2009 : M (CG)
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- すずり おおはし
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1 2009 M
2 2009 : M (CG)
3 A A I
4 A A II
5 1 1.1 CG [1] [2] [3] [4] [5] CG [6] [7] [8] 1
6 : CG L*a*b [9] L*a*b 2
7
8 2 2.1 [10] [11][12] [13][14]
9 2.1: 2.2: 2 [13]
10 2.3: [15][16] [17] [18] 6
11 : 2.5(b) 2.5(a) 7
12 (a) 木目に対し平行な着色 (b) 木目に対し垂直な着色 図 2.5: 木材に対して着色の角度を変える 塗料が繊維方向に流れる距離には差がある 塗料が付着した量が多いところほ ど 滲みが起こる範囲は広くなり 量が少なければ滲みの起こる範囲は狭くなる また 2.1 節で述べたように 早材と晩材の材質の違いによっても差が生じる 早材 は滲みの範囲が狭く 晩材は滲みの範囲が広くなると言うのがその特徴である このことから 滲みが長く発生するのは 塗料が多い箇所であり 晩材部であ るとき また木目に対し垂直かそれに近い角度で境目が存在するときである 8
13 3 2 2 A A A 9
![3.1 2.1 2 2 2 [19] 2 (3.1) Y Y = (0.30 R + 0.59 G + 0.](/docs-images/92/111030507/images/14-1.jpg "11 B) 3 (3.1) R G B NTSC [20] 2 2 2 3.1 (a) (b) 2 3.")
14 [19] 2 (3.1) Y Y = (0.30 R G B) 3 (3.1) R G B NTSC [20] (a) (b) 2 3.1: 10
15 3.2(a) (a) 2 (b) 3.2: 2 (x,y) 8 (8 ) (x-1,y- 1),(x-1, y),(x-1,y+1),(x, y+1),(x+1,y+1),(x+1, y),(x+1,y-1),(x, y-1) x y y=x y=-x 8 3 (x,y) (x,y) 8 3 (x,y) 11
16 (x,y) 3.2(b) 3.2 A 2.3 A 1 A A 2 A 2 2 A 3.3(a) 2 A 3.3(b) (a) A (b) 2 3.3: A A 2 A 12
17 s t t s s s s s 13
18 3.3.2 P P (s t) P s t θ s t s x t y s t P s t s t s t 1 2s P x U U s y V V t P 2s 2s A P A P A P A U V (3.2) P A = αu + βv (3.2) U V α β P A A s t α β (3.3)(3.4) s 2 α s 2 t 2 β t 2 (3.3) (3.4) s t (3.3)(3.4) A 14
19 A n n U P 3.4 A A A P(x, y) P 8 (x+1, y-1),(x-1, y),(x-1, y+1),(x, y+1),(x+1,y+1), (x+1,y),(x+1,y- 1),(x,y-1) 8 5 (x+1, y-1),(x-1, y),(x-1, y+1),(x, y+1)(x,y-1) P y 90 P P 15
20 A 2 A R R U R R R R 2.3 R 3.1 A R R 3.4 R 3.4: 16
21 R R U R O U R R + O R : U U 3.6 U O 17
22 3.6: R R R s 1 R x 1 R x R y 1 x x R j x y j y j y j R x y R 18
23 4 3 GUI 2D Lily C++ GUI Library[21] OpenGL 3DCG FK Toolkit System[22]
24 4.1: A 4.2 A : A A 4.3(a)
25 (a) (b) 4.3: 4.3(b) (a) 4.3(b) 21
26 (a)
27 4.4: 23
28 5 CG
29 25
30 [1], CG,. CAD 78, 1 7 (1995). [2] Cassidy J.Curits, Sean E.Andreson, Joshua E. Seims, Kurt W.Fleischer, David H.Salesin, Computer-Generated Watercolor, SIGGRAPH 97, (1997). [3] Aaron Hertzmann, Painterly Rendering with Curved Brush Strokes of Multiple Sizes, SIGGRAPH 98, (1998). [4],,,, LIC, 1(3), (2002). [5],,,, -, -,. CAD 35, (2000). [6],,,. CAD 70, (1999). [7],,, CG,. 107, (1999). 26
31 [8],, PhD thesis, (2008). [9],, L*a*b, 15, (1991). [10],. [11], [ ], (1982). [12], [ ], (1992). [13] XYLADECOR. [14] ( ),. [15], [ ], (1983). [16], ( ),. 30, (1979). [17],,, 36, (1965). [18],, ( 5 ),,. 47, (1995). [19], [ - CG-] (2001). [20], [ ] (1997). 27
32 [21], Lily Library. [22], FK Tool Kit System. 28
2010 : M
2010 M0107288 2010 : M0107288 1 1 1.1............................ 1 1.2............................... 3 2 4 2.1................ 4 2.2....................... 6 3 9 3.1......................... 9 3.1.1..............................
