学習内容と日常生活との関連性の研究－第２部－第４章－１
|
|
|
- かずし ひきぎ
- 5 years ago
- Views:
Transcription
1
2 69
3 V A V + A V A 2A
4 (1) (1996)35 7 pp (2) (1994) 71
5 ()
6 (1) (2) 73
7 meta- + PRD RT RT DJ boy tall girl dog cat book VT sees reads VI runs ADJ black difficult small ADV quickly skillfully ART a the PRD The boy sees a black dog. The cat runs quickly. A boy reads the difficult book skillfully. 74
8 2005Vol.75, No.1 (2005-4) pp
9 r0.1 a B = A( 1+ r) a a BA A 2 = (1 + r) a x 100r 2 (1 + x r = r) x r = {(1 + r) } a 1 r ( 1+ r) r e = e x x
10 2003pp pp pp
11 ( x, y) ( 0, c) A m/ θ x y t ( x, y) x = At cosθ, g 2 sinθ 2 y = gt + At sinθ + c y = x + x + c = ax + bx + c t ( Acosθ ) cosθ x y 2 ax + bx + c ( R, S) 2 R S = -ax + bx + c 1 ( b ± b 4 ( ) ) 2 + a c S 2a R 1 ( b b 4 ( ) ) 2 + a c S A 2 cos g 2 θ sinθ cosθ 2a sin θ 2g( c S) + A = 2 2 cos θ A cos θ + cosθ sinθ + g sin 2 2 2g( c S) θ + A R A θ c S 2 2 A cosθ 2sinθ A sin 2θ R = R A g g 18m, 30m, 50m, 60m, 70m, 90m () 78
12 1 2 2 b 2 b ax + bx + c = a( x ) + c + 2a 4a b A sin θ tanθ c + = c + c+ R 4a 2g 4 A 50m / A θ 79
13 80
14 81
15 82
16 83
17 3 3 = = 3 3 = 2 3 = 6 3, 2, 6, 4, 5, 1 p r r p 1 1mod p 5 S.K. K.W. a = 7 M = X i = 7 X i 1 mod (2 31 1) mod (2 31 1) IBM RANDU 1970 X i = 65539X i 1 mod 2 31 () 84
18 3 1 = = 9 2mod7 3 3 = 2 3 = = 6 3 = 18 4mod7 3 5 = 4 3 = 12 5mod7 3 6 = 5 3 = 15 1mod7 3, 2, 6, 4, 5, 1 5 S.K. K.W. a = 7 M = X i = 7 X i 1 mod (2 31 1) IBM RANDU 1970 X i = 65539X i 1 mod 2 31 (1) S. K. Park and K. W. Miller, 1988, Random number generators: Good ones are hard to find, Communication of the ACM, Vol.31, pp
19 2 2 C
20 D 87
21 x + 100y x + y s y = 2x
22 y 1 1 x + y = x + 100y = 1000 x 89
23 x ,15 k=30+15=45 90
24 91
25 () 92
26 93
27 x y = 2 94
28
29 96
30 JPEG cos 97
31 asin( ωt) asin( ωt) a ω φ t asin( ω t + φ) a' sin( ω t + φ' ) a' φ' asin( ωt + φ) a'sin( ωt + φ') = asin( ωt + φ) asin( ωt + φ') + ( a a')sin( ωt + φ') φ + φ' φ φ' = a cos( ωt + )sin( ) + ( a a')sin( ωt + φ') 2 2 φ = φ' a = a' φ = φ' a = a' 98
32 99
33 440Hz () 100
34 y = sin x y = 0.8x y = sin x+ sin 0.8x 1998 pp
35 B L L ( L+ 2 ) x x = L+ 2 = = log( x + 1) = log 2 = 81 log 2 = = < x + 1<
36 103
37 3 5 f(x) 3 5 f(x) 3 5f(x) 3 5 yf(x) a a 0 f ( an ) 2an 5 a a n = + f ( a ) 3 3a n 2 n a a2 = + = = = 1.75, a a 4 = = = = = = a 5 a 6 104
38 105
39 106
40 107
41 108
42 109
43 14 N 12 C 14 N 12 C 14 C 12 CO2 14 CO2 CO 2 12 CO2 14 CO 14 C 14 C 2 12 C 14 C t y() t t t+ dt y() t kk ( > 0) y( t+ dt) y( t) = ky( t) dt dt dt 0 y () t = ky() t y = Ce kt t = 0 A y = Ae kt kt 1 t 0 e 0 = t o 14 C
44 1998 pp
45 d dt N( t) α λ t 2 L = N( t) N( ) N t) = 1+ αt me ( 112
46 L N( t) = 1+ αt me L S t L N(t) L ,085, dn dt = b( 1 an( t)) N( t) dn dn adn ( 1 an ) N = bdt N + 1 an = bdt log N log(1 an) = bt C C bt log( 1 an ) / N = bt + C 1/ N a = e C bt bt N = 1/( a + e ) = L /(1 + me ) L = 1/ a, m 1 e a C = 113
47 2 2 = 4 (1)(3) = AB + AB AB A B A A B B Exclusive ORXOR () 114
