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- きみつぐ いさし
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2 1. 2/ 121
3 a x = a t 3/ 121
4 a x = a t 4/ 121
5 a > 0 t a t = a t t {}}{ a a a t 5/ 121
6 a t+s = = t+s {}}{ a a a t s {}}{{}}{ a a a a = a t a s (a t ) s = s {}}{ a t a t = a ts 6/ 121
7 a > 0 t a 0 t t = 0 + t a t = a 0+t = a 0 a t a t 0 a 0 = 1 7/ 121
8 t 0 = t + ( t) 1 = a 0 = a t+( t) = a t a t a t 0 a t = 1 a t 8/ 121
9 a = = = = = = / 121
10 t = m n m n m = m n n ( ) a m = a m n n = a m n n n a m n = n a m 10/ 121
11 a = = = 3 3 = = = = / 121
12 t a = 3 t (1) (2) (3) 12/ 121
13 x = 3 t x 3 1 = = = O t 13/ 121
14 x = 3 t x 3 1 = = = O t 14/ 121
15 x = 3 t 3 0 = 1 x 3 1 = = = O t 15/ 121
16 x = 3 t 3 0 = 1 x 3 1 = = = O t 16/ 121
17 x = 3 t = 3 x = O 3 t 17/ 121
18 x = 3 t = 3 x = O 3 t 18/ 121
19 x = 3 t 20 x t O t 19/ 121
20 x x = a t x = a t a t+s = a t a s 1 (a t ) s = a ts O t 20/ 121
21 2. 21/ 121
22 t x x t 22/ 121
23 x x = f(t) t = f(t 2) f(t 1 ) t 2 t 1 t 1 = t t 2 = t + h x t = f(t + h) f(t) (t + h) t = f(t + h) f(t) h 23/ 121
24 x = f(t) = a t x t = f(t + h) f(t) h = at+h a t = at a h a t h h = at (a h 1) = a t ah 1 h h 24/ 121
25 a t ah 1 h ah 1 h h 0 x = a t t = 0 a h x = a h 1 1 x O x = a t t = h 25/ 121 h t
26 a t ah 1 h ah 1 h h 0 x = a t t = 0 x x = a t a h 1 x = a h 1 t = h O h 26/ 121 t
27 a t ah 1 h ah 1 h h 0 x = a t t = 0 1 x Oh x = a t 27/ 121 t
28 x x = a t a h 1 h 1 O a h 1 t = 0 h t 28/ 121
29 x x = a t a h 1 h 1 O a h 1 t = 0 h t 29/ 121
30 x x = a t a h 1 h 1 O a h 1 t = 0 h t 30/ 121
31 x x = a t 1 O a h 1 t = 0 h t = 0 t 31/ 121
32 x x = a t 1 O x = a t a h 1 t = 0 h t = 0 1 a t 32/ 121
33 x x = a t 1 O a h 1 t = 0 h a h 1 t = 0 1 h t a e 33/ 121
34 x x = e t 1 O e h 1 t = 0 = 1 h x = e t t = 0 1 e t 34/ 121
35 x = e t x = e t x t = et+h e t h = et (e h 1) h = e t eh 1 h e t 1 = e t = x 35/ 121
36 x = e t x = e t x t = et+h e t h = et (e h 1) h = e t eh 1 h e t 1 = e t = x 36/ 121
37 x = e t x = e t dx dt = et = x 37/ 121
38 John Napier e = e (Euler) e 38/ 121
39 John Napier e = e (Euler) e 39/ 121
40 3. 40/ 121
41 x = e t x = e t +C 41/ 121
42 x = e t e t e t e t + C e t 42/ 121
43 x = e t e t e t e t + C e t 43/ 121
44 x = e t x = e t e t e t e t e t + C 44/ 121
45 x = e t x = e 2t x x = e 2t x = e 2t t = O x = e t 1 t 45/ 121
46 x = e t x x = e 2t x = e t t = 0 1 e 1 x = e t 1 O t 46/ 121
47 t = 0 x x = e 2t x = e 2t 1 x = e t 1 O t 47/ 121
48 t = 0 x x = e 2t x = e 2t 1 x = e t 1 O t 48/ 121
49 x = e kt ke kt 49/ 121
50 x = e 2t k = 2 ke kt 2e 2t x = e 2t t = 0 2e 2t t = / 121
51 x = e kt e kt ke kt e kt 1 k ekt + C 51/ 121
52 4. 52/ 121
53 (1) = 3 5+( 3) = 3 2 = 9 a t a s = a t+s (2) ( 2 3) 2 = 2 3 ( 2) = 2 6 = 64 (a t ) s = a t s 53/ 121
54 (1) = 3 5+( 3) = 3 2 = 9 a t a s = a t+s (2) ( 2 3) 2 = 2 3 ( 2) = 2 6 = 64 (a t ) s = a t s 54/ 121
