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「産業上利用することができる発明」の審査の運用指針（案）

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Microsoft Word - 触ってみよう、Maximaに2.doc

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Japan Research Review 1998年7月号

Japan Research Review 1998.7 Perspectives ****************************************************************************************** - 1 - Japan Research Review 1998.7-2 - Japan Research Review 1998.7-3

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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,

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チュートリアル：ノンパラメトリックベイズ

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sekibun.dvi

a d = a + a+ (a ), e d = e, sin d = cos, (af() + bg())d = a d = log, cosd = sin, f()d + b g()d d 3 d d d d d d d ( + 3 + )d ( + )d ( 3 )d (e )d ( sin 3 cos)d g ()f (g())d = f(g()) e d e d ( )e d cos d

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y x = α + β + ε =,, ε V( ε) = E( ε ) = σ α $ $ β w ( 0) σ = w σ σ y α x ε = + β + w w w w ε / w ( w y x α β ) = α$ $ W = yw βwxw $β = W ( W) ( W)( W) w x x w x x y y = = x W y W x y x y xw = y W = w w

UCT探索を用いた大貧民クライアント

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N cos s s cos ψ e e e e 3 3 e e 3 e 3 e

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橡CyberView User's Guide for Macintosh

for Macintosh Mac - 1 Mac - 2 Mac - 2 Mac - 3 PrimeFilm 1800i CyberView 1. PrimeFilm 1800i 2. 3.CyberView 4. CyberView Mac - 4 1. PrimeFilm 1800i 2. 3.CyberView CyberView 1. PrimeFilm 1800i 2. 3. 4. 1.

1 0/1, a/b/c/ {0, 1} S = {s 1, s 2,..., s q } S x = X 1 X 2 X 3 X n S (n = 1, 2, 3,...) n n s i P (X n = s i ) X m (m < n) P (X n = s i X n 1 = s j )

(Communication and Network) 1 1 0/1, a/b/c/ {0, 1} S = {s 1, s 2,..., s q } S x = X 1 X 2 X 3 X n S (n = 1, 2, 3,...) n n s i P (X n = s i ) X m (m < n) P (X n = s i X n 1 = s j ) p i = P (X n = s i )

() 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)

0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

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26 11 5 1 ( 2 2 2 3 5 3.1...................................... 5 3.2....................................... 5 3.3....................................... 6 3.4....................................... 7

ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y

01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y

(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {

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10 log 10 W W 10 L W = 10 log 10 W 10 12 10 log 10 I I 0 I 0 =10 12 I = P2 ρc = ρcv2 L p = 10 log 10 p 2 p 0 2 = 20 log 10 p p = 20 log p 10 0 2 10 5 L 3 = 10 log 10 10 L 1 /10 +10 L 2 ( /10 ) L 1 =10

dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp

+ P (x)y = Q(x) (1) = P (x)y + Q(x) P (x), Q(x) y Q(x) 0 homogeneous = P (x)y 1 y = P (x) log y = P (x) + C y = C exp{ P (x) } = C e R P (x) 5.1 + P (x)y = 0 (2) y = C exp{ P (x) } = Ce R P (x) (3) αy

arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =

arctan arctan arctan arctan 2 2000 π = 3 + 8 = 3.25 ( ) 2 8 650 π = 4 = 3.6049 9 550 π = 3 3 30 π = 3.622 264 π = 3.459 3 + 0 7 = 3.4085 < π < 3 + 7 = 3.4286 380 π = 3 + 77 250 = 3.46 5 3.45926 < π < 3.45927

Chap9.dvi

.,. f(),, f(),,.,. () lim 2 +3 2 9 (2) lim 3 3 2 9 (4) lim ( ) 2 3 +3 (5) lim 2 9 (6) lim + (7) lim (8) lim (9) lim (0) lim 2 3 + 3 9 2 2 +3 () lim sin 2 sin 2 (2) lim +3 () lim 2 2 9 = 5 5 = 3 (2) lim

( ) x y f(x, y) = ax

013 4 16 5 54 (03-5465-7040) [email protected] hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy

main.dvi

4 DFT DFT Fast Fourier Transform: FFT 4.1 DFT IDFT X(k) = 1 n=0 x(n)e j2πkn (4.1) 1 x(n) = 1 X(k)e j2πkn (4.2) k=0 x(n) X(k) DFT 2 ( 1) 2 4 2 2(2 1) 2 O( 2 ) 4.2 FFT 4.2.1 radix2 FFT 1 (4.1) 86 4. X(0)

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( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

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II 2014 2 (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1

II 2014 1 1 I 1.1 72 r 2 72 8 72/8 = 9 9 2 a 0 1 a 1 a 1 = a 0 (1+r/100) 2 a 2 a 2 = a 1 (1 + r/100) = a 0 (1 + r/100) 2 n a n = a 0 (1 + r/100) n a n a 0 2 n a 0 (1 + r/100) n = 2a 0 (1 + r/100) n = 2

(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

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP

09 II 09/11/ y = e x y = log x = log e x 2. ( e x ) = e x 3. ( ) log x = 1 x 1 Warming Up 1 u = log a M a u = M log a 1 a 0 a 1 a r+s 0 a r

09 II 09/11/16 1 5.6 1. y = e x y = log x = log e x 2. e x ) = e x 3. ) log x = 1 x 1 Warming Up 1 u = log a M a u = M log a 1 a 0 a 1 a r+s 0 a r a s 1 a 2 f g) = f g + f g 1. fx) = x e x f x) = 2. fx)

( ) 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) (

6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b

1 n i i 1 i n index = 0; [ index] = 1; = = ; if ( == ) { // [ index ]++; } else if () { if( index == 0 ) { // // // // [ index ]++; = ; } else { // //

7 1 2008 12 22 1 Copyright c 2008 The Japanese Committee for International Olympiad in Informatics 1 n i i 1 i n index = 0; [ index] = 1; = = ; if ( == ) { // [ index ]++; } else if () { if( index == 0