数学演習:微分方程式
|
|
- よしじろう こしの
- 5 years ago
- Views:
Transcription
1 ( ) 1 / 21
2 ( ) 2 / 21
3 x(t)? ẋ + 5x = 0 ( ) 3 / 21
4 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 ( ) 3 / 21
5 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 x(t) = sin 5t? ẋ = 5 cos 5t, ẋ + 5x = 5 cos 5t + 5 sin 5t 0 ( ) 3 / 21
6 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 x(t) = sin 5t? ẋ = 5 cos 5t, ẋ + 5x = 5 cos 5t + 5 sin 5t 0 x(t) = e 5t? ẋ = 5e 5t, ẋ + 5x = 5e 5t + 5e 5t = 0 ( ) 3 / 21
7 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 x(t) = sin 5t? ẋ = 5 cos 5t, ẋ + 5x = 5 cos 5t + 5 sin 5t 0 x(t) = e 5t? ẋ = 5e 5t, ẋ + 5x = 5e 5t + 5e 5t = 0 x(t) = 3e 5t? ẋ = 15e 5t, ẋ + 5x = 15e 5t + 15e 5t = 0 ( ) 3 / 21
8 x(t)? ẋ + 5x = 0 x(t) = Ce 5t (C ) ( ) 4 / 21
9 x(t)? ẋ + 5x = 0 x(0) = 2 ( ) 5 / 21
10 x(t)? ẋ + 5x = 0 x(0) = 2 x(t) = Ce 5t x(0) = Ce 5 0 = C = 2 x(t) = 2e 5t ( ) 5 / 21
11 x(t)? ẋ + 5x = 0 ( ) 6 / 21
12 ẋ + 5x = 0 x(t)? x(t) = e λt ẋ = λe λt λe λt + 5e λt = 0 λ + 5 = 0 λ = 5 x(t) = e 5t ( ) 6 / 21
13 ẋ + 5x = 0 x(t)? x(t) = e λt ẋ = λe λt λe λt + 5e λt = 0 λ + 5 = 0 λ = 5 x(t) = e 5t ( ) 6 / 21
14 x(t)? ẍ + 4ẋ + 3x = 0 ( ) 7 / 21
15 ẍ + 4ẋ + 3x = 0 x(t)? x(t) = e λt ẋ = λe λt ẍ = λλe λt = λ 2 e λt λ 2 e λt + 4λe λt + 3e λt = 0 λ 2 + 4λ + 3 = 0 λ = 3, 1 x(t) = e 3t x(t) = e t ( ) 7 / 21
16 ẍ + 4ẋ + 3x = 0 x(t)? x(t) = e λt ẋ = λe λt ẍ = λλe λt = λ 2 e λt λ 2 e λt + 4λe λt + 3e λt = 0 λ 2 + 4λ + 3 = 0 λ = 3, 1 x(t) = e 3t x(t) = e t ( ) 7 / 21
17 ẍ + 4ẋ + 3x = 0 x(t)? x(t) = C 1 e 3t x(t) = C 2 e t x(t) = C 1 e 3t + C 2 e t ẋ = 3C 1 e 3t C 2 e t, ẍ = 9C 1 e 3t + C 2 e t ẍ + 4ẋ + 3x = (9C 1 e 3t + C 2 e t ) + 4( 3C 1 e 3t C 2 e t ) + 3(C 1 e 3t + C 2 e t ) = C 1 ( )e 3t + C 2 ( )e t = 0 ( ) 8 / 21
18 x(t)? ẍ + 4ẋ + 3x = 0 x(t) = C 1 e 3t + C 2 e t (C 1, C 2 ) ( ) 9 / 21
19 ẍ + 4ẋ + 3x = 0 x(0) = 6, ẋ(0) = 2 x(t)? ( ) 10 / 21
20 ẍ + 4ẋ + 3x = 0 x(t)? x(0) = 6, ẋ(0) = 2 x(t) = C 1 e 3t + C 2 e t x(0) = C 1 + C 2 = 6 ẋ(0) = 3C 1 C 2 = 2 C 1 = 2, C 2 = 8 x(t) = 2e 3t + 8e t ( ) 10 / 21
21 ẍ + 9x = 0 ( ) 11 / 21
22 ẍ + 9x = 0 x(t) = e λt ẋ = λe λt ẍ = λλe λt = λ 2 e λt λ 2 e λt + 9e λt = 0 λ = 0 λ = 3i, 3i x(t) = e 3it x(t) = e 3it x(t) = C 1 e 3it + C 2 e 3it ( ) 11 / 21
23 x(0) = 4, ẍ + 9x = 0 ẋ(0) = 6 ( ) 12 / 21
24 ẍ + 9x = 0 x(0) = 4, ẋ(0) = 6 x(0) = C 1 e 3i 0 + C 2 e 3i 0 = C 1 + C 2 = 4 ẋ(0) = (3i)C 1 e 3i 0 + ( 3i)C 2 e 3i 0 = 3i (C 1 C 2 ) = 6 C 1 = 2 + i, C 2 = 2 i x(t) = (2 + i)e 3it + (2 i)e 3it = (2 + i)(cos 3t + i sin 3t) + (2 i)(cos 3t i sin 3t) = 4 cos 3t 2 sin 3t ( ) 12 / 21
25 ẍ + 4ẋ + 13x = 0 ( ) 13 / 21
26 ẍ + 4ẋ + 13x = 0 x(t) = e λt ẋ = λe λt ẍ = λλe λt = λ 2 e λt λ 2 e λt + 4λe λt + 13e λt = 0 λ 2 + 4λ + 13 = 0 λ = 2 + 3i, 2 3i x(t) = e ( 2+3i)t x(t) = e ( 2 3i)t x(t) = C 1 e ( 2+3i)t + C 2 e ( 2 3i)t ( ) 13 / 21
