 + N( + 1) = p N e qr. p = 2 e qr qr 16 + N( + 1) ( N e qr) = 32 qr + N + 2N. p = 32 qr + N + 2N (2 e 1)p = N e qr (2 e 1)( 32 qr + N( + 2) = N e qr. 32(2 e 1) (2 e 1)qr + (2 e 1)N( + 2) = N e qr (2 e 1)qr 32(2 e 1) + (2 e 1)N( + 2) = N e qr + (2 e 1)qr = N Nqr. 15
16 32(2 e 1) = 16 2(2 e 1) = 16(N 1) 16(N 1) + (2 e 1)N( + 2) = N Nqr. N 16 + (2 e 1)( + 2) = 1 + qr. = q + r,η = 2 e 1, q 0 = q η, r 0 = r η,, 15 + η( + 2) = qr. Θ = η 2 + 2η 15. q 0 r 0 = qr η + η 2 = η 2 + 2η e = 2. η = 2 e 1 = 3, N = 7, q 0 = q 3, r 0 = r 3, Θ = η 2 + 2η 15 = 0. q 0 > r 0, r 0 = 0, r = 3. B = 6q, p = 32 qr + N + 2N = 32 3q + 7 (q + 3) = 4q = 4q + 3., a = 2 2 p, b = 2 2 q r = 6q. 2. e = 3. η = 2 e 1 = 7, N = 15, q 0 = q 7, r 0 = r 7, Θ = η 2 + 2η 15 = 7(7 + 9) 15 = = 48. q 0 r 0 = Θ = 48, r 0 = 4, q 0 = 12. r = 11, q = 19. p = 32 qr + N + 2N = ( ) + 30 = 239., a = , b = 2 3 q r = e = 4. η = 2 e 1 = 15, N = 31, q 0 = q 15, r 0 = r 15, Θ = η 2 + 2η 15 = = q 0 r 0 = Θ = , r 0 = 3 2 = 6, q 0 = 5 4 = 20. r = 21;. r 0 = 5 2 = 10, q 0 = 3 8 = 24. r = 25;
17 5 m 3 (a, b) m. a = 2 e p, b = 2 e qr, (p, q, r < q):, a = 2 e p, N = 2 e+1 1 F (a) = σ(a) a + m = N(p + 1) 2 e p + m = Np 2 e p + N + m = 2 e p p + N + m. F (a) = b = 2 e qr, 2 e p p + N + m = 2 e qr. 2 e p = p N m + 2 e qr. (2 e 1)p = N m + 2 e qr. b = 2 e qr, = q + r F (b) = σ(b) b + m = N(q + 1)(r + 1) 2 e qr + m = Nqr 2 e qr + m + N( + 1) = 2 e qr qr + N( + 1) + m = a = 2 e p. 2 e qr qr + m + N( + 1) = 2 e p. 2 e p = p N m + 2 e qr 2 e qr qr + m + N( + 1) = p N m + 2 e qr. p = 2 e qr qr + m + N( + 1) ( N m + 2 e qr) = 2m qr + N + 2N. p = 2m qr + N + 2N (2 e 1)p = N m + 2 e qr (2 e 1)(2m qr + N( + 2) = N m + 2 e qr. 2m(2 e 1) (2 e 1)qr + (2 e 1)N( + 2) = N m + 2 e qr (2 e 1)qr 2m(2 e 1) + (2 e 1)N( + 2) = N e qr + (2 e 1)qr = N + m + Nqr. 2m(2 e 1) = 2m(2 e 1) = m(n 1) m(n 1) + (2 e 1)N( + 2) = N m + Nqr. 17
18 N m + (2 e 1)( + 2) = 1 + qr. = q + r,η = 2 e 1, q 0 = q η, r 0 = r η,, m η( + 2) = qr. q 0 r 0 = qr η + η 2 = η 2 + 2η + m + 1. Θ = η 2 + 2η + m + 1 = (2 e 1)(2 e + 1) + m + 1 = 2 2e + m q 0 r 0 = 2 2e + m. 1. e = 1. Θ = 4 + m. m = 4 η = 2 e 1 = 1, q 0 r 0 = 0, q 0 > r 0, r 0 = 0, r = e = 2. η = 2 e 1 = 3, N = 7, q 0 = q 3, r 0 = r 3, Θ = 16 + m. m = 16 q 0 > r 0, r 0 = 0, r = 3. B = 6q, p = +2m qr + N + 2N = +2m 3q + 7 (q + 3) = 4q + 2m + 35 = 4q + 3., a = 2 2 p, b = 2 2 q r = 6q. 2. e = 3. η = 2 e 1 = 7, N = 15, q 0 = q 7, r 0 = r 7, Θ = 16 + m. m = 16 r 0 = r 7 = 0., q > 7: B = qr = 7q. Proof. a = 2 e p, N = 2 e+1 1 F (a) = σ(a) a 16 = N(p + 1) 2 e p 16 = Np 2 e p + N 16 = 2 e p p + N 16. F (a) = b = 2 e qr, 2 e p p + N 16 = 2 e qr. 2 e p = p N e qr.. 6 (a, b) m 9: k = 20, 18 k = k =
19 10: k = 16 k =
20 11: k = 14, 12, 10 k = k = k =
21 12: k = 8, 6, 4 k = k = k =
22 13: k = 2, 0 k = k =
23 14: k = 2, 4 k = k =
24 15: k = 6, 8, 10 k = k = k =
25 16: k = 44 k = a = 14p, b = 30q, ((p > 2 7, q > 5): ).. F (a) = σ(a) a 44, F (a) = F (14p) = σ(14p) 14p 44 = 24(p + 1) 14p 44 = 10p 20 = b = 30q; p 2 = 3q. F (b) = F (30q) = σ(30q) 30q 44 = 72(q + 1) 30q 44 = 42q + 28 = 7(6q + 4) = 14(3q + 2) = 14p = a. (q, p = 2 + 3q). p = 29, p = 27 = 2 + 3q, q =
26 17: k = 34 k = a = 6p, b = 28q, ((p > 3, q 2, 7): ). F (a) = F (6p) = 12(p + 1) 6p 34 = 6p 22 = b = 28q; 3p 11 = 14q. F (b) = F (28q) = 56(q +1) 28q 34 = 28q +22 = 2(14q +11) = a = 6p;14q +11 = 3p. p, q:, 14q + 11 = 3p. 26
27 18: k = 12, 14 k = k =
28 19: k = 16, 18, 20 k = k = k =
29 7 k = 64, 20: k = 64, a = 2 3 p, b = 2 3 7r
30 21: k =
31 22: k =
32 23: k =
33 24: k =
34 25: k =
35 26: k =
36 27: k =
37 28: k = 64 k = k = /april/25 a = 44p = p, b = 20 = q, ((p > 3, p 11, q 2, 5): ). D. F (a) = F (44p) = 12 7(p + 1) 44p 64 = 40p + 20 = b = 20q; 2p + 1 = q.. F (b) = F (20q) = 42(q + 1) 28q 64 = 22q 22 = 44p;q 1 = 2p. a = 8p = 2 3 p, b = 56q = q, ((p > 3, p 11, q 2, 5): ). F (a) = F (8p) = 15(p + 1) 8p 64 = 7p + 49 = b = 56q; p + 7 = 8q. F (b) = F (56q) = 120(q + 1) 56q 64 = 64q 56 = 8p;8q 7 = p. (p = 8q 7, q). a = 8p, b = 56q, ((p > 3, p 11, q 2, 5): ). 37
38 8 k = 0, F 0 (a) = σ(a) a, a, b, c, F 0 (a) = b, F 0 (b) = c, F 0 (c) = a (sociable number)... k = 2 a = 6 = 2 3, b = 8 = 2 3, c = 9 =
39 29: k a b c k = k = k = k = k = k = k = k = k =
Super perfect numbers and Mersenne perefect numbers /2/22 1 m, , 31 8 P = , P =
Super perfect numbers and Mersenne perefect numbers 3 2019/2/22 1 m, 2 2 5 3 5 4 18 5 20 6 25 7, 31 8 P = 5 35 9, 38 10 P = 5 39 1 1 m, 1: m = 28 m = 28 m = 10 height48 2 4 3 A 40 2 3 5 A 2002 2 7 11 13
Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n
Part2 47 Example 161 93 1 T a a 2 M 1 a 1 T a 2 a Point 1 T L L L T T L L T L L L T T L L T detm a 1 aa 2 a 1 2 + 1 > 0 11 T T x x M λ 12 y y x y λ 2 a + 1λ + a 2 2a + 2 0 13 D D a + 1 2 4a 2 2a + 2 a
Go a σ(a). σ(a) = 2a, 6,28,496, = 2 (2 2 1), 28 = 2 2 (2 3 1), 496 = 2 4 (2 5 1), 8128 = 2 6 (2 7 1). 2 1 Q = 2 e+1 1 a = 2
Go 2016 8 26 28 8 29 1 a σ(a) σ(a) = 2a, 6,28,496,8128 6 = 2 (2 2 1), 28 = 2 2 (2 3 1), 496 = 2 4 (2 5 1), 8128 = 2 6 (2 7 1) 2 1 Q = 2 e+1 1 a = 2 e Q (perfect numbers ) Q = 2 e+1 1 Q 2 e+1 1 e + 1 Q
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
GO
GO 2016 8 26 1 a σ(a) σ(a) = 2a, 6,28,496,8128 6 = 2 (2 2 1), 28 = 2 2 (2 3 1), 496 = 2 4 (2 5 1), 8128 = 2 6 (2 7 1) 2 1 Q = 2 e+1 1 a = 2 e Q (perfect numbers ) Q = 2 e+1 1 Q 2 e+1 1 e + 1 Q = 2 e+1
1
005 11 http://www.hyuki.com/girl/ http://www.hyuki.com/story/tetora.html http://www.hyuki.com/ Hiroshi Yuki c 005, All rights reserved. 1 1 3 (a + b)(a b) = a b (x + y)(x y) = x y a b x y a b x y 4 5 6
高等学校学習指導要領解説 数学編
5 10 15 20 25 30 35 5 1 1 10 1 1 2 4 16 15 18 18 18 19 19 20 19 19 20 1 20 2 22 25 3 23 4 24 5 26 28 28 30 28 28 1 28 2 30 3 31 35 4 33 5 34 36 36 36 40 36 1 36 2 39 3 41 4 42 45 45 45 46 5 1 46 2 48 3
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
1 + 1 + 1 + 1 + 1 + 1 + 1 = 0? 1 2003 10 8 1 10 8, 2004 1, 2003 10 2003 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 ( )?, 1, 8, 15, 22, 29?, 1 7, 1, 8, 15, 22,
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
, 1. x 2 1 = (x 1)(x + 1) x 3 1 = (x 1)(x 2 + x + 1). a 2 b 2 = (a b)(a + b) a 3 b 3 = (a b)(a 2 + ab + b 2 ) 2 2, 2.. x a b b 2. b {( 2 a } b )2 1 =
x n 1 1.,,.,. 2..... 4 = 2 2 12 = 2 2 3 6 = 2 3 14 = 2 7 8 = 2 2 2 15 = 3 5 9 = 3 3 16 = 2 2 2 2 10 = 2 5 18 = 2 3 3 2, 3, 5, 7, 11, 13, 17, 19.,, 2,.,.,.,?.,,. 1 , 1. x 2 1 = (x 1)(x + 1) x 3 1 = (x 1)(x
() 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
3 m = [n, n1, n 2,..., n r, 2n] p q = [n, n 1, n 2,..., n r ] p 2 mq 2 = ±1 1 1 6 1.1................................. 6 1.2......................... 8 1.3......................... 13 2 15 2.1.............................
