平成 22 年度 ( 第 32 回 ) 数学入門公開講座テキスト ( 京都大学数理解析研究所, 平成 ~8 22 月年 58 日開催月 2 日 ) V := {(x,y) x n + y n 1 = 0}, W := {(x,y,z) x 3 yz = x 2 y z 2

Similar documents

III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2 lim. (x,y) (1,0) x 2 + y 2 lim (x,y) (0,0) lim (x,y) (0,0) lim (x,y) (0,0) 5x 2 y x 2 + y 2. xy x2 + y

1

kb-HP.dvi

1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)

() 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 (

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

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

「産業上利用することができる発明」の審査の運用指針(案)

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

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.

II 2 II

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

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

I II III 28 29

生活設計レジメ

44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)


(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,

6. Euler x

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

³ÎΨÏÀ

n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1)

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

2.4 ( ) ( B ) A B F (1) W = B A F dr. A F q dr f(x,y,z) A B Γ( ) Minoru TANAKA (Osaka Univ.) I(2011), Sec p. 1/30

TOP URL 1

pdf

i

2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta

Chap10.dvi

v er.1/ c /(21)


A

振動と波動

40 6 y mx x, y 0, 0 x 0. x,y 0,0 y x + y x 0 mx x + mx m + m m 7 sin y x, x x sin y x x. x sin y x,y 0,0 x 0. 8 x r cos θ y r sin θ x, y 0, 0, r 0. x,

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

入試の軌跡

II Time-stamp: <05/09/30 17:14:06 waki> ii

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

Gmech08.dvi

5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (

応力とひずみ.ppt

18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb

A

i


Wide Scanner TWAIN Source ユーザーズガイド

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

ii

2011de.dvi

( ) ( )

Part () () Γ Part ,

第1部 一般的コメント

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

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2

slide1.dvi


II III I ~ 2 ~

中堅中小企業向け秘密保持マニュアル


PR映画-1

- 2 -


1 (1) (2)

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a

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

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

第1章 国民年金における無年金


untitled

表1票4.qx4

福祉行財政と福祉計画[第3版]

= π2 6, ( ) = π 4, ( ). 1 ( ( 5) ) ( 9 1 ( ( ) ) (

橡ミュラー列伝Ⅰ.PDF

i


C:/KENAR/0p1.dvi

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x

simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =

203 x, y, z (x, y, z) x 6 + y 6 + z 6 = 3xyz ( 203 5) a 0, b 0, c 0 a3 + b 3 + c 3 abc 3 a = b = c 3xyz = x 6 + y 6 + z 6 = (x 2 ) 3 + (y 2 ) 3

ランダムウォークの境界条件・偏微分方程式の数値計算

分散分析・2次元正規分布

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

keisoku01.dvi

7-12.dvi

Acrobat Distiller, Job 128

Chap11.dvi

, = = 7 6 = 42, =

A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2.


73

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

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

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>

Transcription:

3 90 2006 1. V := {(x,y) x n + y n 1 = 0}, W := {(x,y,z) x 3 yz = x 2 y z 2 = xz y 2 = 0} V (x,y) n = 1 n = 2 (x,y) V n = 1 n = 2 (3/5,4/5),(5/13,12/13)... n 3 V (0,±1),(±1,0) ( ) n 3 x n + y n = z n, xyz 0 (x,y,z)

y = sinx V 1 p 1 + + 1 0 Q,R,C C := {(x,y) xy + x 3 + y 3 = 0}, D := {(x,y) xy = 0} C,D V xy W xyz k n A n := {(x 1,...,x n ) x 1,...,x n k} n f 1 (x 1,...,x n ),..., f r (x 1,...,x n ) V := {(x 1,...,x n ) A n f 1 (x 1,...,x n ) = = f r (x 1,...,x n ) = 0} n P n P n := {[x 0 :... : x n ] x 0,...,x n k 0} k n + 1 [x 0 :... : x n ] x i 0 k 0 t [x 0 :... : x n ] = [tx 0 :... : tx n ] P n x 0 0 U 0 U 0 [x 0 :... : x n ] 1/x 0 [1 : x 1 /x 0 :... : x n /x 0 ] U 0 [1 : x 1 /x 0 :... : x n /x 0 ] (x 1 /x 0,...,x n /x 0 ) A n

