S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.

Size: px
Start display at page:

Download "S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1."

Transcription

1 () (1) L K (i) 0 K 1 K (ii) x, y K x + y K, x y K (iii) x, y K xy K (iv) x K \ {0} x 1 K K L L K ( 0 L 1 L ) L K L/K (2) K M L M K L 1.1. C C 1.2. R K = {a + b 3 i a, b Q} Q( 2, 3) = Q( 2 + 3) 1.3. L K i (i I) i I K i L 1.2. K L L S K S L K(S) K S S = {s 1, s 2,..., s n } K(S) K(s 1, s 2,..., s n ) α L K(α) K 1.4. n K(S) (S = {s 1, s 2,..., s n } L) { } f(s1,..., s n ) K(S) = g(s 1,..., s n ) f, g K[X 1,..., X n ], g(s 1,..., s n ) 0 1

2 S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1. K M L (1) {α 1, α 2,..., α m } M K {β 1, β 2,..., β n } L M {α i β j 1 i m, 1 j n} L K (2) () L/M M/K L/K [L : K] = [L : M][M : K]. (1) () {β 1, β 2,..., β n } M L x y j M(1 j n) x = y j β j 1 j n y j {α 1, α 2,..., α m } M K z ij K(1 i m) y j = z ij α i 1 i m {α i β j 1 i m, 1 j n} K L () c ij K c ij α i β j = 0 1 i m 1 j n

3 () 3 1 j n 1 i m c ij α i β j = 0 i c ijα i M {β j } n j=1 M j c ij α i = 0. 1 i m {α i } m i=1 K c ij = 0 (2) (1) 1.6. K L K L K, L 1.7. M 1, M 2 K L [M 1 : K] [M 2 : K] M 1 M 2 = K K L α L I α = {f(x) K[X] f(α) = 0} K[X] (i) I α = (0) α K (ii) I α (0) α K I α 1 α K Irr(α, K, X) 1.8. K α K α α K 1.9. Q Q Q (1) 2 (2) ω (1 3 ) ω 1 3 (1) ω Q( 3 2) (2) [Q( 3 2, ω) : Q] (3) Q( 3 2, ω) Q α = 3 1 (1) α (2) α L/K L K L/K L/K

4 4 Q e π C/R (1) α K K(α)/K (2) α, β K α ± β, αβ, αβ 1 (β 0) K (3) L K K Q 1.2. L K L K. ( ) [L : K] [L : K] = 1 [L : K] 2 K L α [L : K(α)] < [L : K] L K(α) β 1, β 2,..., β n L = K(α, β 1, β 2,..., β n ) α β i (1 i n) K ( ) n n = 1 L = K(α 1,..., α n 1, α n ) [K(α 1,..., α n 1 ) : K] < L/K(α 1,..., α n 1 ) Irr(α n, K, X) 1.1 [L : K] (1) K, K f : K K f f ( ) K K K = K (2) L/K, L /K ϕ : L L f = ϕ K K K ϕ f K = K id K K ϕ ϕ(x) = x (x K) ϕ K L = L ϕ K f : K K (1) K x f( x) = f(x) (2) K x 0 f(x 1 ) = f(x) 1 (3) Q K ( K 0) f Q (1) L K α L K f(x) L K ϕ : K(α) L ϕ(α) f(x) (2) Q ϕ : Q(α) C (3) ϕ( 3 2) = 3 2 ω L = Q( 3 2) (4) L = Q( 3 2) Q

5 () K f(x) K L K L[X] K[X] f(x) = γ(x α 1 )(X α 2 )... (X α n ); α 1, α 2,..., α n, γ L, γ 0 L f(x) 2.1. f(x) K[X] f(x) L K. f(x) n n = 1 K n > 1 f(x) = f 1 (X)f 2 (X) f 1 (X), f 2 (X) K[X] f 1 (X) L 1 K f 2 (X) L 2 L 1 L 2 K f(x) f(x) K 1 = K[X]/(f(X)) K X K 1 f(x) K 1 [X] f(x) f(x) 1 n 1 f(x) 2.2. L K f(x) K[X] α 1, α 2,..., α n L f(x) L f(x) K(α 1, α 2,..., α n ) f(x) K f(x) K[X] 2.2. σ K 1 K 2 f(x) K 1 L 1 K 1 f(x) L 2 K 2 σf(x) σ L 1 L 2. n = [L 1 : K 1 ] n = 1 σ n > 1 K 1 [X] f(x) 2 g(x) K 2 [X] σg(x) σf(x) g(x) α σg(x) β σ : K 1 (α) K 1 [X]/(g(X)) K 2 [X]/(σg(X)) K 2 (β) σ L 1 K 1 (α) f(x) L 2 K 2 (β) σf(x) L 1 L 2 σ σ 2.1. (1) Q X (2) Q X 3 2 L [L : Q] (3) L = Q( i) Q X 4 + 2X

