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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

熊本県数学問題正解

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Chap9.dvi

.,. f(),, f(),,.,. () lim 2 +3 2 9 (2) lim 3 3 2 9 (4) lim ( ) 2 3 +3 (5) lim 2 9 (6) lim + (7) lim (8) lim (9) lim (0) lim 2 3 + 3 9 2 2 +3 () lim sin 2 sin 2 (2) lim +3 () lim 2 2 9 = 5 5 = 3 (2) lim

1 u t = au (finite difference) u t = au Von Neumann

1 u t = au 3 1.1 (finite difference)............................. 3 1.2 u t = au.................................. 3 1.3 Von Neumann............... 5 1.4 Von Neumann............... 6 1.5............................

1 I

1 I 3 1 1.1 R x, y R x + y R x y R x, y, z, a, b R (1.1) (x + y) + z = x + (y + z) (1.2) x + y = y + x (1.3) 0 R : 0 + x = x x R (1.4) x R, 1 ( x) R : x + ( x) = 0 (1.5) (x y) z = x (y z) (1.6) x y =

(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

i I II I II II IC IIC I II ii 5 8 5 3 7 8 iii I 3........................... 5......................... 7........................... 4........................ 8.3......................... 33.4...................

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 =

1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1

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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 +

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 ),

Macdonald, ,,, Macdonald. Macdonald,,,,,.,, Gauss,,.,, Lauricella A, B, C, D, Gelfand, A,., Heckman Opdam.,,,.,,., intersection,. Macdona

Macdonald, 2015.9.1 9.2.,,, Macdonald. Macdonald,,,,,.,, Gauss,,.,, Lauricella A, B, C, D, Gelfand, A,., Heckman Opdam.,,,.,,., intersection,. Macdonald,, q., Heckman Opdam q,, Macdonald., 1 ,,. Macdonald,

1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 5 3. 4. 5. A0 (1) A, B A B f K K A ϕ 1, ϕ 2 f ϕ 1 = f ϕ 2 ϕ 1 = ϕ 2 (2) N A 1, A 2, A 3,... N A n X N n X N, A n N n=1 1 A1 d (d 2) A (, k A k = O), A O. f

Excel ではじめる数値解析 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

Excel ではじめる数値解析 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009631 このサンプルページの内容は, 初版 1 刷発行時のものです. Excel URL http://www.morikita.co.jp/books/mid/009631 i Microsoft Windows

2000年度『数学展望 I』講義録

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微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)

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.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 +

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1. 1 BASIC PC BASIC BASIC BASIC Fortran WS PC (1.3) 1 + x 1 x = x = (1.1) 1 + x = (1.2) 1 + x 1 = (1.

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linearal1.dvi

19 4 30 I 1 1 11 1 12 2 13 3 131 3 132 4 133 5 134 6 14 7 2 9 21 9 211 9 212 10 213 13 214 14 22 15 221 15 222 16 223 17 224 20 3 21 31 21 32 21 33 22 34 23 341 23 342 24 343 27 344 29 35 31 351 31 352

さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n

1 1.1 1.1.1 A 2 P Q 3 R S T R S T P 80 50 60 Q 90 40 70 80 50 60 90 40 70 8 5 6 1 1 2 9 4 7 2 1 2 3 1 2 m n m n m n n n n 1.1 8 5 6 9 4 7 2 6 0 8 2 3 2 2 2 1 2 1 1.1 2 4 7 1 1 3 7 5 2 3 5 0 3 4 1 6 9 1

A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P

1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A

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

1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A 2 1 2 1 2 3 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 4 P, Q R n = {(x 1, x 2,, x n ) ; x 1, x 2,, x n R}

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ

() 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

数値計算：常微分方程式

( ) 1 / 82 1 2 3 4 5 6 ( ) 2 / 82 ( ) 3 / 82 C θ l y m O x mg λ ( ) 4 / 82 θ t C J = ml 2 C mgl sin θ θ C J θ = mgl sin θ = θ ( ) 5 / 82 ω = θ J ω = mgl sin θ ω J = ml 2 θ = ω, ω = g l sin θ = θ ω ( )

all.dvi

72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G

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 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,

ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2

BASICとVisual Basic

Visual Basic BASIC Visual Basic BASICBeginner's All purpose Symbolic Instruction Code Visual Basic Windows BASIC BASIC Visual Basic Visual Basic End Sub .Visual Basic Visual Basic VB 1-1.Visual Basic

(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

III Kepler ( )

III 9 8 3....................................... 3.2 Kepler ( ).......................... 0 2 3 2.................................. 3 2.2......................................... 7 3 9 3..........................................

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

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,

x V x x V x, x V x = x + = x +(x+x )=(x +x)+x = +x = x x = x x = x =x =(+)x =x +x = x +x x = x ( )x = x =x =(+( ))x =x +( )x = x +( )x ( )x = x x x R

V (I) () (4) (II) () (4) V K vector space V vector K scalor K C K R (I) x, y V x + y V () (x + y)+z = x +(y + z) (2) x + y = y + x (3) V x V x + = x (4) x V x + x = x V x x (II) x V, α K αx V () (α + β)x

limit&derivative

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21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

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numb.dvi

11 Poisson kanenko@mbkniftycom alexeikanenko@docomonejp http://wwwkanenkocom/ , u = f, ( u = u+f u t, u = f t ) 1 D R 2 L 2 (D) := {f(x,y) f(x,y) 2 dxdy < )} D D f,g L 2 (D) (f,g) := f(x,y)g(x,y)dxdy (L

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