1 911 9001030 9:00 A B C D E F G H I J K L M 1A0900 1B0900 1C0900 1D0900 1E0900 1F0900 1G0900 1H0900 1I0900 1J0900 1K0900 1L0900 1M0900 9:15 1A0915 1B0915 1C0915 1D0915 1E0915 1F0915 1G0915 1H0915 1I0915
2010 : M CG 3DCG 3 3
2010 M0107432 2010 : M0107432 3 3 CG 3DCG 3 3 1 1 1.1............................ 1 1.2.............................. 2 2 3 2.1................................. 3 2.2 3........................ 4 2.3.................................
研修コーナー
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2009 : M DCG 3 4 3
2009 M0106121 2009 : M0106121 3DCG 3 4 3 1 1 1.1............................. 1 1.2.............................. 4 2 5 2.1.............................. 5 2.2............................. 6 2.2.1...........................
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add1 2 β β - conversion (λx.x + 1(2 β x + 1 x λ f(x, y = 2 x + y 2 λ(x, y.2 x + y 1 λy.2 x + y λx.(λy.2 x + y x λy.2 x + y EXAMPLE (λ(x, y.2
output: 2011,11,10 2.1 λ λ β λ λ - abstraction λ λ - binding 1 add1 + add1(x = x + 1 add1 λx.x + 1 x + 1 add1 function application 2 add1 add1(2 g.yamadatakahiro@gmail.com 1 add1 2 β β - conversion (λx.x
Taro13-第６章（まとめ）.PDF
% % % % % % % % 31 NO 1 52,422 10,431 19.9 10,431 19.9 1,380 2.6 1,039 2.0 33,859 64.6 5,713 10.9 2 8,292 1,591 19.2 1,591 19.2 1,827 22.0 1,782 21.5 1,431 17.3 1,661 20.0 3 1,948 1,541 79.1 1,541 79.1
2009 3DCG : M0106423 3DCG,,,, 3DCG 2D 3DCG 2D 3DCG 3DCG
2009 3DCG M0106423 2009 3DCG : M0106423 3DCG,,,, 3DCG 2D 3DCG 2D 3DCG 3DCG 1 1 1.1................................. 1 1.2................................. 1 1.3............................... 3 1.4.................................
平成20年5月 協会創立50年の歩み 海の安全と環境保全を目指して 友國八郎 海上保安庁 長官 岩崎貞二 日本船主協会 会長 前川弘幸 JF全国漁業協同組合連合会 代表理事会長 服部郁弘 日本船長協会 会長 森本靖之 日本船舶機関士協会 会長 大内博文 航海訓練所 練習船船長 竹本孝弘 第二管区海上保安本部長 梅田宜弘
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2006 3D M
2006 3D M0103190 2006 3D : M0103190 3DCG 3DCG 3DCG 3D 3 3D SKETCH 3D Teddy 1 1 2 3DModelingTool Dice 5 2.1................................... 5 2.2.............................. 6 2.2.1.............. 6
2007 3DCG : M DCG 3DCG 3DCG 3D (huristic method) C++
2007 3DCG M0104402 2007 3DCG : M0104402 3DCG 3DCG 3DCG 3D (huristic method) C++ 1 1 1.1............................ 1 1.2.............................. 3 2 4 2.1......................... 4 2.2....................