27 x(0) = 4, ẍ + 4ẋ + 13x = 0 ẋ(0) = 2 ( ) 14 / 21
28 ẍ + 4ẋ + 13x = 0 x(0) = 4, ẋ(0) = 2 x(0) = C 1 + C 2 = 4 ẋ(0) = ( 2 + 3i)C 1 + ( 2 3i)C 2 = 2 C 1 = 2 i, C 2 = 2 + i x(t) = (2 i)e ( 2+3i)t + (2 + i)e ( 2 3i)t = e 2t {(2 i)(cos 3t + i sin 3t) + (2 + i)(cos 3t i sin 3t)} = e 2t (4 cos 3t + 2 sin 3t) ( ) 14 / 21
29 ẍ + 6ẋ + 9x = 0 ( ) 15 / 21
30 ẍ + 6ẋ + 9x = 0 x(t) = e λt λ 2 + 6λ + 9 = 0 λ = 3 ( ) x(t) = e 3t x(t) = Ce 3t ( ) 15 / 21
31 ẍ + 6ẋ + 9x = 0 C C(t) x(t) = C(t)e 3t ẋ = Ce 3t + C( 3)e 3t = ( C 3C)e 3t ẍ = ( C 3 C)e 3t + ( C 3C)( 3)e 3t = ( C 6 C + 9C)e 3t ( C 6 C + 9C) + 6( C 3C) + 9C = 0 C = 0 x(t) = te 3t C(t) = t ( ) 15 / 21
32 ẍ + 6ẋ + 9x = 0 x(t) = C 1 e 3t + C 2 te 3t (C 1, C 2 ) ( ) 15 / 21
33 1. (1) ẍ + 9ẋ + 20x = 0 x(0) = 1, ẋ(0) = 7 (2) ẍ + 2ẋ + 5x = 0 x(0) = 2, ẋ(0) = 2 ( π ) 2 (3) ẍ + x = 0 2 x(0) = 3, x(1) = 5 (4) ẋ + y = 0 ẏ x = 0 x(0) = 2, y(0) = 0 ( ) 16 / 21
34 2. (1) x(t) = 8e 7t (2) x(t) = 3e 4t, 5e 6t (3) x(t) = 5 sin 4t (4) x(t) = 2te t ( ) 17 / 21
35 ẍ + (5t)ẋ + 2e t x = 0 ẍ + 5ẋx + 3x = 0 ẍ + 5ẋx + 3x = 0 ẍ + (5t)ẋ + 2e t x = 0 ( ) 18 / 21
36 ( ) ẍ + 5ẋ + 2x = 0 ( ) ẍ + 5ẋ + 2x = 4 sin 6t ( ) 19 / 21
37 ẋ + 5x = 0 ẍ + 4ẋ + 3x = 0... x 6ẍ + 11ẋ 6x = 0 ẋ + 5x = 0 ẋ + 5x = 2e 3t ẍ + 5ẋ + 2x = 4 sin 6t ( ) 20 / 21
38 Ce λt C 1 e λ1t + C 2 e λ 2t λ 1, λ 2 ( ) 21 / 21
08-Note2-web
r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)
More information平成17年度 マスターセンター補助事業
- 1 - - 2 - - 3 - - 4 - - 5 - - 6 - - 7 - - 8 - - 9 - - 10 - - 11 - - 12 - - 13 - - 14 - - 15 - - 16 - - 17 - - 18 - - 19 - - 20 - - 21 - - 22 - - 23 - - 24 - - 25 - - 26 - - 27 - - 28 - IC IC - 29 - IT
More information1 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
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 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
More information重力方向に基づくコントローラの向き決定方法
( ) 2/Sep 09 1 ( ) ( ) 3 2 X w, Y w, Z w +X w = +Y w = +Z w = 1 X c, Y c, Z c X c, Y c, Z c X w, Y w, Z w Y c Z c X c 1: X c, Y c, Z c Kentaro Yamaguchi@bandainamcogames.co.jp 1 M M v 0, v 1, v 2 v 0 v
More information.. p.2/5
IV. p./5 .. p.2/5 .. 8 >< >: d dt y = a, y + a,2 y 2 + + a,n y n + f (t) d dt y 2 = a 2, y + a 2,2 y 2 + + a 2,n y n + f 2 (t). d dt y n = a n, y + a n,2 y 2 + + a n,n y n + f n (t) (a i,j ) p.2/5 .. 8
More informationx A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
More informationohp_06nov_tohoku.dvi
2006 11 28 1. (1) ẋ = ax = x(t) =Ce at C C>0 a0 x(t) 0(t )!! 1 0.8 0.6 0.4 0.2 2 4 6 8 10-0.2 (1) a =2 C =1 1. (1) τ>0 (2) ẋ(t) = ax(t τ) 4 2 2 4 6 8 10-2 -4 (2) a =2 τ =1!! 1. (2) A. (2)
More information2.2 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)
