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
|
|
- ことこ はしかわ
- 5 years ago
- Views:
Transcription
1 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 = αx + βx (2) α(x + y) =αx + αy (3) α(βx) =(αβ)x (4) x = x (2) (3) (4) x + = + x = x, x +( x) =( x)+x = V x, y x y = x +( y), V = + = =
2 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 n R n x R n x 2 = x,x 2,,x n R x n R n x y x + y x αx x 2 y 2 x 2 + y 2 x 2 αx 2 + =, α = x n y n x n + y n x n αx n R n R (I)(II) = t (,,,), t (x,x 2,,x n ) t ( x, x 2,, x n ) R n n Euclidean space 2 R 2 3 R 3 R m n M(m, n; R) a a n M(m, n; R) = a ij R, i m, j n a m a mn M(m, n; R) R 2
3 2 2 R R F f,g F α R f + g αf x R (f + g)(x) =f(x)+g(x), (αf)(x) =αf(x) F R (I)()(2) f (x) f x R f(x) = f(x) f (II) f = f, ( )f = f f f f (x) =(x) f f = f F f x f(x) f(x) =(x) 3 3= 3 3= f f = F R 2 P P = {{a n } n=,2, a n R,n N} {a n }, {b n } P α R {a n } + {b n } α{a n } {a n } + {b n } = {a n + b n }, α{a n } = {αa n } P R V W V W V vector subspace W ( φ) (I)(II) x, y W = x + y W, x W, α R = αx W 3
4 W V (I)()(2) (II)() (4) (I)(3)(4) V W x W x V x W 2 α = x = W α = ( )x = x W V {} V 3 () n M(n; R) =M(n, n; R) S, A, U M(n; R) (2) F 2 C = {f F f(x) R } C = {f C f(x) R } C F C C F C = {f F f(x) } F (3) P 2 G λ = {{a n } P a n+ = λa n (n N)} G λ P A λ = {{a n } P a n+ = a n + λ (n N)} λ P a,a 2,,a n (a,a 2,,a n ) x (,,,) R R n x 2 = x,x 2,,x n R x n 4
5 x Hc n = x 2 R n a x + a 2 x a n x n = c x n H n c Hn c x y x 2 y 2, H c n α R x n y n a (x + y )+a 2 (x 2 + y 2 )+ + a n (x n + y n )=2c a (αx )+a 2 (αx 2 )+ + a n (αx n )=α(a x + a 2 x a n x n )=αc Hc n c = H 3 c R3 H n c Rn hyperplane H 2 c R 2 H n Hc n (c ) 4 R R 3 R 3 x x () W = y x + y + z = (2) W 2 = y x, y, z z z x (3) W 3 = y xyz = z x (4) W 4 = y z x 2 = y 3 = z 4 x (5) W 5 = y x 2 + y 2 + z 2 = z x (7) W 7 = y x =y = z z x (6) W 6 = y x = y = z z x (8) W 8 = y x 2 + y 2 + z 2 < z 5
6 5 V R a, a 2,,a m V, λ,λ 2,,λ m R λ a + λ 2 a λ m a m a, a 2,,a m linear combination a, a 2,,a m V S S = {λ a + λ 2 a λ m a m λ,λ 2,,λ m R} S V 5 S a, a 2,,a m span[a, a 2,,a m ] G(a, a 2,,a m ), {{a, a 2,,a m }}, [a, a 2,,a m ] 6 R 3 a =, a 2 =, a 3 = λ + λ 3 µ span[a, a 2, a 3 ]= λ 2 λ 3 λ,λ 2,λ 3 R = µ 2 λ 2 + λ 3 µ 2 x = y y + z = = span[a, a 2 ] z µ,µ 2 R a, a 2, a 3 a, a 2 a 3 = a + a 2 a, a 2 a 3 span[a, a 2 ] R 3 a a 2 a, a 2 6 V R V a, a 2,,a m span[a, a 2,,a m ] b 6
7 span[a, a 2,,a m, b] = span[a, a 2,,a m ] V a, a 2,,a m m a, a 2,,a m linearly dependent linearly independent () a, a 2,,a m λ a + λ 2 a λ m a m = λ,λ 2,,λ m (λ,λ 2,,λ m ) (,,,) (2) a, a 2,,a m λ a + λ 2 a λ m a m = (λ,λ 2,,λ m )=(,,,) 7 R 2 R 2 a, b a = 2, b = 7+k 7 k 4 2α +(7+k)β =, αa + βb = 4 7+k (7 k)α +4β =, 48α + (28 + 4k)β =, (k 2 )α = (49 k 2 )α + (28 + k)β = k ± α = β = k = 3α +2β = k = 2α + β = (α, β) (, ) α, β αa + βb = k = 2a +3b =, k = a +2b = a, b k ± k = ± a, b R 2 αa + βb = α a = β b, β α b = α a (α, β) (, ) a, b β a, b a, b a, b R 2 a, b R 2 a, b R 2 a, b R 2 7
