2000年度『数学展望 I』講義録

Similar documents
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.

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

数学Ⅱ演習(足助・09夏)

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

() 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 (

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

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

Part () () Γ Part ,

1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x

Chap9.dvi

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

°ÌÁê¿ô³ØII

(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)


18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

70 : 20 : A B (20 ) (30 ) 50 1

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

Untitled

I

untitled

2011de.dvi

29

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π


201711grade1ouyou.pdf

高校生の就職への数学II

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y

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

基礎数学I

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

数学の基礎訓練I

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google

II R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

2009 I 2 II III 14, 15, α β α β l 0 l l l l γ (1) γ = αβ (2) α β n n cos 2k n n π sin 2k n π k=1 k=1 3. a 0, a 1,..., a n α a

i

1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3

all.dvi


newmain.dvi

A S- hara/lectures/lectures-j.html r A = A 5 : 5 = max{ A, } A A A A B A, B A A A %

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt


II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

入試の軌跡

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

Z: Q: R: C: sin 6 5 ζ a, b

85 4

TOP URL 1

Note.tex 2008/09/19( )

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

Z: Q: R: C:

2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

I y = f(x) a I a x I x = a + x 1 f(x) f(a) x a = f(a + x) f(a) x (11.1) x a x 0 f(x) f(a) f(a + x) f(a) lim = lim x a x a x 0 x (11.2) f(x) x

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

1 I

ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α


i 18 2H 2 + O 2 2H 2 + ( ) 3K

DVIOUT

2 R U, U Hausdorff, R. R. S R = (S, A) (closed), (open). (complete projective smooth algebraic curve) (cf. 2). 1., ( ).,. countable ( 2 ) ,,.,,

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou

Gmech08.dvi

y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x

A

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

chap1.dvi

2014 S hara/lectures/lectures-j.html r 1 S phone: ,

最新耐震構造解析 ( 第 3 版 ) 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, 第 3 版 1 刷発行時のものです.

(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z

koji07-01.dvi

2 (2016 3Q N) c = o (11) Ax = b A x = c A n I n n n 2n (A I n ) (I n X) A A X A n A A A (1) (2) c 0 c (3) c A A i j n 1 ( 1) i+j A (i, j) A (i, j) ã i


2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP

熊本県数学問題正解

ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

(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

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2

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

6. Euler x

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0

医系の統計入門第 2 版 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, 第 2 版 1 刷発行時のものです.

ii 3.,. 4. F. (), ,,. 8.,. 1. (75% ) (25% ) =9 7, =9 8 (. ). 1.,, (). 3.,. 1. ( ).,.,.,.,.,. ( ) (1 2 )., ( ), 0. 2., 1., 0,.

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)

B

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

Transcription:

2000 I

I IV I II 2000 I I IV I-IV. i

ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000

A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53 B. Borsuk Ulam... 60 C....65 iii

iv

1 A I f x I f(x) (x, y)- (x, f(x)) x I (x, f(x)) (x, y)- f f f f x (x,f(x)) y=f(x) x A.1: f I [a, b] a, b f f(a) f(b) < 0 (1.1) f(a) f(b) f(a) < 0 f(b) > 0 f(a) > 0 f(b) < 0 y = f(x) 1

2 1 (a x b) 1 x- x- 2 x- x- A.2 f f x- x- ca <c<b) f(c) =0 (1.1) f(a) f(b) =0 f(a) f(b) a, b c f(c) =0 a c b a c b (a) (b) A.2: f(a) < 0 f(b) > 0 f(a) > 0 f(b) < 0 x- 1.1 (). [a, b] f f(a) f(b) 0 f(c) =0 c [a, b] f(x) =0a x b

A. 3 1.2. 1 A.3: 1.3. A B 1 A B

4 1 A B A B A.4: B 1.2 1.2 1.2 1.2 1 1.2 1.2

B. 5 Ω 1 Ω 1 1.2 Ω 1 2 Ω 2 1 Ω 1 1.2 1.4. 1.4 1.4 Ω l O 1.4 l Ω Ω l B.1: l (u, v) u- l l v = t t

6 1 t v = t t Ω v = t Ω (u, v)- t v = t Ω t Ω v = t v = t A A + B.2 ) Ω v = t v A + u A- v=t O B.2: A =0 Ω v = t A + =0 A + A t f(t) f(t) =A + A. f(t) t 1 <t 2 f(t 1 ) f(t 2 ) S Ω t f(t) =S>0 t f(t) = S <0 f(t) t 1.4 f f(t 0 )=0 t 0 f t 0 t 0 v = t 0 Ω f 1.4 1.2 Ω 1 Ω 2

B. 7 f (t) t 0 S O t -S B.3: f(t) u. 1 u O xy x- θ 0 θ 2π 1.4 1 Ω 1 u u Ω 1 u x- θ l θ l θ 1 Ω 1 2 Ω 2 θ g(θ) l θ Ω 2 l θ Ω 2 l θ u A R A L g(θ) =A R A L g(θ) l θ u A L l u 1 A R 2 B.4: Ω 2 g(θ) =(Ω 2 )