U 0 A n U 0 A n U 0 A n x i 0 P n U i U i A n [x 0 :... : x n ] x i 0 P n U 0,...,U n P n = U i 0 i n P n n + 1 n = 2 V 2 := {(x,y) A 2 x 2 + y 2 1 = 0} C u = x + y 1, v = x y 1 V 2 = {(u,v) A 2 uv 1 = 0} V 2 (u,v) u A 1 V 2 A 1 o A 1 \{o} V 2 V 2 P 2 V 2 := {[U : V : W] P 2 UV W 2 = 0} V 2 (u,v) [u : v : 1] V 2 P 1 V 2 [U : V : W] [U : W] [W : V ] P 1 C V 2 C u = x + y 1, v = x y 1 R C C 2. 1 2 W := {(x,y,z) x 3 yz = x 2 y z 2 = xz y 2 = 0}

(t 3,t 4,t 5 ) 1 Y := {(x,y,z,w) xy zw = 0} 4 3 W A 1 t (t 3,t 4,t 5 ) W C := {(x,y) xy + x 3 + y 3 = 0} B B := { (x,y),[x : Y ] A 2 P 1 xy yx = 0} A 2 P 1 (x,y),[x : Y ] [X : Y ] (x,y) B A 2 π B (x,y),[x : Y ] (x,y) A 2 A 2 o x,y π B\π 1 (o) A 2 \{o} x = y = 0 [X : Y ] xy yx = 0 o π 1 (o) P 1 B A 2 P 1 A 2 C B C C C π C C π : B A 2 A 1 W B A 2 3. X

X X X 2 P 1 P 1 X P 1 X X X P 1 X ( ) C C P 1 (C 2 ) = 1 X C,D C D C,D (C D) X Z 1 (X) := { r i C i r i R} (C 2 ) C C C C C C D C C D (C C +C ) = (C D) = 0 (C 2 ) = (C C ) < 0 n X C (C D) D n 1 X n 1 Z 1 (X) := { d i D i d i R} Z 1 (X) Z 1 (X) R N 1 (X) = Z 1 (X)/ N 1 (X) = Z 1 (X)/ N 1 (X) N 1 (X) R C X X K X X X n x 1,...,x n n dx 1 dx n

f (x 1,...,x n ) ω = f dx 1 dx n dx dx = dx dt dt ω ω X P 1 = {[X : Y ]} P = [0 : 1] Y 0 x := X/Y ω = dx Q = [1 : 0] ω Q y := Y /X x = y 1 ω = dx dy dy = y 2 dy Q 2 ( 2 ) y 2 ω P 1 K P 1 = 2Q X ω D i n i divω := n i D i X K X ω P 1 dy K P 1 = 2P X f f D i f i div f := f i D i C (K X C) ω K X C (K X C) X X X (i) X K X K X X X (ii) X S f : X S f K X X P 1 f 4. N 1 (X) NE(X) NE(X) K X P 1

( ) NE(X) = NE KX 0(X) + R 0 [C i ] C i (K X C i ) < 0 R 0 [C i ] π i : X Y i R 0 [C i ] X K X K X X Y Y 3 Y 3 x x 3 X π : X Y π X E Y K Y π X Y K X = π K Y + ae E π E a K Y E π K X E a π : X Y Y K Y π : X Y π + : X + Y π K X + X + X,X + Q- 2 3 Y := {(x,y,z,w) xy zw = 0} Q := {[X : Y : Z : W] XY ZW = 0} B Q {l t : [X : Z] = [W : Y ] = t} t P 1 {l t + : [X : W] = [Z : Y ] = t} t P 1 B X B X + X Y, X + Y

