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- ちかこ ごみぶち
- 4 years ago
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2 (u(x)v(x)) = u (x)v(x) + u(x)v (x) ( ) u(x) = u (x)v(x) u(x)v (x) v(x) v(x) 2 y = g(t), t = f(x) y = g(f(x)) dy dx dy dx = dy dt dt dx., y, f, g y = f (g(x))g (x). ( (f(g(x)). ). [ ] y = e ax+b (a, b ) ax + b = t y = e t, t = ax + b dy dx = dy dt dt dx = d d dt et dx (ax + b) = et a = ae ax+b.. (.) f(x) dx x = g(t) x t, f(x) f(g(t)), dx dx dt dt f(x)dx = f(g(t)) dx dt dt. t a b x g(a) g(b) g(b) g(a) f(x)dx = b a f(g(t)) dx dt dt. 1
3 [ 1] e ax+b dx (a 0, b ). ax + b = t e ax+b = e t, x = t b a e ax+b dx = e t 1 a dt = 1 a eax+b. dx = dx dt dt = 1 dt a dt [ 2] 1 + t. 2 x = tan 1 t dx dt = 1 dx = dx 1 + t 2 dt dt = dt 1 + t = dx = x = tan 1 t. 2 dt 1 + t2 u vdx = uv uv dx, b u vdx = [uv] b a b a a uv dx. e a+b = e a e b, e a b = ea e b, eab = (e a ) b. log AB = log A + log B, log A B = log A log B, log Ak = k log A, A, B 0. a = e b b = log a e log a = a, log e b = b. (e x ) = e x, (log x ) = 1 x. 1 e x dx = e x, dx = log x, x y = e ax log x dx = x log x x. y 1 O a > 0 a < 0 x 2
4 sin 2 x + cos 2 x = 1, 1 + tan 2 x = 1 cos 2 x, cos(α + β) = cos α cos β sin α sin β, sin(α + β) = sin α cos β + cos α sin β ( (sin x) = sin x + π ) = cos x 2 ( (cos x) = cos x + π ) = sin x 2 (tan x) = 1 cos 2 x. ( cos xdx = sin x = cos sin xdx = cos x = sin sin 1 x = y sin y = x 1 < x < 1, π 2 < y < π 2. cos 1 x = y cos y = x 1 < x < 1, 0 < y < π. tan 1 x = y tan y = x < x <, π 2 < y < π 2. (sin 1 x) = x > x 2, (cos 1 x) = x n m = m x n, x n 1 m = m, (n, m = 1, 2,... ) x n x 1 2 = x, x 1 2 = 1 x. (x α ) = αx α 1 (α; ). α = 1 2, ( ) 1 x = 2 x. x α dx = xα+1 (α 1; ). α + 1 dx x 1 dx = = log x. x 1, 1 x 2 (tan 1 x) = x. 2 x π 2 ). ( x π 2 ). 3
5 0 0.1 x, x 2 3x + 2 = 0 x,, x = 1 x = 2., ( x) ( y(x)), y (x) = 3y(x). x, y(x).,.. y(x) = Ce 3x (C )., ( x, y) ( u(x, y)) (u x (x, y), u y (x, y), ) (1) y (x) = 4x x, y = y(x). y = 2x 2 + C, C. (2) y (t) = 4y(t) t, y = y(t). y = Ce 4t, C. (3) u xx (x, y) + u yy (x, y) = 0. x y. u = u(x, y). u = log(x 2 + y 2 ). 4
6 t [s] 0 m [kg] mg t y(t) [m] t v(t) [m/s] t a(t) [m/s 2 ] y(t). (y, v.) mg v(t) = dy dt a(t) = d2 y dt 2. F (t) F (t) = ma(t) a(t). Newton. F F = mg y(t) my (t) = mg y (t) = g. y(t) = 1 2 gt2 + v(0)t + y(0) [m]., y(0), v(0) 0 y(t) = 1 2 gt2 [m/s] [ ]. [ ] t = 1, 5, 10, 20, 30. 5
7 ,. k kv(t), F = mg kv(t). my (t) = mg ky (t) my (t) + ky (t) mg = 0. y = mg (( v(0) m ) ( ) ) 1 e k m t + t + y(0) [m] k g k., y(0), v(0) 0 y = mg ( t m )) (1 e k m t [m] k k. [ ]. t.. ( ) [mm] [m/s]
8 t [s], ky (t) m [kg], t y(t) [m]. (,.) 0 y(t), ky(t) [N]. k > 0. y(t) my (t) = ky(t). k y = C 1 cos m t + C 2 sin. k t m y = A sin (C 1, C 2, A, φ y(0), y (0) ) ( ) k m t + φ [ ] q q t I(t) L C I(t) I (t) + LCI(t) = 0. (.) I = C 1 cos LCt + C 2 sin LCt ( ) y = A sin LCt + φ (C 1, C 2, A, φ I(0), I (0) ). 7