55 (1) = 3 5+( 3) = 3 2 = 9 a t a s = a t+s (2) ( 2 3) 2 = 2 3 ( 2) = 2 6 = 64 (a t ) s = a t s 55/ 121
56 (1) = 3 5+( 3) = 3 2 = 9 a t a s = a t+s (2) ( 2 3) 2 = 2 3 ( 2) = 2 6 = 64 (a t ) s = a t s 56/ 121
57 (1) = 3 5+( 3) = 3 2 = 9 a t a s = a t+s (2) ( 2 3) 2 = 2 3 ( 2) = 2 6 = 64 (a t ) s = a t s 57/ 121
58 (1) e 4t 4e 4t e kt ke kt (2) e kt + at 2 + bt + c ke kt + 2at + b (3) e 8t 1 8 e8t + C e kt 1 k ekt + C 58/ 121
59 (1) e 4t 4e 4t e kt ke kt (2) e kt + at 2 + bt + c ke kt + 2at + b (3) e 8t 1 8 e8t + C e kt 1 k ekt + C 59/ 121
60 (1) e 4t 4e 4t e kt ke kt (2) e kt + at 2 + bt + c ke kt + 2at + b (3) e 8t 1 8 e8t + C e kt 1 k ekt + C 60/ 121
61 (1) e 4t 4e 4t e kt ke kt (2) e kt + at 2 + bt + c ke kt + 2at + b (3) e 8t 1 8 e8t + C e kt 1 k ekt + C 61/ 121
62 (1) e 4t 4e 4t e kt ke kt (2) e kt + at 2 + bt + c ke kt + 2at + b (3) e 8t 1 8 e8t + C e kt 1 k ekt + C 62/ 121
63 (1) e 4t 4e 4t e kt ke kt (2) e kt + at 2 + bt + c ke kt + 2at + b (3) e 8t 1 8 e8t + C e kt 1 k ekt + C 63/ 121
64 a > 0, b > 0 (1) ( a 2 a 3) ( ) 4 = a 20 (5) a b 0 = a 3 (2) = 9 (3) = 5 (4) a a 3 = a 6 (6) = 2 ( ) (7) a b 6 = a 4 b 5 (8) ( ) = / 121
65 a > 0, b > 0 (1) ( a 2 a 3) ( ) 4 = a 20 (5) a b 0 = a 3 (2) = 9 (3) = 5 (4) a a 3 = a 6 (6) = 2 ( ) (7) a b 6 = a 4 b 5 (8) ( ) = / 121
66 (1) e kt ke kt (2) e 3t + 2t + 4 3e 3t + 2 (3) e 5t 1 5 e5t + C (4) e kt 1 k e kt + C 66/ 121
67 (1) e kt ke kt (2) e 3t + 2t + 4 3e 3t + 2 (3) e 5t 1 5 e5t + C (4) e kt 1 k e kt + C 67/ 121
68 5. 68/ 121
69 x = e t t x x x = e t x t x 1 O t t 69/ 121
70 x > 0 x t x x = e t x x > 0 x t 1 O t t 70/ 121
71 x > 0 x x = e t t x t x x = e t x 1 O t t x t = log x 71/ 121
72 t = log x x = e t t = log x = log e t x = e t = e log x 72/ 121
73 t = log x = log e t t = 1 e 1 = e log e = 1 t = 0 e 0 = 1 log 1 = 0 73/ 121
74 74/ 121
75 t = log x s = log y x = e t y = e s u = log xy xy = e u e u = xy = e t e s = e s+t u = s + t log xy = log x + log y 75/ 121
76 t = log x x = e t s = log x m x m = e s e s = x m = (e t ) m = e m t s = m t log x m = m log x 76/ 121
77 log e = 1 log 1 = 0 log xy = log x + log y log x m = m log x 77/ 121
78 6. 78/ 121
79 79/ 121
80 dt dx = 1 dx dt 80/ 121
81 x = e t dx dt = et = x t = log x dt dx = 1 dx dt = 1 e t = 1 x 81/ 121
82 x = e t dx dt = et = x t = log x dt dx = 1 dx dt = 1 e t = 1 x 82/ 121
83 x = e t dx dt = et = x t = log x dt dx = 1 dx dt = 1 e t = 1 x 83/ 121
84 log x 1 x log x + C 1 x 84/ 121
85 log x 1 x log x + C 1 x 85/ 121
86 1 x log x 1 x 1 x log x + C 86/ 121
87 7. 87/ 121
88 log 2 = a, log 3 = b a, b (1) log 24 = log = log 2 3 +log 3 = 3 log 2+log 3 = 3a+b (2) log 1 4 = log = log 2 2 = 2 log 2 = 2a (3) log 2e 3 = log 2 + log e 3 = a / 121
89 log 2 = a, log 3 = b a, b (1) log 24 = log = log 2 3 +log 3 = 3 log 2+log 3 = 3a+b (2) log 1 4 = log = log 2 2 = 2 log 2 = 2a (3) log 2e 3 = log 2 + log e 3 = a / 121