(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n
3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)
2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
独立性の検定・ピボットテーブル
II L04(2016-05-12 Thu) : Time-stamp: 2016-05-12 Thu 12:48 JST hig 2, χ 2, V Excel http://hig3.net ( ) L04 II(2016) 1 / 20 L03-Q1 Quiz : 1 { 0.95 (y = 10) P (Y = y X = 1) = 0.05 (y = 20) { 0.125 (y = 10)
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
76 3 B m n AB P m n AP : PB = m : n A P B P AB m : n m < n n AB Q Q m A B AQ : QB = m : n (m n) m > n m n Q AB m : n A B Q P AB Q AB 3. 3 A(1) B(3) C(
3 3.1 3.1.1 1 1 A P a 1 a P a P P(a) a P(a) a P(a) a a 0 a = a a < 0 a = a a < b a > b A a b a B b B b a b A a 3.1 A() B(5) AB = 5 = 3 A(3) B(1) AB = 3 1 = A(a) B(b) AB AB = b a 3.1 (1) A(6) B(1) () A(
2012 A, N, Z, Q, R, C
2012 A, N, Z, Q, R, C 1 2009 9 2 2011 2 3 2012 9 1 2 2 5 3 11 4 16 5 22 6 25 7 29 8 32 1 1 1.1 3 1 1 1 1 1 1? 3 3 3 3 3 3 3 1 1, 1 1 + 1 1 1+1 2 2 1 2+1 3 2 N 1.2 N (i) 2 a b a 1 b a < b a b b a a b (ii)
10_11p01(Ł\”ƒ)
q q q q q q q q q q q q q q q q q q q q q q q q q qq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q
レイアウト 1
1 1 3 5 25 41 51 57 109 2 4 Q1 A. 93% 62% 41% 6 7 8 Q1-(1) Q2 A. 24% 13% 52% Q3 Q3 A. 68% 64 Q3-(2) Q3-(1) 9 10 A. Q3-(1) 11 A. Q3-(2) 12 A. 64% Q4 A. 47% 47% Q5 QQ A. Q Q A. 13 QQ A. 14 Q5-(1) A. Q6
1.500 m X Y m m m m m m m m m m m m N/ N/ ( ) qa N/ N/ 2 2
1.500 m X Y 0.200 m 0.200 m 0.200 m 0.200 m 0.200 m 0.000 m 1.200 m m 0.150 m 0.150 m m m 2 24.5 N/ 3 18.0 N/ 3 30.0 0.60 ( ) qa 50.79 N/ 2 0.0 N/ 2 20.000 20.000 15.000 15.000 X(m) Y(m) (kn/m 2 ) 10.000
untitled
0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.
( )
18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................
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
I II
I II I I 8 I I 5 I 5 9 I 6 6 I 7 7 I 8 87 I 9 96 I 7 I 8 I 9 I 7 I 95 I 5 I 6 II 7 6 II 8 II 9 59 II 67 II 76 II II 9 II 8 II 5 8 II 6 58 II 7 6 II 8 8 I.., < b, b, c, k, m. k + m + c + c b + k + m log
1: n n e = φ(n) e 2 [2] 1 [1] 3 [3] 2 [2] 4 [2,2] 2 [2] 5 [5] 4 [2,2] 6 [2,3] 2 [2] 7 [7] 6 [2,3] 8 [2,2,2] 4 [2,2] 9 [3,3] 6 [2,3] 10 [2,5] 4 [2,2] 1