48 (1) (2) (3) (4) (1) (2) (4) (3) A B A B (1) (2) A B A B (3) (4) (1)(1986)pp
49 a + b + c + d = 7 1 a, b, c, d 4 6! = (6 3)!3! 4 H 3 = C3 = 6C3 =
50 (2)(1985)JIS X
51 () 118
52 1 (1) 1993pp
53 50% 6 1mm % 52% 30 29% %60% 120
54 (1) (2) (1) (2) 121
55 / 122
56 2005Vol.75, No.1 (2005-4) pp
57 124
58 A A A A A A A A 2005Vol.75, No.1 (2005-4) pp
59 126
60 127
61 128 Internet Explorer
62 HTML This is our Home Page! / / Here we can write any massage or information contents. / /HTML 2005Vol.75, No.1 (2005-4) pp
63 syntactic role 130
64 2005Vol.75, No.1 (2005-4) pp
65 HP Home Page D1,D2,D3,,Dn D1,D2,D3,,Dn K1,K2K13K 1,K2K3Kn1,Kn2Kn3 Di F F Di F Ki1,Ki2,Ki3, F Di Ki1,Ki2,Ki3, F Di Ki1,Ki2,Ki3, F F -1 K Di K F -1 Di, Dj, Dk, F -1 F -1 F F inverted file 132
66 D1 K1 K2 K D2 K1 K2 K D3 K1 K2 K3 Dn Kn1 Kn2 Kn K1 D1 D2 D K2 D1 D2 D K3 D1 D2 D3 Kn Dn1 Dn2 Dn F -1 HP1 HP6 HP1 HP1 HP HP6 HP2 HP HP HP HP HP HP HP HP5 HP1 HP HP HP HP HP 2005Vol.75, No.1 (2005-4) pp
67 134
68 135
69 raw data run-length 136
70 or or or or or 0 or 0 or 0 or 0 or 0 or 0 or 0 or 0 or 2005Vol.75, No.1 (2005-4) pp
71 138
72 139
73
74 i i i i -1 i -1 i 2005Vol.75, No.1 (2005-4) pp
75 r= (%) =T ), S X S X ( 1+ r) S m m k X = (( ( T (1 + r) S)(1 + r) S)(1 + r) S) = T (1 + r) S(1 + r) m 1 k = 0 m k 1 (1 + r) S m m S m S(1 + r) = S = {(1 + r) 1} X = T ( 1+ r) {(1 + r) 1} 1 (1 + r) r r 1. S S = 2. m m= m 1 k = 0 X S X ( 1+ r) S Excel y, (1+r) S (1+r) S S () 142
76 (T) (T)(T)() (T)()(S) T(1+r)S T(1+r)SS T(1+r) 2 S(1+r)S m (T(1+r)S)(S) T(1+r) S() S *** T=, r=, S= Excel *** Excel A1 T B1 C1 S A2 A2 A3 A2 Excel A2 1 S A3 Excel A B C D =A1+B$1) C$1 =A2+B$1) C$1 =A+B$1) C$1 =A+B$1) C$1 =A+B$1) C$1 =A+B$1) C$
77
78 2005Vol.75, No.1 (2005-4) pp
() 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 ()
() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y
![() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y](/thumbs/85/92979039.jpg "() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y")
5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
S 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..............
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 (
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 +
S 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....
chap1.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
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
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
[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
![[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s](/thumbs/94/118579915.jpg "[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s")
[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
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
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
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................
December 28, 2018
e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................
