n 2 + π2 6 x [10 n x] x = lim n 10 n n 10 k x 1.1. a 1, a 2,, a n, (a n ) n=1 {a n } n=1 1.2 ( ). {a n } n=1 Q ε > 0 N N m, n N a m
|
|
|
- れいが とどろき
- 5 years ago
- Views:
Transcription
1 n 2 + π2 6 x [10 n x] x = lim n 10 n n 10 k x 1.1. a 1, a 2,, a n, (a n ) n=1 {a n } n=1 1.2 ( ). {a n } n=1 Q ε > 0 N N m, n N a m a n < ε 1
2 1. ε = 10 1 N m, n N a m a n < ε = 10 1 N a m, a n ε = 10 2 N m, n N a m a n < ε = 10 2 N a m, a n ε = 10 3 N m, n N a m a n < ε = 10 3 N a m, a n ε 10 n n r a n = r, r = 1, 2, {a n } n=1 2. ε > 0 N N Nε > 0 n > m N a n = , n = 1, 2, n2 a n a m = < 1 (m + 1) (m + 2) n 2 1 m(m + 1) n(n + 1) = 1 n 1 m + 1 < 1 n < 1 N 2
3 ε > 0 N N Nε > 1 n > m N a n a m = a n a m < ε n, m N a n a m < ε 1.4 ( ). {a n } n=1 Q 1. ε > 0 N N m, n N a m a n < ε (1) 2. ε > 0 N N m, n > N a m a n < ε (2) 3. ε > 0 N N m, n N a m a n ε (3) 4. ε > 0 N N m, n > N a m a n ε (4) 4. () (1) = (2), (3) = (4) (4) (1) (1) ε > 0 (4) ε > 0 1 ε (4) N N 2 n, m > N a n a m 1 2 ε N M 3
4 M N n, m > M a n a m 1 2 ε (1) N = M + 1 n N n > M 1 2 ε < ε n, m N a n a m < ε 1.5 ( N, ε ). {a n } n=1 Q 1. ε > 0 N N m, n N a m a n < ε (5) 2. ε > 0 N N m, n N a m a n < 2ε (6) 3. ε > 0 N N m, n N + 1 a m a n < ε (7) 4. ε > 0 N N m, n N a m a n < 1 2 ε (8) 4 4
5 N, ε. (3) (7) (1) (5) (6) (8) (1) (8) = (4) (1) = (6) (6) (8) (8) ε > 0 (6) 1 ε (6) N 4 (8) m, n N a m a n < ε {a n } n=1 α {a n + 10 n } n=1 α 1.6 (). {a n } n=1 {b n} n=1 [ ] ε > 0 N N n N a n b n < ε {a n } n=1, {b n} n=1, {c n} n=1 1. {a n } n=1 {a n} n=1 2. {a n } n=1 {b n} n=1 {b n} n=1 {a n} n=1 3. {a n } n=1 {b n} n=1 {b n} n=1 {c n} n=1 {a n } n=1 {c n} n= {a n } n=1 {b n} n=1 {b n } n=1 {c n} n=1 [ ] ε > 0 N N n N a n b n < ε 5
6 [ ] ε > 0 N N n N b n c n < ε ε > 0 [ ] N 1 N n N 1 a n b n < 1 2 ε [ ] N 2 N n N 2 b n c n < 1 2 ε N = max(n 1, N 2 ) n N a n b n < 1 2 ε, b n c n < 1 2 ε α + β α + β n N a n c n a n b n + b n c n < 1 2 ε ε = ε (). {a n } n=1 M a n M, n N {a n } n=1 ε = 1 > 0 N m, n N a n a m 1 a n M, n N M = max( a 1, a 2,, a N 1, a N + 1) {a n } n=1, {b n} n=1 1. {a n + b n } n=1 6
7 2. {a n b n } n=1 3. {a n b n } n=1 4. {b n } n=1 0 ε 0 N n N b n > 1 2 ε 0 > 0 5. {b n } n=1 0 b n 0, n = 1, 2, {a n /b n } n=1 4,5 4. {b n } n=1 0 ε 0 > 0 N n N b n > ε 0 {b n } n=1 N n, m N b n b m > 1 2 ε N n N b n > ε 0 N = N n N b n > ε 0 m N b m b n b m b n > ε ε 0 = 1 2 ε 0 > 0 ε 0 N n N b n > 1 2 ε 0 > 0 5. {a n /b n } n=1 M > 0 a n, b n M 4 ε 0 N n N b n > 1 2 ε 0 > 0 b 1, b 2,, b N 1 0 ( κ = min b 1, b 2,, b N 1, 1 ) 2 ε 0 b n κ > 0, n N m, n N a n a m b n b m = a nb m a m b n b n b m a nb m a m b m + a m b m a m b n b n b m a nb m a m b m + a m b m a m b n κ 2 M κ 2 ( a n a m + b m b n ) 7