6 K α, β K K 1. a + b i (a, b R) a b i R Q 2.3. L f(x) K[X] (1) f(x) α β K (2) σ(α) = β K σ : L L. (2) (1). g(x) = Irr(α, K, X) g(x) K g(β) = g(σ(α)) = σ(g(α)) = 0 β g(x) (1) (2). K σ : K(α) = K(β) L K K(α) f(x) K(α) = L K(β) L [K(β) : K] = [L : K] K(β) = L f(x) K(α) 2 f 1 (X) K(α) f 1 (X) α 1 K(β) f σ 1 (X) β 1 σ K K(α, α 1 ) = K(α)[X]/(f 1 (X)) = K(β)[X]/(f σ 1 (X)) = K(β, β 1 ) K(α, α 1,... ) L 2.3. f(x) K α 1, α 2,..., α t f(x) L f(x) K σ : L L α 1, α 2,..., α t 2.4. Q X 3 2 L σ : L L Q (1) L = Q( 3 2, ω) (2) σ( 3 2) σ(ω) (3) τ( 3 2) = 3 2 ω, τ(ω) = ω Q τ : L L τ X L/K (1) f(x) K[X] L L (2) α L α K L (3) α L α K L L/K

7 () (1) Q( 2)/Q (2) Q( 3 2)/Q 3.2. (1) L/K L, K M L/M (2) M, N L/K M/K N/K MN/K (3) M, N L/K M/K MN/N 3.2. L/K (1) L/K (2) L M K σ : M M σ(l) L. (1) (2). L = K(α 1, α 2,..., α n ) α i K f i (X) f i (σ(α i )) = σ(f i (α i )) = 0 σ(α i ) α i K σ(α i ) L (2) (1). α L f(x) α K M f(x) 2.3 α K β σ(α) = β M K β = σ(α) σ(l) L β L 3.3. L/K L K. ( ) 1.4 L K α i (i = 1, 2,..., n) α i f i (X) f(x) = f 1 (X)f 2 (X) f n (X) L/K L f(x) K ( ) L f(x) K f(x) α 1, α 2,..., α n L = K(α 1, α 2,..., α n ) L K σ f(σ(α i )) = σ(f(α i )) = 0 σ(α i ) f(x) σ(l) L 3.2 L/K 3.3. Q( 4 2, i) Q( 4 2) Q (1) Q( 4 2, i)/q (2) Q( 4 2, i)/q( 4 2) (3) Q( 4 2)/Q 3.4. Q( 4 2)/Q( 2), Q( 2)/Q Q( 4 2)/Q

8 (1) K α Irr(α, K, X) K (2) L/K L K f(x) = a n X n + a n 1 X n a 1 X + a 0 K[X] f (X) f (X) = na n X n 1 + (n 1)a n 1 X n a 1 f(x), g(x) K[X] a, b K (4.1) (4.2) (af(x) + bg(x)) = af (X) + bg (X) (f(x)g(x)) = f (X)g(X) + f(x)g (X) 4.1. (1) (4.1) (2) (4.2) ( f(x), g(x) ) 4.2. K f(x), g(x) (1) (g(x) k ) = kg(x) k 1 g (X) (0 k Z) (2) (f(g(x))) = f (g(x))g (X) f(x) g(x) g(α) 0 f(x) = (X α) e g(x), f(x) α e 2 α e = 1 f (X) = g(x) + (X α)g (X) f (α) = g(α) 0 e 2 f (X) = e(x α) e 1 g(x) + (X α) e g (X) f (α) = 0 α f(x) f(α) = f (α) = K 0 L/K. α L f(x) f(x) 1 K 0 f (X) 0 f (X) f(x) f(x) f (X) f(x) (4.3) a(x)f(x) + b(x)f (X) = 1 a(x), b(x) K[X] K[X] f(α) = f (α) = 0 (4.3)

9 () K L L = L \ {0} 1 L K K L K L K 2 L = K(α, β) α, β K f(x), g(x) α, β K α 1 = α, α 2,..., α m f(x) β 1 = β, β 2,..., β n g(x) f 1 (X) = f((α 1 cβ 1 ) + cx) g(x) β 1 0 c K f 1 (X) β 1 β 1 + α i α 1 c (i = 2, 3,..., m) (i, j) : 2 i m, 2 j n β 1 + α i α 1 c β j c 0 K c θ = α cβ(= α 1 cβ 1 ) h(x) β K(θ) h(x) f 1 (X) g(x) f 1 (X) g(x) β h(x) = X β h(x) K(θ) β K(θ) α = θ + cβ K(θ) K(α, β) K(θ) K(θ) K(α, β) 5. L Aut(L) L 5.1. Aut(L) 5.1. (1) L Aut(L) G F(G) = {a L σ(a) = a σ G} L G (2) L/K L K G(L/K) = {σ Aut(L) σ(a) = a a K} 5.2. (1) F(G) L (2) G(L/K) Aut(L) 1