III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2 lim. (x,y) (1,0) x 2 + y 2 lim (x,y) (0,0) lim (x,y) (0,0) lim (x,y) (0,0) 5x 2 y x 2 + y 2. xy x2 + y
III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2. (x,y) (1,0) x 2 + y 2 5x 2 y x 2 + y 2. xy x2 + y 2. 2x + y 3 x 2 + y 2 + 5. sin(x 2 + y 2 ). x 2 + y 2 sin(x 2 y + xy 2 ). xy (i) (ii) (iii) 2xy x 2 +
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 29 30 ( ) ( ) ( ) 31 32 ( ) ( ) 33 34 35 36 37 38 39 40 41 42 43 44 ) i ii i ii 45 46 47 2 48 49 50 51 52 53 54 55 56 57 58
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i ii (1) (1) (2) (1) (3) (1) (1) (2) (1) (3) (1) (1) (2) (1) (3) (2) (3) (1) (2) (3) (1) (1) (1) (1) (2) (1) (3) (1) (2) (1) (3) (1) (1) (1) (2) (1) (3) (1) (1) (2) (1) (3)
23 15961615 1659 1657 14 1701 1711 1715 11 15 22 15 35 18 22 35 23 17 17 106 1.25 21 27 12 17 420,845 23 32 58.7 32 17 11.4 71.3 17.3 32 13.3 66.4 20.3 17 10,657 k 23 20 12 17 23 17 490,708 420,845 23
平成１８年度「商品先物取引に関する実態調査」報告書
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ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4
20 20.0 ( ) 8 y = ax 2 + bx + c 443 ax 2 + bx + c = 0 20.1 20.1.1 n 8 (n ) a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 444 ( a, b, c, d
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
2010 : M0107189 3DCG 3 (3DCG) 3DCG 3DCG 3DCG S
2010 M0107189 2010 : M0107189 3DCG 3 (3DCG) 3DCG 3DCG 3DCG S 1 1 1.1............................ 1 1.2.............................. 4 2 5 2.1............................ 5 2.2.............................
A, B, C. (1) A = A. (2) A = B B = A. (3) A = B, B = C A = C. A = B. (3)., f : A B g : B C. g f : A C, A = C. 7.1, A, B,. A = B, A, A A., A, A
91 7,.,, ( ).,,.,.,. 7.1 A B, A B, A = B. 1), 1,.,. 7.1 A, B, 3. (i) A B. (ii) f : A B. (iii) A B. (i) (ii)., 6.9, (ii) (iii).,,,. 1), Ā = B.. A, Ā, Ā,. 92 7 7.2 A, B, C. (1) A = A. (2) A = B B = A. (3)
24 21 21115025 i 1 1 2 5 2.1.................................. 6 2.1.1........................... 6 2.1.2........................... 7 2.2...................................... 8 2.3............................
独立性の検定・ピボットテーブル
II L04(2016-05-12 Thu) : Time-stamp: 2016-05-12 Thu 12:48 JST hig 2, χ 2, V Excel http://hig3.net ( ) L04 II(2016) 1 / 20 L03-Q1 Quiz : 1 { 0.95 (y = 10) P (Y = y X = 1) = 0.05 (y = 20) { 0.125 (y = 10)
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2018 M
2018 M0115001 2018 9 2018 : M0115001 1 3D 3D 1 3D 3D 1 1 1.1......................................... 1 1.2......................................... 3 2 4 2.1...................................... 4 2.2................................
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +
2003 : 00P249,,, CG
2003 Web3D 00P249 2003 : 00P249,,, CG 1 1 1.1................................. 1 1.2................................. 3 1.3.............................. 3 2 4 2.1............................... 4 2.2.....................