90 log 2 = a, log 3 = b a, b (1) log 24 = log = log 2 3 +log 3 = 3 log 2+log 3 = 3a+b (2) log 1 4 = log = log 2 2 = 2 log 2 = 2a (3) log 2e 3 = log 2 + log e 3 = a / 121
91 log 2 = a, log 3 = b a, b (1) log 24 = log = log 2 3 +log 3 = 3 log 2+log 3 = 3a+b (2) log 1 4 = log = log 2 2 = 2 log 2 = 2a (3) log 2e 3 = log 2 + log e 3 = a / 121
92 (1) log x = 3t x = e 3t 1 (2) log 3x log x x 1 log 3x = log x + log 3 x (3) 3 x 1 x log x + C 3 1 x 3 log x + C = log x3 + C 92/ 121
93 (1) log x = 3t x = e 3t 1 (2) log 3x log x x 1 log 3x = log x + log 3 x (3) 3 x 1 x log x + C 3 1 x 3 log x + C = log x3 + C 93/ 121
94 (1) log x = 3t x = e 3t 1 (2) log 3x log x x 1 log 3x = log x + log 3 x (3) 3 x 1 x log x + C 3 1 x 3 log x + C = log x3 + C 94/ 121
95 (1) log x = 3t x = e 3t 1 (2) log 3x log x x 1 log 3x = log x + log 3 x (3) 3 x 1 x log x + C 3 1 x 3 log x + C = log x3 + C 95/ 121
96 (1) log x = 3t x = e 3t 1 (2) log 3x log x x 1 log 3x = log x + log 3 x (3) 3 x 1 x log x + C 3 1 x 3 log x + C = log x3 + C 96/ 121
97 (1) log x = 3t x = e 3t 1 (2) log 3x log x x 1 log 3x = log x + log 3 x (3) 3 x 1 x log x + C 3 1 x 3 log x + C = log x3 + C 97/ 121
98 log 2 = a, log 3 = b (1) log 144 = 4a + 2b (2) log 1 6 = a b (1) log x 2 = 4t x = e 2t (2) log x 2 2 x (3) k x k log x + C = log xk + C 98/ 121
99 log 2 = a, log 3 = b (1) log 144 = 4a + 2b (2) log 1 6 = a b (1) log x 2 = 4t x = e 2t (2) log x 2 2 x (3) k x k log x + C = log xk + C 99/ 121
100 8. 100/ 121
101 dx dt = kx 101/ 121
102 dx dt = kx 1 x t dx x = kdt 102/ 121
103 dx dt = kx 2 1 x dx = ( k) dt 1 dx x = kdt log x = kt + C 0 C 0 103/ 121
104 dx dt = kx 3 x = x = e kt+c 0 = e C 0 e kt 1 2 dx x = kdt 1 x dx = ( k) dt = Ce kt log x = kt + C 0 e C 0 = C 104/ 121
105 dx dt = kx x = Ce kt 1 dx x = kdt 2 1 x dx = ( k) dt log x = kt + C 0 C 3 x = e kt+c 0 = e C 0 e kt = Ce kt 105/ 121
106 dx dt = kx x = Ce kt 1 dx x = kdt 2 1 x dx = ( k) dt log x = kt + C 0 C 3 x = e kt+c 0 = e C 0 e kt = Ce kt 106/ 121
107 dx dt = kx + g 107/ 121
108 dx dt = kx + g 1 dx dt = k ( x g k ) 108/ 121
109 dx dt = kx + g 2 1 ( dx dt = k x g ) k X = x g k g k 0 109/ 121
110 dx dt = dx dt 1 ( dx dt = k x g ) k 2 dx dt = dx dt = kx X = x g k 110/ 121
111 dx dt = kx + g 3 dx = kx dt X = Ce kt X = x g k x = Ce kt + g k 1 dx dt = k ( x g ) k 2 X = x g k dx dt = dx dt = kx 111/ 121
112 dx dt = kx+g x = Ce kt + g k dx dt = k ( x g ) k X = x g k dx = dx dt dt dx = kx dt X = Ce kt x = Ce kt + g k C 112/ 121
113 dx dt = kx+g x = Ce kt + g k dx dt = k ( x g ) k X = x g k dx = dx dt dt dx = kx dt X = Ce kt x = Ce kt + g k C 113/ 121
114 dv dt = k m v + g t = 0 v = 0 v m g k 114/ 121
115 dv dt = k ( v mg ) m k V = v mg k dv = dv dt dt 115/ 121
116 dv dt = k m V 1 V dv = k m dt 116/ 121