10 10 Aut(L) G 1, G 2 L K 1, K 2 G 1 G 2 F(G 1 ) F(G 2 ) K 1 K 2 G(L/K 1 ) G(L/K 2 ) 5.3. (1) σ : C C; a + b i a b i (a, b R) R G(C/R) = {id, σ} (2) G(Q( 2)/Q) 5.1. (1) L Aut(L) G (a) (b) (2) L/K (c) G(L/F(G)) G F(G(L/F(G))) = F(G) F(G(L/K)) K (d) G(L/F(G(L/K))) = G(L/K). (a) (c) (b) (a) F F(G(L/F(G))) F(G) (c) K F(G) F(G(L/F(G))) F(G) (d) (b) 5.4. (1) G(Q( 3 2)/Q) = {id} (2) F(G(Q( 3 2)/Q)) Q L/K G(L/K) L/K 6.1. L M K L/K L/M 6.1. L/K G (1) L (2) [L : K] = G (3) F(G) = K.(1) 4.2 L = K(α) α L L/K f(x) = Irr(α, K, X) L L f(x) K L/K f(x) (2) L = K(α) L K σ σ(α) 2.3

11 () 11 α K β σ(α) = β K σ G α K f(x) (3) L F(G) K L/F(G) 5.1(d) G(L/F(G)) = G (2) K F(G) [L : F(G)] = G [L : K] = [L : F(G)][F(G) : K] [F(G) : K] = 1 2 K f(x) L f(x) α 1, α 2,..., α n L/K G ( ) α1... α σ i... α n σ(α 1 )... σ(α i )... σ(α n ) α 1, α 2,..., α n 6.2. Q( d)/q d 6.3. L = Q( 2, i) (1) L Q (2) σ L Q σ( 2) = 2, σ(i) = i τ L Q τ( 2) = 2, τ(i) = i σ τ G = σ, τ G (3) G(L/Q) = G (4) F(G) = Q 6.2. L/K {[K(α) : K] α L} m [K(µ) : K] = m µ L L = K(µ). α L L 4.2 K(µ, α) = K(ν) ν L K(µ) K(ν) [K(µ) : K] [K(µ) : K] = [K(ν) : K] K(µ) = K(ν) α K(µ) = L 6.3. L G Aut(L) K = F(G) (1) L/K (2) [L : K] = G (3) G(L/K) = G. (1) α L {α 1, α 2,..., α r } = {σ(α) σ G} ( ) f(x) = (X α 1 )(X α 2 )... (X α r ) σ G σ(f) = f f(x) K[X] Irr(α, K, X) f(x) L L/K 2 (X a) n

12 12 {[K(α) : K] α L} G = deg f(x) 6.2 L/K (6.1) [L : K] G (2) K (6.2) G G(L/K) (1) 6.1(2) G G(L/K) = [L : K] (6.1) G = G(L/K) = [L : K] (3) (6.2) ((2) ) 6.4. Q X 3 2 L (1) L = Q( 3 2, ω) ω = ( i)/2 (2) σ G(L/Q) { 3 2, 3 2 ω, 3 2 ω 2 } (3) L/Q G(L/Q) 3 S 3 (4) σ L Q σ( 3 2) = 3 2 ω, σ(ω) = ω τ L Q τ( 3 2) = 3 2, τ(ω) = ω 2 σ, τ (5) H = {ρ G(L/Q) ρ( 3 2) = 3 2} K = F(H) (6) L/K (7) K/Q 6.5. Q( i)/q (1) α = i Q α f(x) = Irr(α, Q, X) (2) f(x) L = Q( i) (3) f(x) α 1 = α, α 2, α 3, α 4 (4) ( ). L/K G (1) L/K M G H M G(L/M) F(H) H (2) M H L/M H [L : M] = H, [M : K] = G : H (3) M H M/K H G G(M/K) = G/H.. (1),(2)

13 () 13 (3) ( ) M/K u M σ G σ(u) M τ H τ(σ(u)) = σ(u) σ 1 τσ G(L/M) = H H G ( ) τ H, σ G σ 1 τσ H τσ(u) = σ(u) σ(u) M M/K M/K M/K G(L/K) G(M/K); σ σ M x 3 + ax 2 + bx + c = 0 x = y a/3 (8.1) y 3 + 3py + q = p = b/3 a 2 /9, q = 2a 3 /27 ab/3 + c y = u + v u 3 + v 3 + 3(u + v)(uv + p) + q = 0 (8.2) u 3 + v 3 = q (8.3) uv = p u, v y = u + v (8.1) (8.3) 3 (8.4) u 3 v 3 = p 3 (8.2),(8.4) u 3, v 3 x 2 + qx p 3 = 0 u 3, v 3 = q ± q 2 + 4p 3. 2 y = u + v + u 3 (8.3) (u, v) 3 q + q 2 + 4p 3, 3 q q 2 + 4p 3, q + q 2 + 4p 3 ω, 3 q q 2 + 4p 3 ω 2, q + q 2 + 4p 3 ω 2, 3 q q 2 + 4p 3 ω 2 2