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 +
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
(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n
. 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n
) Binary Cubic Forms / 25
2016 5 2 ) Binary Cubic Forms 2016 5 2 1 / 25 1 2 2 2 3 2 3 ) Binary Cubic Forms 2016 5 2 2 / 25 1.1 ( ) 4 2 12 = 5+7, 16 = 5+11, 36 = 7+29, 1.2 ( ) p p+2 3 5 5 7 11 13 17 19, 29 31 41 43 ) Binary Cubic
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
Microsoft Word - ２１年仕様書案0905011600.doc
Computer System in General Information Processing Ce nter 21 4 April,2009 ShinshuUniversity 1 ( ) ( ) ( 2 3 - 4 - - 5 - - 6 - - 7 - - 8 - - 9 - - 10 - - 11 - 1-4 1-4-1 1-4-1-1 Thin 1-4-1-2 1-4-2 1-4-2-1
s s U s L e A = P A l l + dl dε = dl l l
P (ε) A o B s= P A s B o Y l o s Y l e = l l 0.% o 0. s e s B 1 s (e) s Y s s U s L e A = P A l l + dl dε = dl l l ε = dε = l dl o + l lo l = log l o + l =log(1+ e) l o Β F Α E YA C Ο D ε YF B YA A YA
.1 A cos 2π 3 sin 2π 3 sin 2π 3 cos 2π 3 T ra 2 deta T ra 2 deta T ra 2 deta a + d 2 ad bc a 2 + d 2 + ad + bc A 3 a b a 2 + bc ba + d c d ca + d bc +
.1 n.1 1 A T ra A A a b c d A 2 a b a b c d c d a 2 + bc ab + bd ac + cd bc + d 2 a 2 + bc ba + d ca + d bc + d 2 A a + d b c T ra A T ra A 2 A 2 A A 2 A 2 A n A A n cos 2π sin 2π n n A k sin 2π cos 2π
, = = 7 6 = 42, =
http://www.ss.u-tokai.ac.jp/~mahoro/2016autumn/alg_intro/ 1 1 2016.9.26, http://www.ss.u-tokai.ac.jp/~mahoro/2016autumn/alg_intro/ 1.1 1 214 132 = 28258 2 + 1 + 4 1 + 3 + 2 = 7 6 = 42, 4 + 2 = 6 2 + 8
09 II 09/12/ (3D ) f(, y) = 2 + y 2 3D- 1 f(0, 0) = 2 f(1, 0) = 3 f(0, 1) = 4 f(1, 1) = 5 f( 1, 2) = 6 f(0, 1) = z y (3D ) f(, y) = 2 + y
09 II 09/12/21 1 1 7 1.1 I 2D II 3D f() = 3 6 2 + 9 2 f(, y) = 2 2 + 2y + y 2 6 4y f(1) = 1 3 6 1 2 9 1 2 = 2 y = f() f(3, 2) = 2 3 2 + 2 3 2 + 2 2 6 3 4 2 = 8 z = f(, y) y 2 1 z 8 3 2 y 1 ( y ) 1 (0,
3D VR CAD 3D CAD CAD [1] CAD 3DCG [2] [3] CAD 3D NC CG [4] Ccurve XY C curve α C curve [5], [6], [7], [8], [9] 2 [10] 1 [11], [12] 2.2 [13] Tcu
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1,a) 2 2 2011 10 21, 2012 5 12 A Study of Simulating Log-aesthetic Curved Surfaces under Various Light Source Environments with Augmented Reality Ryo Hirano 1,a) Toshinobu Harada 2 Kohe Tokoi 2 Received:
f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) B p.1/14
B p.1/14 f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) B p.1/14 f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) f(x 1,...,x n ) (x 1 x 0,...,x n 0), (x 1,...,x n ) i x i f xi
24 201170068 1 4 2 6 2.1....................... 6 2.1.1................... 6 2.1.2................... 7 2.1.3................... 8 2.2..................... 8 2.3................. 9 2.3.1........... 12
II 1 3 2 5 3 7 4 8 5 11 6 13 7 16 8 18 2 1 1. x 2 + xy x y (1 lim (x,y (1,1 x 1 x 3 + y 3 (2 lim (x,y (, x 2 + y 2 x 2 (3 lim (x,y (, x 2 + y 2 xy (4 lim (x,y (, x 2 + y 2 x y (5 lim (x,y (, x + y x 3y
