f(x) = e x2 25 d f(x) 0 x d2 dx f(x) 0 x dx2 f(x) (1 + ax 2 ) 2 lim x 0 x 4 a 3 2 a g(x) = 1 + ax 2 f(x) g(x) 1/2 f(x)dx n n A f(x) = Ax (x R

Similar documents
36 3 D f(z) D z f(z) z Taylor z D C f(z) z C C f (z) C f(z) f (z) f(z) D C D D z C C 3.: f(z) 3. f (z) f 2 (z) D D D D D f (z) f 2 (z) D D f (z) f 2 (

高等学校学習指導要領

高等学校学習指導要領

() 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)

- II

, 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,,,,

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

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

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

[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )

1 4 1 ( ) ( ) ( ) ( ) () 1 4 2


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

() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.

body.dvi


1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 +

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

A

4 R f(x)dx = f(z) f(z) R f(z) = lim R f(x) p(x) q(x) f(x) = p(x) q(x) = [ q(x) [ p(x) + p(x) [ q(x) dx =πi Res(z ) + Res(z )+ + Res(z n ) Res(z k ) k

2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+

, 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


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

知能科学:ニューラルネットワーク

知能科学:ニューラルネットワーク

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

all.dvi

[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29

2

II ( : )

Chap11.dvi

P /fi¡ficfiNŒç

Microsoft Word - 触ってみよう、Maximaに2.doc

7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a

main.dvi

A

II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K

応力とひずみ.ppt

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

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

6.1 (P (P (P (P (P (P (, P (, P.


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

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

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

資料5:聖ウルスラ学院英智小・中学校 提出資料(1)

x i [, b], (i 0, 1, 2,, n),, [, b], [, b] [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ] ( 2 ). x 0 x 1 x 2 x 3 x n 1 x n b 2: [, b].,, (1) x 0, x 1, x 2,, x n

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 =

untitled

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

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


6.1 (P (P (P (P (P (P (, P (, P.101

1 yousuke.itoh/lecture-notes.html [0, π) f(x) = x π 2. [0, π) f(x) = x 2π 3. [0, π) f(x) = x 2π 1.2. Euler α

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

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

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

Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0

°ÌÁê¿ô³ØII

FX ) 2

FX自己アフリエイトマニュアル

DVIOUT

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


MF 型

1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx

mugensho.dvi

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

all.dvi

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

II K116 : January 14, ,. A = (a ij ) ij m n. ( ). B m n, C n l. A = max{ a ij }. ij A + B A + B, AC n A C (1) 1. m n (A k ) k=1,... m n A, A k k

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


all.dvi

(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t

Gmech08.dvi

function2.pdf

2S III IV K A4 12:00-13:30 Cafe David 1 2 TA 1 appointment Cafe David K2-2S04-00 : C

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 =

CD納品用.indd

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

Acrobat Distiller, Job 128

tomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p.

f : R R f(x, y) = x + y axy f = 0, x + y axy = 0 y 直線 x+y+a=0 に漸近し 原点で交叉する美しい形をしている x +y axy=0 X+Y+a=0 o x t x = at 1 + t, y = at (a > 0) 1 + t f(x, y

(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (

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

B [ 0.1 ] x > 0 x 6= 1 f(x) µ 1 1 xn 1 + sin sin x 1 x 1 f(x) := lim. n x n (1) lim inf f(x) (2) lim sup f(x) x 1 0 x 1 0 (

1

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

III ϵ-n ϵ-n lim n a n = α n a n α 1 lim a n = 0 1 n a k n n k= ϵ-n 1.1

109

(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)


2 Husserl Husserl 1 Husserl [1966] Hua. Bd. 10 Husserl 2 Husserl Husserl

18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (

no35.dvi

1


Transcription:

29 ( ) 90 1 2 2 2 1 3 4 1 5 1 4 3 3 4 2 1 4 5 6 3 7 8 9

f(x) = e x2 25 d f(x) 0 x d2 dx f(x) 0 x dx2 f(x) (1 + ax 2 ) 2 lim x 0 x 4 a 3 2 a g(x) = 1 + ax 2 f(x) g(x) 1/2 f(x)dx 11 0 24 n n A f(x) = Ax (x R n ) 25 A 2 = A A 2 A 2 = A t A = A A t A A 3 n = 2 A 2 = O A f Ker(f) m(f) 1

n (n > 2) x 1, x 2,, x n G A A k (i, j) (A k ) ij 3 G G 50 G A x l x l 3 A = 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 2 G A 3 (A) ij (A 2 ) ij 0 i, j G 2

