知能科学：ニューラルネットワーク

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知能科学：ニューラルネットワーク

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =


6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

高等学校学習指導要領

高等学校学習指導要領

85 4

() 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)

4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X

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

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資料5：聖ウルスラ学院英智小・中学校　提出資料（1）


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

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日本内科学会雑誌第96巻第11号

平成20年5月 協会創立50年の歩み　海の安全と環境保全を目指して 友國八郎 海上保安庁 長官 岩崎貞二 日本船主協会 会長 前川弘幸 JF全国漁業協同組合連合会 代表理事会長 服部郁弘 日本船長協会 会長 森本靖之 日本船舶機関士協会 会長 大内博文 航海訓練所 練習船船長 竹本孝弘 第二管区海上保安本部長 梅田宜弘

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重力方向に基づくコントローラの向き決定方法

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Q & A Q A p

c y /2 ddy = = 2π sin θ /2 dθd /2 [ ] 2π cos θ d = log 2 + a 2 d = log 2 + a 2 = log 2 + a a 2 d d + 2 = l

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1


66 σ σ (8.1) σ = 0 0 σd = 0 (8.2) (8.2) (8.1) E ρ d = 0... d = 0 (8.3) d 1 NN K K 8.1 d σd σd M = σd = E 2 d (8.4) ρ 2 d = I M = EI ρ 1 ρ = M EI ρ EI


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

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,

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

1 y(t)m b k u(t) ẋ = [ 0 1 k m b m x + [ 0 1 m u, x = [ ẏ y (1) y b k m u


( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

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さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1

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9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =


B

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(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37

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II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

Gmech08.dvi

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, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

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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

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数学演習：微分方程式

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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.

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

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Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

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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

7

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

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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 (

Gmech08.dvi

2.5 (Gauss) (flux) v(r)( ) S n S v n v n (1) v n S = v n S = v S, n S S. n n S v S v Minoru TANAKA (Osaka Univ.) I(2012), Sec p. 1/30

高等学校学習指導要領解説　数学編


main.dvi

1

(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 {

(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

zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {

(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 ( ) ( ) (1) C z = a ρ. f(z) dz = C = = (z a) n dz C n= p 2π (ρe iθ ) n ρie iθ dθ 0 n= p { 2πiA 1 n = 1 0 n 1 (2) C f(z) n.. n f(z)dz = 2πi Re

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4 Mindlin -Reissner 4 δ T T T εσdω= δ ubdω+ δ utd Γ Ω Ω Γ T εσ (1.1) ε σ u b t 3 σ ε. u T T T = = = { σx σ y σ z τxy τ yz τzx} { εx εy εz γ xy γ yz γ

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

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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 =

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

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

Transcription:

2 3 4

(Neural Network) (Deep Learning) (Deep Learning) (

x

x

= ax + b x

x

x

? x

x

x w σ b = σ(wx + b) x w b w b

.2.8.6 σ(x) = + e x.4.2 -.2 - -5 5

x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x, x 2, x 3 w, w 2, w 3 b w, w 2, w 3 b

x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x, x 2, x 3 w, w 2, w 3 b w, w 2, w 3 b

x w x2 w2 σ x3 w3 b = σ(w T x + b) x x = x 2 x 3 w = w w 2 w 3

( ) x

(3 2 ) x x2 x3 2

(3 2 ) x x2 x3 2

(3 2 ) x x2 x3 2

(3 2 ) x x2 x3 2

(3 2 ) x x2 x3 2

(3 2 ) x x2 x3 2

x x2 x3 2, 2 [, ]

x w x2 w2 Σ x3 w3 b = w x + w 2 x 2 + w 3 x 3 + b x, x 2, x 3 w, w 2, w 3 b w, w 2, w 3 b

σ σ x σ σ x σ Σ σ σ [, ] (, )

x -2-3 5 2 6 5 u = σ(5x + 5).8 4 4 σ -3 u 2 = σ( 3x + 6) u 3 = σ( 2x + 4) = σ(2u + u 2 + 4u 3 3).6.4.2 - -5 5 x

x 5-3 -2 5 2 6 4 4 Σ -3 u = σ(5x + 5) u 2 = σ( 3x + 6) u 3 = σ( 2x + 4) = 2u + u 2 + 4u 3 3 4 3 2 - -2-3 - -5 5 x

x x x2 x3 2

= σ(wx + b) w =. b =..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) w = 5. b =..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) w =. b =..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) w = 5. b =..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) w = 5. b =..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) w = 5. b = 5..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) w = 5. b =..2.8.6.4.2 -.2 - -5 5

= σ(wx + b) wx + b = b/w = 2.2.8.6.4.2 -.2 - -5 5

.2.8.6.4.2 -.2 - -5 5.2 + =.2.8.6.4.2.8.6.4 (-) -.2 - -5 5.2 -.2 - -5 5

.2.8.6.4.2 -.2 - -5 5.2.8.6.4 + 7 (-7) = 9 8 7 6 5 4 3 2 - - -5 5.2 -.2 - -5 5

x 5 σ 7-5 5 σ -7 Σ = 9 8 7 6 5 4 3 2 - - -5 5

.2.8.6.4.2 -.2 - -5 5.2.8.6.4 + 4 (-4) = 9 8 7 6 5 4 3 2 - - -5 5.2 -.2 - -5 5

x 5-5 σ 4-5 σ -4 Σ = 9 8 7 6 5 4 3 2 - - -5 5

9 8 7 6 5 4 3 2 - - -5 5 + 9 8 7 6 5 4 3 2 - - -5 5 = 9 8 7 6 5 4 3 2 - - -5 5

σ x 5-5 7 5 σ -7-5 5 σ 4 5 - -4 Σ = 9 8 7 6 5 4 3 2 - - -5 5 σ

x

x x

x x + + + x x

4 3 2 - -2 (, ) [, ] -3 - -5 5 x σ = =.8.6.4.2 σ - -5 5 x

4 3 2 - -2 (, ) [, ] -3 - -5 5 x x 5-2 -3 5 2 6 4 4 Σ -3 σ = =.8.6.4.2 σ - -5 5 x