8 7 R 2 a = a, a 2 = b, c d () a, a 2, ad bc (2) ad bc, R 2 e, e 2 a, a 2 8 R 3 R 3 a, b, c k a =, b =, c = αa + βb + γc = γk =,α = β,α + β + γ = k γ =, α = β = k = γ = 2α, α = β, (α, β, γ) (,, ) α, β, γ αa + βb + γc = a + b 2c = a, b, c k k = a, b, c R 3 αa + βb + γc = α a = β α b γ c, α a b c a b c a, b, c β,γ a, b, c R 3 a, b, c R 3 a, b, c R 3 a, b, c R 3 8 R () a =, b = 3, c = 3 (2) a =, b =, c = (3) a =, b =, c = (4) a =, b =, c =, d = 8
9 2 (5) a =, b =, c = 3 k (k ) 9 R n R n a, a 2,,a n λ a + λ 2 a λ n a n = (λ,λ 2,,λ n ) = (,,,) a, a 2,,a n n A n λ = t (λ,λ 2,,λ n ) λ Aλ = λ = (A) 9 (A) n A n a, a 2,,a n det A = (B) R n k a,,a k n P a,,a k P a,,pa k V R V a, a 2 aa + ba 2,ca + da 2 a, b, c, d aa + ba 2,ca + da 2 α(aa + ba 2 )+ β(ca + da 2 )= α = β = a, a 2 αa + βc = α, β α = β = αb + βd = ad bc V R (A) a, b, c V () a, b, c 2 (2) a + b, b + c, c + a (3) a, a + b + c, a b c (4) x = a + b 2c, y = a b c, z = a + c (5) u = a + b 3c, v = a +3b c, w = b + c 9
10 (B) a, a 2,,a m V, c 2,c 3,,c m R a = a + c 2 a 2 + c 3 a c m a m, a, a 2,,a m a, a 2,,a m (C) a, a 2,,a k V a, a 2,,a k V c a + c 2 a c k a k = c a + c 2 a c k a k = c = c,c 2 = c 2,,c k = c k, 9(B) A rank A =(A ) =(A ) n A =(a, a 2,,a n ) rank A = n a, a 2,,a n n A =(a, a 2,,a n ) () A n (2) rank A = n (3) A n a, a 2,,a n (4) det A (5) x R n Ax = b b R n (6) x R n Ax = (7) A
11 V a, a 2,,a n V () span[a, a 2,,a n ]=V (2) a, a 2,,a n a, a 2,,a n V basis a, a 2,,a n R R n = x x 2 x n x,x 2,,x n R e =, e 2 =,,e n = R n x = x x 2 x n () x = x e + x 2 e x n e n (2) x e + x 2 e x n e n = x x 2 x n = e, e 2,,e n e, e 2,,e n R n () R n a =, a 2 =,,a n =
12 (2) R C (3) C C (4) R V a, b a + b, a b 2 (4) V n V n dim V = n (2)(3) R C 2 C C C V V x P f (x) =,f (x) =x, f 2 (x) =x 2,,f n (x) =x n, P 2 x () W = y R3 x = y = z z x x 2 (2) H = R 4 x x + x 2 + x 3 + x 4 = 3 x 4 (3) R C 2 (4) C C 2 n V W a, a 2,,a k n k V b, b 2,,b n k a, a 2,,a k, b, b 2,,b n k V V k V 2
13 (I) A, B M(m, n; R) A + B M(m, n; R) ()(2) (3) m n O m,n, (4)A A (II) A M(m, n; R) α R αa M(m, n; R) () (4) 2 (I)() ({a n } + {b n })+{c n } = {a n + b n } + {c n } = {(a n + b n )+c n } = {a n +(b n + c n )} = {a n } + {b n + c n } = {a n } +({b n } + {c n }) (2) (3) {z n } : z =,z 2 =, (4){a n } {a n } = { a n } {a n } {z n } = {}, {a n } = {a n } 3 () S, A, U M(n; R) S M(n; R) A, U (2) C C F C F C C f F C f(x) ( x R), f F C F C F (3) G λ P {a n }, {b n } G λ a n+ + b n+ = λa n + λb n = λ(a n + b n ) {a n } + {b n } = {a n + b n } G λ α R αa n+ = αλa n = λ(αa n ) α{a n } = {αa n } G λ G λ A λ P {a n }, {b n } A λ a n+ + b n+ = a n + λa n + b n + λ = a n + b n +2λ λ {a n } + {b n } = {a n + b n } A λ, λ = {a n } + {b n } A α R αa n+ = α(a n + λ) =αa n + αλ λ α{a n } = {αa n } A λ, λ = α{a n } A A P A λ (λ ) 2 4 (4)(6) W 4 = t 3 t R, W 6 = t t R 4 3