A (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
高等学校学習指導要領解説 数学編
5 10 15 20 25 30 35 5 1 1 10 1 1 2 4 16 15 18 18 18 19 19 20 19 19 20 1 20 2 22 25 3 23 4 24 5 26 28 28 30 28 28 1 28 2 30 3 31 35 4 33 5 34 36 36 36 40 36 1 36 2 39 3 41 4 42 45 45 45 46 5 1 46 2 48 3
(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 )
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
4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 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 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0
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)
( ) 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
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)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
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
sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V
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
8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,
(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
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
SO(2)
TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6
http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg
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
: January 14, 28..,. 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, A. lim k A k = A. A k = (a (k) ij ) ij, A k = (a ij ) ij, i,
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
DVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
DE-resume
- 2011, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 21131 : 4 1 x y(x, y (x,y (x,,y (n, (1.1 F (x, y, y,y,,y (n =0. (1.1 n. (1.1 y(x. y(x (1.1. 1 1 1 1.1... 2 1.2... 9 1.3 1... 26 2 2 34 2.1,... 35 2.2
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
θ i ) AB θ ) A = B = sin θ = sin θ A B sin θ) ) < = θ < = Ax Bx = θ = sin θ ) abc θ sin 5θ = sin θ fsin θ) fx) = ax bx c ) cos 5 i sin 5 ) 5 ) αβ α iβ) 5 α 4 β α β β 5 ) a = b = c = ) fx) = 0 x x = x =
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
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 =
( ) 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
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) = 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)
Note.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
数学Ⅱ演習（足助・09夏）
II I 9/4/4 9/4/2 z C z z z z, z 2 z, w C zw z w 3 z, w C z + w z + w 4 t R t C t t t t t z z z 2 z C re z z + z z z, im z 2 2 3 z C e z + z + 2 z2 + 3! z3 + z!, I 4 x R e x cos x + sin x 2 z, w C e z+w
LCR 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
function2.pdf
2... 1 2009, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 38 : 5) i) [], : 84 85 86 87 88 89 1000 ) 13 22 33 56 92 147 140 120 100 80 60 40 20 1 2 3 4 5 7.1 7 7.1 1. *1 e = 2.7182 ) fx) e x, x R : 7.1)
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
) 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
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y
No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ
I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT
I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345
untitled
- k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =
No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
( ) ,
II 2007 4 0. 0 1 0 2 ( ) 0 3 1 2 3 4, - 5 6 7 1 1 1 1 1) 2) 3) 4) ( ) () H 2.79 10 10 He 2.72 10 9 C 1.01 10 7 N 3.13 10 6 O 2.38 10 7 Ne 3.44 10 6 Mg 1.076 10 6 Si 1 10 6 S 5.15 10 5 Ar 1.01 10 5 Fe 9.00
85 4
85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V
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
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
I
I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
Microsoft Word - 触ってみよう、Maximaに2.doc
i i e! ( x +1) 2 3 ( 2x + 3)! ( x + 1) 3 ( a + b) 5 2 2 2 2! 3! 5! 7 2 x! 3x! 1 = 0 ",! " >!!! # 2x + 4y = 30 "! x + y = 12 sin x lim x!0 x x n! # $ & 1 lim 1 + ('% " n 1 1 lim lim x!+0 x x"!0 x log x
IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t
![4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t](/thumbs/70/62705042.jpg "4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t")
1 1.1 sin 2π [rad] 3 ft 3 sin 2t π 4 3.1 2 1.1: sin θ 2.2 sin θ ft t t [sec] t sin 2t π 4 [rad] sin 3.1 3 sin θ θ t θ 2t π 4 3.2 3.1 3.4 3.4: 2.2: sin θ θ θ [rad] 2.3 0 [rad] 4 sin θ sin 2t π 4 sin 1 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 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
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 =
5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =
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
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
3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (
振動工学に基礎
Ky Words. ω. ω.3 osω snω.4 ω snω ω osω.5 .6 ω osω snω.7 ω ω ( sn( ω φ.7 ( ω os( ω φ.8 ω ( ω sn( ω φ.9 ω anφ / ω ω φ ω T ω T s π T π. ω Hz ω. T π π rad/s π ω π T. T ω φ 6. 6. 4. 4... -... -. -4. -4. -6.
all.dvi
38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t
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
1 (1) ( i ) 60 (ii) 75 (iii) 15 () ( i ) (ii) 4 (iii) 7 1 ( () r, AOB = θ 0 < θ < ) OAB A OB P ( AB ) < ( AP ) (4) 0 < θ < sin θ < θ < tan θ 0 x, 0 y (1) sin x = sin y (x, y) () cos x cos y (x, y) 1 c
( 9 1 ) 1 2 1.1................................... 2 1.2................................................. 3 1.3............................................... 4 1.4...........................................