117 1 V dv = ( k ) dt m log V = k m t + C 0 C 0 117/ 121
118 V = e k m t+c 0 = e C 0 e k m t C = e C 0 V = Ce k m t V = v mg k 118/ 121
119 dv dt = k m v + g v = Ce k m t + mg k 119/ 121
120 t = 0 v = 0 v = mg k e 0 = Ce 0 + mg k C = mg k k m t + mg k = mg k (1 e k m t ) 120/ 121
121 v mg k v = mg k (1 e k m t ) O t t v mg k 121/ 121
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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 +
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
< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3)
< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3) 6 y = g(x) x = 1 g( 1) = 2 ( 1) 3 = 2 ; g 0 ( 1) =
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(1) / 130 g 25 g 520% 170 g 30 g 560% 70 mg 600 mg 11.6% 0 10.5 mg 0% (1) (2) / 50100 g 25 g 200400% 50100 g 30 g 167333% 5001000 mg 600 mg 83167% 1020 mg 10.5 mg 95190% (2) / (1) 45.6 g 30 g 152% (2)
2.2 ( 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
(u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b )
m(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
v_-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)
( ) 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 +
(.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d
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1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
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 =
, 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,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 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,
1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0
A c 2008 by Kuniaki Nakamitsu 1 1.1 t 2 sin t, cos t t ft t t vt t xt t + t xt + t xt + t xt t vt = xt + t xt t t t vt xt + t xt vt = lim t 0 t lim t 0 t 0 vt = dxt ft dft dft ft + t ft = lim t 0 t 1.1
EPSON エプソンプリンタ共通 取扱説明書 ネットワーク編
K L N K N N N N N N N N N N N N L A B C N N N A AB B C L D N N N N N L N N N A L B N N A B C N L N N N N L N A B C D N N A L N A L B C D N L N A L N B C N N D E F N K G H N A B C A L N N N N D D
ありがとうございました
- 1 - - 2 - - 3 - - 4 - - 5 - 1 2 AB C A B C - 6 - - 7 - - 8 - 10 1 3 1 10 400 8 9-9 - 2600 1 119 26.44 63 50 15 325.37 131.99 457.36-10 - 5 977 1688 1805 200 7 80-11 - - 12 - - 13 - - 14 - 2-1 - 15 -
EPSON エプソンプリンタ共通 取扱説明書 ネットワーク編
K L N K N N N N N N N N N N N N L A B C N N N A AB B C L D N N N N N L N N N A L B N N A B C N L N N N N L N A B C D N N A L N A L B C D N L N A L N B C N N D E F N K G H N A B C A L N N N N D D
公務員人件費のシミュレーション分析
47 50 (a) (b) (c) (7) 11 10 2018 20 2028 16 17 18 19 20 21 22 20 90.1 9.9 20 87.2 12.8 2018 10 17 6.916.0 7.87.4 40.511.6 23 0.0% 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2.0% 4.0% 6.0% 8.0%
Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4 A B (A/B) 1 1,185 17,801 6.66% 2 943 26,598 3.55% 3 3,779 112,231 3.37% 4 8,174 246,350 3.32% 5 671 22,775 2.95% 6 2,606 89,705 2.91% 7 738 25,700 2.87% 8 1,134
橡hashik-f.PDF
1 1 1 11 12 13 2 2 21 22 3 3 3 4 4 8 22 10 23 10 11 11 24 12 12 13 25 14 15 16 18 19 20 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 144 142 140 140 29.7 70.0 0.7 22.1 16.4 13.6 9.3 5.0 2.9 0.0
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197 198 199 200 201 202 A B C D E F G H I J K L 203 204 205 A B 206 A B C D E F 207 208 209 210 211 212 213 214 215 A B 216 217 218 219 220 221 222 223 224 225 226 227 228 229 A B C D 230 231 232 233 A
ネットショップ・オーナー２ ユーザーマニュアル
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1
1 2 3 4 5 (2,433 ) 4,026 2710 243.3 2728 402.6 6 402.6 402.6 243.3 7 8 20.5 11.5 1.51 0.50.5 1.5 9 10 11 12 13 100 99 4 97 14 A AB A 12 14.615/100 1.096/1000 B B 1.096/1000 300 A1.5 B1.25 24 4,182,500
05[ ]戸田(責)村.indd
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147 2 62 4 3.2.1.16 3.2.1.17 148 63 1 3.2.1.F 3.2.1.H 3.1.1.77 1.5.13 1 3.1.1.05 2 3 4 3.2.1.20 3.2.1.22 3.2.1.24 3.2.1.D 3.2.1.E 3.2.1.18 3.2.1.19 2 149 3.2.1.23 3.2.1.G 3.1.1.77 3.2.1.16 570 565 1 2
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1. 2018/9/ ) 1) 8 9) 2) 6 14) + 14 ) 1 4 8a 8b) 2 a + b) 4) 2 : 7 = x 8) : x ) x ) + 1 2 ) + 2 6) x + 1)x + ) 15 2. 2018/9/ ) 1) 1 2 2) 4) 2 + 6 5) ) 2x + y 42x + y + 1) 4) : 6 = x 5) : x 2) x 2 15 12
Gmech08.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
a b b / / - CD 68
CD 67 a b b / / - CD 68 CD A B C D A B 69 a b A a b c d e f g h i j 70 a a - a b c a b 71 a b c / d e f g h i j k 72 a - a b a c / - d e f c a b a 73 a / b c a b c d e f g b 74 a a b / b a b c a 75 b a
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
![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](/thumbs/95/125540821.jpg "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")
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 =
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A大扉・騒音振動.qxd
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C:/KENAR/0p1.dvi
2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2 54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N 2{3. 55 (b) x
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limit&derivative
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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,
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)
D d dx 1 1.1 n d n y a 0 dx n + a d n 1 y 1 dx n 1 +... + a dy n 1 dx + a ny = f(x)...(1) dk y dx k = y (k) a 0 y (n) + a 1 y (n 1) +... + a n 1 y + a n y = f(x)...(2) (2) (2) f(x) 0 a 0 y (n) + a 1 y
[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
![[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](/thumbs/101/152334931.jpg "[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 [.]