4 3 2 - -2 (, ) [, ] -3 - -5 5 x x 5-2 -3 5 2 6 4 4 Σ -3 σ = =.8.6.4.2 σ x - -5 5 x 5-2 -3 5 2 6 4 4 σ -3

= σ(a x + a 2 x 2 + c) σ(a x + a 2 x 2 c) 2.5.5 -.5 - -.5-2 -2 -.5 - -.5.5.5 2 a = r cos θ a 2 = r sin θ r = 2 θ = c =

= σ(a x + a 2 x 2 + c) σ(a x + a 2 x 2 c) 2.5.5 -.5 - -.5-2 -2 -.5 - -.5.5.5 2 a = r cos θ a 2 = r sin θ r = 2 θ = π/4 c =

= σ(a x + a 2 x 2 + c) σ(a x + a 2 x 2 c) 2.5.5 -.5 - -.5-2 -2 -.5 - -.5.5.5 2 a = r cos θ a 2 = r sin θ r = 2 θ = 2π/4 c =

= σ(a x + a 2 x 2 + c) σ(a x + a 2 x 2 c) 2.5.5 -.5 - -.5-2 -2 -.5 - -.5.5.5 2 a = r cos θ a 2 = r sin θ r = 2 θ = 3π/4 c =

N = 8 2.5.5 -.5 2 3 - -.5-2 -2 -.5 - -.5.5.5 2

N = 6 2.5.5 2 4 6 -.5 - -.5-2 -2 -.5 - -.5.5.5 2

N = 32 2.5.5 4 8 -.5 - -.5-2 -2 -.5 - -.5.5.5 2

N = 36 2.5.5 4 2 -.5 2 - -.5-2 -2 -.5 - -.5.5.5 2

b w b2 u w 2 x w 2 b3 u2 w 2 2 b w 3 u3 w 2 3 u = σ(w x + b ) u 2 = σ(w2 x + b 2 ) u 3 = σ(w3 x + b 3 ) = σ(w 2 u + w2 2 u 2 + w3 2 u 3 + b)

x b w b2 w 2 w 3 b3 u u2 u3 w 2 w 2 2 b w 2 3 + - t E t E = ( t)2 2

x b w b2 w 2 w 3 b3 u u2 u3 w 2 w 2 2 b w 2 3 + - t E E

f (x, ) (x, ) min f (x, ) (x n, n ) (x n+, n+ ) x n+ = x n α f x (x n, n ) n+ = n α f (x n, n ) α

σ(x) = + e x σ(x) = e x + e x σ (x) = ( e x ) ( + e x ) 2 = + e e x x + e x = σ(x)( σ(x))

x w σ b = σ( + wx + + b) x = σ ( ) w = σ( ) { σ( )} w = ( )w

x w σ b = σ( + wx + + b) w = σ ( ) x = ( )x b = σ ( ) = ( )

x w σ b = σ( + wx + + b) = ( )w x = ( )x w b = ( )

E w 2 = ( t) w 2 = σ( + w 2 u + ) w 2 = ( )u E w 2 = ( t)( )u

E b = ( t) b = σ( + b) b = ( ) E b = ( t)( )

E = ( t) w w w u u w = u u w = σ( + w 2 u + ) u = ( )w 2

u = σ( + w x + ) u w = u ( u )x E w = ( t)( )w 2 u ( u )x = E w 2 w 2 ( u )x

E = ( t) b b b u u = u b u b = σ( + w 2 u + ) u = ( )w 2

u = σ( + b) u b = u ( u ) E b = ( t)( )w 2 u ( u ) = E b w 2 u ( u )

E w 2 k E b = ( t)( )u k = ( t)( ) E = E w 2 wk wk 2 k ( u k )x E = E b k b w k 2 u k ( u k ) (Back Propagation)

w 2 k := w 2 k α E w 2 k b := b α E b w k := w k α E w k b k := b k α E b k (Back Propagation)

u u2 + - u3 t

u w 2 x u2 w 2 2 + - u3 w 2 3

x 2 2 t.5.2.6.9.3 5 α =.

.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

2.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

5.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

.9.8.7.6.5.4.3.2. -2 -.5 - -.5.5.5 2

x b u w b2 w 2 u2 v w 2 c w 3 w 2 2 w 2 2 b w 2 22 w 3 2 v2 c2 + - t E

x b u w b2 w 2 u2 v w 2 c w 3 w 2 2 w 2 2 b w 2 22 w 3 2 v2 c2 + - t E

x b u w b2 w 2 u2 v w 2 c w 3 w 2 2 w 2 2 b w 2 22 w 3 2 v2 c2 + - t E

x b u w b2 w 2 u2 v w 2 c w 3 w 2 2 w 2 2 b w 2 22 w 3 2 v2 c2 + - t E

8

2 3 4 5 6 7 8 9 2 77 5 3 85 8 6 4 73 5 2 5 9 6 8 5 7 9 5 2 89 8 86.5 %

(Back Propagation)

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