8 1 g(θ) = (Ω 2 ) g(θ) (0 θ<2π) g(θ + π) = g(θ), 0 θ<π (1.2) (1.2) l θ l θ+π Ω 1 1.4 l θ, l θ+π g(θ) g(θ) g(θ + π) g(θ) = g(θ + π) l θ l θ+π l θ l θ+π l θ l θ+π g(θ) (1.2) (1.2) g(0) g(π) 0 g(θ) g(θ) g(θ 0 )=0 θ 0 θ 0 l θ0 1 Ω 1 2 Ω 2 l θ g(θ) 1.2 f g 1.3 Newton

B. 9 θ π/2 θ π/2) B.5 θ θ h(θ) h( ) A B B.5: h(θ) h( π/2) = π/2 h(π/2) = π/2 h h h(θ 0 )=0 θ 0 θ 0

10 1 1.3 1 C A I A 1 2 1.2 1.3 f, g, h 3 A 3 2 3 R n R m 1 A B 1 1 30 0.1 89

C. 11 n- R n m- R m (m + n)- R m+n m + n>3 2 1.5. f I a I f a {a n } n a n I a lim f(a n)=f(a) (1.3) n a I f f I (1.3) a = lim n a n lim f(a n)=f( lim a n), n n f lim n I 2 ɛ δ ɛ δ ɛ δ ɛ δ

12 1 f(x) = { x +1 x<0, x 1 x 0 x =0 C.1 1.5 a n = 1/n (n =1, 2, ) {a n } a n < 0 a n 0(n ) f( lim a n)= 1 lim f(a n)=1 n n lim f(a n) f( lim a n) f x =0 n n y O x C.1:

D. 13 1.2 1.3 3 f, g, h f g g l θ h D 1.5 1.5 1.6. {a n } (1) a 0 a 1 a n a n+1 {a n } a 0 a 1 a n a n+1 {a n } (2) K n =0, 1, a n K {a n } K n =0, 1, a n K

14 1 {a n } a n = n {a n } {b n } b n = 2 n {a n } a 0 <a 1 < <a n <a n+1 < {a n } a 0 >a 1 > >a n >a n+1 > 2 1.7 (). {a n } a n a n n K {a n }

D. 15 a 1 a 2 a 0 a 3 K D.1: 1.8. {a n } {a n } N {a n } a N = a N+1 = a N+2 = 1.8 1.8 1.8 1.8 1.8 1.8 1.7

16 1 1.7 {a n } {a n } {a n } b n = a n {b n } {a n } n =0, 1, a n = M n + r n M n 0 r n < 1 r n, r n =0.d n1 d n2 d n1,d n2, 0 9 10 {a n } {M n } 1.8 N 0 M n M {d n1 } {a n } a n M n N 0 {d n1 } N 0 1.8 N 1 N 1 N 0 ) d n1 d 1. d 2,d 3, D.2

E. 17 21745. 392... M n d n1 d n2 d n3... D.2: r d 1,d 2, r =0.d 1 d 2 a a = M + r a n a E 1.1. f(a) f(b) =0f(a) =0 f(b) =0 c = a c = b f(a) f(b) < 0 f(a) f(b) < 0 f(a) < 0 f(b) > 0 f(a) > 0 f(b) < 0 2 1 f f 1 1 f(a) < 0 f(b) > 0 {a n }, {b n } a 0 = a, b 0 = b

18 1 a 0 b 0 a n b n a n+1 b n+1 ( an + b ) n f 2 ( an + b ) n f 2 0 a n+1 = a n, b n+1 = a n + b n, 2 < 0 a n+1 = a n + b n, b n+1 = b n 2 E.1 {a n } {b n } an =a n+1 an m n =an+1 m n =b n+1 bn bn =bn+1 (a) (b) E.1: m n =(a n + b n )/2 f (a) a n+1 = a n, b n+1 = m n (b) a n+1 = m n, b n+1 = b n a a n b, a b n b; (1.4) b n a n = b a 2 n ; (1.5) f(a n ) < 0, f(b n ) 0; (1.6) {a n } {b n } ; (1.7) (1.4) (1.7) {a n } {b n } (1.5) b n a n 0 n {a n }, {b n } c c = lim a n = lim b n. n n (1.4) a c b c f(c) =0 c = lim n a n f(a n ) < 0 f(c) =f( lim n a n) = lim n f(a n) 0.

E. 19 c = lim n b n f(b n ) 0 f(c) =0 f(c) =f( lim n b n) = lim n f(b n) 0. intermediate intermediate intermediate 2 10 = 1024 intermediate intermediate 512 intermediate 1 512 256 intermediate 256 512 intermediate f(c) =0 c 1024 f [a, b] 512 intermediate [a, b] (a + b)/2 =(a 0 + b 0 )/2 f 512 intermeditate 1 512