9 0.3.4 t [h] ( ) y = y(t) [ ] u(t, t) = y(t + t) y(t) y(t). u(t, t) t t + t ( t ). u(t, t) t t u( t) (u(0) = 0 ) y(t + t) y(t) t = u( t) u(0) y(t) t. t 0 y(t) y (t) = u (0)y(t). u (0) = k y = y(0)e kt. k t = 0,. [ ].. ( ) ( )
10 . k ly(t) (l ) y (t) = (k ly(t))y(t) y = y(0) ( 1 l y(0)) e k kt + l y(0) k. [ ]. [ ], , ,. 9
11 1 1 x. x y = y(x) y (x), y (x),... F (x, y, y, y, ) = 0 y..,., y, y, y 1,., y, y, y, x., ( ) f(x, y) 2, y (x) = f(x, y) (1) 1. x x 0, y(x 0 ), y (x 0 ),. { y = f(x, y) (2) y(x 0 ) = a 0. (2). y = y(x) ( ) (x, y) f(x, y). (x, y) (1, f(x, y)) ( )., (x, y) f(x, y). (x, y). (x, y) f(x, y).,.,. x. 10
12 1.0 f(x, y), f y (x, y) xy D, (x 0, a 0 ) D. (x 0, a 0 ) (2). [ ] f(x, y), f y (x, y). [ ] (x 0, a 0 ). f(x, y) = y y 2, ( )..(. ) y = y y y = y y 2. y = y 1 (x), y = y 2 (x) (2) 2. x > x 0. d dx (y 1(x) y 2 (x)) 2 = 2(y 1(x) y 2(x))(y 1 (x) y 2 (x)) = 2(f(x, y 1 ) f(x, y 2 ))(y 1 (x) y 2 (x)), f(x, y 1 ) f(x, y 2 ) = f y (x, ξ) (ξ y 1 (x) y 2 (x) ), f x (x, y) (x 0, a 0 ) C < 2C(y 1(x) y 2 (x)) 2 11
13 e 2Cx e 2Cx d dx (y 1(x) y 2 (x)) 2 2Ce 2Cx (y 1 (x) y 2 (x)) 2 < 0 = d ( e 2Cx (y 1 (x) y 2 (x)) 2) dx e 2Cx (y 1 (x) y 2 (x)) 2. e 2Cx (y 1 (x) y 2 (x)) 2 < e 2Cx 0 (y 1 (x 0 ) y 2 (x 0 )) 2 = 0 x > x 0 y 1 (x) = y 2 (x) (2) ( ) a y (x) + ay(x) = 0 (3) 1. 1 (1) f(x, y) = ay (1.1) y(x) = Ce ax (C : ) (4). [ ] (4), (e ax ) = ae ax. (4). 1.0, a, x 0, y 0, y (x) + ay(x) = 0, (3) y(x 0 ) = y 0 (5) 12
14 y(x) = y 0 e a(x x 0) (6). [ ] (3) 1.1 y(x) = Ce ax (4),, (5) C. (4) x = x 0, y(x 0 ) = Ce ax 0 (5) y 0 = Ce ax 0 C = y 0 e ax 0. C (4) (6) ( ) x, y(x), a, r(x). y (x) + ay(x) = r(x) (7) 1. f(x, y) = ay + r(x). (1) 1.3. r(x). (7) ( ) y(x) = r(x) e ax dx + C e ax (C : ) (8). [ ] (8) ( ( ) y (x) = r(x) e ax dx + C) e ax + r(x) e ax dx + C (e ax ) ( ) = (r(x) e ax ) e ax + r(x) e ax dx + C ( ae ax ) = r(x) a y(x) (8) (7). 13
15 (8), (7) 1.0,. [ 1.4] y (x) + 2y(x) = 2x. e 2x,. e 2x e 2x y (x) + 2e 2x y(x) = 2xe 2x. = ( e 2x y(x) ) e 2x y(x) = 2xe 2x dx + C. ( ) 1 2xe 2x dx = 2x 2 e2x, y = x Ce 2x. ( ) e2x dx + C = xe 2x 1 2 e2x + C y(x), f(x), r(x). y (x) + f(x)y(x) = r(x). (9) 1. r(x) 0, r(x) 0. (1) f(x, y) = f(x)y + r(x) f(x). 1 y(x) = C e F (x) (C : ) (10). F (x) f(x) f(x), r(x). 1 ( ) y(x) = r(x) e F (x) dx + C e F (x) (C : ) (11) 14
16 . F (x) f(x). e F (x) (10). ( [ 1.5 ] (11) (9). ) e F (x) = f(x) e F (x) (11) ( ( ) (e y (x) = r(x) e F (x) dx + C) e F (x) + r(x) e F (x) F (x) dx + C ) ( ) ( f(x) = r(x) e F (x) e F (x) + r(x) e F (x) dx + C ) e F (x) = r(x) f(x) y(x) (9). (11) (9) 1.0,. [ 1.8.] y (x) + 2 x y(x) = x, y(1) = 0. y 2x 2x dx = x 2 + C (C ) e x2. e x2 y (x) + 2xe x2 y(x) = xe x2 ( ). e x2 = 2xe x 2 = ( ) e x2 y(x) e x2 y(x) = ( x)e x2 dx + C C. x 2 = t ( x)e x2 dx = 1 2 ex2 + C C. e x2 y(x) = Ce x2 C,. x = 1 y(1) = 0 C = e 2 y(x) = e1 x ,
