1. A0 A B A0 A : A1,...,A5 B : B1,...,B

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1 1. A0 A B A0 A : A1,...,A5 B : B1,...,B

3 A0 A B f : A B 4 (i) f (ii) f (iii) C 2 g, h: C A f g = f h g = h (iv) C 2 g, h: B C g f = h f g = h 4 (1) (i) (iii) (2) (iii) (i) (3) (ii) (iv) (4) (iv) (i) 1

4 A1 f R n (n ) R n V W f(v ) W f(w ) V (1) dim V dim W f (2) V + W = R n f(v ) = W f(w ) = V f (3) V W = {0} α R \ {0} α f V W A2 (1) x = 0 f(x) = x 3 x 2 + x + 1 g g (1), g (1), g (1) (2) x = 0 F (x) = sin 3 x sin 2 x + sin x + 1 G G (1), G (1), G (1) 2

5 A3 S S U, V U = {U S U = S U = S U } V = {V S V = S V = S V } (1) U, V S ( ) (2) (S, U) (S, V) (3) f : (S, U) (S, V) g : (S, V) (S, U) f(x) = x, g(x) = x (x S) f g A4 0 < p < 1 X P (X = k) = { p(1 p) k 1 k = 1, 2,... 0 X p Ge(p) X 1 X 2 Ge(p 1 ) Ge(p 2 ) p 1 p 2 (1) Y = X 1 + X 2 P (Y = k) (2) Y (3) Z = min(x 1, X 2 ) P (Z = k) (4) Z 3

6 A5 Pascal, program sieve(input, output); const maxind = 200; var table: array[0..maxind] of boolean; function ti(n: integer): integer; begin ti := (n div 7) * 2 + (n mod 7) div 5 end; n: integer; function fi(i: integer): integer; begin if i mod 2 = 0 then fi := (i div 2) * else fi := (i div 2) * end; procedure mktab(maxnum: integer); var n, m, d, dm, i: integer; begin for i := 0 to maxind do table[i] := true; n := 6; d := 2; while n <= maxnum do begin if table[ti(n)] then begin m := n; dm := d; while m <= maxnum div n do begin table[ti(n*m)] := false; m := m+dm; dm := 7 dm end; n := n+d; d := 7 d end end; end begin readln(n); if (ti(n) <= maxind) and (n = fi(ti(n))) then begin mktab(n); writeln(n, table[ti(n)]) end end. (1) 34, (.) (2) n, 4

7 B1 p G { g p g G } G G (1) G G G (2) G H G p 1 f : H/H G/G ; f(hh ) = hg (3) (2) G f B2 K 4 S = K[x 1, x 2, x 3, x 4 ] y = x 1 x 4 + x 2 x 3, z = x 3 x 4 R = K[y, x 2, x 3, x 4 ] T = K[y, x 3, x 4 ] R T { 1, z, z 2,... } R z T z a K σ a : S S K x 3, x 4 σ a (x 1 ) = x 1 ax 3, σ a (x 2 ) = x 2 + ax 4 (1) a K σ a (y) (2) S R z (3) K f S a K σ a (f) = f f T z 5

8 B3 a f a : R 2 R 2 f a (x, y) = (x, y 3 + xy + ay) (1) a = 0 f 0 df 0 (p) : T p R 2 T f0 (p)r 2 p R 2 ( f 0 0 p R 2 ) C 0 R 2 C 0 R 2 (2) C 0 f a f a (C 0 ) R 2 a B4 D, E 2 C {z C z 1}, {z C z = 1} D, E, C D E D E n f n : D C g n : D E f n (z) = z n, g n (z) = z n D E g n Y n = D gn E Y n D E D x g n (x) E Y n = (D E)/ D C f n X n = D fn C (1) X 1, Y 1 H q (X 1, Z), H q (Y 1, Z) (q = 0, 1, 2) (2) X 0, Y 0 H q (X 0, Z), H q (Y 0, Z) (q = 0, 1, 2) (3) X 2, Y 2 H q (X 2, Z), H q (Y 2, Z) (q = 0, 1, 2) 6

9 B5 C R (R > 0) { C R = z C \ {0} Re 1 z = 1 } R Re (1) C R (2) n C R z n 1 e 1/z dz C R (C R.) B6 y, z, w x, y, z, w. (1) y + xy = 0. (2) y + xy = x 3. (3) y z w x 0 0 y x 3 = z + x w x y(x), z(x), w(x). y(0) = 2, z(0) = 1 9, w(0) = 1 3 7

10 B7 V (, ). x V x = (x, x). (1) Cauchy-Schwartz. (x, y) x y, x, y V. (2) f : V R x = 1 f(x) M M., N 1 x 1, x 2,..., x N, { 1 i = j (x i, x j ) = 0 i j. N f(x i ) 2 M 2 i=1 8

11 B8 X M(X) P ( X > M(X) ) 1/2 P ( X < M(X) ) 1/2 (1) M(X) 1 (2) E[X] <, E[X 2 ] < M(X) E[X] + 2 Var(X) Var(X) X (3) X 1 p(x) Ce λx, 0 < x < 1/3, p(x) = 0, x 0 1/3 x 2/3 1 x, Ce (1 λ)(x 1), 2/3 < x < 1. λ 0 λ 1 C M(X 1 ) 1 λ B9 B(n, p) m X 1, X 2,..., X m X = 1 m X k n p m k=1 (1) X/n p (2) p (3) X/n p (4) m p 1 α. 9

12 B10 E {0, 1} k n- Π E : {0, 1} k {0, 1} n {0, 1} n {0, 1} 2k n- Π F F (K 1 K 2, M) def = E(K 2, E(K 1, M)), ( K 1, K 2 {0, 1} k ) 2 ((m 1, c 1 ) (m 2, c 2 )) Π Π ( ) B11 p, q,... A ν(a) {0, 1} ν, ν( ) = 0 ν(a B) = max{1 ν(a), ν(b)} A A : (A ). Γ Γ 0 A Γ 0 (ν 0 (A) = 1) ν 0, Γ G Γ 0 Γ 1 (1) Γ G Γ ( ) A A, A Γ (2) Γ ( ) Γ 1 ( ) 10

13 B12 Scheme, (define (r x) (lambda (z) x)) (define (b g f) (lambda (z) ((f (g z)) z))) (define t1 (b (lambda (z) (* 2 z)) r)) (define t2 (b car (lambda (u) (b cadr (lambda (v) (r (+ u v))))))) (define (tn f a) (if (null? f) (r a) (b (car f) (lambda (u) (tn (cdr f) (cons u a)))))) (1) (t1 2), (2) (t2 l) 0 l, (3) l, ((tn (list cadr car caddr) (0)) l), 11

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)