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

3 A0 (1) A, B A B f K K A ϕ 1, ϕ 2 f ϕ 1 = f ϕ 2 ϕ 1 = ϕ 2 (2) N A 1, A 2, A 3,... N A n X N n X N, A n N n=1 1

4 A1 d (d 2) A (, k A k = O), A O. f : R d R d f(x) = Ax.. (1) A p(x) X d. (2) A. (3) a n = rank(a n ), d = a 0 a 1 a d = 0. (4) Im(f) = Ker(f), {a n } A. A2 1 < x < 1, f(x) =. (1) f(x). (2) F (x). x 0 dt t 4 1 F (x) = (3) F (x) Taylor. x 0 f(t)dt (4) D : x 2 + y 2 < 1 2, 2 φ(x, y) = f(x2 + y 2 ). 2

5 A3 C C O : A O A = A C O (1) f(z) = z 2 f : C C (2) B = {z C z < 1} (3) Z 2 = { n + m 1 } n, m Z A4 N(µ, σ 2 ) f(x) = 1 (x µ)2 e 2σ 2 2πσ 2.. (1) X N(µ, σ 2 ), Z = X µ σ. (2) a 1, a 2. X 1 X 2, N(µ 1, σ 2 1), N(µ 2, σ 2 2), Y = a 1 X 1 + a 2 X 2. (3) n 2. X k (k = 1,..., n) N(µ, σ 2 ), W = 1 n X k. n k=1 3

6 A5 Pascal ( ) const n = max; var a : array [1..n] of integer; p, c : integer; function src(t : integer) : boolean; var b, e, m : integer; begin if t <= 0 then begin src := false; p := 0; c := 0 end else begin b := 1; e := n; c := 1; while b <= e do begin m := (b + e) div 2; c := c + 1; if t < a[m] then e := m 1 else b := m + 1 end; p := e; { * } src := (t = a[e]) end end; (1) max max = 8 a[1] a[n] 2, 3, 5, 7, 11, 13, 17, 19 src(10) p (2) max a[1] a[n] (1) m src(m) m (3) (2) { * } (4) max t a[1] < a[2] < < a[n] a src(t) c O(log max) 4

7 B1 G = A 7 7 σ = ( ) σ(1) = 2, σ(2) = 3,..., σ(7) = 1 S = σ G X N G (X) = { g G gx = Xg } C G (X) = { g G gx = xg ( x X) } (1) C G (S) N G (S) (2) N G (S) (3) ρσρ 1 = σ 2 ρ ( G) ρ C G (S) (4) ρσρ 1 = σ m (1 m 7) ρ ( G) m (5) N G (N G (S)) = N G (S) B2 K = Q( 4 2),. (1) K Q. (2) K Q L. (3) Gal(L/Q) 8. (4) L. 5

8 B3 f : R 4 R f(x, y, z, w) = x 2 + y 2 + z 2 + w 2 1 ( f (1) f 1 (0) f J(f) = x, f y, f z, f ) w (2) M = f 1 (0) F : R 4 R 2 M = F 1 (0) F (x, y, z, w) = ( x 2 + y 2 + z 2 + w 2 1, x 2 + y 2 z 2 w 2) (3) M F J(F ) (4) M (5) φ : M R φ(x, y, z, w) = x + y + z + w φ φ M ξ i φ/ ξ i = 0 ( i) B4 R 3 A, B, S A = { (x, 0, 0) x 1 } B = { (x, y, 0) x + y 1 } S = { (x, y, z) x + y + z = 1 } X = A S, Y = B S (1) X Y (2) H q (X; Z), H q (Y ; Z) (q = 0, 1, 2) (3) i : X Y i : H q (X; Z) H q (Y ; Z) i (H q (X; Z)) (q = 0, 1, 2) 6

9 B5 C f(z) (1) f(z) z z C (2) k M f(z) M z k z C (3) k M f(z) M z k z > 1 z C (4) f(0) = f (0) = 0, f(1) = 1 z C f (z) 2 z B6 f CR 1 [0, 1] [0, 1] 1 ( )., f, g H H = {f C 1 R[0, 1] : f(0) = 0} f, g = 1 0 f (t)g (t)dt, f = f, f. (1) f H x [0, 1]. f(x) f x (2) {f n } H Cauchy, 2 h {f n },. g(x) = x 0 h(t)dt (x [0, 1]) 7

10 B7 p(x), q(x) I R 2 y + p(x)y + q(x)y = 0 (E) (E) y 1 (x), y 2 (x) W (y1,y 2 )(x) W (y1,y 2 )(x) = y 1 (x) y 2 (x) y 1(x) y 2(x) = y 1(x)y 2(x) y 1(x)y 2 (x) (1) y 1 (x), y 2 (x) (E) I W (y1,y 2 )(x) = 0 (x I) (2) y 1 (x), y 2 (x) (E) W (y1,y 2 )(x) I (3) y = y(x) (E) a, b I (a < b) y(x) y(a) = y(b) = 0 y(x) 0 (a < x < b). y (a)y (b) < 0 (4) y 1 (x), y 2 (x) (E) y 1 (x) y 2 (x) 8

11 B8 X 1, X 2,... (Ω, F, P ), p(x) (x Z)., p(x) = p( x) (x Z).. (1) X 1 + X 2 p(x) (x Z). (2) x Z. P (X 1 + X 2 = 0) P (X 1 + X 2 = x) (3) n x Z. ( 2n ) ( 2n P X j = 0 P X j = x j=1 j=1 ) B9 X 1, X 2,..., X n, µ, σ 2 D µ, σ 2 > 0 n 2 X = 1 n n i=1 X i, S 2 = 1 n 1 n (X i X) 2 i=1 (1) X 2 E[X 2 ] (2) S 2 σ 2 (3) D N(µ, σ 2 ) X k E[(X µ) k ] k 9

12 B10 A : N 2 N ( N 0 ) (1) σ : N 3 N A(0, y) = y + 1 A(x + 1, 0) = A(x, 1) A(x + 1, y + 1) = A(x, A(x + 1, y)) σ(x, y, z) = { 1 if A(x, y) = z 0 otherwise A A(x, y) > x + y A(x, y) < A(x, y + 1) A(x, y) < A(x + 1, y) (2) B : N N B(A(x, y)) = p(x, y ) A(x, y) = A(x, y ) p : N 2 N p 1 (p(x, y)) = x, p 2 (p(x, y)) = y p 1, p 2 B11 p, q n = pq e 1, e 2 φ(n) f f(x, e) = x e mod n m n n n, e 1, e 2, f(m, e 1 ), f(m, e 2 ) m φ x, y xz 1 mod y z I 10

13 B12 Scheme (define (enumerate-tree tree) (cond ((null? tree) ()) ((not (pair? tree)) (list tree)) (else (append (enumerate-tree (car tree)) (enumerate-tree (cdr tree)))))) (define t0 (list 1 (list 2 (list 3 4) 5))) (1) (enumerate-tree t0) (2) (enumerate-tree t0) cons (a) enumerate-tree list cons (b) enumerate-tree append cons list append 1 cons (3) t ( ) l prepend-leaves (equal? (append (enumerate-tree t) l) (prepend-leaves t l)) #t (prepend-leaves t ()) cons (enumerate-tree t) list cons prepend-leaves set-car! set-cdr! 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)