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
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1 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(a, b) = m 1 + m 2 CM(a, b), qm 1 CM(a, b) m 1, m 2 CM(a, b), m 1 > m 2 = m 1 m 2 CM(a, b). CM(a, b) LCM(a, b), (Least common multiple). CD(a, b) GCD(a, b), (Greatest common divisor). a = bq + r, r 0 CD(a, b) = CD(b, r) ( ) 1
2 1 m CM(a, b) = L m, (L = LCM(a, b)) Proof m L, m = ql + r, r < L. L, m CM(a, b) r = m ql CM(a, b)., r = 0. LCM GCD. ( ) 2 d CD(a, b) = d δ, (δ = GCD(a, b)) Proof d CD(a, b), δ = GCD(a, b), d δ LCM d, δ LCM. L 0 = LCM(d, δ).,l 0 δ. L 0 = δ.l 0 δ. d, δ CD(a, b) a = a d, b = b d, a = a δ, b = b δ. a = a d, a = a δ = a CM(d, δ) 1 a LCM(d, δ). a 0, a = a 0 L 0 b 0, b = b 0 L 0 a = a 0 L 0, b = b 0 L 0 = L 0 CD(a, b), L 0 δ = GCD(a, b). L 0 = δ = GCD(a, b). d δ 3 ab = LCM(a, b) GCD(a, b) L = LCM(a, b), δ = GCD(a, b), a L L = ab, b L L = ba. ab CM(a, b) ab L. ab = LD D. ab = LD = ab D, ab = ba D = b = b D, a = a D 2
3 D CD(a, b). 2,D δ. δ = De e. b = b δ = Db, δ = De b De = Db ; b e = b. L = ab = ab e, L = a b = ba e. L/e = L 0, L 0 = a b, L 0 = b a. L 0 CM(a, b). 1 L 0 L. L L 0 = fl L L 0 = L; L = L 0, e = 1. δ = D ab = LD = Lδ. 4 a, b:, a bc = a c Proof bc = ak k Z. GCD(a, b) = 1 ab = LCM(a, b). X = bc = ak CM(a, b). X ab = LCM(a, b). X = abs, (s Z). X = bc = abs c = as; a c. 1.1,, 0 a, b c a = bc b a (divisor), a b (multiple). b a ( (factor)). Z (ring of integers).,,. Z b bz (b) b (ideal). (b) = bz b (b) (generator). b = 1 (1) = Z. a b a (b). k, a = bk. b a, b a. 1.2 d a, b, d = ax + by x, y.. 3
4 a = 11, b = 8 d 1. 1 = 11x+8y x, y. y = 1, 2, 3, 1 8y : 1 = 11x + 8y x, y y y , y = 7, 11x = 1 8y = 55, x = 5. 1 = = 11( 5) , 1 : 1 = , d a, b d = ax + by.,.,. 2 J d = ax + by x, y, d ax + by, Z J. J = {ax + by x, y Z}. J α, β J α + β J, 2. z Z, α J zα J. 1 a, b a = bq + r, 0 r < b r. 4
5 J (Z ) (ideal). (b) ( principal ideal). J {ax + by x, y Z} a, b az + bz. (a, b). (2) 1 J z Z z J. J = Z. (a, b). (a, b),,. 2.1 J J α = ax + by. a, b d. a = a d, b = b d a, b α = ax + by = a dx + b dy = (a x + b y)d. α = (a x + b y)d d. J + = {α > 0 α J}, a, b d J +. d J +. J +,. δ. α δ q, r α = qδ + r, 0 r < δ. r = α + ( q)δ (2) δ J ( q)δ J. α J (1) α+( q)δ J. r = α+( q)δ J. 0 r J. δ J + r < δ r = 0. α = qδ. α (δ). α < 0 α J α (δ). J (δ). δ J, (δ) J. J = (δ). a, b J = (δ) δ a, b. δ a, b. d 1 a, b a = a 1 d 1, b = b 1 d 1 a 1, b 1. δ J δ = ax 0 + by 0 δ = ax 0 + by 0 = a 1 d 1 x 0 + b 1 d 1 y 0 = (a 1 x 0 + b 1 y 0 )d 1 5
6 δ d 1. δ a, b d., δ = d. J = (d). J d. d J d J +. x, y a, b d d = ax + by x, y 1.3. a, b 3. v 0 = (1, 0, a), v 1 = (0, 1, b). a b q 1, r 1 a = bq 1 + r 1. a = bq 1 + r 1 (b, r 1 ) (a, b) (b, r 1 ). r 1 = a + ( q)b (a, b) (b, r 1 ) (a, b). (b, r 1 ) = (a, b). v 2 v 2 = (1, q 1, r 1 ). v 2 = v 0 q 1 v 1 2: d = ax + by v a v 1 q b v 2 1 q 1 r 1 r 1 > 0 b r 1 q 2, r 2 b = r 1 q 2 + r 2. v 3 v 3 = v 1 q 2 v 2,. b > r 1 >. r h 1 > r h = 0 h > 0. v h 1 = (x, y, d) d = r h 1 a, b (a, b) = (b, r 1 ) = = (r h 1, r h ) = (r h 1 ). x, y ax + by = d. 6
7 3: a = 11, b = a = b = x = 3 y = 4 d = 1 a = 11, b = 8. v 0, v 1,,. x = 3, y = 4, d = 1., ( 4) = 1. 7
8 d = ax + by. w w = (a, b, 1). v 0 w = (1, 0, a) (a, b, 1) = 1 a a = 0, v 1 w = 1 b b = 0 v 2 w = v 0 w q 1 v 1 w = 0, v 3 w = v 1 w q 2 v 2 w = 0, v h 1 w = 0. v h 1 = (x, y, d) v h 1 w = ax + by d. d = ax + by. 3 a, b, c a, b d 1 a b (relatively prime). a b (a, b) 1 (a, b) = Z. a, b a, b, a, b. 1 = ax + by x, y. a, b, a, b 0. a, b, c. 1 a, b, c a, b. a bc a c. Proof a bc ak = bc k. a, b 1 = ax + by x, y. c c = acx + bcy = acx + aky = a(cx + ky). f = cx + ky c = af. a c. a, b (a, b) = (1).. (a, b) = (1), bc (a) c (a). 8
