1.1 1 A
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- たつや あさぶき
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1 . A..2
2 2 2. () (xyz) ( xyz) ( xy z) = (x x)yz ( xy z) = yz ( xy z) = y(z ( x z)) = y((z x)(z z)) = y( x z) (2) (3) M aj (x, y, M aj ( x, ȳ, z)) = xy ȳm aj ( x, ȳ, z) M aj ( x, ȳ, z)x M aj (x, y, z) x = xy ȳ(xȳ y z z x) ( x y)(ȳ z)( z x)x = xy xȳ xȳz ( xȳ xz yz)( z x)x = xy xȳ xȳz ( xȳ z xyz)x = xy xȳ xȳz xyz = x(y ȳz) x(ȳ yz) = x(y z) x(ȳ z) = xy xȳ xz xz = x y z = M aj (0, y, z) M aj (, y, z) = (y z z) (ȳ y z) = (y z z)(ȳ y z) (y z z)(ȳ y z) = (y z z)y(ȳ z) (ȳ z) z(ȳ y z) = (y z z)yz ȳ z(ȳ y z) = yz ȳ z
3 2.2 ()F (x, y, z) = x y z 3 F (x, y, z) = xyz xȳ z xy z xȳz F (x, y, z) = ( x ȳ z)(x y z)(x ȳ z)( x y z) (2)F (x, y, z) = (x y)(ȳ z) F (x, y, z) = (x y)(ȳ z) = xȳ xz yz = xyz xȳz xyz xȳ z F (x, y, z) = (xy z) ( xȳ z) ( xȳz) ( xy z) = ( x ȳ z)(x y z)(x y z)(x ȳ z) (3)F (x, y, z) = (xȳ) (y z) (x z) F (x, y, z) = (xȳ) (y z) (x z) = xȳ y z x z
4 4 = ( x y)(ȳ z)( x z) = xȳz xz xȳ xyz yz = xȳz xyz xȳ z xyz F (x, y, z) = ( xy z) (xȳz) (xy z) (xȳ z) = (x ȳ z)( x y z)( x ȳ z)( x y z) (4)F (x, y, z) = (x ȳ) xz F (x, y, z) = (x ȳ) xz = (x ȳ) xz (x ȳ)( xz) = xy xz ((x ȳ)(x z)) = xyz (x ȳ z) = xyz xyz xȳz xy z xȳ z xȳ z F (x, y, z) = ( xȳz) ( xy z) = (x y z)(x ȳ z)
5 5 2.3 () x yz = xyz xȳz xy z xȳ z () (2) (xy ȳz) z = x ȳ z () (3) x(y z) = xȳ zxyz (NAND ) (4) xȳ xz = x z y 0 x 0 y 0 z (NOR ) 2.4 () x y z x y z = F (x, y, z) F (x, y, z) = x y z = xȳ z F (0, y, z) = ȳ z F (, y, z) = 0 z = z F (0, y, z) F (, y, z) = ȳ z x z, y ȳ z = z y y, z (y, z) = (0, 0), (0, ), (, ) x α x = F (0, y, z) F (, y, z)α = 0 zα = zα (x, y, z) = (0, 0, 0), (0, 0, ), (, 0, ), (0,, ), (,, ) (2) x y = xz y (x y)(ȳ xz) = F (x, y, z)
6 6 F = (x y)(ȳ xz) F (0, y, z) = y(ȳ z) = yz F (, y, z) = ȳ F (0, y, z) F (, y, z) = yz ȳ = z ȳ x z, y z ȳ = z y x α x = F (0, y, z) F (, y, z)α = yz ȳα = ȳ z ȳα = ȳ z z y ȳ z x = ȳ y, z z ȳ = (y, z) = (0, 0), (0, ), (, ) (x, y, z) = (, 0, 0), (, 0, ), (0,, ) (3) x ȳz = F (x, y, z) F (x, y, z) = x ȳz F (0, y, z) = ȳz F (, y, z) = y z F (0, y, z) F (, y, z) = z, y x x α
7 7 x = F (0, y, z) F (, y, z)α = y z (y z)α = y z (x, y, z) = (, 0, 0), (0, 0, ), (,, 0), (,, )
8 8 3. F 3 (x 5, x 4, x 3, x 2, x ) F 2 (x 5, x 4, x 3, x 2, x ) AND F 3 (x 5, x 4, x 3, x 2, x ) OR F 2 (x 5, x 4, x 3, x 2, x ) 3.2 (xy ȳz) z x, y, z 0 x = 0 ȳz z = ȳ zx = x y = 0 y = x z y z = 0 z = xy ȳ z (xy ȳz) z 3.3 x, y, z 3 0,,2, xy xz yz, xȳ z xy z xȳz xyz, xyz xȳz xy z xȳ z, ȳ z x z xȳ 3.4 f(x 3, x 2, x ) = a 3 x 3 a 2 x 2 a x a 0
9 9 f( x 3, x 2, x ) = a 3 (x 3 ) a 2 (x 2 ) a (x ) a 0 = a 3 (x 3 ) a 2 (x 2 ) a (x ) a 0 = a 3 x 3 a 2 x 2 a x a 3 a 2 a a f(x, y, z) ()xy z (x y) z (2)x ȳ z (3)x y z (4) f(0, 0, 0) = f(, 0, 0) = f(, 0, ) = f(,, 0) = 0 yz x(y z) (5)xyz (6)x y z 87) xy yz xz
10 0 x3 x2 x x5 x4 A4.. M aj (x 5, x 4, x 3, x 2, x ) x3 x2 x x5 x4 A4..2 H 2 (x 5, x 4, x 3, x 2, x ) 4. () M aj (x 5, x 4, x 3, x 2, x ) 4A.. NOT-AND-OR M aj (x 5, x 4, x 3, x 2, x ) = x 3 x 2 x x 4 x 2 x x 5 x 2 x x 4 x 3 x x 5 x 3 x x 5 x 4 x x 4 x 3 x 2 x 5 x 3 x 2 x 5 x 4 x 2 x 5 x 4 x 3 (2) H 2 (x 5, x 4, x 3, x 2, x ) H 2d (x 5, x 4, x 3, x 2, x ) ( 4 ) A4..2 A4..3 NOT-AND- OR AND OR H 2 (x 5, x 4, x 3, x 2, x ) NOT-OR-AND H 2d (x 5, x 4, x 3, x 2, x ) = x 4 x 3 x 2 x x 5 x 3 x 2 x x 5 x 4 x 2 x x 5 x 4 x 3 x x 5 x 4 x 3 x 2