14 ()(3)(5)(7) (2)(8) x x y α R α y z z a = 2 (2) a W 2 3a W 2 (8) a W 8 2a W 8 5 S x, y x = λ a + + λ m a m, y = µ a + + µ m a m x + y =(λ + µ )a + +(λ m + µ m )a m S α R αx =(αλ )a + +(αλ m )a m S 6 span[a, a 2,,a m, b] span[a, a 2,,a m ] x span[a, a 2,,a m ] x = λ a + λ 2 a λ m a m = λ a + λ 2 a λ m a m +b x span[a, a 2,,a m, b] x span[a, a 2,,a m, b] b span[a, a 2,,a m ] b = λ a + λ 2 a λ m a m x = µ a + + µ m a m + µ m+ b =(µ + µ m+ λ )a + +(µ m + µ m+ λ m )a m x span[a, a 2,,a m ] span[a, a 2,,a m, b] span[a, a 2,,a m ] span[a, a 2,,a m, b] = span[a, a 2,,a m ] aα + bβ =, 7 () αa + βa 2 = α, β cα + dβ = a b α = c d β a, a 2 ad bc = α = β = ad bc 4
15 ad bc a c a, a 2 b α = d β 9(A) R n (2) αa +βa 2 = a b α = = e α = c d β β d e = ad bc c ad bc (da ca 2 ) e 2 = ad bc d c b = a ad bc ( ba + aa 2 ) α + β +5γ =, 8 () αa + βb + γc = α +3β +3γ =, 3 β +2γ =, β =2γ 2 α =9γ 6γ = γ = α = β = (2) (3) (4) (5)k k = 9 (A) α,α 2,,α n R, α a + + α n a n = A α α n = ( ) det A, A, (α,,α n ) ( ) α = = α n = a,,a n det A =, ( ) (α,,α n ) (,,) a,,a n (B) α (P a )+ + α k (P a k )=P(α a + + α k a k )= P α a + + α k a k = a,,a k α = = α k = P a,,pα k (A) α, β, γ R () a, b, c 2, a, b,, a = αb a, b, c (2) α(a + b)+β(b + c)+γ(c + a) =(α + γ)a +(α + β)b +(β + γ)c = a, b, c, α + γ = α + β = β + γ =, α = β = γ = (3) αa + β(a + b + c)+γ(a b c) =(α + β + γ)a +(β γ)b +(β γ)c = a, b, c, α = 2β,γ = β 5
16 (4) αx + βy + γz =(α + β + γ)a +(α β)b +( 2α β + γ)c =, a, b, c, α + β + γ = α β = 2α β + γ =, α = β = γ = x, y, z (5) αu + βv + γw =(α + β)a +(α +3β + γ)b +( 3α β + γ)c =, a, b, c, α + β = α +3β + γ = 3α β + γ =, β = α, γ =2α, u, v, w (B) b,b 2,,b m F, b a + b 2a b m a m = b a +(b c 2 + b 2 )a 2 + +(b c m + b m )a m = a, a 2,,a m, b = b c 2 + b 2 = = b c m + b m =, b = b 2 = = b m = a, a 2,,a m (C) c a + c 2 a c k a k = c a + c 2a c ka k (c c )a +(c 2 c 2 )a 2 + +(c k c k )a k = a, a 2,,a k c c = c 2 c 2 = = c k c k = c = c,c 2 = c 2,,c k = c k c = c 2 = = c k = a, a 2,,a k x () R n x 2 x = e, e 2,,e n x n n x = x i e i a i = e i, = (i =,,n) i= ( n n n nk= ) x k x = x i a i + x k = i= k= i= n x i a i a, a 2,,a n n n n x i a i = x i = x i e i i= i= i= n x i = x = x 2 = = x n x = x 2 = = x n = a, a 2,,a n i= a, a 2,,a n R n (2) C x α,α 2 R e =, e 2 = i x = α e + α 2 e 2 e, e 2 =,i R C (3) C x α = x C e = x = αe e = C C 6
17 (4) V x α, β R x = αa + βb λ = α + β,µ = α β αa + βb = λ(a + b) +µ(a b) x 2 2 a + b, a b λ(a + b)+µ(a b) = a, b λ + µ = λ µ = λ = µ = a + b, a b a + b, a b 2 () a = W = {ta t R} W a R3 (2) a =, b =, c = H = {ra + sb + tc r, s, t R} H a, b, c R 4 3 (3) e =, e 2 =, e 3 = i, e 4 = R C 2 i z R a, b, c, d z = ae + be 2 + ce 3 + de 4 C 2 z 2 z 2 e, e 2, e 3, e 4 4 (4) e =, e 2 = C C 2 z C a = z,b= z 2 z = ae + be 2 C 2 e, e 2 2 z 2 z 2 7