20 1 f(a) < 0, f(b) > 0 f((a + b)/2) 0 a 1 = a 0 = a, b 1 =(a 0 + b 0 )/2 =(a + b)/2 f((a + b)/2) < 0 a 1 =(a 0 + b 0 )/2 =(a + b)/2, b 1 = b 0 = b f(c) =0 c [a 1,b 1 ] a=a 0 m b=b 0 a=a 0 m b=b 0 (a) (b) E.2: [a, b] m =(a + b)/2 [a, m] [m, b] f m [a 1,b 1 ] f(a 1 ) < 0, f(b 1 ) 0 [a 1,b 1 ] f(x) =0 [a 1,b 1 ] [a 1,m 1 ], [m 1,b 1 ] m 1 =(a 1 + b 1 )/2 [a 2,b 2 ] a 2 f b 2 f [a 2,b 2 ] f(x) =0 f(x) =0

2 Brouwer A 2.1. f [0, 1] 0 f(x) 1, (0 x 1) f(x) =x x (0 x 1) f(x) =x x f y = f(x) y = x x- 1 y x = y y = f (x) O 1 x A.1:. [0, 1] g g(x) =f(x) x (x [0, 1]) g g(0) = f(0) 0, 21

22 2 BROUWER g(1) = f(1) 1 0 g(x) =0 x [0, 1] x f(x) =g(x)+x = x X Y 1 X x Y y X Y ϕ X Y ϕ : X Y X, Y ϕ (domain) (range) ϕ : X Y x X Y X x ϕ Y y = ϕ(x) A.2: X x Y y = ϕ(x) X Y ϕ : X Y ϕ(x) ϕ x X = Y = {m : m } m X m X Y m X Y 2.1 1 2 2.2. X ϕ ϕ(p )=P 1 X Y 2

A. 23 P X ϕ 2.1 [0, 1] [0, 1] R 2 = {(x, y) :x, y R} 3 (x 0,y 0 ) R 2 x- x 0 y- y 0 (x, y)- R 2 R 2 4 D = {P =(x, y) R 2 : x 2 + y 2 1}. 2.3 (Brouwer ). D Brouwer D 3 X, Y X Y X Y = {(x, y) :x X, y Y } (x, y), (x,y ) X Y (x, y) =(x,y ) x = x y = y X X X X 2 R R 2 4 X, Y x X = x Y X Y X Y X Y Z R

24 2 BROUWER y O 1 x D A.3: D n x n 5 n x n {x n } n P n {P n } P n =(1/n, 1/n 2 ) P n R 2 X {P n } X {P n } n P n P {P n } P lim P n = P P n P (n ) n {P n } P P n P 0 (n ) a n = P n P P n P {a n } P n =(1/n, 1/n 2 ) {P n } P n O (n ) {P n } P P n (x n,y n ) P (x, y) {P n } P x n x y n y (n ) lim P n = P lim x n = x lim y n = y. n n n 5

A. 25 X R 2 2.4. f : X R f {P n } X X P {f(p n )} f(p ) {P n } X f( lim n P n) = lim n f(p n) 1 X R 2 ϕ X R 2 X P ϕ ϕ(p ) x- y- P ϕ(p ) f(p ) x- y- P f(p ), g(p ) ϕ(p )=(f(p ),g(p )) (P X) f g X f, g f, g ϕ ϕ : X R 2 X {P n } X lim ϕ(p n)=ϕ( lim P n). n n ϕ : D D X = D R 2 2.3

26 2 BROUWER B Brouwer 1 2 2.5. A B 6 A B 7 B A B.1: X R 2 V : X R 2 X V X P V (P ) OV (P ) P P P X X P V (P ) 6 7

B. 27 V(P) P O X B.2: V : X R 2 P X OV (P ) P X X 8 P X V (P ) V (P ) 2m/ 2m X R 2 V : X R 2 V X R 2 V V (P ) X P V V D D = {P =(x, y) R 2 : x 2 + y 2 =1} V P D V V P D ɛ>0 P + ɛv (P ) D D D {P =(x, y) R 2 : x 2 + y 2 < 1} B.3 8

28 2 BROUWER (a) (b) B.3: (a) (b) D Brouwer 2.6. D V D V D D V 2.6. V D P D P V (P ) l = {P + tv (P ):t>0} V D l D ϕ(p ) l D λ(p ) ϕ(p ) ϕ(p )=P + λ(p ) 2 V (P ) V (P ) ϕ(p ) D λ D ϕ : D D

B. 29 ϕ(p) V(P) λ(p) P l B.4: ϕ Brouwer ϕ(p )=P P D ϕ P λ(p )=0 λ V 2.7. X, Y ϕ X Y (1) x 1 x 2 x 1,x 2 X ϕ(x 1 ) ϕ(x 2 ) ϕ (2) y Y ϕ(x) =y x X ϕ (3) ϕ ϕ ϕ : X Y b Y ϕ(x) =b x X x X b Y 9 ϕ 9

30 2 BROUWER b Y ϕ f : I R I R y = f(x) y = b b b R f b R f B.5 y y = f(x) y b y = b y = b b y = f(x) O x O x (a) (b) B.5: (a) y = b y = f(x) (b) y = b y = f(x) 2.8. R R (1) f 1 (x) =sinx; (2) f 2 (x) =x 3 x; (3) f 3 (x) =e x (4) f 4 (x) =x 3 : x R