1 16 10 5 1 2 2.1 a a a 1 1 1 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) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a
More informationK E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................
More information1 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
( ) LPV( ) 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 m 1 m m 2, b 1 b b 2, k 1 k k 2 (2) [m b k ( ) k 0 b m ( ) 2 ẋ = Ax, x(0) 0 (3) (x(t) 0) ( ) V (x) V (x) = x T P x >
More informationFdData社会地理
[ [ 1(3 ) [ 2(3 ) A C [ [ [ 3(2 ) (1) X Y Z (2) X Y Z 3,000m [ 4(3 ) [ [ [ 5(2 ) ( ) 1 [ [ 6( ) (1) A (2) (1) B [ 7(3 ) (1) A (2) A (3) A 2 [ 8(2 ) [ 9(3 ) 2 [ 10(2 ) A H [ [ 11( ) A H 3 3 [ 12(2 ) [ (
More information消防力適正配置調査報告
8 5 5 20 11 22 4 25 1.1 1 1.2 1 1.3 2 2.1 6 2.2 6 2.3 8 2.4 8 2.5 9 3.1 10 3.2 10 3.3 13 4.1 15 4.2 17 4.3 19 4.4 21 4.5 23 (1) - 1 - (2) (1) ()1 ( ) 8 1 1 143 116 (2) 1-2 - 26 24 19 24 6 21 24 4 19 24
More informationMicrosoft Word - 01Ł\”ƒ.doc
226821,416* 13,226 22 62,640 46,289 13,226 28.6 * 8,030 4,788 408 13,226 2,249 2,868 55 5,173 2,153 716 93 2,962 3,628 1,204 260 5,092 173 10 361 25.5% 40 220 112 50.9% 4,922 804 16.3% 3040 141 54 38.3%
More information2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t) = x 0 + v 0 t 1 2 gt2 v(t) = v 0 gt t x = x 0 + v2 0 2g v2 2g 1.1 (x, v) θ
1 1 1.1 (Isaac Newton, 1642 1727) 1. : 2. ( ) F = ma 3. ; F a 2 t x(t) v(t) = x (t) v (t) = x (t) F 3 3 3 3 3 3 6 1 2 6 12 1 3 1 2 m 2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t)
More informationchap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
More information) 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)
4 4 ) 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) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
More information5. [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 { (
5 5.1 [ ] ) d f(t) + a d f(t) + bf(t) : f(t) 1 dt dt ) u(x, t) c u(x, t) : u(x, t) t x : ( ) ) 1 : y + ay, : y + ay + by : ( ) 1 ) : y + ay, : yy + ay 3 ( ): ( ) ) : y + ay, : y + ay b [],,, [ ] au xx
More information6 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
6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
More informationB B 10 7 581 10 8 582 10 9 583 B B 10 11 585 10 12 586 B 10 10 584 B
10 1 575 10 12 586 B B 10 1 575 10 2 576 B B 10 4 578 10 5 579 10 3 577 B 10 6 580 B B B 10 7 581 10 8 582 10 9 583 B B 10 11 585 10 12 586 B 10 10 584 B 11 1 587 11 12 598 B B 11 1 587 11 2 588 11 3 589
More information(資料2)第7回資料その1(ヒアリング概要)
2 3 4 5 6 7 8 9 10 11 12 13 1 1 1 1 5 1 6 533 4 505 722 13 3325 475 1 2 3 13 10 31 1 1 1 (1) 1 (2) 2 (3) 3 (4) 4 5 5 6 7 8 8 8 9 11 11 12 13 14 15 16 19 (1) (2) (3) (1) (5 ) 1 (10 ) ( ) (2) 2 4 (3) 3 3,100