14 14 ω 1 y = u + v x = y a/ (1) X 3 + 3X 1 = 0 (2) X 3 3X 1 = 0 (3) X 3 + 6X 7 = A, B A = , B = (1) X 3 + X 2 = 0 (2) A + B X 3 + X 2 = 0 A + B (3) X 3 + X 2 = X 3 +ax 2 +bx+c = 0 a, b, c f(x) = X 3 + ax 2 + bx + c α 1, α 2, α 3 a, b, c K = Q(a, b, c) f(x) L L = K(α 1, α 2, α 3 ) = Q(a, b, c, α 1, α 2, α 3 ) = Q(α 1, α 2, α 3 ) Aut(L) G {α 1, α 2, α 3 } {1, 2, 3} G F(G) = K K(α 1, α 2, α 3 )/K G = S 3 F(G) = K α 1, α 2, α 3 α 1 + α 2 + α 3 = a, α 1 α 2 + α 2 α 3 + α 3 α 1 = b, α 1 α 2 α 3 = c A 3 = {id, (1, 2, 3), (1, 3, 2)} F(A 3 ) = (α 1 α 2 )(α 1 α 3 )(α 2 α 3 ) D = 2 f(x) = 0 D = 27(4p 3 + q 2 ) K F(A 3 ) = K( ) K( ) K M = K( ) S 3 > A 3 > {id} K M L L M 3 A 3 ω K ω p, q, u, v p, q, ω K, uv = p = (1 ω) 3 (u 3 v 3 ) = (1 ω) 3 3 q 2 + 4p 3 M = K( ) = K( 3 q 2 + 4p 3 ), L = M(u) = M(v)

15 () f(x) = X 3 + 3pX + q Q L Q f(x) { A 3 D 2 G(L/Q) = D 2 S 3 D = 27(4p 3 + q 2 ) f(x) = Q f(x) = X 3 3X + 1 L f(x) L/Q σ S n f(x 1, X 2,..., X n ) f(x σ(1), X σ(2),..., X σ(n) ) K(X 1,..., X n ) S n σ S n σf = f f K(X 1,..., X n ) n () n (X + X 1 )(X + X 2 )... (X + X n ) n k T k T k = ( ) X i U i U U k {1, 2,..., n} T k X 1, X 2,..., X n k (k = 1, 2,..., n) 2. a, b, c a, b, c, d a + b + c, ab + ac + bc, abc a + b + c + d, ab + ac + ad + bc + bd + cd, abc + abd + acd + bcd, abcd 8.2. T 1, T 2,..., T n X 1, X 2,..., X n L = K(X 1, X 2,..., X n ) M = K(T 1, T 2,..., T n ) L/M S n. S n L K F F(S n ) = F L/F S n L F M F = M T k X 1, X 2,..., X n k f(x) = X n T 1 X n ( 1) n T n f(x) = (X X 1 )(X X 2 )... (X X n )

16 16 L f(x) M [L : M] n! L/F [L : F ] = S n = n! ( 1.1) [L : M] 1 [F : M] = [L : F ] 1 F = M 8.4. (1) f/g f g (2) ( ) G G = G 0 G 1 G 2... G i G i 1 (i = 1, 2, 3,... ) G G = G 0 G 1 G 2 G n = {e} G i 1 /G i G 3. G H K K G 4 S 4 ρ = ( ) σ = (1 2)(3 4) G G = {id, ρ, ρ 2, ρ 3, σ, σρ, σρ 2, σρ 3 } ( 4 ) ρ 4 = σ 2 = id ρσ = σρ 1 G G A 4 H H = {id, ρ 2, σ, σρ 2 } K = {id, σ} G H K K G 8.5. G H G K G K H H 8.3. K 0 L f(x) K[X] f(x) = 0 L/K L f(x) F [X] G(L/F ) n S n