B. 31 f 1 f 2 f 3 f 4 ϕ : X Y Y Im ϕ = {ϕ(x) :x X} ϕ (image) 10 f(x) =x 2 R R [0, ) ϕ : X Y Im ϕ = Y 2.8 (1) f 1 R [ 1, 1] f 1 ϕ : X Y Y ϕ : X Y Im ϕ ϕ : X Y Y =Imϕ 2.6 2.9. ϕ : D D D P D ϕ(p )=P (2.1) ϕ 11 2 3 12 10 11 12

32 2 BROUWER 2.9. ϕ D \ Im ϕ 13 14 Q D V V (P )=Q ϕ(p ), P D D P D ϕ(p ) Q V (2.1) ϕ(p) Im ϕ V(P) Q B.6: D Im ϕ Q D Q D (2.1) V D V 2.6 V D V 2.5 2.5. 1 1 1 D D 13 X, Y X \ Y = {x X : x Y } Q R \ Q 14 φ

C. 33 1 1 D B.7: A D P B D Q Q P P Q = ϕ(p ) ϕ : D D A P D ϕ(p )=P 2.9 ϕ ϕ(p ) D P P C X R 2 X X 2.3 D A = {(x, y) R 2 :1 x 2 + y 2 4} C.1 90 y O 1 2 x C.1: A :1 x 2 + y 2 4

34 2 BROUWER A 3 Brouwer D C.2: 3 ϕ : X Y y Y ϕ(x) =y x X y Y ϕ(x) =y x Y X ϕ ϕ 1 ϕ 1 : Y X. f :[0, ) R f(x) =x 2 f 1 (x) = x X x ϕ 1 ϕ y Y C.3:

C. 35 2.10. X, Y R 2 X Y X Y X Y X Y D x 2 +4y 2 =1 E = {(x, y) R 2 : x 2 +4y 2 1} C.4: D 2.11. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

36 2 BROUWER. 9 1 : A R 2 : B 3 : C G I J L M N S U V W Z 4 : D O 5 : E F T Y 6 : H K 7 : P 8 : Q 9 : X 2.12. ϕ X Y ψ Y Z ϕ : X Y, ψ : Y Z. x X ϕ y = ϕ(x) Y ψ Z ψ(y) =ψ(ϕ(x)) x X ψ(ϕ(x)) X Z ϕ, ψ ψ ϕ ψ ϕ : X Z. f,g : R R f(x) = sin x, g(x) =e x (g f)(x) =e sin x,(f g)(x) = sin e x 2.12. X, Y R 2 H : X Y X

C. 37 ψ ϕ x ϕ ψ y = ϕ(x) ψ(y) =ψ(ϕ(x)) X Y Z C.5: Y ψ Y ϕ = H 1 ψ H X 15 X H X Y ϕ X H ψ Y ϕ(p )=P P X Y Q = H(P ) ψ(q) =Q Y ψ 3 D Brouwer 2.3 2.13. V V P ɛ> P + ɛv (P ) V C.6 2.13 15 X, Y, Z R 2 φ : X Y, ψ : Y Z ψ φ : X Z H 1 ψ H

38 2 BROUWER (a) (b) (c) C.6: (a) (b) (c) 2.14. V V 2.6 2.14 = 2.13. ϕ : P V (P )=ϕ(p ) P V P ϕ(p) V(P) P C.7: V (P ) V P 2.14 P ϕ ϕ(p )=P V

D. 39 D 2.14 Brouwer {x n } k n (n =0, 1, )k 0 <k 1 < {k n } y 0 = x k0,y 1 = x k1 y n = x kn (n =0, 1, ) {y n } {y n } {x n } {x n }, {y n } x n =2 n, y n =4 n =2 2n (n =0, 1, ) R 2 {P n } 2.15. (1) X R x n X (n =0, 1, ) {x n } X X (2) R 2 X X X X {x n } x 2m =1 2 m, x 2m+1 = 1+2 m (m =0, 1, ) y n = x 2n (n =0, 1, ) {y n } 1 R [0, )

40 2 BROUWER x n = n {x n } x n [0, ) (n =0, 1, ) {x n } n + {x n } [0, ) (0, 1] x n =1/n {x n } 0 (0, 1] d y c O a b x D.1: [a, b] [c, d] a, b, c, d a b, c d [a, b] [c, d] ={P =(x, y) R 2 : a x b, c y d} 4 x = a, x = b, y = c, y = d D.1 2.16. R 2 X X X [a, b] [c, d] a, b, c, d a b, c dx R 2 D(P ; r) ={Q R 2 : Q P <r} D(P ; r) ={Q R 2 : Q P <r}

D. 41 y y P r P r O x O x (a) D(P ; r) (b) D(P ; r) y O 1 x (c) {(x, y) R 2 :0 x 1} D.2: (a), (b) (c) {(x, y) R 2 :0 x 1} R 2 {(x, y) R 2 :0 x 1} 2.17. R 2 X X R 2 {P n } X lim n P n X R 2 D(P ; r) ={Q R 2 : Q P r} D(P ; r) ={Q R 2 : Q P <r} 2.18. R 2