More informationIT 180 181 1) 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 (a) (b) (c) (d) (e) (f) (a) (a) (b) 214 215 216 (a) (a) (a)
More information-------------------------- ----------------------------------------------------- -------------------------------------------------------------- ----------------------------------------------------- --------------------------------------------------------------
More information<4D F736F F D DEC8BC A95BD90AC E A982BA81698AB A B B4790DF90AB8EBE8AB FC89408A4F816A82CC93AE8CFC82C98AD682B782E9838C837C815B D
27 29 2 IT 1,234 1,447 2,130 1,200 3,043 4 3 75 75 70-74 -10 J00 J101 J110 J111 J118 J300 J302-304 J301 26,475,118 155,290,311 1,234 14,472,130 75,784,748 12,003,043 79,505,563 1 1.00% 0.62% 1.31% 9 12
More information, , ,210 9, ,
2006 5 642 7 2,671 35 732 1,727 602 489 386 74 373 533 305 1,210 9,786 2004 1,024 43.7 16.4 2004 978.6 40.2 2003 1 2006 5 1997 1998 1999 774 3,492 11 2,603 35 843 5,118 1,686 476 358 2000 738 3,534 11
More information( ) ( )
20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
More informationII 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
II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
More information曲面のパラメタ表示と接線ベクトル
L11(2011-07-06 Wed) :Time-stamp: 2011-07-06 Wed 13:08 JST hig 1,,. 2. http://hig3.net () (L11) 2011-07-06 Wed 1 / 18 ( ) 1 V = (xy2 ) x + (2y) y = y 2 + 2. 2 V = 4y., D V ds = 2 2 ( ) 4 x 2 4y dy dx =
More information() 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 ()
More informationGmech08.dvi
145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
More informationω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
More informationS 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
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
More information1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O
: 2014 4 10 1 2 2 3 2.1...................................... 3 2.2....................................... 4 2.3....................................... 4 2.4................................ 5 2.5 Free-Body
More informationLCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)
338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x
More information#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 =
#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/
More information( ) 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
More information1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0
: 2016 4 1 1 2 1.1......................................... 2 1.2................................... 2 2 2 2.1........................................ 2 2.2......................................... 3 2.3.........................................