17 () p p f(x) Q 2 p 2 f(x) Q L G = G(L/Q) p S p. G G Q f(x) Q p p [L : Q] = G G p p p p S p p G = S p ( ) 5. f(x) = X 5 5X + 1 Q[X] Q S 5 f(x) = (1) n f(x) = X n + a n 1 X n a 0 K = Q(a n 1,..., a 0 ) K f(x) L L/K n S n (2) A n 8.7. A 5 a a n = 1 n a ord(a) a n = 1 n a 8.6. a, b (1) a n = 1 ord(a) n. (2) ord(a) = mn ord(a m ) = n. (3) a, b m = ord(a) n = ord(b) ord(ab) = mn.. (1) ( ). m = ord(a) q, r n = qm + r, 0 r < m 1 = a n = a qm+r = (a m ) q a r = a r m r = 0 ( ). m = ord(a) n = mq q a n = (a m ) q = 1. (2) p = ord(a m ) (a m ) n = a mn = 1 (1) p n a mp = (a m ) p = 1 (1) mn = ord(a) mp n p (3). p = ord(ab) m, n (ab) mn = (a m ) n (b n ) m = 1

18 18 (1) p mn mn p b np = (b n ) p = 1 a np = a np b np = (ab) np = ((ab) p ) n = 1 (1) m np m n m p n p p m, n mn 8.7. G. G n n = 1 n > 1 n n n = p e 1 1 pe pes s p 1, p 2,..., p s i p ei i n G 8.6(2) p e i i a i 8.6(3) ord(a 1 a 2... a s ) = n G n i (1 i s) a G p e pe i 1 i 1 pe i 1 i p e i+1 i+1... pe s s = n/p i a n/p i = 1 X n/p i = 1 n/p i G a, b a 1 b 1 ab [a, b] a, b G H, K [h, k](h H, k K) [h, k] h H, k K [H, K] [G, G] G [a, b] (a, b G) G g g 1 [a, b]g = [g 1 ag, g 1 bg] [a i, b i ] (a i, b i G, 1 i n) g 1 [a 1, b 1 ][a 2, b 2 ]... [a n, b n ]g = g 1 [a 1, b 1 ]gg 1 [a 2, b 2 ]g... g 1 [a n, b n ]g = [g 1 a 1 g, g 1 b 1 g][g 1 a 2 g, g 1 b 2 g]... [g 1 a n g, g 1 b n g] [G, G] G G H (1) G/H (2) bah = abh (a, b G) (3) a 1 b 1 ab H (a, b G) (4) [G, G] H G [G, G] G/H H

19 () 19 G D 0 (G) = G G D i (G) = [D i 1 (G), D i 1 (G)] (i > 0) D 0 (G) D 1 (G) D 2 (G) G G. G = G 0 G 1 G 2 G n = {e} G i 1 /G i 8.8 G i D i (G) D n (G) = {e} A n (n 3) 3. (i j), (k l) 0 3 ( 0 1 ) {i, j} {k, l} {i, j} {k, l} 1 () j = l (i j)(k j) = (i j k) 3 3 (i j)(k j) = (i k j)(i k l) 6. i, j, k, l, m (m l k) 1 (i j k) 1 (m l k)(i j k) = (j k l) n 5 j, k, l i, m 3 3 D(A n ) = A n A n p S p τ p σ. S n (1 2) ( n) τ (1 2) σ 1 k 2 σ k σ k 1 p σ k 1 = ( ) 3,..., p τ σ k 1 S n τ σ S p [1] ( )1983 [2] ( )1989 [3] John M. Howie Fields and Galois Theory (Springer)2006

16 B

16 B 16 B (1) 3 (2) (3) 5 ( ) 3 : 2 3 : 3 : () 3 19 ( ) 2 ax 2 + bx + c = 0 (a 0) x = b ± b 2 4ac 2a 3, 4 5 1824 5 Contents 1. 1 2. 7 3. 13 4. 18 5. 22 6. 25 7. 27 8. 31 9. 37 10. 46 11. 50 12. 56 i 1 1. 1.1..

More information

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

More information

newmain.dvi

newmain.dvi 数論 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/008142 このサンプルページの内容は, 第 2 版 1 刷発行当時のものです. Daniel DUVERNEY: THÉORIE DES NOMBRES c Dunod, Paris, 1998, This book is published

More information

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C 0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,

More information

,2,4

,2,4 2005 12 2006 1,2,4 iii 1 Hilbert 14 1 1.............................................. 1 2............................................... 2 3............................................... 3 4.............................................

More information

1

1 1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................