17 = ( ) 1 (9). r(x) 0 (9) F (x) y(x) = C e (F (x) : f(x), C : ), C u(x) F (x) y(x) = u(x) e r(x) 0 (9) (7) r(x),,., ( ) 1 y (x) + f(x)y(x) = r(x). (9) F (x) y(x) = y p (x) + C e. y p (x) (9), F (x) f(x), C. C e F (x) (9) r(x) 0 y (x) + f(x)y(x) = 0 (12). [ ] y(x), y p (x) (9) y (x) + f(x)y(x) = r(x), (y p (x)) + f(x)y p (x) = r(x) 16
18 . (y(x) y p (x)) + f(x)(y(x) y p (x)) = 0, y(x) y p (x) (12). 1.4 F (x) y(x) y p (x) = C e. 1.6 (9) y p (12). f(x) a ( ) r(x),,,. [r(x) = n ] { n (a 0 ) y p (x) = n + 1 (a = 0 ). [ 1.10] y + 3y = x 2 1 (a). y p = αx 2 + βx + γ (a) y (y p ) + 3y p = (2αx + β) + 3(αx 2 + βx + γ) = 3αx 2 + (2α + 3β)x + β + γ = x α = 1 2α + 3β = 0 β + γ = 1. α = 1 3, β = 2 9, γ = 7 27 y p = 1 3 x2 2 9 x 7 27., y + 3y = 0 (b) y = Ce 3x a y = 1 3 x2 2 9 x Ce 3x. 17
19 [r(x) = cos ωx, sin ωx ] y p (x) = α cos ωx + β sin ωx α, β. [ 1.12] y + 2y = cos x (a). y p = α cos x + β sin x (a) y (y p ) +2y p = ( α sin x+β cos x)+2(α cos x+β sin x) = (2α+β) cos x+(2β α) sin x = cos x. 2α + β = 1 α + 2β = 0. α = 2 5, β = 1 5 y p = 2 5 cos x + 1 sin x. 5, y + 2y = 0 (b) y = Ce 2x (a) y = 2 5 cos x sin x + Ce 2x. [r(x) = ke qx (q a) ] y p (x) = αe qx α. [ 1.13] y y = 2e 2x (a). y p = αe 2x (a) y (y p ) y p = (2α α)e 2x = 2e 2x. α = 2., y y = 0 (b) y = Ce x (a) y = 2e 2x + Ce x. 1.5, y (x) = f(x)g(y(x)) (f,g. ) (13), ( ). (13). 18
20 [ ] (Step 1) g(y 0 ) = 0 y 0., y(x) y 0 (13). (Step 2) y(x) = y 0. ((13) x y(x) y 0 x y(x) y 0.) (Step 2-1) (13) g(y(x)) : 1 dy g(y(x)) dx = f(x). (Step 2-2) x : 1 dy g(y(x)) dx dx = f(x) dx + C. (Step 2-3) x y dy g(y) = f(x) dx + C.. [ ] y (x) = y(x)(2 y(x)) (a). (a) (13) f(x) = 1, g(y) = y(2 y). y = 0, y = 2 (a). y 0, 2 y(y 2) 1 dy y(y 2) dx = 1. x 1 dy = y(y 2) dx dx = ( 1 = 2(y 2) 1 ) 2y = 1 2 log y 2 y = x + C. y y = 2Ce2x Ce 2x 1 1 dx = x + C. (C ). 1 y(y 2) dy dx = 1 2 log y 2 y + C.. y = 0 y =
21 1.6 ( ), ( ) y(x) y (x) = f x (f. ) [ ] u = y y u u, x y = (xu) = u + xu u = f(u) u x. [ ] y (x) = x (a). y(x) f(u) = 1 u u = 1 + u2 xu, u = ± C x 2 x x 2 + y 2 = C y (x) = f(ax + by(x) + c) (f, a, b, c. ) [ ] u = ax + by + c y u u,. [ ] y (x) = 1 x + y(x) (a). u = x + y u = u u log u = x + C y log(x + y + 1) = C. 20
22 (2.1) y + ay + by = 0, a, b : 2., λ 1 = cos q 1 + i sin q 1, λ 2 = cos q 2 + i sin q 2 λ 1 λ 2 = (cos q 1 + i sin q 1 )(cos q 2 + i sin q 2 ) = cos(q 1 + q 2 ) + i sin(q 1 + q 2 ). λ 1 λ 2 λ 2 λ 1 q 1 q 2 q 1 O O O q 2. λ = p + iq (p, q ) e λ = e p (cos q + i sin q). e λ 1+λ 2 = e λ 1 e λ 2., λ = p + iq, x, x x e λx, d dx eλx = λ e λx, e λx dx = 1 λ eλx. λ 0., i. 21