8 ε > 0 N 1, N 2 n, m N 1 = a n a m < κ2 ε 2M, n, m N 2 = b n b m < κ2 ε 2M N = max(n 1, N 2 ) n, m N = a n a m < κ2 ε 2M, b n b m < κ2 ε 2M n, m N a n a m b n M κ 2 ( a n a m + b m b n ) ε {a n /b n } n= {a n } n=1, {b n} n=1 b m 1. {a n } n=1 + {b n} n=1 {a n + b n } n=1 2. {a n } n=1 {b n} n=1 {a n b n } n=1 3. {a n } n=1 {b n} n=1 {a nb n } n=1 4. {b n } n=1 0 {a n} n=1 {b n} n=1 {a n/b n } n= A B A := B, A B {a n } n=1, {b n} n=1, {a n} n=1, {b n} n=1 {a n} n=1 {a n} n=1 {b n } n=1 {b n} n=1 1. {a n } n=1 + {b n} n=1 {a n} n=1 + {b n} n=1 2. {a n } n=1 {b n} n=1 {a n} n=1 {b n} n=1 3. {a n } n=1 {b n} n=1 {a n} n=1 {b n} n=1 4. {b n } n=1 {b n} n=1 0 {a n} n=1 {b n } n=1 {a n} n=1 {b n} n= M a n + a n + b n + b n M {b n } n=1 {b n} n=1 0 κ > 0 b n, b n κ 8
9 ε > 0 N 1, N 2 n N 1 = a n a n < κ2 2M ε n N 2 = b n b n < κ2 2M ε a n a n b n = a nb n a nb n b n b n b n a nb n a nb n + a nb n a nb n b n b n a nb n a nb n + a nb n a nb n κ 2 M κ 2 ( a n a n + b n b n ) n N = max(n 1, N 2 ) a n a n b n M κ 2 ( a n a n + b n b n ) < ε b n {a n } n=1 {b n} n=1 {a n} n=1 {b n} n= {a n } n=1 {b n} n=1 {a n} n=1 {b n} n=1 ε > 0 N n N b n a n > ε 2.2. {a n } n=1, {b n} n=1, {a n} n=1, {b n} n=1 {a n} n=1 {a n} n=1 {b n } n=1 {b n} n=1 {a n} n=1 {b n} n=1 {a n} n=1 {b n} n=1. {a n } n=1 {b n} n=1 ε > 0 N n N b n a n > 1 3 ε {a n} n=1 {a n} n=1 {b n } n=1 {b n} n=1 N 1, N 2 n N 1 = a n a n < 1 3 ε n N 2 = b n b n < 1 3 ε 9
10 n N = max(n 1, N 2, N) b n a n = b n b n + b n a n + a n a n > 1 3 ε 1 3 ε 1 3 ε = ε {a n} n=1 {b n} n= {a n } n=1, {b n} n=1, {c n} n=1 1. {a n } n=1 {a n} n=1 2. {a n } n=1 {b n} n=1 {b n} n=1 {a n} n=1 {a n} n=1 = {b n} n=1 3. {a n } n=1 {b n} n=1 {b n} n=1 {c n} n=1 {a n} n=1 {c n} n=1 4. {a n } n=1 {b n} n=1 {b n} n=1 {a n} n=1 5. {a n } n=1 {b n} n=1 {a n} n=1 + {c n} n=1 {b n} n=1 + {c n} n= {a n } n=1 {b n} n=1 {c n} n=1 {a nb n } n=1 {a nc n } n=1. 4 {a n } n=1 {b n} n=1 ε 0 > 0 N n 0 N b n0 a n0 ε 0 {a n } n=1 {b n} n=1 ε > 0 N 1, N 2 n, m N 1 = a n a m < 1 3 ε 0 n, m N 2 = b n b m < 1 3 ε 0 N = max(n 1, N 2 ) n 0 N a n0 b n0 ε 0 m N b m a m = (b m b n0 ) + (b n0 a n0 ) + (a n0 a m ) < 1 3 ε 0 ε ε 0 = 1 3 ε 0 ε 0 N m N = b m a m < 1 3 ε 0 < 0 ε > 0 n N = a n b n > 1 3 ε 0 > ε {a n } n=1 {b n} n=1 10