![1: n n e = φ(n) e 2 [2] 1 [1] 3 [3] 2 [2] 4 [2,2] 2 [2] 5 [5] 4 [2,2] 6 [2,3] 2 [2] 7 [7] 6 [2,3] 8 [2,2,2] 4 [2,2] 9 [3,3] 6 [2,3] 10 [2,5] 4 [2,2] 1](/thumbs/92/108896862.jpg "1: n n e = φ(n) e 2 [2] 1 [1] 3 [3] 2 [2] 4 [2,2] 2 [2] 5 [5] 4 [2,2] 6 [2,3] 2 [2] 7 [7] 6 [2,3] 8 [2,2,2] 4 [2,2] 9 [3,3] 6 [2,3] 10 [2,5] 4 [2,2] 1")
(3) 25 7 17 1 φ n φ(n) n k n. k < n k n k φ(n). 10 1 10, 3 10, 7 10, 9 10 φ(10) = 4. φ(100) = 40, φ(1000) = 400. 100.,. φ.. 1. n p p p φ(p) = p 1. 2. n p 2 p p 2 φ(p 2 ) = p 2 p = p(p 1). 3. φ(p e ) =
1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C
0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,
xyz,, uvw,, Bernoulli-Euler u c c c v, w θ x c c c dv ( x) dw uxyz (,, ) = u( x) y z + ω( yz, ) φ dx dx c vxyz (,, ) = v( x) zθ x ( x) c wxyz (,, ) =
,, uvw,, Bernoull-Euler u v, w θ dv ( ) dw u (,, ) u( ) ω(, ) φ d d v (,, ) v( ) θ ( ) w (,, ) w( ) θ ( ) (11.1) ω φ φ dθ / dφ v v θ u w u w 11.1 θ θ θ 11. vw, (11.1) u du d v d w ε d d d u v ω γ φ w u
(yx4) 1887-1945 741936 50 1995 1 31 http://kenboushoten.web.fc.com/ OCR TeX 50 yx4 e-mail: yx4.aydx5@gmail.com i Jacobi 1751 1 3 Euler Fagnano 187 9 0 Abel iii 1 1...................................
note4.dvi
10 016 6 0 4 (quantum wire) 4.1 4.1.1.6.1, 4.1(a) V Q N dep ( ) 4.1(b) w σ E z (d) E z (d) = σ [ ( ) ( )] x w/ x+w/ π+arctan arctan πǫǫ 0 d d (4.1) à ƒq [ƒg w ó R w d V( x) QŽŸŒ³ džq x (a) (b) 4.1 (a)
B
B YES NO 5 7 6 1 4 3 2 BB BB BB AA AA BB 510J B B A 510J B A A A A A A 510J B A 510J B A A A A A 510J M = σ Z Z = M σ AAA π T T = a ZP ZP = a AAA π B M + M 2 +T 2 M T Me = = 1 + 1 + 2 2 M σ Te = M 2 +T
![Œ{Ł¶/1ŒÊ −ªfiª„¾ [ 1…y†[…W ]](/thumbs/50/26134150.jpg "Œ{Ł¶/1ŒÊ −ªfiª„¾ [ 1…y†[…W ]")
平成20年5月 協会創立50年の歩み 海の安全と環境保全を目指して 友國八郎 海上保安庁 長官 岩崎貞二 日本船主協会 会長 前川弘幸 JF全国漁業協同組合連合会 代表理事会長 服部郁弘 日本船長協会 会長 森本靖之 日本船舶機関士協会 会長 大内博文 航海訓練所 練習船船長 竹本孝弘 第二管区海上保安本部長 梅田宜弘
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%)
,2,4
2005 12 2006 1,2,4 iii 1 Hilbert 14 1 1.............................................. 1 2............................................... 2 3............................................... 3 4.............................................