9 4 D.A.(, ). p a, b : ab 0 mod p a 0 b 0 mod p. ab 0 mod p a 0 b 0 mod p., a, p. b. p b Q r p = bq + r, r < b. ab = pm p = bq + r a ap = abq + ar = pmq + ar. p(a mq) = ar m = a mq ar = pm, 0 r < b. b r = 0. p = bq. p Q = 1; p = b. :b 0 mod p. p J = pz.. a J = pz p. a p 1 1 = am + pn n, m. a J., J,.. 5. σ(a) a, σ(a). P, m σ(p e ) + m q a = P e q m P ( ).. P σ(a) P a = (P 2)Maxp(a) m(p 1). (1) Maxp(a) a. a m P ( ). 9
10 6 φ φ(a). φ(p e ), (e > 1)., 1 φ(p e ) + 1 q a = P e q P φ. P = 2. 4: P = 2 φ e a φ(a) ,a 3,5,17,257,65537.,, P = 2 q = φ(p e ) + 1 = 2 e e 1 = 2 m q. 7 P, m q = P e m a = P e q P, m, ( subperfect number), q ( subprime number)... a = P e q P σ(a) = P σ(p e q) = (P e+1 1)(q + 1) = P a (q + 1 P e ) q = P e m q + 1 P e = m P σ(a) = P a m.. 10
11 P, m ( subperfect number with translation parameter m).. P > 2, m = 0 P e m. σ(p e ) q = σ(p e ) 1 + m a = P e q., m P e m. 7.1 a = P f Q(Q : ). P σ(a) = (P f+1 1)(Q + 1) = P a + P f+1 (Q + 1) m = P σ(a) P a = P f+1 (Q + 1). Q = P f+1 + m 1. P, m. 7.2 P = 3, m = 3 P = 3 Q = 3 e m. m. m = 1, 7, 10 Q. m = 3. 5, , 2.. A = B =
12 5: [P = 3, m = 3], e a A B 7.3 a = P f rq(r < q : ). P σ(a) = (P f+1 1)(r + 1)(q + 1), P a m = P f+1 rq m. N = P f+1 1, A = (r + 1)(q + 1), B = rq, = r + q NA = (P f+1 1)(r + 1)(q + 1), P a m = P f+1 rq m = (N + 1)B m. A = B NB + N( + 1) = P f+1 rq m = (N + 1)B m. N( + 1) = B m. q 0 = q N, r 0 = r N, B 0 = q 0 r 0 B 0 = B N + N 2. N( + 1) = B m = B 0 + N N 2. D = N(N + 1) + m, B 0 = D.. f m N = P f+1 1, D = N(N + 1)+m D, B 0 = D q = q 0 +N, r = r 0 +N., a = P f rq. 12
13 8 P = 3, m = 0 m = 0 q = 3 e m. 2σ(a) = 3a... 9 P :, m = 0 P σ(a) = P a. P = 2, P, m = 0. 1 P = 3, a = 2 P :, m = 0. Proof., σ(a) P,σ(a) = P ε L(L P ). P σ(a) = P P ε L = P a., a = P ε 1 LP. N = P ε 1, M = P L a = P ε 1 M, P a = (N + 1)M. M P, σ(a) = σ(m)σ(p ε 1 ). P σ(a) = P a, P σ(a) = σ(m)p σ(p ε 1) = Nσ(M). Nσ(M) = (N + 1)M. N N + 1 = M σ(m). N, M = kn, σ(m) = k(n + 1) k N + 1., σ(m) M = k. M = kn, k M, σ(m) = M + k M :,k = 1. 13
14 M :, P = 3, L = 1, a = 4. M = P L. ( ) 14
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1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.
1 1 n 0, 1, 2,, n 1 1.1 n 2 a, b a n b n a, b n a b (mod n) 1 1. n = 10 1567 237 (mod 10) 2. n = 9 1567 1826578 (mod 9) n II Z n := {0, 1, 2,, n 1} 1.2 a b a = bq + r (0 r < b) q, r q a b r 2 1. a = 456,
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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
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(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
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15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = N N 0 x, y x y N x y (mod N) x y N mod N mod N N, x, y N > 0 (1) x x (mod N) (2) x y (mod N) y x
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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)
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数学概論I
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1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります
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空き容量一覧表（154kV以上）
1/3 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量 覧 < 留意事項 > (1) 空容量は 安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 熱容量を考慮した空き容量を記載しております その他の要因 ( や系統安定度など ) で連系制約が発 する場合があります (3) 表 は 既に空容量がないため
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meiji_resume_1.PDF
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1/8 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( や系統安定度など ) で連系制約が発生する場合があります (3)
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