4 C C i, N i, P i, S i 1 4 50 C = 10 i 1 2 3 4 N i FUN 1 FUN 2 FUN 3 FUN 4 P i 8 9 12 15 S i 3 5 7 8 2 C ( ) C (j = 1, 2,, C) j (V 4 j ) 4 4 (V i j ) j = 1, 2,, 10 i = 1,, 4 2 2 j 1 2 3 4 5 6 7 8 9 10 Vj 1 0 Vj 2 0 Vj 3 0 Vj 4 0 3

2 3 4 i = 1 N 1 j(= 1, 2,,C) (V 1 j ) L j ( ) V 1 j 0 L j i = 2 N 1 N 2 j(= 1, 2,, C) (V 2 j ) L j 2 i = 3 N 1 N 2 N 3 j(= 1, 2,, C) (V 3 j ) L j 2 i = 4 N 1 N 2 N 3 N 4 j(= 1, 2,, C) (V 4 j ) L j 2 C 2 Vj i = Vj i 1 j j S i 0 P i + V i j S i j Vj i V j i P i + Vj S i i L j L j Si N i V k 0 = 0 (k = 1,, 4) 4

50 00 20 T1 10 T2 T1 T2 2 3 4 5

29 ( ) CT 90 1 2 2 2 1 3 4 1 5 1 4 3 3 4 2 1 4 5 6 3 7 8 9

f(x) = e x2 25 d f(x) 0 x d2 dx f(x) 0 x dx2 f(x) (1 + ax 2 ) 2 lim x 0 x 4 a 3 2 a g(x) = 1 + ax 2 f(x) g(x) 1/2 f(x)dx 11 0 24 n n A f(x) = Ax (x R n ) 25 A 2 = A A 2 A 2 = A t A = A A t A A 3 n = 2 A 2 = O A f Ker(f) m(f) 1

n (n > 2) x 1, x 2,, x n G A A k (i, j) (A k ) ij 3 G G 50 G A x l x l 3 A = 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 2 G A 3 (A) ij (A 2 ) ij 0 i, j G 2

4 C C i, N i, P i, S i 1 4 50 C = 10 i 1 2 3 4 N i FUN 1 FUN 2 FUN 3 FUN 4 P i 8 9 12 15 S i 3 5 7 8 2 C ( ) C (j = 1, 2,, C) j (V 4 j ) 4 4 (V i j ) j = 1, 2,, 10 i = 1,, 4 2 2 j 1 2 3 4 5 6 7 8 9 10 Vj 1 0 Vj 2 0 Vj 3 0 Vj 4 0 3

2 3 4 i = 1 N 1 j(= 1, 2,,C) (V 1 j ) L j ( ) V 1 j 0 L j i = 2 N 1 N 2 j(= 1, 2,, C) (V 2 j ) L j 2 i = 3 N 1 N 2 N 3 j(= 1, 2,, C) (V 3 j ) L j 2 i = 4 N 1 N 2 N 3 N 4 j(= 1, 2,, C) (V 4 j ) L j 2 C 2 Vj i = Vj i 1 j j S i 0 P i + V i j S i j Vj i V j i P i + Vj S i i L j L j Si N i V k 0 = 0 (k = 1,, 4) 4

50 00 20 T1 10 T2 T1 T2 2 3 4 5

29 ( ) 90 1 2 1 2 1 3 1 4 5 1 4 3 3 4 2 1 4 5 6 3 5 7 8 9

50 00 2 50 1

50 Atkinson Shiffrin (1968) 150 2 Craik Lockhart (1972) 250 Atkinson, R. C., & Shiffrin, R. M. (1968). Human memory: A proposed system and its control processes. n K. W. Spence & J. T. Spence (Eds.), The psychology of learning and motivation, Vol. 2. New York: Academic Press, pp. 89-195. Craik, F.. M., & Lockhart, R. S. (1972). Levels of processing: A framework for memory research. Journal of Verbal Learning and Verbal Behavior, 11, pp. 671-684. 2

2008 3 50 2 50 2008 pp. 19-21. 3

4 C C i, N i, P i, S i 1 4 50 C = 10 i 1 2 3 4 N i FUN 1 FUN 2 FUN 3 FUN 4 P i 8 9 12 15 S i 3 5 7 8 2 C ( ) C (j = 1, 2,, C) j (V 4 j ) 4 4 (V i j ) j = 1, 2,, 10 i = 1,, 4 2 2 j 1 2 3 4 5 6 7 8 9 10 Vj 1 0 Vj 2 0 Vj 3 0 Vj 4 0 4

2 3 4 i = 1 N 1 j(= 1, 2,,C) (V 1 j ) L j ( ) V 1 j 0 L j i = 2 N 1 N 2 j(= 1, 2,, C) (V 2 j ) L j 2 i = 3 N 1 N 2 N 3 j(= 1, 2,, C) (V 3 j ) L j 2 i = 4 N 1 N 2 N 3 N 4 j(= 1, 2,, C) (V 4 j ) L j 2 C 2 Vj i = Vj i 1 j j S i 0 P i + V i j S i j Vj i V j i P i + Vj S i i L j L j Si N i V k 0 = 0 (k = 1,, 4) 5