23 2.1.2 C 1, C 2 y 1 (x), y 2 (x) (2.1) = C 1 y 1 (x) + C 2 y 2 (x) (2.1).. [ ] C 1 y 1 (x) + C 2 y 2 (x) (2.1) (C 1 y 1 (x) + C 2 y 2 (x)) + a(c 1 y 1 (x) + C 2 y 2 (x)) + b(c 1 y 1 (x) + C 2 y 2 (x)) = C 1 (y 1(x) + ay 1(x) + by 1 (x)) + C 2 (y 2(x) + ay 2(x) + by 2 (x)) = 0 (2.1) , a = OA, b = OB { a, b} ( ), O, A, B. k 1, k 2 k 1 a + k 2 b = 0 = k 1 = k 2 = 0., a = (a 1, a 2 ), b = (b 1, b 2 ) { a, b} a 1 b 1 a 2 b {y 1 (x), y 2 (x),, y m (x)} k 1, k 2,, k m x k 1 y 1 (x) + k 2 y 2 (x) + + k m y m (x) = 0 = k 1 = k 2 = = k m = 0.. [ ] 1. {1, x, x 2,, x m }. 2. {e a 1x, e a 2x,, e amx }, (a 1,, a m. (2.1) 1. 22
24 (2.2) λ 2 + aλ + b = 0 (2.1),. a 2 4b D. [1] (2.1). [1-1] D > 0. λ 1, λ 2 ( 2 ) {e λ 1x, e λ 2x }. [1-2] D = 0. λ ( ) {e λx, x e λx }. [1-3] D < 0. λ 1, λ 2 ( 2 ) {e λ 1x, e λ 2x } [2] {y 1 (x), y 2 (x)} (2.1), (2.1) y(x) = k 1 y 1 (x) + k 2 y 2 (x), (k 1, k 2 )... A. [1] λ y = e λx (2.1). [2] λ, y = xe λx (2.1). [.] [1] y = e λx (2.1) (e λx ) + a(e λx ) + b(e λx ) = λ 2 e λx + aλe λx + be λx = (λ 2 + aλ + b)e λx = 0 (2.1). [2] λ λ = a 2, y = xeλx (2.1) (xe λx ) + a(xe λx ) + b(xe λx ) = (λ 2 + aλ + b)xe λx + (2λ + a)e λx = 0 23
25 y = xe λx. [ ] [1]. (2.5) { y + ay + by = 0, (1) y(0) = A, y (0) = B (2) 2 (2.1). (2). (2 2.) B. (2.5). y(0) = 0, y (0) = 0 y(x) 0. [.] y 1 (x), y 2 (x) (2.5). ( (y 1 y 2 ) 2 + (y 1 y 2) 2 ) = 2(y1 y 2 )(y 1 y 2) + 2(y 1 y 2)(y 1 y 2) = 2{(y 1 y 2 ) a(y 1 y 2) b(y 1 y 2 )}(y 1 y 2) ( ) < K (y 1 y 2 ) 2 + (y 1 y 2) 2 (K ), (y 1 (x) y 2 (x)) 2 + (y 1(x) y 2(x)) 2 < ekx ( (y 1 (0) y 2 (0)) 2 + (y 1(0) y 2(0)) 2 ). 0 y 1 (x) y 2 (x). [ ] C. {y 1 (x), y 2 (x)} {( ) ( )} y 1 (0) y 2 (0), y 1(0) y 2(0) [ ]., k 1 y 1 (x) + k 2 y 2 (x) 0, (k 1, k 2 ) (0, 0) k 1, k 2. k 1 y 1(x) + k 2 y 2(x) 0 ( ) ( ) ( ) y 1 (x) y 2 (x) 0 k 1 + k y 1(x) 2 = y 2(x) 0 x =
26 ( ) ( ) ( ) y 1 (0) y 2 (0) 0 k 1 + k y 1(0) 2 =, (k y 2(0) 1, k 2 ) (0, 0) 0 k 1, k 2. Y (x) = k 1 y 1 (x) + k 2 y 2 (x), Y (0) = 0, Y (0) = 0 (2.1) B Y (x) 0. {y 1 (x), y 2 (x)}. [ ] [1-1] [1-3] D. {y 1 (x), y 2 (x)} (2.1) (2.1) y(x) k 1, k 2 y(x) = k 1 y 1 (x) + k 2 y 2 (x), (k 1, k 2 ). [ ] {( ) ( )} y 1 (0) y 2 (0), y 1(0) y 2(0) k 1, k 2. ( ) ( ) ( ) y 1 (0) y 2 (0) y(0) k 1 + k y 1(0) 2 = y 2(0) y (0) Y (x) = k 1 y 1 (x) + k 2 y 2 (x), Y (0) = y(0), Y (0) = y (0) (2.1) B Y (x) y(x). y(x) = k 1 y 1 (x) + k 2 y 2 (x). [ ] [2]. 25
27 2.1.5 D < 0, D < 0 y(x) = k 1 e λ 1x + k 2 e λ 2x, (k 1, k 2 ).,. a, b λ 1 = A+iB, λ 2 = A ib (A, B ). A, B {e Ax cos Bx, e Ax sin Bx} y = C 1 e Ax cos Bx + C 2 e Ax sin Bx (C 1, C 2 ).. [ ] 1 2 (eλ 1x + e λ 2x ) = e Ax cos Bx (= y 1 (x) ), 1 2i (eλ 1x e λ 2x ) = e Ax sin Bx (= y 2 (x) ) ( )2., y 1 (0) y 2 (0) y 1(0) y 2(0) = 1 0 D A B = B = 0 2. [2]. 26