K X,K X + 0 (x,y,z,w) ( x,y,z, w) X Y X + S Q- X S (i) K X X (ii) K X K X π : X Y (ii-1) Y X π X (ii-2) π Y S X Y (ii-3) π Y / S π + : X + Y X + S X X + ( ) N 1 (X) ( ) 5. 3 3 3 1 3 3 80 3 X (X, ) X Q- R- R- 0 (X, ) K X K X 1

D H 0 (X,D) := { f div f + D 0} π : X Y Y P r π : X P r X f 0,..., f r [ f 0 :... : f r ] P r H = P r 1 π H π H 0 (X,π H) f 0,..., f r V K X H i (X,K X + ) ( ) X A H i (X,K X + A) = 0 (i 1) X S (X,S + B) S (S,B S ) K X + S + B S = K S + B S B 0 H 0 (X,K X + B) H 0 (X,K X + S + B) H 0 (S,K S + B S ) H 1 (X,K X + B) (X,S +B) (S,B S ) 3 90 6. 3 pl (X,S +B) 21 200 4 pl 2005 pl ( ) n 1 n pl (X,S + B) (S,B S ) H 0 (X,m(K X + S + B)) H 0 (S,m(K S + B S ))

m(k X + S + B) m pl B Q- (X,S + B) R X := m 0H 0 (X,m(K X + S + B)) R X H 0 (X,m(K X + S + B)) m H 0 (X,m(K X + S + B)) H 0 (X,m (K X + S + B)) H 0 (X,(m + m )(K X + S + B)) R S := m 0H 0 (X,m(K S + B S )) ρ : R X R S pl R X ρ(r X ) R S ρ R X m X,S X m,s m S m + B m H 0 (X,ml(K X + S + B)) = H 0 (X m,ml(k Xm + S m + B m )), H 0 (X m,ml(k Xm + S m + B m )) H 0 (S m,ml(k Sm + B m,sm )) l l(k X + S + B) m X m m S m m S m T R l X := H 0 (X,ml(K X + S + B)) R l T := m 0 m 0H 0 (T,ml(K T + B m,t )) R l T ρ(rl X ) ρ(r l X ) ρ(r X) 7. R- R- R- R-

( ) (X, ) (i) ( ) K X + R- R- (X, ) (ii) ( ) (X, ) Y Y (iii) ( ) K X + R- R- R- R- (i) (iii) K X + ( ) (ii) (X, = A + B ) A R- B pl X X S X Y X Y (i) ( ) X S ( ) (X, ) Q- m 0 H 0 (X,m(K X + )) ( ) (pl n ) n pl (i n ), (ii n ), (iii n ) (i), (ii), (iii) n (ii) (ii ) (i) (i n 1 ), (ii n 1 ), (iii n 1 ) (pl n )

(i n 1 ), (ii n 1 ), (iii n 1 ) (ii) (ii n 1 ) (ii n) (ii n) (X,S + B ) (S,B S ) (iii) (pl n ), (ii n) (i n ) pl (ii n) (iv) (iii n 1 ), (i n ), (ii n) (iii n ) R- Q- (v) (i n ), (iii n ) (ii n ) [0,1] [0,1] R- X Q- (X, ) C t (X, tc) K X + tc R- R- (X, ) C (i) K X + tc t (ii) t = 0 K X + X (iii) t > 0 NE(X) K X + K X + tc 0 π (iii-1) π (iii-2) π (X, ) C C ( ) C (X, ) C (X, ) K X + ( ) K X + C (X, ) C

8. R- A R- B A ( ) K X + (X, ) ( ) (X, ) 5 H 0 (X,m(K X + )) Φ m : X P r m m(k X + ) m Φ m ( ) K X + R- R- 3 0 ( ) 3 9. J. Kollár

61 (2009), 162-186 C. Birkar, P. Cascini, C. Hacon and J. McKernan, Existence of minimal models for varieties of log general type, J. Am. Math. Soc. 23 (2009), 405-468 C. Hacon and J. McKernan, Existence of minimal models for varieties of log general type II, J. Am. Math. Soc. 23 (2009), 469-490