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X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2

CG38.PDF

............3...3...6....6....8.....8.....4...9 3....9 3.... 3.3...4 3.4...36...39 4....39 4.....39 4.....4 4....49 4.....5 4.....57...64 5....64 5....66 5.3...68 5.4...7 5.5...77...8 6....8 6.....8 6.....83

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r

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) du (L) = f (9.3) dx (9.) P

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)

64 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

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

本文／目次（裏白）

液晶の物理１：連続体理論（弾性，粘性）

The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers

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..............

I y = f(x) a I a x I x = a + x 1 f(x) f(a) x a = f(a + x) f(a) x (11.1) x a x 0 f(x) f(a) f(a + x) f(a) lim = lim x a x a x 0 x (11.2) f(x) x

11 11.1 I y = a I a x I x = a + 1 f(a) x a = f(a +) f(a) (11.1) x a 0 f(a) f(a +) f(a) = x a x a 0 (11.) x = a a f (a) d df f(a) (a) I dx dx I I I f (x) d df dx dx (x) [a, b] x a ( 0) x a (a, b) () [a,

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP

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

: , 2.0, 3.0, 2.0, (%) ( 2.

2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................

II R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k

II 231017 1 1.1. R n k +1 v 0,, v k k v 1 v 0,, v k v 0 1.2. v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ kσ dimσ = k 1.3. k σ {v 0,...,v k } {v i0,...,v il } l σ τ < τ τ σ 1.4.

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

A

A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................

, 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,

,. 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,

70 : 20 : A B (20 ) (30 ) 50 1

70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................

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( 1) 1 ( [1] ) [] ( ) (AC) [3] [4, 5, 6] 3 (i) AC (ii) (iii) 3 AC [3, 7] [4, 5, 6] 1.1 ( e; e>0) Ze r v [ 1(a)] v [ 1(a )] B = μ 0 4π Zer v r 3 = μ 0 4π 1 Ze l m r 3, μ 0 l = mr v ( l s ) s μ s = μ B s

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)

DVIOUT-講

005-10-14 1 1 [1] [] [3] [4] (a + b) = a +ab + b [5] (a + b) 3 a 3 +a b + ab + a b +ab + b 3 a 3 +3a b +3ab + b 3 [6] (a + b) 4 (a + b) 5 [7] technology expand((a+b) n n =?) [8] technology n =6, 7, 8,

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ

圧延理論概論（中級）

圧延理論 -(Orowan の理論 板プロフィルの形成機構 ) 東京大学生産技術研究所 柳本潤. はじめに圧延理論 -では Orowan の圧延理論とその数値解析について詳しく述べる ついで 板プロフィルの形成機構とその解析に利用されている理論 ( 分割モデル スリットモデル ) についてもふれる なお 方程式番号 参考文献番号は初級コースからの通番となっているので 是非初級コースのテキストもご一読頂きたい.Orowan

A

A04-164 2008 2 13 1 4 1.1.......................................... 4 1.2..................................... 4 1.3..................................... 4 1.4..................................... 5 2

高等学校学習指導要領

高等学校学習指導要領

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

.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T

NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977

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

z 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

f f x, y, u, v, r, θ r > = x + iy, f = u + iv C γ D f f D f f, Rm,. = x + iy = re iθ = r cos θ + i sin θ = x iy = re iθ = r cos θ i sin θ x = + = Re, y = = Im i r = = = x + y θ = arg = arctan y x e i =

画像工学特論

.? (x i, y i )? (x(t), y(t))? (x(t)) (X(ω)) Wiener-Khintchine 35/97 . : x(t) = X(ω)e jωt dω () π X(ω) = x(t)e jωt dt () X(ω) S(ω) = lim (3) ω S(ω)dω X(ω) : F of x : [X] [ = ] [x t] Power spectral density

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July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i

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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

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IA 2013 : :10722 : 2 : :2 :761 :1 23-27) : : 1 1.1 / ) 1 /, ) / e.g. Taylar ) e x = 1 + x + x2 2 +... + xn n! +... sin x = x x3 6 + x5 x2n+1 + 1)n 5! 2n + 1)! 2 2.1 = 1 e.g. 0 = 0.00..., π = 3.14..., 1

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9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) Ĥ0 ψ n (r) ω n Schrödinger Ĥ 0 ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ0 + Ĥint (

9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) 2. 2.1 Ĥ ψ n (r) ω n Schrödinger Ĥ ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ + Ĥint (t)] ψ (r, t), (2) Ĥ int (t) = eˆxe cos ωt ˆdE cos ωt, (3)

#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/

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