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
More informationn ( (
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.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 informationad 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 informationII 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 informationkoji07-01.dvi
2007 I II III 1, 2, 3, 4, 5, 6, 7 5 10 19 (!) 1938 70 21? 1 1 2 1 2 2 1! 4, 5 1? 50 1 2 1 1 2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 3 1, 2 1, 3? 2 1 3 1 2 1 1, 2 2, 3? 2 1 3 2 3 2 k,l m, n k,l m, n kn > ml...?
More information/ 2 n n n n x 1,..., x n 1 n 2 n R n n ndimensional Euclidean space R n vector point R n set space R n R n x = x 1 x n y = y 1 y n distance dx,
1 1.1 R n 1.1.1 3 xyz xyz 3 x, y, z R 3 := x y : x, y, z R z 1 3. n n x 1,..., x n x 1. x n x 1 x n 1 / 2 n n n n x 1,..., x n 1 n 2 n R n n ndimensional Euclidean space R n vector point 1.1.2 R n set
More informationax 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 informationver 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 informationx () 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.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 information1990 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 information190 87 28 1 212 77 1777 77 219 1 171 28 201 1 1 16 102 17 10 1 16 99 1 1 1 1 960 1 1 1 1 1 2 168 1 12 2 18 100 2 1 6 1 61 7 16 18 20 2 961 2 11 6 2 6 6 0 17 86 1 2 16 1 1 9 2 1 1 1 1 1 1 0 2 17 16 6 1
More informationx, 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 information20 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 information1 n A a 11 a 1n A =.. a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = 0 ( x 0 ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 1.1 Th
1 n A a 11 a 1n A = a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = ( x ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 11 Th9-1 Ax = λx λe n A = λ a 11 a 12 a 1n a 21 λ a 22 a n1 a n2
More informationII 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漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト
https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,
More informationi 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 informationJanuary 27, 2015
e-mail : kigami@i.kyoto-u.ac.jp January 27, 205 Contents 2........................ 2.2....................... 3.3....................... 6.4......................... 2 6 2........................... 6
More informationA(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 information1. 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数学Ⅱ演習(足助・09夏)
II I 9/4/4 9/4/2 z C z z z z, z 2 z, w C zw z w 3 z, w C z + w z + w 4 t R t C t t t t t z z z 2 z C re z z + z z z, im z 2 2 3 z C e z + z + 2 z2 + 3! z3 + z!, I 4 x R e x cos x + sin x 2 z, w C e z+w