More information

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C 8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,

More information

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

More information

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 = 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

More information

AI n Z f n : Z Z f n (k) = nk ( k Z) f n n 1.9 R R f : R R f 1 1 {a R f(a) = 0 R = {0 R 1.10 R R f : R R f 1 : R R 1.11 Z Z id Z 1.12 Q Q id

AI n Z f n : Z Z f n (k) = nk ( k Z) f n n 1.9 R R f : R R f 1 1 {a R f(a) = 0 R = {0 R 1.10 R R f : R R f 1 : R R 1.11 Z Z id Z 1.12 Q Q id 1 1.1 1.1 R R (1) R = 1 2 Z = 2 n Z (2) R 1.2 R C Z R 1.3 Z 2 = {(a, b) a Z, b Z Z 2 a, b, c, d Z (a, b) + (c, d) = (a + c, b + d), (a, b)(c, d) = (ac, bd) (1) Z 2 (2) Z 2? (3) Z 2 1.4 C Q[ 1] = {a + bi

More information

2012 A, N, Z, Q, R, C

2012 A, N, Z, Q, R, C 2012 A, N, Z, Q, R, C 1 2009 9 2 2011 2 3 2012 9 1 2 2 5 3 11 4 16 5 22 6 25 7 29 8 32 1 1 1.1 3 1 1 1 1 1 1? 3 3 3 3 3 3 3 1 1, 1 1 + 1 1 1+1 2 2 1 2+1 3 2 N 1.2 N (i) 2 a b a 1 b a < b a b b a a b (ii)

More information

II Time-stamp: <05/09/30 17:14:06 waki> ii

II Time-stamp: <05/09/30 17:14:06 waki> ii II waki@cc.hirosaki-u.ac.jp 18 1 30 II Time-stamp: ii 1 1 1.1.................................................. 1 1.2................................................... 3 1.3..................................................

More information

, = = 7 6 = 42, =

, = = 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

More information

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a 3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (

More information

r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B

r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B 1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n

More information

( )

( ) 18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................

More information

熊本県数学問題正解

熊本県数学問題正解 00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (

More information

.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 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

More information

名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト

名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト 名古屋工業大の数学 年 ~5 年 大学入試数学動画解説サイト http://mathroom.jugem.jp/ 68 i 4 3 III III 3 5 3 ii 5 6 45 99 5 4 3. () r \= S n = r + r + 3r 3 + + nr n () x > f n (x) = e x + e x + 3e 3x + + ne nx f(x) = lim f n(x) lim

More information

1 Abstract 2 3 n a ax 2 + bx + c = 0 (a 0) (1) ( x + b ) 2 = b2 4ac 2a 4a 2 D = b 2 4ac > 0 (1) 2 D = 0 D < 0 x + b 2a = ± b2 4ac 2a b ± b 2

1 Abstract 2 3 n a ax 2 + bx + c = 0 (a 0) (1) ( x + b ) 2 = b2 4ac 2a 4a 2 D = b 2 4ac > 0 (1) 2 D = 0 D < 0 x + b 2a = ± b2 4ac 2a b ± b 2 1 Abstract n 1 1.1 a ax + bx + c = 0 (a 0) (1) ( x + b ) = b 4ac a 4a D = b 4ac > 0 (1) D = 0 D < 0 x + b a = ± b 4ac a b ± b 4ac a b a b ± 4ac b i a D (1) ax + bx + c D 0 () () (015 8 1 ) 1. D = b 4ac

More information

IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a

IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a 1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =

More information

2000年度『数学展望 I』講義録

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

More information

1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.

1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1. 1 1 n 0, 1, 2,, n 1 1.1 n 2 a, b a n b n a, b n a b (mod n) 1 1. n = 10 1567 237 (mod 10) 2. n = 9 1567 1826578 (mod 9) n II Z n := {0, 1, 2,, n 1} 1.2 a b a = bq + r (0 r < b) q, r q a b r 2 1. a = 456,

More information

II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K

II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F

More information

2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+

2 7 V 7 {fx fx 3 } 8 P 3 {fx fx 3 } 9 V 9 {fx fx f x 2fx } V {fx fx f x 2fx + } V {{a n } {a n } a n+2 a n+ + a n n } 2 V 2 {{a n } {a n } a n+2 a n+ R 3 R n C n V??,?? k, l K x, y, z K n, i x + y + z x + y + z iv x V, x + x o x V v kx + y kx + ky vi k + lx kx + lx vii klx klx viii x x ii x + y y + x, V iii o K n, x K n, x + o x iv x K n, x + x o x

More information

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 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.

More information

ver Web

ver Web ver201723 Web 1 4 11 4 12 5 13 7 2 9 21 9 22 10 23 10 24 11 3 13 31 n 13 32 15 33 21 34 25 35 (1) 27 4 30 41 30 42 32 43 36 44 (2) 38 45 45 46 45 5 46 51 46 52 48 53 49 54 51 55 54 56 58 57 (3) 61 2 3

More information

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

I A A441 : April 15, 2013 Version : 1.1 I   Kawahira, Tomoki TA (Shigehiro, Yoshida ) I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17

More information

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

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

More information

2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a

More information

A S- hara/lectures/lectures-j.html r A = A 5 : 5 = max{ A, } A A A A B A, B A A A %