42 2 BROUWER D 3 F 2.14 2.19 (Sperner ). 3 3 3 α, β, γ 1 3 3 A, B, C α, β, γ 2 3 AB α β BC β γ CA γ α 16 3 3 α, β, γ Sperner 2.18 2.14 2.14. V n 3 P V α, β, γ x- 16 3 ABC

D. 43 C A B D.3: V (P ) 0 θ 0 θ<2π 0 θ<2π/3 α, 2π/3 θ<4π/3 β, 4π/3 θ<2π γ. V Sperner V(P) O x D.4: P α Sperner 3 α, β, γ 3 3 α A n β B n γ C n {m n } {A n }, {B n }, {C n } {A n }, {B n }, {C n } A n = A mn, B n = B mn, C n = C mn (n =0, 1, ) 3 {A mn }, {B mn }, {C nm }

44 2 BROUWER 1 {A n } A {A n} A n A (n ). {A n} {A n } {k n } A n = A kn (n =0, 1, ) {B n } {B n} B n = B kn (n =0, 1, ) {B n } B {B n} B n B (n ). {B n} {k n } {l n } B n = B ln (n =0, 1, ) {C n } {C n} C n = C ln (n =0, 1, ) {C n } C C n C (n ). {C n } {l n } {m n } C n = C mn (n =0, 1, ) {A n }, {B n } {A n }, {B n } A n = A mn, B n = B mn (n =0, 1, ) {m n } {l n } {l n } {k n } {A n }, {B n } {A n} = {A kn }, {B n} = {B ln } {A n}, {B n} A, B {A n }, {B n } A, B A n A, B n B (n ).

E. SPERNER 45 ( 3 ) 3 ABC 1 L 3 A n B n C n 1 L/n 3 n 3 A n B n C n 1 n 3 {A n }, {B n }, {C n } A, B, C P V P = lim n A n = lim n B n = lim n C n. V (P ) = lim n V (A n) = lim n V (B n) = lim n V (C n) A n, B n, C n α, β, γ Λ 1 Λ 2 O x Λ 3 D.5: x- V (A n ), V (B n ), V (C n ) 0 θ<2π/3, 2π/3 θ<4π/3, 4π/3 θ<2π V (P ) 3 Λ 1 = {(r cos θ, r sin θ) :r 0, 0 θ 2π 3 }, Λ 2 = {(r cos θ, r sin θ) :r 0, 2π 3 θ 4π 3 }, Λ 3 = {(r cos θ, r sin θ) :r 0, 4π 3 θ 2π} 3 V (P )=0 V V E Sperner Sperner Sperner 1

46 2 BROUWER 2.20. [0, 1] n P 0 = 0,P 1,,P n 1,P n =1 P k α β P 0, P n P k 1 P k P 0 = 0 P 1 P 2 P 3 P n = 1 E.1: 1 Sperner k =0, 1,,n h k 0 P k α; h k = 1 P k β h k P k P 0 =0,P n =1 2 [0, 1] δ k = h k h k 1 (k =1, 2,,n) 2 P k 1 P k 1, 0, +1 δ k = 1 δ k =+1 k =1,,n δ 1 + + δ n 1 n δ k =(δ k =1 k ) (δ k = 1 k ) k=1 δ k = 1, 0 1 δ k n n δ k = (h k h k 1 )=h n h 0 k=1 k=1 P 0, P n ±1 (δ k =1 k ) (δ k = 1 k ) =±1. 2.20

E. SPERNER 47 1 0 E.2: Sperner 2.19. 3 α, β 3 E.3 (a) 1 (b) 2 (c) 3 E.3: 3 3 3 1 3 λ 2 3 µ, ν 1 3 α, β 2 3 3 λ +2(µ + ν) α, β 3 2 α, β l m λ +2(µ + ν) =l +2m 2 2(µ + ν) 2 2m λ l 2.20 l λ

48 2 BROUWER C A B E.4: F 2.18 17 2.21. [a, b]. {x n } [a, b] m =0, 1, a m, b m a m b m [a m,b m ] x (m) 0,x (m) 1, m =0a 0 = a, b 0 = b, x (0) n = x n (n =0, 1, ) a m, b m a m b m [a m,b m ] {x n (m) } (i), (ii) 18 (i) (ii) x (m) n x (m) n [a m, a m + b m ] n 2 [ a m + b m,b m ] n 2 17 18 (i), (ii) (i), (ii)

F. 49 (i) a m+1 = a m, b m+1 = a m + b m 2 a m+1, b m+1 {k n } x (m) k n [a m+1,b m+1 ]=[a m, a m + b m ] (n =0, 1, ) 2 x (m+1) n = x (m) k n, n =0, 1, {x (m+1) n } (i) a m+1 = a m + b m, b m+1 = b m 2 a m+1, b m+1 (ii) {k n } x (m) k n [a m+1,b m+1 ]=[ a m + b m,b m ], n =0, 1, 2 x (m+1) n {x (m+1) n } = x (m) k n n =0, 1, {a n }, {b n } c c = lim n a n = lim n b n. c [a, b] {x (m) n } (m =0, 1, ) {y n } y 0 = x (0) 0,y 1 = x (1) 1, 19 m =1, 2, {x (m+1) n } {x n (m) } {y n } {x (0) n } = {x n } 19 {x (m) n } x (0) 0,x(0) 1,x(0) 2, x (1) 0,x(1) 1,x(1) 2, x (2) 0,x(2) 1,x(2) 2, {y n}