More information第86回日本感染症学会総会学術集会後抄録(I)
κ κ κ κ κ κ μ μ β β β γ α α β β γ α β α α α γ α β β γ μ β β μ μ α ββ β β β β β β β β β β β β β β β β β β γ β μ μ μ μμ μ μ μ μ β β μ μ μ μ μ μ μ μ μ μ μ μ μ μ β
More information2 (2016 3Q N) c = o (11) Ax = b A x = c A n I n n n 2n (A I n ) (I n X) A A X A n A A A (1) (2) c 0 c (3) c A A i j n 1 ( 1) i+j A (i, j) A (i, j) ã i
[ ] (2016 3Q N) a 11 a 1n m n A A = a m1 a mn A a 1 A A = a n (1) A (a i a j, i j ) (2) A (a i ca i, c 0, i ) (3) A (a i a i + ca j, j i, i ) A 1 A 11 0 A 12 0 0 A 1k 0 1 A 22 0 0 A 2k 0 1 0 A 3k 1 A rk
More informationA, B, C. (1) A = A. (2) A = B B = A. (3) A = B, B = C A = C. A = B. (3)., f : A B g : B C. g f : A C, A = C. 7.1, A, B,. A = B, A, A A., A, A
91 7,.,, ( ).,,.,.,. 7.1 A B, A B, A = B. 1), 1,.,. 7.1 A, B, 3. (i) A B. (ii) f : A B. (iii) A B. (i) (ii)., 6.9, (ii) (iii).,,,. 1), Ā = B.. A, Ā, Ā,. 92 7 7.2 A, B, C. (1) A = A. (2) A = B B = A. (3)
More information.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
More informationDVIOUT-HYOU
() P. () AB () AB ³ ³, BA, BA ³ ³ P. A B B A IA (B B)A B (BA) B A ³, A ³ ³ B ³ ³ x z ³ A AA w ³ AA ³ x z ³ x + z +w ³ w x + z +w ½ x + ½ z +w x + z +w x,,z,w ³ A ³ AA I x,, z, w ³ A ³ ³ + + A ³ A A P.
More information2001 年度 『数学基礎 IV』 講義録
4 A 95 96 4 1 n {1, 2,,n} n n σ ( ) 1 2 n σ(1) σ(2) σ(n) σ σ 2 1 n 1 2 {1, 2,,n} n n! n S n σ, τ S n {1, 2,,n} τ σ {1, 2,,n} n τ σ σ, τ τσ σ n σ 1 n σ 1 ( σ σ ) 1 σ = σσ 1 = ι 1 2 n ι 1 2 n 4.1. 4 σ =
More information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 information4 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 information2S III IV K A4 12:00-13:30 Cafe David 1 2 TA 1 appointment Cafe David K2-2S04-00 : C
2S III IV K200 : April 16, 2004 Version : 1.1 TA M2 TA 1 10 2 n 1 ɛ-δ 5 15 20 20 45 K2-2S04-00 : C 2S III IV K200 60 60 74 75 89 90 1 email 3 4 30 A4 12:00-13:30 Cafe David 1 2 TA 1 email appointment Cafe
More informationO x y z O ( O ) O (O ) 3 x y z O O x v t = t = 0 ( 1 ) O t = 0 c t r = ct P (x, y, z) r 2 = x 2 + y 2 + z 2 (t, x, y, z) (ct) 2 x 2 y 2 z 2 = 0
9 O y O ( O ) O (O ) 3 y O O v t = t = 0 ( ) O t = 0 t r = t P (, y, ) r = + y + (t,, y, ) (t) y = 0 () ( )O O t (t ) y = 0 () (t) y = (t ) y = 0 (3) O O v O O v O O O y y O O v P(, y,, t) t (, y,, t )
More information取扱説明書 -詳細版- 液晶プロジェクター CP-AW3019WNJ
B A C D E F K I M L J H G N O Q P Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C LAN RS-232C LAN LAN BE EF 03 06 00 2A D3 01 00 00 60 00 00 BE EF 03 06 00 BA D2 01 00 00 60 01 00 BE EF 03 06 00 19 D3 02 00
More informationA
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
More informationBD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B
2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :
More informationuntitled
0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.