11 3 3.1 ( ). {{a m,n } n=1 } m=1 m N {a m,n } n=1 3.2 ( ). {α n } m=1 = {{a m,n } n=1 } m=1 ε = {ε n} n=1 0 N n n, m > N ε < α m α n < ε 3.3 ( ). α = {α n } n=1 {α n } m=1 = {{a m,n} n=1 } m=1 α ε = {ε n } n=1 0 N n n, m > N ε < α m α < ε α {α n } m=1 = {{a m,n} n=1 } m=1 {α n } m=1 = {{a m,n} n=1 } m=1 3.4 ( ). α = {a n } n=1 α m = {a m } n=1 α Proof. ε = {ε n } n=1 0 ε 0 N 0 n N 0 ε n ε 0 > 0 ε > 0 {a n } n=1 N m N 0 lim m α m = α n, m N = a n a m < min(ε, ε 0 ) α α m < min(ε, ε 0 ) < ε {α m } m=1 = {{a m,n} n=1 } m=1 α m = {a m,n } n=1 N(m) n 1, n 2 N(m) 11
12 a m,n1 a m,n2 < 2 m {α m } m=1 L N M(L) m 1, m 2 M(L) 2 L 2 < α m1 α m2 < 2 L 2 m 1, m 2 M(L) Ñ(m 1, m 2 ) m 1, m 2 M(L), n > Ñ(m 1, m 2 ) 2 L 1 < α m1,n α m2,n < 2 L 2 2 L 1 < α m1,n(m 1 )+N(m 2 )+Ñ(m 1,m 2 ) α m 2,N(m 1 )+N(m 2 )+Ñ(m 1,m 2 ) < 2 L 1 2 m 1 < α m1,n(m 1 ) α m1,n(m 1 )+N(m 2 )+Ñ(m 1,m 2 ) < 2 m 1 2 m 2 < α m2,n(m 2 ) α m2,n(m 1 )+N(m 2 )+Ñ(m 1,m 2 ) < 2 m 2 m 1, m 2 L L < α m1,n(m 1 ) α m2,n(m 2 ) < 2 L ε > 0 2 L < ε L L m 1, m 2 L + 2 ε < α m1,n(m 1 ) α m2,n(m 2 ) < ε α = {a L,N(L) } L=1 lim m α m = α ε > 0 L N 2 L < ε α m α = {α m,n α n,n(n) } n=1 m n N(m) α m,n α n,n(n) = α m,n α m,n(m) + α m,n(m) α n,n(n) m, n L L < α m,n(m) α n,n(n) < 2 L N(m) n N(m) 2 m α m,n α m,n(m) < 2 m m L + 2 n L L+1 < α m,n α n,n(n) < 2 L+1 m L L+1 < α m α < 2 L+1 lim α m = α m 12
13 3.6. K = {k n } n=1 {α m} = {{a m,n } n=1 } α 1 α 2 α m α m+1 K α = {α m } m=1. α ε > 0 M m 1, m 2 m 1 > m 2 M α m1 α m2 > ε {α m } m=1 ε α m m K 3.7. A M a A a M M M 0 M 0 1. a A a M 0 2. m < M 0 a A a > m. {M n } n=1 M 1 M 0 n, n = 0, 1, 2, M 1, M 2,, M n M n+1 1. a A a M n 2 n M n+1 = M n 2 n 2. M n+1 = M n M n 2 n M n+1 M n {M n } n=1 M 1 1 M n M = lim n M n a M a M n, n = 1, 2,, a A 13
14 4 Q (a) R i α = α + 0. ii. α α α+( α) = ( α)+α = 0. iii. α, β, γ α + (β + γ) = (α + β) + γ. iv. α, β α + β = β + α (b) i. 1 α1 = 1α ii. α, β, γ α(βγ) = (αβ)γ. iii. α, β, γ α(β+γ) = αβ+αγ (α+β)γ = αγ + βγ. iv. α, β αβ = βα. (c) i. 0 α αα 1 = α 1 α = 0. ii (a) i. α α. ii. α β, β α α = β. iii. α β, β γ α γ. iv. α β β α (b) α β α + γ β + γ. (c) α β, 0 γ αγ βγ. 1. R 2. Q Q Q π 14
第86回日本感染症学会総会学術集会後抄録（I）
κ κ κ κ κ κ μ μ β β β γ α α β β γ α β α α α γ α β β γ μ β β μ μ α ββ β β β β β β β β β β β β β β β β β β γ β μ μ μ μμ μ μ μ μ β β μ μ μ μ μ μ μ μ μ μ μ μ μ μ β
koji07-02.dvi
007 I II III 1,, 3, 4, 5, 6, 7 5 4 1 ε-n 1 ε-n ε-n ε-n. {a } =1 a ε N N a a N= a a
1 I
1 I 3 1 1.1 R x, y R x + y R x y R x, y, z, a, b R (1.1) (x + y) + z = x + (y + z) (1.2) x + y = y + x (1.3) 0 R : 0 + x = x x R (1.4) x R, 1 ( x) R : x + ( x) = 0 (1.5) (x y) z = x (y z) (1.6) x y =