A S-   hara/lectures/lectures-j.html r A = A 5 : 5 = max{ A, } A A A A B A, B A A A % A S- http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html r A S- 3.4.5. 9 phone: 9-8-444, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office

More information

2016 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 16 2 1 () X O 3 (O1) X O, O (O2) O O (O3) O O O X (X, O) O X X (O1), (O2), (O3) (O2) (O3) n (O2) U 1,..., U n O U k O k=1 (O3) U λ O( λ Λ) λ Λ U λ O 0 X 0 (O2) n =

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

More information

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 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

More information

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)

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

More information

() 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 (

() 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

More information

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

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 ),

More information

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.   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)

More information

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59

More information

基礎数学I

基礎数学I I & II ii ii........... 22................. 25 12............... 28.................. 28.................... 31............. 32.................. 34 3 1 9.................... 1....................... 1............

More information

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ = 1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A

More information

dynamics-solution2.dvi

dynamics-solution2.dvi 1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj

More information

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

http://know-star.com/ 3 1 7 1.1................................. 7 1.2................................ 8 1.3 x n.................................. 8 1.4 e x.................................. 10 1.5 sin

More information

8 i, III,,,, III,, :!,,,, :!,,,,, 4:!,,,,,,!,,,, OK! 5:!,,,,,,,,,, OK 6:!, 0, 3:!,,,,! 7:!,,,,,, ii,,,,,, ( ),, :, ( ), ( ), :... : 3 ( )...,, () : ( )..., :,,, ( ), (,,, ),, (ϵ δ ), ( ), (ˆ ˆ;),,,,,,!,,,,.,,

More information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

医系の統計入門第 2 版 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, 第 2 版 1 刷発行時のものです.

医系の統計入門第 2 版 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, 第 2 版 1 刷発行時のものです. 医系の統計入門第 2 版 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/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987

More information

?

? 240-8501 79-2 Email: nakamoto@ynu.ac.jp 1 3 1.1...................................... 3 1.2?................................. 6 1.3..................................... 8 1.4.......................................

More information

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

More information

D 24 D D D

D 24 D D D 5 Paper I.R. 2001 5 Paper HP Paper 5 3 5.1................................................... 3 5.2.................................................... 4 5.3.......................................... 6

More information

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1 ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD

More information

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

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

More information

ad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign(

ad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign( I n n A AX = I, YA = I () n XY A () X = IX = (YA)X = Y(AX) = YI = Y X Y () XY A A AB AB BA (AB)(B A ) = A(BB )A = AA = I (BA)(A B ) = B(AA )B = BB = I (AB) = B A (BA) = A B A B A = B = 5 5 A B AB BA A

More information

a (a + ), a + a > (a + ), a + 4 a < a 4 a,,, y y = + a y = + a, y = a y = ( + a) ( x) + ( a) x, x y,y a y y y ( + a : a ) ( a : a > ) y = (a + ) y = a

a (a + ), a + a > (a + ), a + 4 a < a 4 a,,, y y = + a y = + a, y = a y = ( + a) ( x) + ( a) x, x y,y a y y y ( + a : a ) ( a : a > ) y = (a + ) y = a [] a x f(x) = ( + a)( x) + ( a)x f(x) = ( a + ) x + a + () x f(x) a a + a > a + () x f(x) a (a + ) a x 4 f (x) = ( + a) ( x) + ( a) x = ( a + a) x + a + = ( a + ) x + a +, () a + a f(x) f(x) = f() = a

More information

直交座標系の回転

直交座標系の回転 b T.Koama x l x, Lx i ij j j xi i i i, x L T L L, L ± x L T xax axx, ( a a ) i, j ij i j ij ji λ λ + λ + + λ i i i x L T T T x ( L) L T xax T ( T L T ) A( L) T ( LAL T ) T ( L AL) λ ii L AL Λ λi i axx

More information

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..........

More information

untitled

untitled 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

More information

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0 1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

More information

O E ( ) A a A A(a) O ( ) (1) O O () 467

O E ( ) A a A A(a) O ( ) (1) O O () 467 1 1.0 16 1 ( 1 1 ) 1 466 1.1 1.1.1 4 O E ( ) A a A A(a) O ( ) (1) O O () 467 ( ) A(a) O A 0 a x ( ) A(3), B( ), C 1, D( 5) DB C A x 5 4 3 1 0 1 3 4 5 16 A(1), B( 3) A(a) B(b) d ( ) A(a) B(b) d AB d = d(a,

More information

n ( (

n ( ( 1 2 27 6 1 1 m-mat@mathscihiroshima-uacjp 2 http://wwwmathscihiroshima-uacjp/~m-mat/teach/teachhtml 2 1 3 11 3 111 3 112 4 113 n 4 114 5 115 5 12 7 121 7 122 9 123 11 124 11 125 12 126 2 2 13 127 15 128

More information

ii

ii ii iii 1 1 1.1..................................... 1 1.2................................... 3 1.3........................... 4 2 9 2.1.................................. 9 2.2...............................