untitled
20 7 1 22 7 1 1 2 3 7 8 9 10 11 13 14 15 17 18 19 21 22 - 1 - - 2 - - 3 - - 4 - 50 200 50 200-5 - 50 200 50 200 50 200 - 6 - - 7 - () - 8 - (XY) - 9 - 112-10 - - 11 - - 12 - - 13 - - 14 - - 15 - - 16 -
untitled
19 1 19 19 3 8 1 19 1 61 2 479 1965 64 1237 148 1272 58 183 X 1 X 2 12 2 15 A B 5 18 B 29 X 1 12 10 31 A 1 58 Y B 14 1 25 3 31 1 5 5 15 Y B 1 232 Y B 1 4235 14 11 8 5350 2409 X 1 15 10 10 B Y Y 2 X 1 X
さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1
... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
13 0 1 1 4 11 4 12 5 13 6 2 10 21 10 22 14 3 20 31 20 32 25 33 28 4 31 41 32 42 34 43 38 5 41 51 41 52 43 53 54 6 57 61 57 62 60 70 0 Gauss a, b, c x, y f(x, y) = ax 2 + bxy + cy 2 = x y a b/2 b/2 c x
xyr x y r x y r u u
xyr x y r x y r u u y a b u a b a b c d e f g u a b c d e g u u e e f yx a b a b a b c a b c a b a b c a b a b c a b c a b c a u xy a b u a b c d a b c d u ar ar a xy u a b c a b c a b p a b a b c a
株式会社日清製粉グループ本社 第158期中間事業報告書
C O N T E N T S...1...3...5...7...9...11...12...13...14 1 2 3 4 3.7% 5.8% 8.5% 70,100kL 81.2% 0.8% 25 20 15 10 5 0 9.18 9.54 9.74 9.62 9.65 9.71 21.04 21.97 22.44 22.23 8.54 22.31 22.45 20.41 15 12 9 6
89 91 93 95 97 99 101 103 105 107 109 111 113 115 H 117 119 l l 121 l l 123 125 127 129 l l l l 131 kl kl kl kl 133 135 137 139 141 143 145 147 149 151 153 155 157 159
IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a
1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =
1. A0 A B A0 A : A1,...,A5 B : B1,...,B
1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A, B Z Z m, n Z m n m, n A m, n B m=n (1) A, B (2) A B = A B = Z/ π : Z Z/ (3) A B Z/ (4) Z/ A, B (5) f : Z Z f(n) = n f = g π g : Z/ Z A, B (6)
ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2
働く女性の母性健康管理、母性保護に関する法律のあらまし
17 1 3 3 12 3 13 10 19 21 22 22 23 26 28 33 33 35 36 38 39 1 I 23 2435 36 4/2 4/3 4/30 12 13 14 15 16 (1) 1 2 3 (2) 1 (1) (2)(1) 13 3060 32 3060 38 10 17 20 12 22 22 500 20 2430m 12 100 11 300m2n 2n
4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X
4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0
LCM,GCD LCM GCD..,.. 1 LCM GCD a b a b. a divides b. a b. a, b :, CD(a, b) = {d a, b }, CM(a, b) = {m a, b }... CM(a, b). q > 0, m 1, m 2 CM
LCM,GCD 2017 4 21 LCM GCD..,.. 1 LCM GCD a b a b. a divides b. a b. a, b :, CD(a, b) = {d a, b }, CM(a, b) = {m a, b }... CM(a, b). q > 0, m 1, m 2 CM(a, b) = m 1 + m 2 CM(a, b), qm 1 CM(a, b) m 1, m 2
Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)
![Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)](/thumbs/94/118373946.jpg "Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)")
3 3 22 Z[i] Z[i] π 4, (x) π 4,3 (x) x (x ) 2 log x π m,a (x) x ϕ(m) log x. ( ). π(x) x (a, m) = π m,a (x) x modm a π m,a (x) ϕ(m) π(x) ϕ(m) x log x ϕ(m) m f(x) g(x) (x α) lim f(x)/g(x) = x α mod m (a,
n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz
1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x
() 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 ()