_0212_68<5A66><4EBA><79D1>_<6821><4E86><FF08><30C8><30F3><30DC><306A><3057><FF09>.pdf
2010 IA ε-n I 1, 2, 3, 4, 5, 6, 7, 8, ε-n 1 ε-n ε-n? {a n } n=1 1 {a n } n=1 a a {a n } n=1 ε ε N N n a n a < ε
00 IA ε-n I,, 3, 4, 5, 6, 7, 8, 9 4 6 ε-n ε-n ε-n? {a } = {a } = a a {a } = ε ε N N a a < ε ε-n ε ε N a a < ε N ε ε N ε N N ε N [ > N = a a < ε] ε > 0 N N N ε N N ε N N ε a = lim a = 0 ε-n 3 ε N 0 < ε
「産業上利用することができる発明」の審査の運用指針（案）
1 1.... 2 1.1... 2 2.... 4 2.1... 4 3.... 6 4.... 6 1 1 29 1 29 1 1 1. 2 1 1.1 (1) (2) (3) 1 (4) 2 4 1 2 2 3 4 31 12 5 7 2.2 (5) ( a ) ( b ) 1 3 2 ( c ) (6) 2. 2.1 2.1 (1) 4 ( i ) ( ii ) ( iii ) ( iv)
44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)
(1) I 44 II 45 III 47 IV 52 44 4 I (1) ( ) 1945 8 9 (10 15 ) ( 17 ) ( 3 1 ) (2) 45 II 1 (3) 511 ( 451 1 ) ( ) 365 1 2 512 1 2 365 1 2 363 2 ( ) 3 ( ) ( 451 2 ( 314 1 ) ( 339 1 4 ) 337 2 3 ) 363 (4) 46
i ii i iii iv 1 3 3 10 14 17 17 18 22 23 28 29 31 36 37 39 40 43 48 59 70 75 75 77 90 95 102 107 109 110 118 125 128 130 132 134 48 43 43 51 52 61 61 64 62 124 70 58 3 10 17 29 78 82 85 102 95 109 iii
数学概論I
{a n } M >0 s.t. a n 5 M for n =1, 2,... lim n a n = α ε =1 N s.t. a n α < 1 for n > N. n > N a n 5 a n α + α < 1+ α. M := max{ a 1,..., a N, 1+ α } a n 5 M ( n) 1 α α 1+ α t a 1 a N+1 a N+2 a 2 1 a n
(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)
1. 1.1...,. 1.1.1 V, V x, y, x y x + y x + y V,, V x α, αx αx V,, (i) (viii) : x, y, z V, α, β C, (i) x + y = y + x. (ii) (x + y) + z = x + (y + z). 1 (iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y
(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
π, R { 2, 0, 3} , ( R),. R, [ 1, 1] = {x R 1 x 1} 1 0 1, [ 1, 1],, 1 0 1,, ( 1, 1) = {x R 1 < x < 1} [ 1, 1] 1 1, ( 1, 1), 1, 1, R A 1
![π, R { 2, 0, 3} , ( R),. R, [ 1, 1] = {x R 1 x 1} 1 0 1, [ 1, 1],, 1 0 1,, ( 1, 1) = {x R 1 < x < 1} [ 1, 1] 1 1, ( 1, 1), 1, 1, R A 1](/thumbs/91/105315987.jpg "π, R { 2, 0, 3} , ( R),. R, [ 1, 1] = {x R 1 x 1} 1 0 1, [ 1, 1],, 1 0 1,, ( 1, 1) = {x R 1 < x < 1} [ 1, 1] 1 1, ( 1, 1), 1, 1, R A 1")
sup inf (ε-δ 4) 2018 1 9 ε-δ,,,, sup inf,,,,,, 1 1 2 3 3 4 4 6 5 7 6 10 6.1............................................. 11 6.2............................... 13 1 R R 5 4 3 2 1 0 1 2 3 4 5 π( R) 2 1 0
O1-1 O1-2 O1-3 O1-4 O1-5 O1-6
O1-1 O1-2 O1-3 O1-4 O1-5 O1-6 O1-7 O1-8 O1-9 O1-10 O1-11 O1-12 O1-13 O1-14 O1-15 O1-16 O1-17 O1-18 O1-19 O1-20 O1-21 O1-22 O1-23 O1-24 O1-25 O1-26 O1-27 O1-28 O1-29 O1-30 O1-31 O1-32 O1-33 O1-34 O1-35