50 2 BROUWER m =0, 1, x (m) n [a m,b m ](n =0, 1, ) a n y n b n n {a n }, {b n } c {y n } c [a, b] 2.21 2.21 2 2.22. [a, b] [c, d]. {P n } [a, b] [c, d] P n (x n,y n ) 2.14 2 2.21 {k n } x n = x kn, ȳ n = y kn (n =0, 1, ) { x n }, {ȳ n } [a, b], [c, d] {k n } {P n } {P n } P n = P kn (n =0, 1, ) P n =( x n, ȳ n )(n =0, 1, ) {P n } [a, b] [c, d] [a, b] [c, d] {P n } [a, b] [c, d] {a n } {a n } R 2 {P n } P n [a, b] [c, d] (n =0, 1, ) [a, b] [c, d] {P n } 2.21 2.22

F. 51 2.23. (1) (2) 2.18. X R 2 X {P n } X X X {P n } 2.23 {P n } {Q n } Q Q X X {Q n } X X X X X a, b, c, d (a b, c d) P [a, b] [c, d] P X a = c = n, b = d = n n =0, 1, P P n X {P n } X X X X X {P n } R 2 P P n P (n ) P X {P n } P P X X 2.18 R 2 2.24. X R

52 2 BROUWER (1) a, b x X a x b X (2) X {x n } lim n x n X X 2 2.25. R

3 A C 3.1 (). a n z n + + a 1 z + a 0 =0 (n 1, a 0,,a n C, a n 0) S 1 1 S 1 = {P R 2 : P =1}. f : S 1 S 1 P S 1 1 P f f(p ) S 1 f (mapping degree) deg f f(p ) 2 deg f = 2 f(p ) 5 3 5 3=2 1 S 1 D D 53

54 3 P f (P) A.1: P S 1 1 f(p ) S 1 +1 3.2. ρ :[0, 1] R A.2 f : S 1 S 1 f(cos 2πθ, sin 2πθ) = (cos 2πρ(θ), sin 2πρ(θ)), 0 θ<1 f ρ(θ) 1.8 1 O 1 θ 1 1.2 A.2:. deg f =1.8 ( 1.2) = 3 2 2.9

A. 55 3.3. ϕ : D D ϕ( D) D ϕ D = S 1 2 f = ϕ D : S 1 S 1 deg f 0 ϕ 2.9 ϕ : D D D D 3 1 2.9 3.3 C 3.3 z C z = x + iy (x, y R) z (x, y) R 2 C R 2 y z = x +iy O x A.3: z = x + iy (x, y R) (x, y)- (x, y) C 3.1 2 ϕ : X Y X X ψ(x) =ϕ(x), x X ψ : X Y ϕ X ϕ X 3 X id(x) =x, x X id : X X X

56 3 a n z n + a n 1 z n 1 + + a 1 z + a 0 =0 (a 0,,a n C, a n 0) a n z n + a n 1 a n z n 1 + + a 1 a n z + a 0 a n =0 z n + b n 1 z n 1 + + b 1 z + b 0 =0 b 0,b 1,,b n 1 C F (z) =z n + b n 1 z n 1 + + b 1 z + b 0, G(z) =z n M = b 0 + b 1 + + b n 1 4 z = R 1 z C F (z) G(z) = bn 1 z n 1 + + b 1 z + b 0 b n 1 R n 1 + + b 1 R + b 0 b n 1 R n 1 + + b 1 R n 1 + b 0 R n 1 = MR n 1 (3.1) A.4S 1 (R) R S 1 (R) ={z C : z = R}. G(z) =z n S 1 (R) R n S 1 (R n ) z S 1 (R) 1 4 z = x + iy (x, y R) z z = x 2 + y 2 z z + w z + w, zw = z w, 1 z = 1 z

A. 57 G (z ) F (z ) z O O 1 n S (R ) 1 S (R) A.4: z S 1 (R) F, G F (z), G(z) F (z) G(z) MR n 1 R 2 R n G(z) S 1 (R n ) n 5 F (z) S 1 (R) z S 1 (R n ) F (z) (z S 1 (R)) S 1 (R n ) G(z) MR n 1 (3.1) R >0 MR n 1 S 1 (R n ) R n z C F (z) 0 O F (z) O S 1 (R n ) ϕ(z) ϕ : C C A.5 6 z S 1 (R) OG(z) Oϕ(z) θ 0 θ π sin θ M R (3.1) A.6 R>0 θ 2π 1000 (3.2) 5 z = r(cos θ + i sin θ) (r>0, θ R) z n = r n (cos nθ + sin nθ), z n = z n, arg(z n )=n arg z 6 ϕ(z) =F (z)/ F (z) (z C)