28 2.3 2 (2.12) y + ay + by = r(x), r(x) : (2.12) y p (x) (2.12).. y = y p (x) + (2.1) [ ] y(x) (2.12). Y (x) = y(x) y p (x) Y (x)+ay (x)+by (x) = (y (x)+ay (x)+by(x)) (y p(x)+ay p(x)+by p (x)) = r(x) r(x) = 0 Y (x) (2.1) [ 1] 2 y i (x) (i = 1, 2) y + ay + by = r i (x), i = 1, 2., C 1, C 2 Y = C 1 y 1 (x) + C 2 y 2 (x) Y + ay + by = C 1 r i (x) + C 2 r 2 (x). [ ]. = (C 1 y 1 (x) + C 2 y 2 (x)) + a(c 1 y 1 (x) + C 2 y 2 (x)) + b(c 1 y 1 (x) + C 2 y 2 (x)) = C 1 (y 1 + ay 1 + by 1 ) + C 2 (y 2 + ay 2 + by 2 ) = C 1 r 1 + C 2 r 2 = 27
29 [ 2] (i) λ, y p (x) = A e λx (a) y + ay + by = ke λx (b) A. (ii) λ, D 0, y p (x) = Ax e λx (c) (b) A. [ ] (i) (a) (b) = (A e λx ) + a(a e λx ) + b(a e λx ) = A(λ 2 + aλ + b)e λx λ 2 + aλ + b 0. (i) (c) (b) = (Ax e λx ) + a(ax e λx ) + b(ax e λx ) = A{(2λ + a)e λx + (λ 2 + aλ + b)xe λx } λ 2 + aλ + b = 0, 2λ + a = D 0. [ 3] y(x), r(x), Y (X) = Re y(x), Z(X) = Im y(x). y(x) y + ay + by = r(x), (a), Y (x), Z(x) Y + ay + by = Re r(x) Z + az + bz = Im r(x). [ ] y = Y + iz, r(x) = Re r(x) + iim r(x) (a) = (Y + iz) + a(y + iz) + b(y + iz) = (Y + ay + by ) + i (Z + az + bz) (a) = Re r(x) + iim r(x),. 28
30 k, k 1, k 2, α, β. (2.12) y p (x). 1. r(x) = k e λx, (λ ) (a) λ y p (x) = Ae λx. (b) λ, D = a 2 4b 0 y p (x) = Axe λx. 2. r(x) = k cos βx k sin βx, (a) ±iβ y p (x) = A cos βx + B sin βx. (b) ±iβ y p (x) = A x cos βx + B x sin βx. 3. r(x) = k e αx cos βx k e αx sin βx, (a) α ± iβ y p (x) = A e αx cos βx + B e αx sin βx. (b) D 0 α ± iβ y p (x) = A xe αx cos βx + B xe αx sin βx. 4. r(x) = n, (a) a 0, b 0 (b) a 0, b = 0 (c) a = 0, b = 0 y p = n. y p = n + 1. y p = n + 2. (2.12) A, B,. [ ] 1 [ 2]. 3 λ = α + iβ e αx cos βx = Ree λx, e αx sin βx = Ime λx [ 2, 3]
31 2.3.3,.. C. (i). Ids I B r P Ids db Φ = S Ids P db = k Ids r r 3. (Biot-Savard ) C, C B = k Ids r r 3 C, C S.. B S B n ds (n S ). Φ S C. Φ = LI (L )., ( ) Φ, C Φ E. E = dφ dt. (Lenz ). 2 E = L di dt. ( ). 30
32 (ii). +q E q E. ±q E E = q C (C ).. (iii). R E I. E = IR.. E = sin ωt, t I(t). L, C, R,,, E L, E C, E R, q. E L = LI : ( ) E C = q C : ( ) E C E R = IR : ( ) q q I = q, E L L C I R E R E L + E C + E R + E = 0 : ( ) E LI RI + q C = E = sin ωt E, I = q (P ) LI + RI + 1 I = ω cos ωt C. 31
33 (P ), (P ). (P 0 ) LI + RI + 1 C I = 0 Lλ 2 + Rλ + 1 C = 0. D ( D = R 2 4L ) < 0 C.. λ = R 2L ± Di, { e R 2L t cos Dt, e R 2L t sin Dt } (P ) I(t) = I 0 (t) + I p (t),. I 0 (t) = C 1 e R 2L t cos Dt + C 2 e R 2L t sin Dt, I p (t) (P ) ((P 0 ) ). I 0 (t) < ( C 1 + C 2 )e R 2L t 0 (t ) I 0 (t), I p (t). I p (t). Ĩ ( P ) LĨ + RĨ + 1 C Ĩ = ωeiωt. ( P ). ( P ) A Ĩ = Ae iωt ( Lω 2 + irω + 1 C ) e iωt = ωe iωt 32
34 A = ω Lω 2 + irω + 1 C = 1 Lω + ir + 1 Cω ( Z ). 1 Z = 1 π Z ei(φ ( P ) 2 ), φ tan φ = Ĩ = 1 Z eiωt = 1 π Z ei(ωt+φ 2 ) 1 Lω cω R., cos ωt = Ree iωt [ 3] ReĨ I p = 1 Z cos(ωt + φ π 2 ) = 1 sin(ωt + φ) Z (P ). I = E/R. Z = ( 1 Cω Lω)2 + R 2., ω ω 1 LC. L = C = 1 ω I 1. 1 I_1 where L=C=R=1,omega=0.25,0.5,