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
1 911 9001030 9:00 A B C D E F G H I J K L M 1A0900 1B0900 1C0900 1D0900 1E0900 1F0900 1G0900 1H0900 1I0900 1J0900 1K0900 1L0900 1M0900 9:15 1A0915 1B0915 1C0915 1D0915 1E0915 1F0915 1G0915 1H0915 1I0915
i
14 i ii iii iv v vi 14 13 86 13 12 28 14 16 14 15 31 (1) 13 12 28 20 (2) (3) 2 (4) (5) 14 14 50 48 3 11 11 22 14 15 10 14 20 21 20 (1) 14 (2) 14 4 (3) (4) (5) 12 12 (6) 14 15 5 6 7 8 9 10 7
() 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
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an
第１部 一般的コメント
(( 2000 11 24 2003 12 31 3122 94 2332 508 26 a () () i ii iii iv (i) (ii) (i) (ii) (iii) (iv) (a) (b)(c)(d) a) / (i) (ii) (iii) (iv) 1996 7 1996 12
表1票4.qx4
iii iv v 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 22 23 10 11 24 25 26 27 10 56 28 11 29 30 12 13 14 15 16 17 18 19 2010 2111 22 23 2412 2513 14 31 17 32 18 33 19 34 20 35 21 36 24 37 25 38 2614
第１章 国民年金における無年金
1 2 3 4 ILO ILO 5 i ii 6 7 8 9 10 ( ) 3 2 ( ) 3 2 2 2 11 20 60 12 1 2 3 4 5 6 7 8 9 10 11 12 13 13 14 15 16 17 14 15 8 16 2003 1 17 18 iii 19 iv 20 21 22 23 24 25 ,,, 26 27 28 29 30 (1) (2) (3) 31 1 20
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
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,
CVMに基づくNi-Al合金の
CV N-A (-' by T.Koyama ennard-jones fcc α, β, γ, δ β α γ δ = or α, β. γ, δ α β γ ( αβγ w = = k k k ( αβγ w = ( αβγ ( αβγ w = w = ( αβγ w = ( αβγ w = ( αβγ w = ( αβγ w = ( αβγ w = ( βγδ w = = k k k ( αγδ
20 15 14.6 15.3 14.9 15.7 16.0 15.7 13.4 14.5 13.7 14.2 10 10 13 16 19 22 1 70,000 60,000 50,000 40,000 30,000 20,000 10,000 0 2,500 59,862 56,384 2,000 42,662 44,211 40,639 37,323 1,500 33,408 34,472
- 2 -
- 2 - - 3 - (1) (2) (3) (1) - 4 - ~ - 5 - (2) - 6 - (1) (1) - 7 - - 8 - (i) (ii) (iii) (ii) (iii) (ii) 10 - 9 - (3) - 10 - (3) - 11 - - 12 - (1) - 13 - - 14 - (2) - 15 - - 16 - (3) - 17 - - 18 - (4) -
2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4 4 4 2 5 5 2 4 4 4 0 3 3 0 9 10 10 9 1 1
1 1979 6 24 3 4 4 4 4 3 4 4 2 3 4 4 6 0 0 6 2 4 4 4 3 0 0 3 3 3 4 3 2 4 3? 4 3 4 3 4 4 4 4 3 3 4 4 4 4 2 1 1 2 15 4 4 15 0 1 2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4
I? 3 1 3 1.1?................................. 3 1.2?............................... 3 1.3!................................... 3 2 4 2.1........................................ 4 2.2.......................................
