m は nで割り切れない?
. T ::= x f n t 1 t n F n,m (x(t 1 t n )t 1 t m) x, f n n, F n,m n, m-., F n,m (x(t 1 t n )t 1 t m), x, t 1,..., t n, t 1,..., t m. F n,m (x(t 1 t n )
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2.1.,., { n Q[t ±1 ] := a k t k a k Q, m, n N k= m. Z., s Z, n k= m a kt k s := n k= m a kt k+s. : Q[t ±1 ] {t n } n Z Q t 2 Q t 1 Q t 0 Q t Q t 2 (Q-
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flF™m…−…n2011_’Î−Û
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Cinema Kobo Brand index Visual Communication V i s u a l C o m m u n i c a t i o n Visual Communication screen product guide vol-
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$x(n+1)=qx(n)-\displaystyle \sum_{j=0}^{m} a_j(n)f_j(x(n-j))$に対する大域吸引性(関数方程式の解のダイナミクスと数値シミュレーション)
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真空ポンプ106 キヤノンアネルバ株式会社真空機器総合カタログ Vol.8.9 フォンブリンオイル CF3 CF3 - (O-CF-CF2)n-(O - CF2)m O-CF3 フォンブリン Y n > m 概要 フォンブリン は 炭素 フッ素 酸素の 3 原子よりなる完全フッ素化油であり また安定
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n 2 + π2 6 x [10 n x] x = lim n 10 n n 10 k x 1.1. a 1, a 2,, a n, (a n ) n=1 {a n } n=1 1.2 ( ). {a n } n=1 Q ε > 0 N N m, n N a m
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1 1 Lambert Adolphe Jacques Quetelet ( ) [ ] 1 (1 ) n x 1, x 2,..., x n x a 1 a i a m f f 1 f i f m n 1.1 ( ( ))
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+ 1 ( ) I IA i i i 1 n m a 11 a 1j a 1m A = a i1 a ij a im a n1 a nj a nm.....
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Dρ Dt =0, Du Dt = 1 ρ p + ν 2 u + g. (1) m ρ: [kg/m 3 ] u: [m/s] p: [N/m 2 ] ν: [m 2 /s] g: [m/s 2 ] 3 MPS p i = D s n 0 2 u i = 2D s n 0 λ j i
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2 1 1 (1) 1 (2) (3) Lax : (4) Bäcklund : (5) (6) 1.1 d 2 q n dt 2 = e q n 1 q n e q n q n+1 (1.1) 1 m q n n ( ) r n = q n q n 1 r ϕ(r) ϕ (r)
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a : [m] a c-c : [m] b : SWNT [m] c : [m/s] D(ω) : d : SWNT [m] : [Js] f : [N] k : [1/m] k B : [J/K] m : [kg] n : L : SWNT [m] Q : [W] q : [W/m 2 ] R T
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(a) (b) (c) 1 (a) (b) m m = (c) p i p i+1 < = ς m L i : {1,..., n} R SVM p i (i = 1,..., m) n ς Kvarnström [1] (1) p m p 1 < = ς O(mnς) [1] (1) n O(mn
6
42 3 u = (37) MeV/c 2 (3.4) [1] u amu m p m n [1] m H [2] m p = (4) MeV/c 2 = (13) u m n = (4) MeV/c 2 =
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3 3.1 algebraic datatype data k = 1 1,1... 1,n1 2 2,1... 2,n2... m m,1... m,nm 1 m m m,1,..., m,nm m 1, 2,..., k 1 data Foo x y = Alice x [y] B
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1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3
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HydroxypropylcelluloseCellulose 2-hydroxypropyl ether CH 2 OR R=H H O O H CH H 3 OR H CH 2 CH O m H OR m1 n 30,000n 100 1,000,000n 2,500 13) 1 14 C 14
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a m 1 mod p a km 1 mod p k<s 1.6. n > 1 n 1= s m, (m, = 1 a n n a m 1 mod n a km 1 mod n k<sn a 1.7. n > 1 n 1= s m, (m, = 1 r n ν = min ord (p 1 (1 B
14
flF™m…−…n„Efic’æ’¶
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No. No. 4 No f(z) z = z z n n sin x x dx = π, π n sin(mπ/n) x m + x n dx = m, n m < n e z, sin z, cos z, log z, z α 4 4 9
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