11 x3 x2 x x5 x4 A4..3 H 2d (x 5, x 4, x 3, x 2, x ) x3 x2 x ux5 x4 A4..4 u = H 2 (x 5, x 4, x 3, x 2, x ) H 2 (x 5, x 4, x 3, x 2, x ) = (x 4 x 3 x 2 x )(x 5 x 3 x 2 x )(x 5 x 4 x 2 x ) (x 5 x 4 x 3 x )(x 5 x 4 x 3 x 2 ) (3) M aj (x 5, x 4, x 3, x 2, x ) = x 5 M aj (, x 4, x 3, x 2, x ) x 5 M aj (0, x 4, x 3, x 2, x ) (4) u = H 2 (x 5, x 4, x 3, x 2, x ) M aj (x 5, x 4, x 3, x 2, x ) A4..4 A4..4 () M aj (x 5, x 4, x 3, x 2, x )
12 2 M aj (x 5, x 4, x 3, x 2, x ) = x 3 x 2 x x 4 x 2 x x 5 x 2 x x 4 x 3 x x 5 x 3 x x 5 x 4 x x 4 x 3 x 2 x 5 x 3 x 2 x 5 x 4 x 2 x 5 x 4 x A4..5 NOT-AND x 5 x 3 x x 5 x 3 x 2 x 5 x 4 x 3 x 5 x 2 x x 5 x 4 x 2 x 5 x 2 x x 4 x 3 x x 5 x 3 x x 4 x 2 x x 5 x 4 x 3 x 5 x 3 x x 5 x 4 x x 5 x 4 x 2 x 5 x 2 x x 4 x x 3 x 2 0 A A A4.2.3 x 3 x 2, x 5 x 2 x, x 5 x 3 x, x 5 x 2 x 4?? NOT-AND-OR F (x 5, x 4, x 3, x 2, x ) = x 3 x 2 x 5 x 2 x x 5 x 3 x x 5 x 2 x 4.3 (2)y y 2 y, y 2 ( A4.3)
13 3 4.4 y = x 3 x 2 x 3 x 2 y 2 = x 2 x x 3 x 2 ()(2)(3) NOT AND OR NAND 4.5 A4.4 NOT-AND-OR f(x, y, z) = xȳ z xz A4.4 NAND 3 H = x, H 2 = z τ = ȳ z, tau 2 x H τ H 2 tau 2 f(x, y, z) = xȳz z x = xȳxz zxz xz NAND xz NAND f(x, y, z) = xȳ zzx = xȳxzzxz 3 x 3, x 2, x u = x x 2 x 3 u A4.5 y = u x 3 u x 2
14 4 p = p = p = p = p = p = 5 (a) p = p = p = p = p = 4 (b) A4..5 NOT-AND p = p = p = p = 3 (c) p = 0 p = 2 (d)
15 5 A4.2. A
16 6 A x2 x x2 x x2 x x3 x3 x3 A4.3 y, y 2, y y 2 y, y 2
17 7 A4.4 NOTANDOR NAND NAND x2 x u x3 A4.5
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 +
応力とひずみ.ppt
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7-12.dvi
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A_chapter5.dvi
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d dt P = d ( ) dv G M vg = F M = F (4.1) dt dt M v G P = M v G F (4.1) d dt H G = M G (4.2) H G M G Z K O I z R R O J x k i O P r! j Y y O -
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k m m d2 x i dt 2 = f i = kx i (i = 1, 2, 3 or x, y, z) f i σ ij x i e ij = 2.1 Hooke s law and elastic constants (a) x i (2.1) k m σ A σ σ σ σ f i x
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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
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II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )
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剛塑性FEM入門.ppt
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座標変換におけるテンソル成分の変換行列
座標変換におけるテンソル成分の変換行列 座標変換におけるテンソル成分の変換関係は 次元数によらず階数によって定義される変換行列で整理することができる 位置ベクトルの変換行列を D としてそれを示そう D の行列式を ( = D ) とするとき 鏡映や回映といった pseudo rotation に対しては = -1 である が問題になる基底は 対称操作に含まれる pseudo rotation に依存する
7 27 7.1........................................ 27 7.2.......................................... 28 1 ( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a -1 1 6
26 11 5 1 ( 2 2 2 3 5 3.1...................................... 5 3.2....................................... 5 3.3....................................... 6 3.4....................................... 7
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35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
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. p.1/11 [ ] F(x,y,z) = (F 1 (x,y,z),f 2 (x,y,z),f 3 (x,y,z)) = F 1 i+f 2 j+f 3 k div F = F 1 x + F 2 y + F 3 z F (divergence). p.1/11 [ ] F(x,y,z) = (F 1 (x,y,z),f 2 (x,y,z),f 3 (x,y,z)) = F 1 i+f 2 j+f
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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