1 (1) (2)
1 2 (1) (2) (3) 3-78 - 1 (1) (2) - 79 - i) ii) iii) (3) (4) (5) (6) - 80 - (7) (8) (9) (10) 2 (1) (2) (3) (4) i) - 81 - ii) (a) (b) 3 (1) (2) - 82 - - 83 - - 84 - - 85 - - 86 - (1) (2) (3) (4) (5) (6)
koji07-01.dvi
2007 I II III 1, 2, 3, 4, 5, 6, 7 5 10 19 (!) 1938 70 21? 1 1 2 1 2 2 1! 4, 5 1? 50 1 2 1 1 2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 3 1, 2 1, 3? 2 1 3 1 2 1 1, 2 2, 3? 2 1 3 2 3 2 k,l m, n k,l m, n kn > ml...?
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
provider_020524_2.PDF
1 1 1 2 2 3 (1) 3 (2) 4 (3) 6 7 7 (1) 8 (2) 21 26 27 27 27 28 31 32 32 36 1 1 2 2 (1) 3 3 4 45 (2) 6 7 5 (3) 6 7 8 (1) ii iii iv 8 * 9 10 11 9 12 10 13 14 15 11 16 17 12 13 18 19 20 (2) 14 21 22 23 24
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
i ii iii iv v vi vii ( ー ー ) ( ) ( ) ( ) ( ) ー ( ) ( ) ー ー ( ) ( ) ( ) ( ) ( ) 13 202 24122783 3622316 (1) (2) (3) (4) 2483 (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) 11 11 2483 13
III ϵ-n ϵ-n lim n a n = α n a n α 1 lim a n = 0 1 n a k n n k= ϵ-n 1.1
III http://www2.mth.kyushu-u.c.jp/~hr/lectures/lectures-j.html 1 1 1.1 ϵ-n ϵ-n lim n = α n n α 1 lim n = 0 1 n k n k=1 0 1.1.7 ϵ-n 1.1.1 n α n n α lim n = α ϵ Nϵ n > Nϵ n α < ϵ 1.1.1 ϵ n > Nϵ n α < ϵ 1.1.2
¿ô³Ø³Ø½øÏÀ¥Î¡¼¥È
2011 i N Z Q R C A def B, A B. ii..,.,.. (, ), ( ),.?????????,. iii 04-13 04-20 04-27 05-04 [ ] 05-11 05-18 05-25 06-01 06-08 06-15 06-22 06-29 07-06 07-13 07-20 07-27 08-03 10-05 10-12 10-19 [ ] 10-26
178 5 I 1 ( ) ( ) 10 3 13 3 1 8891 8 3023 6317 ( 10 1914 7152 ) 16 5 1 ( ) 6 13 3 13 3 8575 3896 8 1715 779 6 (1) 2 7 4 ( 2 ) 13 11 26 12 21 14 11 21
I 178 II 180 III ( ) 181 IV 183 V 185 VI 186 178 5 I 1 ( ) ( ) 10 3 13 3 1 8891 8 3023 6317 ( 10 1914 7152 ) 16 5 1 ( ) 6 13 3 13 3 8575 3896 8 1715 779 6 (1) 2 7 4 ( 2 ) 13 11 26 12 21 14 11 21 4 10 (
研修コーナー
l l l l l l l l l l l α α β l µ l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l
高校生の就職への数学II
II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................