58 3 R n θ O G (z ) F (z ) ϕ(z) A.5: G (z ) R n r ϕ(z) θ F (z ) O A.6: θ OF(z) G(z) r = MR n 1 sin θ = r/r n = M/R z S 1 (R) S 1 (R) G(z) S 1 (R n ) n ϕ(z) ϕ(z) S 1 (R n ) (3.2) ϕ S 1 (R) ϕ S 1 (R) : S 1 (R) S 1 (R n ) 7 G S 1 (R) n>0 z S 1 (R) z S 1 (R) S 1 (R) 1 1 S 1 (R n ) 1km 7 ϕ S 1 (R) S 1 (R n ) S 1 f : S 1 S 1 ϕ

A. 59 z S 1 (R) G(z) 1 n ϕ(z) 1 1 2 7 (3.2) ϕ(z) z G(z) S 1 (R) A.7: S 1 (R n ) ϕ D(R) ={z C : z R} S 1 (R n ) ϕ D(R) ϕ ϕ : D(R) S 1 (R n ) ϕ D(R) D(R) =S 1 (R) ϕ S 1 (R) : S 1 (R) S 1 (R n ) G S 1 (R) : S 1 (R) S 1 (R n ) deg ϕ S 1 (R) = deg G S 1 (R). deg G S 1 (R) = n>0 deg ϕ S 1 (R) 0 3.3 ϕ : D(R) S 1 (R n ) D(R n ) ϕ D(R n ) ϕ D(R n ) S 1 (R n )= D(R n )

60 3 B Borsuk Ulam 3.3 (x, y, z)- x- y- z- R 3 = {(x, y, z) :x, y, z R 3 } R 3 3 P =(x, y, z) R 3 z c (a, b, c) O b y a x B.1: 3 (x, y, z)- 3 3 R 3 (x, y, z)- P = x 2 + y 2 + z 2 O =(0, 0, 0) R 3 S 2 = {P R 3 : P =1} 1 P =(x, y, z), Q=(u, v, w) R 3 P + Q =(x + u, y + v, z + w), P Q =(x u, y + v, z + w)

B. BORSUK ULAM 61 z S 2 O 1 y x B.2: P Q 2 P, Q 3 R 3 {P n } R 3 P n P n P 0 3 2 X R 3 f : X R X X {P n } lim f(p n)=f( lim P n) n n 3.4 (Borsuk-Ulam ). ϕ : S 2 R 2 ϕ( P )= ϕ(p ) P S 2 f(p )=O P S 2 P = (x, y, z) R 3, Q = (x, y) R 2 P = ( x, y, z), Q =( x, y) 3 2 P, P 2 Q, Q ϕ : S 2 R 2 S 2 2 2

62 3 P P O ϕ ϕ( P ) γ O ϕ(p) B.3: Borsuk Ulum ϕ : S 2 R 2 S 2 2 2 O R 2 ϕ S+ 2 S 2 + = {P =(x, y, z) S 2 : z 0}. ϕ 8 S+ 2 D H(x, y, z) =(x, y) H : S+ 2 D S+ 2 D z S 2 + P O y x D x O H(P) y B.4: S 2 + = {(x, y, z) R 3 : x 2 + y 2 + z 2 =1,z 0} D = {(x, y) R 2 : x 2 + y 2 1} H H(x, y, z) =(x, y) S 2 + D 8 P =(x, y, z) S 2 S 2 = {(x, y, z) S 2 : z 0} P S 2 + ϕ( P ) P ϕ ϕ(p )= ϕ( P )

B. BORSUK ULAM 63 3.4. ϕ(p ) O P S 2 D S 1 ψ ψ(p )= ϕ H 1 (P ) ϕ H 1 (P ), P D ϕ : S 2 R 2 ψ ψ( P )= ψ(p ), P D (3.3) 9 P D P 0 =(1, 0) D P 0 =( 1, 0) ψ(p ) ψ(p 0 ) ψ( P 0 ) ψ(p 0 ) ψ( P 0 )= ψ(p 0 ) ψ(p ) S 1 m + 1 2 m B.5 P D P ψ(p) ψ(p ) 0 P 0 O P 0 O P ψ ( P ) = ψ (P ) 0 0 ψ ( P) B.5: P 0 P 0 ψ(p ) S 1 ψ(p 0 ) ψ(p 0 ) m + 1 2 ψ (3.3) P D D 1 ψ(p ) S 1 (m + 1 2 )+(m + 1 2 )=2m +1 deg ψ D deg ψ D ψ : D S 1 D 3.3 ψ D Im ψ S 1 1 1.2 9 φ (i) S 2 + D S 2 + S 2 + = {(x, y, z) S 2 : z =0} D D (ii) P S 2 + P S 2 +.