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
表紙
A B,BW C Rc No. h Rc C 2 3 4 M2 M3 M4 M5 M6 25 3.2 6.8 3.5 6.4 2.8 2.7 2.4 9.1 15.5 21.6 28.0 34.4 35 41 40 50 60 80 100 120 140 160 180 200 240 4.8 10.2 5.2 10.1 4.3 4.1 3.8 14.23 24.36 34.1 44.2 54.3
ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y
01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y
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
A_chapter3.dvi
: a b c d 2: x x y y 3: x y w 3.. 3.2 2. 3.3 3. 3.4 (x, y,, w) = (,,, )xy w (,,, )xȳ w (,,, ) xy w (,,, )xy w (,,, )xȳ w (,,, ) xy w (,,, )xy w (,,, ) xȳw (,,, )xȳw (,,, ) xyw, F F = xy w x w xy w xy w
n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz
1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x
96 7 1m =2 10 7 N 1A 7.1 7.2 a C (1) I (2) A C I A A a A a A A a C C C 7.2: C A C A = = µ 0 2π (1) A C 7.2 AC C A 3 3 µ0 I 2 = 2πa. (2) A C C 7.2 A A
7 Lorentz 7.1 Ampère I 1 I 2 I 2 I 1 L I 1 I 2 21 12 L r 21 = 12 = µ 0 2π I 1 I 2 r L. (7.1) 7.1 µ 0 =4π 10 7 N A 2 (7.2) magnetic permiability I 1 I 2 I 1 I 2 12 21 12 21 7.1: 1m 95 96 7 1m =2 10 7 N
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5. NMR µ = γ I µ = γ I r H D 0 I r I r { I 3 I r r µ γγ 3( )( ) = } (5..) zb 0 µ 0 γγ I r I r 0( γ z γ z) { I 3 I r r 3( )( ) H = B I + I + } (5..) (5..) z z µ 0 γγ D = ( + + + + + ) 3 r H A B C D F (5..3)
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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 +
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06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
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1 ( 12 11 44 7 20 10 10 1 1 ( ( 2 10 46 11 10 10 5 8 3 2 6 9 47 2 3 48 4 2 2 ( 97 12 ) 97 12 -Spencer modulus moduli (modulus of elasticity) modulus (le) module modulus module 4 b θ a q φ p 1: 3 (le) module
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
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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
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Fermat s Last Theorem Hajime Mashima November 19, 2018 Abstract About 380 years ago, Pierre de Fermat wrote the following idea to Diophantus s Arithme
Fermat s Last Theorem Hajime Mashima November 19, 2018 Abstract About 380 years ago, Pierre de Fermat wrote the following idea to Diophantus s Arithmetica. Cubum autem in duos cubos, aut quadratoquadratum
x y x-y σ x + τ xy + X σ y B = + τ xy + Y B = S x = σ x l + τ xy m S y = σ y m + τ xy l σ x σ y τ xy X B Y B S x S y l m δu δv [ ( σx δu + τ )
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dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp
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Chap11.dvi
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m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d
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II 2014 1 1 I 1.1 72 r 2 72 8 72/8 = 9 9 2 a 0 1 a 1 a 1 = a 0 (1+r/100) 2 a 2 a 2 = a 1 (1 + r/100) = a 0 (1 + r/100) 2 n a n = a 0 (1 + r/100) n a n a 0 2 n a 0 (1 + r/100) n = 2a 0 (1 + r/100) n = 2
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論理回路 ( 基礎 ) 法政大学 情報科学部 大森健児 参考書 論理演算 () AND,OR,NOT,XOR AND OR NOT XOR 論理演算 (2) NAND,NOR NAND NOR 前提 結論 If A then B は A が真のとき B が真であるならば この文は真であり A が偽のときは B が真であろうとなかろうとこの文は真である A が真のとき B が偽であればこの文は偽である
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