2 p T, Q
270 C, 6000 C, 2 p T, Q p: : p = N/ m 2 N/ m 2 Pa : pdv p S F Q 1 g 1 1 g 1 14.5 C 15.5 1 1 cal = 4.1855 J du = Q pdv U ( ) Q pdv 2 : z = f(x, y). z = f(x, y) (x 0, y 0 ) y y = y 0 z = f(x, y 0 ) x x =
A A = a 41 a 42 a 43 a 44 A (7) 1 (3) A = M 12 = = a 41 (8) a 41 a 43 a 44 (3) n n A, B a i AB = A B ii aa
1 2 21 2 2 [ ] a 11 a 12 A = a 21 a 22 (1) A = a 11 a 22 a 12 a 21 (2) 3 3 n n A A = n ( 1) i+j a ij M ij i =1 n (3) j=1 M ij A i j (n 1) (n 1) 2-1 3 3 A A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33
Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)
![Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)](/thumbs/94/118373946.jpg "Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)")
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I 1
I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
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II(c) 1 October. 21, 2009 1 CS53 yamamoto@cs.kobe-u.ac.jp 3 1 7 1.1 : : : : : : : : : : : : : : : : : : : : : : 7 1.2 : : : : : : : : : : : : : : : : 8 1.2.1 : : : : : : : : : : : : : : : : : : : 8 1.2.2
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9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)
KENZOU
KENZOU 2008 8 2 3 2 3 2 2 4 2 4............................................... 2 4.2............................... 3 4.2........................................... 4 4.3..............................
2. 2 I,II,III) 2 x expx) = lim + x 3) ) expx) e x 3) x. ) {a } a a 2 a 3...) a b b {a } α : lim a = α b) ) [] 2 ) f x) = + x ) 4) x > 0 {f x)} x > 0,
![2. 2 I,II,III) 2 x expx) = lim + x 3) ) expx) e x 3) x. ) {a } a a 2 a 3...) a b b {a } α : lim a = α b) ) [] 2 ) f x) = + x ) 4) x > 0 {f x)} x > 0,](/thumbs/91/107555276.jpg "2. 2 I,II,III) 2 x expx) = lim + x 3) ) expx) e x 3) x. ) {a } a a 2 a 3...) a b b {a } α : lim a = α b) ) [] 2 ) f x) = + x ) 4) x > 0 {f x)} x > 0,")
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master.dvi
4 Maxwell- Boltzmann N 1 4.1 T R R 5 R (Heat Reservor) S E R 20 E 4.2 E E R E t = E + E R E R Ω R (E R ) S R (E R ) Ω R (E R ) = exp[s R (E R )/k] E, E E, E E t E E t E exps R (E t E) exp S R (E t E )
a a apier sin 0; 000; 000 = 0 7 sin 0 0; 000; 000 a = 0 7 ;r = 0: = 0 7 a n =0 7 ( 0 7 ) n n =0; ; 2; 3; n =0; ; 2; 3; ; 00 a n+ =0 7 ( 0 7 ) n
apier John apier(550-67) 0 2 3 4 5 6 7 8 9 0 2 4 8 6 32 64 28 256 52 024 4 32 = 28 2+5=7 2 n n 2 n 2 m n + m a 0 ;a ;a 2 ;a 3 ; a = a 0 ; r = a =a 0 = a 2 =a = a 3 =a 2 = n a n a n = ar n a r 2 a m = ar
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1 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
6. Euler x
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