40 6 y mx x, y 0, 0 x 0. x,y 0,0 y x + y x 0 mx x + mx m + m m 7 sin y x, x x sin y x x. x sin y x,y 0,0 x 0. 8 x r cos θ y r sin θ x, y 0, 0, r 0. x,
9.. x + y + 0. x,y, x,y, x r cos θ y r sin θ xy x y x,y 0,0 4. x, y 0, 0, r 0. xy x + y r 0 r cos θ sin θ r cos θ sin θ θ 4 y mx x, y 0, 0 x 0. x,y 0,0 x x + y x 0 x x + mx + m m x r cos θ 5 x, y 0, 0,
.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,
.1.1 Y K L Y = K 1 3 L 3 L K K (K + ) 1 1 3 L 3 K 3 L 3 K 0 (K + K) 1 3 L 3 K 1 3 L 3 lim K 0 K = L (K + K) 1 3 K 1 3 3 lim K 0 K = 1 3 K 3 L 3 z = f(x, y) x y z x-y-z.1 z = e x +xy y 3 x-y ( ) z 0 f(x,
2005 3DCG 3DCG M
2005 3DCG 3DCG M0102244 2005 3DCG : M0102244 3DCG Inverse Kinematics 3DCG 3DCG 1 1 1.1................................. 1 1.2.............................. 2 2 3 2.1............................. 3 2.2............................
(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n
3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4
#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/
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.....................................
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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
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
1 1.1 R(x) = 0 y + P (x)y + Q(x)y = R(x)...(1) y + P (x)y + Q(x)y = 0...(2) 1 2 u(x) v(x) c 1 u(x)+ c 2 v(x) = 0 c 1 = c 2 = 0 c 1 = c 2 = 0 2 0 2 u(x) v(x) u(x) u (x) W (u, v)(x) = v(x) v (x) 0 1 1.2
第86回日本感染症学会総会学術集会後抄録（I）
κ κ κ κ κ κ μ μ β β β γ α α β β γ α β α α α γ α β β γ μ β β μ μ α ββ β β β β β β β β β β β β β β β β β β γ β μ μ μ μμ μ μ μ μ β β μ μ μ μ μ μ μ μ μ μ μ μ μ μ β
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
017 8 10 f : R R f(x) = x n + x n 1 + 1, f(x) = sin 1, log x x n m :f : R n R m z = f(x, y) R R R R, R R R n R m R n R m R n R m f : R R f (x) = lim h 0 f(x + h) f(x) h f : R n R m m n M Jacobi( ) m n
() 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 ()
II
II 16 16.0 2 1 15 x α 16 x n 1 17 (x α) 2 16.1 16.1.1 2 x P (x) P (x) = 3x 3 4x + 4 369 Q(x) = x 4 ax + b ( ) 1 P (x) x Q(x) x P (x) x P (x) x = a P (a) P (x) = x 3 7x + 4 P (2) = 2 3 7 2 + 4 = 8 14 +
O1-1 O1-2 O1-3 O1-4 O1-5 O1-6
O1-1 O1-2 O1-3 O1-4 O1-5 O1-6 O1-7 O1-8 O1-9 O1-10 O1-11 O1-12 O1-13 O1-14 O1-15 O1-16 O1-17 O1-18 O1-19 O1-20 O1-21 O1-22 O1-23 O1-24 O1-25 O1-26 O1-27 O1-28 O1-29 O1-30 O1-31 O1-32 O1-33 O1-34 O1-35
a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552
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カテゴリ変数と独立性の検定
II L04(2015-05-01 Fri) : Time-stamp: 2015-05-01 Fri 22:28 JST hig 2, Excel 2, χ 2,. http://hig3.net () L04 II(2015) 1 / 20 : L03-S1 Quiz : 1 2 7 3 12 (x = 2) 12 (y = 3) P (X = x) = 5 12 (x = 3), P (Y =
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A
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