More informationA11 (1993,1994) 29 A12 (1994) 29 A13 Trefethen and Bau Numerical Linear Algebra (1997) 29 A14 (1999) 30 A15 (2003) 30 A16 (2004) 30 A17 (2007) 30 A18
2013 8 29y, 2016 10 29 1 2 2 Jordan 3 21 3 3 Jordan (1) 3 31 Jordan 4 32 Jordan 4 33 Jordan 6 34 Jordan 8 35 9 4 Jordan (2) 10 41 x 11 42 x 12 43 16 44 19 441 19 442 20 443 25 45 25 5 Jordan 26 A 26 A1
More information研修コーナー
l l l l l l l l l l l α α β l µ l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l
More informationX 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 information17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,
17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ
More information内科96巻3号★/NAI3‐1(第22回試験問題)
µ µ α µ µ µ µ µ µ β β α γ µ Enterococcus faecalis Escherichia coli Legionella pneumophila Pseudomonas aeruginosa Streptococcus viridans α β 正解表正解記号問題 No. 正解記号問題 No. e(4.5) 26 e 1 a(1.2) 27 a 2
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. 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入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
More information, = = 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(2018 2Q C) [ ] 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
(2018 2Q C) [ ] 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 information20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................
More information18 ( ) 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 information2 A id A : A A A A id A def = {(a, a) A A a A} 1 { } 1 1 id 1 = α: A B β : B C α β αβ : A C αβ def = {(a, c) A C b B.((a, b) α (b, c) β)} 2.3 α
20 6 18 1 2 2.1 A B α A B α: A B A B Rel(A, B) A B (A B) A B 0 AB A B AB α, β : A B α β α β def (a, b) A B.((a, b) α (a, b) β) 0 AB AB Rel(A, B) 1 2 A id A : A A A A id A def = {(a, a) A A a A} 1 { } 1
More informationHITACHI 液晶プロジェクター CP-AX3505J/CP-AW3005J 取扱説明書 -詳細版- 【技術情報編】
B A C E D 1 3 5 7 9 11 13 15 17 19 2 4 6 8 10 12 14 16 18 H G I F J M N L K Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C LAN RS-232C LAN LAN BE EF 03 06 00 2A D3 01 00 00 60 00 00 BE EF 03 06 00 BA D2 01
More information4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t
1 1.1 sin 2π [rad] 3 ft 3 sin 2t π 4 3.1 2 1.1: sin θ 2.2 sin θ ft t t [sec] t sin 2t π 4 [rad] sin 3.1 3 sin θ θ t θ 2t π 4 3.2 3.1 3.4 3.4: 2.2: sin θ θ θ [rad] 2.3 0 [rad] 4 sin θ sin 2t π 4 sin 1 1
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 informationiii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................
More informationn 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
1 1 1 + 1 4 + + 1 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 a n < ε 1 1. ε = 10 1 N m, n N a m a n < ε = 10 1 N
More informationu V u V u u +( 1)u =(1+( 1))u =0 u = o u =( 1)u x = x 1 x 2. x n,y = y 1 y 2. y n K n = x 1 x 2. x n x + y x α αx x i K Kn α K x, y αx 1
5 K K Q R C 5.1 5.1.1 V V K K- 1) u, v V u + v V (a) u, v V u + v = v + u (b) u, v, w V (u + v)+w = u +(v + w) (c) u V u + o = u o V (d) u V u + u = o u V 2) α K u V u α αv V (a) α, β K u V (αβ)u = α(βv)
More information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
More information1 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 informationv 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 informationTIJ日本語教育研究会通信40号
XYZ AX B X Y Z A A ) AA B A D ) AD B D E AE B E A B ) A B A B A B A B A ( ) B A B A B C ( ) A B ( ) A B ( ) A B A ( ) B A B A B B A B 21 21 A B ( ) ( ) ( ) ( ) ( ) ) 300 A B BBQ vs vs A B 4 4 RPG 10 A
More informationnewmain.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 information16 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(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 informationI 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 information1 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 informationHITACHI 液晶プロジェクター CP-EX301NJ/CP-EW301NJ 取扱説明書 -詳細版- 【技術情報編】 日本語
A B C D E F G H I 1 3 5 7 9 11 13 15 17 19 2 4 6 8 10 12 14 16 18 K L J Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C RS-232C RS-232C Cable (cross) LAN cable (CAT-5 or greater) LAN LAN LAN LAN RS-232C BE
More informationIMO 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..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A
.. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.