29 ( ) 90 1 2 2 2 1 3 1 4 5 1 4 3 3 4 2 1 4 5 6 3 7 8 9

f(x) = e x2 25 d f(x) 0 x d2 dx f(x) 0 x dx2 f(x) (1 + ax 2 ) 2 lim x 0 x 4 a 3 2 a g(x) = 1 + ax 2 f(x) g(x) 1/2 f(x)dx 11 0 24 n n A f(x) = Ax (x R n ) 25 A 2 = A A 2 A 2 = A t A = A A t A A 3 n = 2 A 2 = O A f Ker(f) m(f) 1

n (n > 2) x 1, x 2,, x n G A A k (i, j) (A k ) ij 3 G G 50 G A x l x l 3 A = 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 2 G A 3 (A) ij (A 2 ) ij 0 i, j G 2

D z = x + iy w(z) = u(x, y) + iv(x, y) w u, v Cauchy-Riemann u x = v y u y = v x u, v, x, y R, i = 1 50 u, v 2 u x 2 + 2 u y 2 = 0 2 v x 2 + 2 v y 2 = 0 2 x = r cos θ, y = r sin θ Cauchy-Riemann u r = 1 r v θ v r = 1 r u θ 3 w u u = x 3 + axy 2 a w a v(1, 1) = 0 w v 3

4 C C i, N i, P i, S i 1 4 50 C = 10 i 1 2 3 4 N i FUN 1 FUN 2 FUN 3 FUN 4 P i 8 9 12 15 S i 3 5 7 8 2 C ( ) C (j = 1, 2,, C) j (V 4 j ) 4 4 (V i j ) j = 1, 2,, 10 i = 1,, 4 2 2 j 1 2 3 4 5 6 7 8 9 10 Vj 1 0 Vj 2 0 Vj 3 0 Vj 4 0 4

2 3 4 i = 1 N 1 j(= 1, 2,,C) (V 1 j ) L j ( ) V 1 j 0 L j i = 2 N 1 N 2 j(= 1, 2,, C) (V 2 j ) L j 2 i = 3 N 1 N 2 N 3 j(= 1, 2,, C) (V 3 j ) L j 2 i = 4 N 1 N 2 N 3 N 4 j(= 1, 2,, C) (V 4 j ) L j 2 C 2 Vj i = Vj i 1 j j S i 0 P i + V i j S i j Vj i V j i P i + Vj S i i L j L j Si N i V k 0 = 0 (k = 1,, 4) 5

29 ( ) 90 1 2 2 2 1 3 1 4 5 1 4 3 3 4 2 1 4 5 6 3 7 8 9

f(x) = e x2 25 d f(x) 0 x d2 dx f(x) 0 x dx2 f(x) (1 + ax 2 ) 2 lim x 0 x 4 a 3 2 a g(x) = 1 + ax 2 f(x) g(x) 1/2 f(x)dx 11 0 24 n n A f(x) = Ax (x R n ) 25 A 2 = A A 2 A 2 = A t A = A A t A A 3 n = 2 A 2 = O A f Ker(f) m(f) 1

n (n > 2) x 1, x 2,, x n G A A k (i, j) (A k ) ij 3 G G 50 G A x l x l 3 A = 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 2 G A 3 (A) ij (A 2 ) ij 0 i, j G 2

3 3 3 50 2 3

4 C C i, N i, P i, S i 1 4 50 C = 10 i 1 2 3 4 N i FUN 1 FUN 2 FUN 3 FUN 4 P i 8 9 12 15 S i 3 5 7 8 2 C ( ) C (j = 1, 2,, C) j (V 4 j ) 4 4 (V i j ) j = 1, 2,, 10 i = 1,, 4 2 2 j 1 2 3 4 5 6 7 8 9 10 Vj 1 0 Vj 2 0 Vj 3 0 Vj 4 0 4

2 3 4 i = 1 N 1 j(= 1, 2,,C) (V 1 j ) L j ( ) V 1 j 0 L j i = 2 N 1 N 2 j(= 1, 2,, C) (V 2 j ) L j 2 i = 3 N 1 N 2 N 3 j(= 1, 2,, C) (V 3 j ) L j 2 i = 4 N 1 N 2 N 3 N 4 j(= 1, 2,, C) (V 4 j ) L j 2 C 2 Vj i = Vj i 1 j j S i 0 P i + V i j S i j Vj i V j i P i + Vj S i i L j L j Si N i V k 0 = 0 (k = 1,, 4) 5