_TZ_4797-haus-local
1.1.................................... 3.3.................................. 4.4......................... 8.5... 10.6.................... 1.7... 14 3 16 3.1 ()........................... 16 3. 7... 17
量子力学 問題
3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,
A
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
31 gh gw
30 31 gh gw 32 33 1406 1421 640 0 (mm) (mm) MAX1513 MIN349 MIN280 MAX900 gh gw 34 gh gh gw gw gh gh gw gw gh gh gw gw 35 175 176 177 178 179 180 181 195 196 197 198 202 203 2 1 L L L2 L2 L2 L 2 2 1 L L
,,,,., = (),, (1) (4) :,,,, (1),. (2),, =. (3),,. (4),,,,.. (1) (3), (4).,,., () : = , ( ) : = F 1 + F 2 + F 3 + ( ) : = i Fj j=1 2
6 2 6.1 2 2, 2 5.2 R 2, 2 (R 2, B, µ)., R 2,,., 1, 2, 3,., 1, 2, 3,,. () : = 1 + 2 + 3 + (6.1.1).,,, 1 ,,,,., = (),, (1) (4) :,,,, (1),. (2),, =. (3),,. (4),,,,.. (1) (3), (4).,,., () : = 1 + 2 + 3 +,
cm H.11.3 P.13 2 3-106-
H11.3 H.11.3 P.4-105- cm H.11.3 P.13 2 3-106- 2 H.11.3 P.47 H.11.3 P.27 i vl1 vl2-107- 3 h vl l1 l2 1 2 0 ii H.11.3 P.49 2 iii i 2 vl1 vl2-108- H.11.3 P.50 ii 2 H.11.3 P.52 cm -109- H.11.3 P.44 S S H.11.3
2010 II / y = e x y = log x = log e x 2. ( e x ) = e x 3. ( ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0
2010 II 6 10.11.15/ 10.11.11 1 1 5.6 1.1 1. y = e x y = log x = log e x 2. e x ) = e x 3. ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0 log a 1 a 1 log a a a r+s log a M + log a N 1 0 a 1 a r
NL09
Information September, 2005 1 2 Japanese Association for Molecular Target Therapy of Cancer News Letter No.9 September, 2005 3 2005 4 Japanese Association for Molecular Target Therapy of Cancer News Letter
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
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 ( )
BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B
2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :
gr09.dvi
.1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {
untitled
i ii iii iv v 43 43 vi 43 vii T+1 T+2 1 viii 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 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 a) ( ) b) ( ) 51
AccessflÌfl—−ÇŠš1
ACCESS ACCESS i ii ACCESS iii iv ACCESS v vi ACCESS CONTENTS ACCESS CONTENTS ACCESS 1 ACCESS 1 2 ACCESS 3 1 4 ACCESS 5 1 6 ACCESS 7 1 8 9 ACCESS 10 1 ACCESS 11 1 12 ACCESS 13 1 14 ACCESS 15 1 v 16 ACCESS
9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =
5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =
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... (
2
1 2 3 4 5 6 7 8 9 10 I II III 11 IV 12 V 13 VI VII 14 VIII. 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 _ 33 _ 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 VII 51 52 53 54 55 56 57 58 59
x V x x V x, x V x = x + = x +(x+x )=(x +x)+x = +x = x x = x x = x =x =(+)x =x +x = x +x x = x ( )x = x =x =(+( ))x =x +( )x = x +( )x ( )x = x x x R
V (I) () (4) (II) () (4) V K vector space V vector K scalor K C K R (I) x, y V x + y V () (x + y)+z = x +(y + z) (2) x + y = y + x (3) V x V x + = x (4) x V x + x = x V x x (II) x V, α K αx V () (α + β)x
x 3 a (mod p) ( ). a, b, m Z a b m a b (mod m) a b m 2.2 (Z/mZ). a = {x x a (mod m)} a Z m 0, 1... m 1 Z/mZ = {0, 1... m 1} a + b = a +
1 1 22 1 x 3 (mod ) 2 2.1 ( )., b, m Z b m b (mod m) b m 2.2 (Z/mZ). = {x x (mod m)} Z m 0, 1... m 1 Z/mZ = {0, 1... m 1} + b = + b, b = b Z/mZ 1 1 Z Q R Z/Z 2.3 ( ). m {x 0, x 1,..., x m 1 } modm 2.4
I II III IV V
I II III IV V N/m 2 640 980 50 200 290 440 2m 50 4m 100 100 150 200 290 390 590 150 340 4m 6m 8m 100 170 250 µ = E FRVβ β N/mm 2 N/mm 2 1.1 F c t.1 3 1 1.1 1.1 2 2 2 2 F F b F s F c F t F b F s 3 3 3
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