35 1 I_1 where L=C=R=1,omega=1,2,
36 f(t) dt = F (t) = f(t) 0 f(t) dt = f(t) (0, ) t, s. F (s) = 0 e st f(t) dt s, F (s) f(t) L(f(t)), L(f), L(f)(s),. (5.1) L(f)(s) = 0 e st f(t) dt. (5.1) L t f(t) s F (s),. f(t) :,. : M, α f(t) < Meαt, (t > 0), α < s s F (s), Laplace L(f(t)) = F (s) t f(t), ( ). L 1 (F (s)) = f(t) L 1. 35
37 3.1.4 [ 1. ] f 1 (t), f 2 (t),f 1 (s), F 2 (s) c 1, c 2 L(c 1 f 1 (t) + c 2 f 2 (t)) = c 1 L(f 1 (t)) + c 2 L(f 2 (t)), L 1 (c 1 F 1 (s) + c 2 F 2 (s)) = c 1 L 1 (F 1 (s)) + c 2 L 1 (F 2 (s)). : [ 1..] a : L(e at ) = 1 s a (s > Re a ) L(1) = 1 s ( ) ( ) 1 1 L 1 = e at L 1 = 1 s a s [ 2..] a L(f(t)) = F (s) (s > α ) L(e at f(t)) = F (s a) (s > α+a ) : 36
38 [ 2..] n = 1, 2, : L(t n ) = n! s n+1, (s > 0 ) n! L(e at t n ) =, (s > a ) (s a) n+1 ( ) ( ) 1 L 1 = tn s n+1 n!, 1 L 1 = eat t n. (s a) n+1 n! [ 3. Heaviside.] { 0 (t < λ ) H λ (t) = 1 (t > λ ) λ > 0 Heaviside. : L(H λ ) = e λs s (s > 0.) 37
39 3.1.5 ( ) (I) f(t) L(f (t)) = s L(f(t)) f(0), (s ). (II) f(t) 2 L(f (t)) = s 2 L(f(t)) s f(0) f (0), (s ). : 38
40 { y (t) + y (t) 6y(t) = 6e t, (1) y(0) = 0, y (0) = 1 (2). 1 (1) L(y (t)) + L(y (t)) 6L(y(t)) = 6L(e t ). L(y (t)) = s 2 L(y(t)) s y(0) y (0), L(y (t)) = s L(y(t)) y(0),, L(e t ) = L(y(t)) = Y (s) (s 2 + s 6)Y (s) sy(0) y (0) y(0) = 2 (2) (s 2 + s 6)Y (s) =. Y. 3 Y (s) Y (s) = 39
41 .. = s s 1 1 s 2 4 (3) y(t) = L 1 (Y (s)) = 1 2 e 3t et e 2t (1), (2). 5. (3) y (t) = 3 2 e 3t et 2 e 2t, y (t) = 9 2 e 3t et 4 e 2t. (1) (3) (1). (2). 40
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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
IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2
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.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
x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (
II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )
W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
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..........................
29
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() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y
![() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y](/thumbs/85/92979039.jpg "() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y")
5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R
![II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R](/thumbs/95/125540821.jpg "II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R")
II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
1W 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
( z = x 3 y + y ( z = cos(x y ( 8 ( s8.7 y = xe x ( 8 ( s83.8 ( ( + xdx ( cos 3 xdx t = sin x ( 8 ( s84 ( 8 ( s85. C : y = x + 4, l : y = x + a,