More informationQMI_10.dvi
75 8 8.1 8.1.1 8.1 V 0 > 0 V ) = 0 < a) V 0 a a ) 0 a0 ft) ft) = e iωt ω = Ē h 8.2) ω 76 8 1 1 2 2m 1 k d
More informationz 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
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
More informationB
B YES NO 5 7 6 1 4 3 2 BB BB BB AA AA BB 510J B B A 510J B A A A A A A 510J B A 510J B A A A A A 510J M = σ Z Z = M σ AAA π T T = a ZP ZP = a AAA π B M + M 2 +T 2 M T Me = = 1 + 1 + 2 2 M σ Te = M 2 +T
More informationS 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.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....
More informationx x x 2, A 4 2 Ax.4 A A A A λ λ 4 λ 2 A λe λ λ2 5λ + 6 0,...λ 2, λ 2 3 E 0 E 0 p p Ap λp λ 2 p 4 2 p p 2 p { 4p 2 2p p + 2 p, p 2 λ {
K E N Z OU 2008 8. 4x 2x 2 2 2 x + x 2. x 2 2x 2, 2 2 d 2 x 2 2.2 2 3x 2... d 2 x 2 5 + 6x 0 2 2 d 2 x 2 + P t + P 2tx Qx x x, x 2 2 2 x 2 P 2 tx P tx 2 + Qx x, x 2. d x 4 2 x 2 x x 2.3 x x x 2, A 4 2
More information1. (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 (
1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +
More information215 11 13 1 2 1.1....................... 2 1.2.................... 2 1.3..................... 2 1.4...................... 3 1.5............... 3 1.6........................... 4 1.7.................. 4
More information1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................
More information知能科学:ニューラルネットワーク
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,
More information知能科学:ニューラルネットワーク
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,
More informationGmech08.dvi
63 6 6.1 6.1.1 v = v 0 =v 0x,v 0y, 0) t =0 x 0,y 0, 0) t x x 0 + v 0x t v x v 0x = y = y 0 + v 0y t, v = v y = v 0y 6.1) z 0 0 v z yv z zv y zv x xv z xv y yv x = 0 0 x 0 v 0y y 0 v 0x 6.) 6.) 6.1) 6.)
More information<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
単純適応制御 SAC 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. http://www.morikita.co.jp/books/mid/091961 このサンプルページの内容は, 初版 1 刷発行当時のものです. 1 2 3 4 5 9 10 12 14 15 A B F 6 8 11 13 E 7 C D URL http://www.morikita.co.jp/support
More information2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................
More informationr d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t
1 1 2 2 2r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t) V (x, t) I(x, t) V in x t 3 4 1 L R 2 C G L 0 R 0
More informationv_-3_+2_1.eps
I 9-9 (3) 9 9, x, x (t)+a(t)x (t)+b(t)x(t) = f(t) (9), a(t), b(t), f(t),,, f(t),, a(t), b(t),,, x (t)+ax (t)+bx(t) = (9),, x (t)+ax (t)+bx(t) = f(t) (93), b(t),, b(t) 9 x (t), x (t), x (t)+a(t)x (t)+b(t)x(t)
More informationhttp://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................
More informationUntitled
http://www.ike-dyn.ritsumei.ac.jp/ hyoo/dynamics.html 1 (i) (ii) 2 (i) (ii) (*) [1] 2 1 3 1.1................................ 3 1.2..................................... 3 2 4 2.1....................................