64 3 3.5 ( ). R 3 3 R 3 3 B.6: 3 1 Borsuk Ulam. 3 Ω 1,Ω 2,Ω 3 S 2 P OP 1 Ω 1 π P 2, 3 Ω 2,Ω 3 π P Ω 2 OP V + (P ) V (P ) π P Ω 3 W + (P ), W (P ) B.7 ϕ : S 2 R 2, ϕ(p )= ( V + (P ) V (P ), W + (P ) W (P ) ) 10 10 ϕ( P )= ϕ(p ) (3.4)

C. 65 + V (P) + W (P) Ω 3 π P Ω 2 Ω 1 V (P) P W (P) O S 2 B.7: P S 2 π P π P π P O P P V + ( P )=V (P ), V ( P )=V + (P ), W + ( P )=W (P ), W ( P )=W + (P ) (3.4) ϕ Borsuk-Ulam ϕ(p 0 )=O P 0 S 2 ϕ π P0 Ω 1 Ω 2,Ω 3 C 3.3 2 V : D R 2 D = {P R 2 : P 1} D = {P R 2 : P =1} V (P ) 0, P D. P D V (P ) 1 V (P )/ V (P )

66 3 V (P )/ V (P ) S 1 = {P R 2 : P =1} f(p ) D = S 1 f f P D V (P ) f V (P ) D V (P ) V (P ) P O f(p ) D S 1 C.1: V D (winding number) wind D V wind V wind D V = wind V = deg f. D V C.2: D V D wind V =2 3.6. V D V D

C. 67 wind V 0 V D C.3: V D D V D wind V V D wind V =1 2 2.6 D V D deg V =1 3.3 3.6 2.6 2 3.7. 3.6 3.3. ϕ : D D ϕ( D) D deg ϕ D 0 ϕ Q D \ Im ϕ D V (P )=Q ϕ(p ), P D wind V = deg φ 0 3.6 V D P D ϕ(p )=Q Q Im ϕ

68 3 ϕ(p) Im ϕ V(P) Q C.4: 3 V wind V = wind V 3.6 3 D 3.8. V wind V 0 V 2 2.14 2.14 Sperner 3.8 Sperner 3.9. 1 1 3 ABC 1 1/n 3 3 α, β, γ ( C.5 ) 3 ABC 1/n 2 α, β 3 ABC α, β M + β, α M M + M 0 1/n 3 3 α, β, γ 1

C. 69 C A B C.5: Sperner M + =3,M =2 3.9 3 α, β, γ 3 3 3 3 3 α, β, γ 1 2 3 1 3 α, β, γ 1 + γ, β, α 1 1 + 3 N + 1 3 N N + N = M + M C.5 N + =2,N =1 (a) 1 + (b) 1 C.6: 2 3

70 3 3.9. 3 1 α, β 3 E α, β +1 1 E 1 α, β 3 C.7 4 1 + 1 3 2 3 3 1 + 3 1 α, β (a) 1 + (b) 1 (c) 2 (d) 3 C.7: 4 3 +1 1 3 1 2 3 (+1) + ( 1)=0 2 3 3 3 E = N + N α, β 3 1 3 3 E E E = M + M

C. 71 C.8: N + N = E = M + M M + M 0 N + 0 N 0 3.8 2 Sperner 2.14 3.8 V : R 2 wind V 0 2.14 3 n 3 3 3 I 1,I 2,,I 3n P I k (k =1, 2,, 3n) V V (P ) 0 P I k V (P )/ V (P ) S 1 = {Q R 2 : Q =1} I k S 1 I k S 1, P V (P ) V (P ) S 1 θ n,k 0 θ n,k < 2π θ n,1,,θ n,3n Θ n Θ n = max{θ n,1,,θ n,3n }. 11 0 Θ n < 2π n Θ n 11 a 1,,a m max{a 1,,a m}

72 3 {Θ n } I 1 I 3n V (P )/ V (P ) I 2 θ n,k P S 1 I k C.9: 3.10. n Θ n 12 3.8 V n Θ n < 2π 3 (3.5) 3.10 n 2.14 α, β, γ 3.9 M +, M 3 α, β α, β M + β, α M wind V = M + M (3.6) P 1 V (P ) V (P )/ V (P ) wind V P V (P ) wind V (3.6) 12

C. 73 wind V P 1 f(p )= V (P ) V (P ) S 1 S 1 f(p ) S 1 wind V S 1 1 Q 0 f(p ) Q 0 L + L wind V = L + L 13 I k (k =1,, 3n) 3 3 I k P Q 0 I k P 0 P 1 f(p ) S 1 C.10: P 1 f(p )=V (P )/ V (P ) S 1 f(p ) S 1 Q 0 L + L wind V = L + L L + =2,L =1 wind V =2 1=1 P 0, P 1 P P 0 P 1 I k f(p ) S 1 Q 0 I k,+ L k, L + = 3n k=1 wind V = L k,+, L = 3n 3n L k, k=1 (L k,+ L k, ) k=1 13 f(p ) Q 0 L +, L

74 3 14 S 1 Q 0 (cos 2π/3, sin 2π/3) α β I k P 0 α P 1 β L k,+ L k, =1 (3.5) (3.5) P P 0 P 1 f(p ) S 1 (3.5) P I k f(p ) S 1 θ n,k < 2π/3 f(p ) S 1 P 0 β P 1 α L k,+ L k, = 1 P 0, P 1 L k,+ L k, 1 =0 (3.6) f(p 0 ) f(p ) Q 0 α β f(p 1 ) θ n,k O S 1 γ C.11: wind V 0 (3.6) M + M 0 3.9 3 3 α, β, γ 3 α A n β β γ C n 2.14 n 3.8 14 I k f Q 0 L k,+, L,k