More information

(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y

(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y (2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b

More information

2011de.dvi

2011de.dvi 211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37

More information

1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x

1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x . P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +

More information

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10% 1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n

More information

function2.pdf

function2.pdf 2... 1 2009, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 38 : 5) i) [], : 84 85 86 87 88 89 1000 ) 13 22 33 56 92 147 140 120 100 80 60 40 20 1 2 3 4 5 7.1 7 7.1 1. *1 e = 2.7182 ) fx) e x, x R : 7.1)

More information

zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {

zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 { 04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory

More information

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3 13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >

More information

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i 1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [

More information

SO(2)

SO(2) TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6

More information

( ) 2002 1 1 1 1.1....................................... 1 1.1.1................................. 1 1.1.2................................. 1 1.1.3................... 3 1.1.4......................................

More information

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

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

More information

,. 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,,.

,. 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,

More information

6.1 (P (P (P (P (P (P (, P (, P.

6.1 (P (P (P (P (P (P (, P (, P. (011 30 7 0 ( ( 3 ( 010 1 (P.3 1 1.1 (P.4.................. 1 1. (P.4............... 1 (P.15.1 (P.16................. (P.0............3 (P.18 3.4 (P.3............... 4 3 (P.9 4 3.1 (P.30........... 4 3.

More information

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

More information

20 9 19 1 3 11 1 3 111 3 112 1 4 12 6 121 6 122 7 13 7 131 8 132 10 133 10 134 12 14 13 141 13 142 13 143 15 144 16 145 17 15 19 151 1 19 152 20 2 21 21 21 211 21 212 1 23 213 1 23 214 25 215 31 22 33

More information

20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................

More information

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,, 01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,

More information

行列代数2010A

行列代数2010A a ij i j 1) i +j i, j) ij ij 1 j a i1 a ij a i a 1 a j a ij 1) i +j 1,j 1,j +1 a i1,1 a i1,j 1 a i1,j +1 a i1, a i +1,1 a i +1.j 1 a i +1,j +1 a i +1, a 1 a,j 1 a,j +1 a, ij i j 1,j 1,j +1 ij 1) i +j a

More information

量子力学 問題

量子力学 問題 3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................

More information

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6 1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67

More information

No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y

No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ

More information

i 6 3 ii 3 7 8 9 3 6 iii 5 8 5 3 7 8 v...................................................... 5.3....................... 7 3........................ 3.................3.......................... 8 3 35

More information

.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(

.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,

More information

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 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

More information

A

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

More information

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

More information

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

: , 2.0, 3.0, 2.0, (%) ( 2. 2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................

More information

25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3

More information

2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................

More information

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π . 4cm 6 cm 4cm cm 8 cm λ()=a [kg/m] A 4cm A 4cm cm h h Y a G.38h a b () y = h.38h G b h X () S() = π() a,b, h,π V = ρ M = ρv G = M h S() 3 d a,b, h 4 G = 5 h a b a b = 6 ω() s v m θ() m v () θ() ω() dθ()

More information

Microsoft Word - 11問題表紙(選択).docx

Microsoft Word - 11問題表紙(選択).docx A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx

More information

/02/18

/02/18 3 09/0/8 i III,,,, III,?,,,,,,,,,,,,,,,,,,,,?,?,,,,,,,,,,,,,,!!!,? 3,,,, ii,,,!,,,, OK! :!,,,, :!,,,,,, 3:!,, 4:!,,,, 5:!,,! 7:!,,,,, 8:!,! 9:!,,,,,,,,, ( ),, :, ( ), ( ), 6:!,,, :... : 3 ( )... iii,,

More information

n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................

More information

6.1 (P (P (P (P (P (P (, P (, P.101

6.1 (P (P (P (P (P (P (, P (, P.101 (008 0 3 7 ( ( ( 00 1 (P.3 1 1.1 (P.3.................. 1 1. (P.4............... 1 (P.15.1 (P.15................. (P.18............3 (P.17......... 3.4 (P................ 4 3 (P.7 4 3.1 ( P.7...........

More information

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx 4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan

More information

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

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 m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m

More information

5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6 cos π 6.7 MP 4 P P N i i i i N i j F j ii N i i ii F j i i N ii li i F j i ij li i i i

5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6 cos π 6.7 MP 4 P P N i i i i N i j F j ii N i i ii F j i i N ii li i F j i ij li i i i i j ij i j ii,, i j ij ij ij (, P P P P θ N θ P P cosθ N F N P cosθ F Psinθ P P F P P θ N P cos θ cos θ cosθ F P sinθ cosθ sinθ cosθ sinθ 5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6

More information