[ ] 8 IC. y d y dx = ( dy dx ( p = dy p y dx ( ( ( 8 ( s8. 3 A A = ( A ( A (3 A P A P AP.3 π y(x = { ( 8 ( s8 x ( π < x x ( < x π y(x π π O π x ( 8 ( s83.4 f (x, y, z grad(f ( ( ( f f f grad(f = i + j
I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
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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
.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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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
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20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
![[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s](/thumbs/94/118579915.jpg "[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s")
[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
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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
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Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n
Part2 47 Example 161 93 1 T a a 2 M 1 a 1 T a 2 a Point 1 T L L L T T L L T L L L T T L L T detm a 1 aa 2 a 1 2 + 1 > 0 11 T T x x M λ 12 y y x y λ 2 a + 1λ + a 2 2a + 2 0 13 D D a + 1 2 4a 2 2a + 2 a
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009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :
9 ( ) 9 5 I II III A B (0 ) 5 I II III A B (0 ), 6 8 I II A B (0 ), 6, 7 I II A B (00 ) OAB A B OA = OA OB = OB A B : P OP AB Q OA = a OB = b () OP a b () OP OQ () a = 5 b = OP AB OAB PAB a f(x) = (log
17 ( ) 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
II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
![II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2](/thumbs/101/149159646.jpg "II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2")
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
simx simxdx, cosxdx, sixdx 6.3 px m m + pxfxdx = pxf x p xf xdx = pxf x p xf x + p xf xdx 7.4 a m.5 fx simxdx 8 fx fx simxdx = πb m 9 a fxdx = πa a =
II 6 ishimori@phys.titech.ac.jp 6.. 5.4.. f Rx = f Lx = fx fx + lim = lim x x + x x f c = f x + x < c < x x x + lim x x fx fx x x = lim x x f c = f x x < c < x cosmx cosxdx = {cosm x + cosm + x} dx = [
.1 z = e x +xy y z y 1 1 x 0 1 z x y α β γ z = αx + βy + γ (.1) ax + by + cz = d (.1') a, b, c, d x-y-z (a, b, c). x-y-z 3 (0,
.1.1 Y K L Y = K 1 3 L 3 L K K (K + ) 1 1 3 L 3 K 3 L 3 K 0 (K + K) 1 3 L 3 K 1 3 L 3 lim K 0 K = L (K + K) 1 3 K 1 3 3 lim K 0 K = 1 3 K 3 L 3 z = f(x, y) x y z x-y-z.1 z = e x +xy y 3 x-y ( ) z 0 f(x,
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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熊本県数学問題正解
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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. 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)