More informationm(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)
2.6 2.6.1 mẍ + γẋ + ω 0 x) = ee 2.118) e iωt Pω) = χω)e = ex = e2 Eω) m ω0 2 ω2 iωγ 2.119) Z N ϵω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j 2.120) Z ω ω j γ j f j f j f j sum j f j = Z 2.120 ω ω j, γ ϵω) ϵ
More information64 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
63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5
More informationii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx
i B5 7.8. p89 4. ψ x, tψx, t = ψ R x, t iψ I x, t ψ R x, t + iψ I x, t = ψ R x, t + ψ I x, t p 5.8 π π π F e ix + F e ix + F 3 e 3ix F e ix + F e ix + F 3 e 3ix dx πψ x πψx p39 7. AX = X A [ a b c d x
More information<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
新 Excel コンピュータシミュレーション 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. http://www.morikita.co.jp/books/mid/084871 このサンプルページの内容は, 初版 1 刷発行当時のものです. Microsoft Excel Excel Visual Basic Visual Basic 2007 Excel Excel
More information表1-表4_No78_念校.indd
mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm Fs = tan + tan. sin(1.5) tan sin. cos Fs ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
More informationp = mv p x > h/4π λ = h p m v Ψ 2 Ψ
II p = mv p x > h/4π λ = h p m v Ψ 2 Ψ Ψ Ψ 2 0 x P'(x) m d 2 x = mω 2 x = kx = F(x) dt 2 x = cos(ωt + φ) mω 2 = k ω = m k v = dx = -ωsin(ωt + φ) dt = d 2 x dt 2 0 y v θ P(x,y) θ = ωt + φ ν = ω [Hz] 2π
More information[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt
3.4.7 [.] =e j(t+/4), =5e j(t+/3), 3 =3e j(t+/6) ~ = ~ + ~ + ~ 3 = e j(t+φ) =(e 4 j +5e 3 j +3e 6 j )e jt = e jφ e jt cos φ =cos 4 +5cos 3 +3cos 6 =.69 sin φ =sin 4 +5sin 3 +3sin 6 =.9 =.69 +.9 =7.74 [.]
More informationc 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
c 28. 2, y 2, θ = cos θ y = sin θ 2 3, y, 3, θ, ϕ = sin θ cos ϕ 3 y = sin θ sin ϕ 4 = cos θ 5.2 2 e, e y 2 e, e θ e = cos θ e sin θ e θ 6 e y = sin θ e + cos θ e θ 7.3 sgn sgn = = { = + > 2 < 8.4 a b 2
More information( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )
n n (n) (n) (n) (n) n n ( n) n n n n n en1, en ( n) nen1 + nen nen1, nen ( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( ) ( n) Τ n n n ( n) n + n ( n) (n) n + n n n n n n n n
More informationZE_Œ{‘‚‡Ì„©Łû
87 5 7 1 2002 8 28 1 2 29 e r w 7 5 q Excel 20022000 2000 2002 2000 2002 STEP UP Check 2000 CONTENTS Excel 20022000 2000 5 1 15 1 1 2 5 7 18 ce 20 22 2 2 28 2002 0 2 8 9 10 11 12 1 1 1 2 8 0 2 2002 8 50
More information2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (
(. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2
More information1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
More information3 filename=quantum-3dim110705a.tex ,2 [1],[2],[3] [3] U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h
filename=quantum-dim110705a.tex 1 1. 1, [1],[],[]. 1980 []..1 U(x, y, z; t), p x ˆp x = h i x, p y ˆp y = h i y, p z ˆp z = h i z (.1) Ĥ ( ) Ĥ = h m x + y + + U(x, y, z; t) (.) z (U(x, y, z; t)) (U(x,
More informationB ver B
B ver. 2017.02.24 B Contents 1 11 1.1....................... 11 1.1.1............. 11 1.1.2.......................... 12 1.2............................. 14 1.2.1................ 14 1.2.2.......................