D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y
5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x
, 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,
Gmech08.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
x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)
2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
3. ( 1 ) Linear Congruential Generator:LCG 6) (Mersenne Twister:MT ), L 1 ( 2 ) 4 4 G (i,j) < G > < G 2 > < G > 2 g (ij) i= L j= N
RMT 1 1 1 N L Q=L/N (RMT), RMT,,,., Box-Muller, 3.,. Testing Randomness by Means of RMT Formula Xin Yang, 1 Ryota Itoi 1 and Mieko Tanaka-Yamawaki 1 Random matrix theory derives, at the limit of both dimension
数学の基礎訓練I
I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
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 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/
1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.
1 1 n 0, 1, 2,, n 1 1.1 n 2 a, b a n b n a, b n a b (mod n) 1 1. n = 10 1567 237 (mod 10) 2. n = 9 1567 1826578 (mod 9) n II Z n := {0, 1, 2,, n 1} 1.2 a b a = bq + r (0 r < b) q, r q a b r 2 1. a = 456,
untitled
1 17 () BAC9ABC6ACB3 1 tan 6 = 3, cos 6 = AB=1 BC=2, AC= 3 2 A BC D 2 BDBD=BA 1 2 ABD BADBDA ABC6 BAD = (18 6 ) / 2 = 6 θ = 18 BAD = 12 () AD AD=BADCAD9 ABD ACD A 1 1 1 1 dsinαsinα = d 3 sin β 3 sin β
keisoku01.dvi
2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.
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
![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](/thumbs/85/91372642.jpg "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")
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
30 (11/04 )
30 (11/04 ) i, 1,, II I?,,,,,,,,, ( ),,, ϵ δ,,,,, (, ),,,,,, 5 : (1) ( ) () (,, ) (3) ( ) (4) (5) ( ) (1),, (),,, () (3), (),, (4), (1), (3), ( ), (5),,,,,,,, ii,,,,,,,, Richard P. Feynman, The best teaching
K 1 mk(
R&D ATN K 1 mk(0.01 0.05 = ( ) (ITS-90)-59.3467 961.78 (T.J.Seebeck) A(+ T 1 I T 0 B - T 1 T 0 E (Thermoelectromotive force) AB =d E(AB) /dt=a+bt----------------- E(AB) T1 = = + + E( AB) α AB a b ( T0
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
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
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
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 (
![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 (](/thumbs/100/143999768.jpg "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 (")
8 ) ) [ ] [ ) 8 5 5 II III A B ),,, 5, 6 II III A B ) ),,, 7, 8 II III A B ) [ ]),,, 5, 7 II III A B ) [ ] ) ) 7, 8, 9 II A B 9 ) ) 5, 7, 9 II B 9 ) A, ) B 6, ) l ) P, ) l A C ) ) C l l ) π < θ < π sin
.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
all.dvi
5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0
chap03.dvi
99 3 (Coriolis) cm m (free surface wave) 3.1 Φ 2.5 (2.25) Φ 100 3 r =(x, y, z) x y z F (x, y, z, t) =0 ( DF ) Dt = t + Φ F =0 onf =0. (3.1) n = F/ F (3.1) F n Φ = Φ n = 1 F F t Vn on F = 0 (3.2) Φ (3.1)
http://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....................
The Physics of Atmospheres CAPTER :
The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(
17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,
17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ
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.
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,
Chap10.dvi
=0. f = 2 +3 { 2 +3 0 2 f = 1 =0 { sin 0 3 f = 1 =0 2 sin 1 0 4 f = 0 =0 { 1 0 5 f = 0 =0 f 3 2 lim = lim 0 0 0 =0 =0. f 0 = 0. 2 =0. 3 4 f 1 lim 0 0 = lim 0 sin 2 cos 1 = lim 0 2 sin = lim =0 0 2 =0.
1 3 1.1.......................... 3 1............................... 3 1.3....................... 5 1.4.......................... 6 1.5........................ 7 8.1......................... 8..............................
1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
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
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
2000年度『数学展望 I』講義録
2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53
main.dvi
5 IIR IIR z 5.1 5.1.1 1. 2. IIR(Infinite Impulse Response) FIR(Finite Impulse Response) 3. 4. 5. 5.1.2 IIR FIR 5.1 5.1 5.2 104 5. IIR 5.1 IIR FIR IIR FIR H(z) = a 0 +a 1 z 1 +a 2 z 2 1+b 1 z 1 +b 2 z 2
高校生の就職への数学II
II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................
2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37