More informationII 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,
More informationnakata/nakata.html p.1/20
http://www.me.titech.ac.jp/ nakata/nakata.html p.1/20 1-(a). Faybusovich(1997) Linear systems in Jordan algebras and primal-dual interior-point algorithms,, Euclid Jordan p.2/20 Euclid Jordan V Euclid
More information6.1 (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 information1 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 =
More informationlimit&derivative
- - 7 )................................................................................ 5.................................. 7.. e ).......................... 9 )..........................................
More informationDecember 28, 2018
e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................
More informationii-03.dvi
2005 II 3 I 18, 19 1. A, B AB BA 0 1 0 0 0 0 (1) A = 0 0 1,B= 1 0 0 0 0 0 0 1 0 (2) A = 3 1 1 2 6 4 1 2 5,B= 12 11 12 22 46 46 12 23 34 5 25 2. 3 A AB = BA 3 B 2 0 1 A = 0 3 0 1 0 2 3. 2 A (1) A 2 = O,
More informationVI VI.21 W 1,..., W r V W 1,..., W r W W r = {v v r v i W i (1 i r)} V = W W r V W 1,..., W r V W 1,..., W r V = W 1 W
3 30 5 VI VI. W,..., W r V W,..., W r W + + W r = {v + + v r v W ( r)} V = W + + W r V W,..., W r V W,..., W r V = W W r () V = W W r () W (W + + W + W + + W r ) = {0} () dm V = dm W + + dm W r VI. f n
More informationS 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.
() 1.1.. 1. 1.1. (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
More informationy π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
More informationさくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a
φ + 5 2 φ : φ [ ] a [ ] a : b a b b(a + b) b a 2 a 2 b(a + b). b 2 ( a b ) 2 a b + a/b X 2 X 0 a/b > 0 2 a b + 5 2 φ φ : 2 5 5 [ ] [ ] x x x : x : x x : x x : x x 2 x 2 x 0 x ± 5 2 x x φ : φ 2 : φ ( )
More information春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an
More informationD 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 information2 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 informationPart () () 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 information1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D
1W II K200 : October 6, 2004 Version : 1.2, kawahira@math.nagoa-u.ac.jp, http://www.math.nagoa-u.ac.jp/~kawahira/courses.htm TA M1, m0418c@math.nagoa-u.ac.jp TA Talor Jacobian 4 45 25 30 20 K2-1W04-00
More informationCVMに基づくNi-Al合金の
CV N-A (-' by T.Koyama ennard-jones fcc α, β, γ, δ β α γ δ = or α, β. γ, δ α β γ ( αβγ w = = k k k ( αβγ w = ( αβγ ( αβγ w = w = ( αβγ w = ( αβγ w = ( αβγ w = ( αβγ w = ( αβγ w = ( βγδ w = = k k k ( αγδ
More information2000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 information2012 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<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
電気電子数学入門 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/073471 このサンプルページの内容は, 初版 1 刷発行当時のものです. i 14 (tool) [ ] IT ( ) PC (EXCEL) HP() 1 1 4 15 3 010 9 ii 1... 1 1.1 1 1.
More information6.1 (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 informationN cos s s cos ψ e e e e 3 3 e e 3 e 3 e
3 3 5 5 5 3 3 7 5 33 5 33 9 5 8 > e > f U f U u u > u ue u e u ue u ue u e u e u u e u u e u N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 > A A > A E A f A A f A [ ] f A A e > > A e[ ] > f A E A < < f ; >
More information1 α X (path) α I = [0, 1] X α(0) = α(1) = p α p (base point) loop α(1) = β(0) X α, β α β : I X (α β)(s) = ( )α β { α(2s) (0 s 1 2 ) β(2s 1) ( 1 2 s 1)
1 α X (path) α I = [0, 1] X α(0) = α(1) = p α p (base point) loop α(1) = β(0) X α, β α β : I X (α β)(s) = ( )α β { α(2s) (0 s 1 2 ) β(2s 1) ( 1 2 s 1) X α α 1 : I X α 1 (s) = α(1 s) ( )α 1 1.1 X p X Ω(p)
More information() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
More information