x x x 2, A 4 2 Ax.4 A A A A λ λ 4 λ 2 A λe λ λ2 5λ + 6 0,...λ 2, λ 2 3 E 0 E 0 p p Ap λp λ 2 p 4 2 p p 2 p { 4p 2 2p p + 2 p, p 2 λ {
K E N Z OU 2008 8. 4x 2x 2 2 2 x + x 2. x 2 2x 2, 2 2 d 2 x 2 2.2 2 3x 2... d 2 x 2 5 + 6x 0 2 2 d 2 x 2 + P t + P 2tx Qx x x, x 2 2 2 x 2 P 2 tx P tx 2 + Qx x, x 2. d x 4 2 x 2 x x 2.3 x x x 2, A 4 2
x ( ) x dx = ax
x ( ) x dx = ax 1 dx = a x log x = at + c x(t) = e at C (C = e c ) a > 0 t a < 0 t 0 (at + b ) h dx = lim x(t + h) x(t) h 0 h x(t + h) x(t) h x(t) t x(t + h) x(t) ax(t) h x(t + h) x(t) + ahx(t) 0, h, 2h,
ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y
01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y
I 1
I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
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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.............................
webkaitou.dvi
( c Akir KANEKO) ).. m. l s = lθ m d s dt = mg sin θ d θ dt = g l sinθ θ l θ mg. d s dt xy t ( d x dt, d y dt ) t ( mg sin θ cos θ, sin θ sin θ). (.) m t ( d x dt, d y dt ) = t ( mg sin θ cos θ, mg sin
y = f(x) y = f( + h) f(), x = h dy dx f () f (derivtive) (differentition) (velocity) p(t) =(x(t),y(t),z(t)) ( dp dx dt = dt, dy dt, dz ) dt f () > f x
I 5 2 6 3 8 4 Riemnn 9 5 Tylor 8 6 26 7 3 8 34 f(x) x = A = h f( + h) f() h A (differentil coefficient) f f () y = f(x) y = f( + h) f(), x = h dy dx f () f (derivtive) (differentition) (velocity) p(t)
chap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4
35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (
6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b
( ) x y f(x, y) = ax
013 4 16 5 54 (03-5465-7040) nkiyono@mail.ecc.u-okyo.ac.jp hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy
II 1 3 2 5 3 7 4 8 5 11 6 13 7 16 8 18 2 1 1. x 2 + xy x y (1 lim (x,y (1,1 x 1 x 3 + y 3 (2 lim (x,y (, x 2 + y 2 x 2 (3 lim (x,y (, x 2 + y 2 xy (4 lim (x,y (, x 2 + y 2 x y (5 lim (x,y (, x + y x 3y
Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0
Tips KENZOU 28 7 6 P no problem 2 w = f(z) z w w z w = (z z ) b b w = log (z z ) z = z 2π 2 z = z w = z /2 z = re iθ θ (z = ) 2π 4π 2 θ θ 2π 4π z r re iθ re i2π = r re i4π = r w r re iθ/2 re iπ = r re
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 =
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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
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9.. x + y + 0. x,y, x,y, x r cos θ y r sin θ xy x y x,y 0,0 4. x, y 0, 0, r 0. xy x + y r 0 r cos θ sin θ r cos θ sin θ θ 4 y mx x, y 0, 0 x 0. x,y 0,0 x x + y x 0 x x + mx + m m x r cos θ 5 x, y 0, 0,
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
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,