(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37
4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin
広報さっぽろ 2016年8月号 厚別区
8/119/10 P 2016 8 11 12 P4 P6 P6 P7 13 P4 14 15 P8 16 P6 17 18 19 20 P4 21 P4 22 P7 23 P6 P7 24 25 26 P4 P4 P6 27 P4 P7 28 P6 29 30 P4 P5 31 P5 P6 2016 9 1 2 3 P4 4 P4 5 P5 6 7 8 P4 9 10 P4 1 b 2 b 3 b
No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y
No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ
ブック 1.indb
20 29 29 18 21 29 10 30 31 10 11 12 30 13 10 30 14 11 30 15 12 16 13 17 14 18 15 19 16 20 17 21 18 10 20 29 82 83 84 85 86 87 88 20 10 89 20 12 11 90 20 13 12 91 20 14 13 92 20 14 14 93 15 15 94 15 16
6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4
35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
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
untitled
351 351 351 351 13.0 0.0 25.8 1.0 0.0 6.3 92.9 0.0 80.5 0.0 1.5 15.9 0.0 3.5 13.1 0.0 30.0 54.8 18.0 0.0 27.5 1.0 0.0 2.5 94.7 0.0 91.7 0.0 1.3 14.7 0.0 3.8 14.4 0.0 25.0 50.5 16.0 0.0 27.5 2.0 0.0 2.5
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/
/02/18
3 09/0/8 i III,,,, III,?,,,,,,,,,,,,,,,,,,,,?,?,,,,,,,,,,,,,,!!!,? 3,,,, ii,,,!,,,, OK! :!,,,, :!,,,,,, 3:!,, 4:!,,,, 5:!,,! 7:!,,,,, 8:!,! 9:!,,,,,,,,, ( ),, :, ( ), ( ), 6:!,,, :... : 3 ( )... iii,,
,. 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,
untitled
. 96. 99. ( 000 SIC SIC N88 SIC for Windows95 6 6 3 0 . amano No.008 6. 6.. z σ v σ v γ z (6. σ 0 (a (b 6. (b 0 0 0 6. σ σ v σ σ 0 / v σ v γ z σ σ 0 σ v 0γ z σ / σ ν /( ν, ν ( 0 0.5 0.0 0 v sinφ, φ 0 (6.
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,
8 i, III,,,, III,, :!,,,, :!,,,,, 4:!,,,,,,!,,,, OK! 5:!,,,,,,,,,, OK 6:!, 0, 3:!,,,,! 7:!,,,,,, ii,,,,,, ( ),, :, ( ), ( ), :... : 3 ( )...,, () : ( )..., :,,, ( ), (,,, ),, (ϵ δ ), ( ), (ˆ ˆ;),,,,,,!,,,,.,,
Step 2 O(3) Sym 0 (R 3 ), : a + := λ 1 λ 2 λ 3 a λ 1 λ 2 λ 3. a +. X a +, O(3).X. O(3).X = O(3)/O(3) X, O(3) X. 1.7 Step 3 O(3) Sym 0 (R 3 ),
1 1 1.1,,. 1.1 1.2 O(2) R 2 O(2).p, {0} r > 0. O(3) R 3 O(3).p, {0} r > 0.,, O(n) ( SO(n), O(n) ): Sym 0 (R n ) := {X M(n, R) t X = X, tr(x) = 0}. 1.3 O(n) Sym 0 (R n ) : g.x := gxg 1 (g O(n), X Sym 0
名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト
名古屋工業大の数学 年 ~5 年 大学入試数学動画解説サイト http://mathroom.jugem.jp/ 68 i 4 3 III III 3 5 3 ii 5 6 45 99 5 4 3. () r \= S n = r + r + 3r 3 + + nr n () x > f n (x) = e x + e x + 3e 3x + + ne nx f(x) = lim f n(x) lim
II
II 16 16.0 2 1 15 x α 16 x n 1 17 (x α) 2 16.1 16.1.1 2 x P (x) P (x) = 3x 3 4x + 4 369 Q(x) = x 4 ax + b ( ) 1 P (x) x Q(x) x P (x) x P (x) x = a P (a) P (x) = x 3 7x + 4 P (2) = 2 3 7 2 + 4 = 8 14 +
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a
![さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a](/thumbs/97/131721994.jpg "さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a")
φ + 5 2 φ : φ [ ] a [ ] a : b a b b(a + b) b a 2 a 2 b(a + b). b 2 ( a b ) 2 a b + a/b X 2 X 0 a/b > 0 2 a b + 5 2 φ φ : 2 5 5 [ ] [ ] x x x : x : x x : x x : x x 2 x 2 x 0 x ± 5 2 x x φ : φ 2 : φ ( )