More informationp06.dvi
I 6 : 1 (1) u(t) y(t) : n m a n i y (i) = b m i u (i) i=0 i=0 t, y (i) y i (u )., a 0 0, b 0 0. : 2 (2), Laplace, (a 0 s n +a 1 s n 1 + +a n )Y(s) = (b 0 s m + b 1 s m 1 + +b m )U(s),, Y(s) U(s) = b 0s
More informationⅠ 調査研究の概要
1999 IP 2005 6 IT IC 2005 3 . 1-1. 1-2. 1-3.. 2-1. 2-2. 2-3. 2-4. 2-5. 2-6. 2-7... 1 2 3 1 4 5 6 TCA 7 8 9 10 11 12 13 14 15 16 17 MCF 2004 18 19 MCF 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
More information(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 {
7 4.., ], ], ydy, ], 3], y + y dy 3, ], ], + y + ydy 4, ], ], y ydy ydy y y ] 3 3 ] 3 y + y dy y + 3 y3 5 + 9 3 ] 3 + y + ydy 5 6 3 + 9 ] 3 73 6 y + y + y ] 3 + 3 + 3 3 + 3 + 3 ] 4 y y dy y ] 3 y3 83 3
More informationA(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
More information(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
More informationẍ = kx, (k > ) (.) x x(t) = A cos(ωt + α) (.). d/ = D. d dt x + k ( x = D + k ) ( ) ( ) k k x = D + i D i x =... ( ) k D + i x = or ( ) k D i x =.. k.
K E N Z OU 8 9 8. F = kx x 3 678 ẍ = kx, (k > ) (.) x x(t) = A cos(ωt + α) (.). d/ = D. d dt x + k ( x = D + k ) ( ) ( ) k k x = D + i D i x =... ( ) k D + i x = or ( ) k D i x =.. k. D = ±i dt = ±iωx,
More informationQMI_09.dvi
63 6 6.1 6.1.1 6.1 V 0 > 0 V ) = 0 < a) V 0 a a ) 0 a0 Ct) Ct) = e iωt ω = Ē h 6.2) ω 64 6 1 1 2 2m 1 k d
More informationA = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B
9 7 A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B x x B } B C y C y + x B y C x C C x C y B = A
More information211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
More information9 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
2009 9 6 16 7 1 7.1 1 1 1 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(cos y y sin y) y dy 1 sin
More informationChap9.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
More information23 7 28 i i 1 1 1.1................................... 2 1.2............................... 3 1.2.1.................................... 3 1.2.2............................... 4 1.2.3 SI..............................
More informationIA
IA 31 4 11 1 1 4 1.1 Planck.............................. 4 1. Bohr.................................... 5 1.3..................................... 6 8.1................................... 8....................................
More information1 c Koichi Suga, ISBN
c Koichi Suga, 4 4 6 5 ISBN 978-4-64-6445- 4 ( ) x(t) t u(t) t {u(t)} {x(t)} () T, (), (3), (4) max J = {u(t)} V (x, u)dt ẋ = f(x, u) x() = x x(t ) = x T (), x, u, t ẋ x t u u ẋ = f(x, u) x(t ) = x T x(t
More informationA (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
More informationK E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................
More information4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k
4.6 (E i = ε, ε + ) T Z F Z = e ε + e (ε+ ) = e ε ( + e ) F = kt log Z = kt loge ε ( + e ) = ε kt ln( + e ) (4.8) F (T ) S = T = k = k ln( + e ) + kt e + e kt 2 + e ln( + e ) + kt (4.20) /kt T 0 = /k (4.20)
More information数値計算:常微分方程式
( ) 1 / 82 1 2 3 4 5 6 ( ) 2 / 82 ( ) 3 / 82 C θ l y m O x mg λ ( ) 4 / 82 θ t C J = ml 2 C mgl sin θ θ C J θ = mgl sin θ = θ ( ) 5 / 82 ω = θ J ω = mgl sin θ ω J = ml 2 θ = ω, ω = g l sin θ = θ ω ( )
More information214 March 31, 214, Rev.2.1 4........................ 4........................ 5............................. 7............................... 7 1 8 1.1............................... 8 1.2.......................
More information213 March 25, 213, Rev.1.5 4........................ 4........................ 6 1 8 1.1............................... 8 1.2....................... 9 2 14 2.1..................... 14 2.2............................
More information振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
More informationGmech08.dvi
51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r
More information