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2 常微分方程式の局所漸近解析 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行当時のものです.
3
4 i Leibniz ydy = y 2 / [6] 100 Bernoulli Riccati 19 Fuchs x [5] 3 2
5 ii y y + f (y) = 0, f (y) = (y y 5/3 ) t x = 5 4 log(t4 6t 2 + 4C 1 t 3) + C 2 y = (t 3 3t + C 1 ) 3/2 (t 4 6t 2 + 4C 1 t 3) 9/8 C 1, C 2 [5] d 2 y dx 2 + ax2 y = 0, a > 0 (1-1-24) y = x [ ( ) ( )] a a AJ 1/4 2 x2 + BY 1/4 2 x2 A, B (1-1-25) 1/4 1-1 (1-1-24) x + x + global analysis local analysis
6 iii A
7 iv I II
8 v I II p c x =
9
10 [5] x + x Painlevé 1-7
11 f (x) g(x) f (x) g(x) lim x x 0 f (x) g(x) = 0 (1-1-1) x x 0 x x 0 f (x) g(x) g(x) f (x) f (x) g(x) x x 0 (1-1-2a) g(x) f (x) x x 0 (1-1-2b) Landauo f (x) = o(g(x)) x x 0 (1-1-3) f (x) g(x) f (x) g(x) lim x x 0 f (x) g(x) = 1 (1-1-4) f (x) g(x) f (x) x x 0 (1-1-5) x x 0 f (x) g(x) f (x) g(x) x x 0 (1-1-6a) g(x) f (x) x x 0 (1-1-6b) *1 f (x) g(x) x = x 0 x K 1 o O
12 1-1 3 f (x) g(x) K (1-1-7) x x 0 f (x) g(x) O f (x) = O(g(x)) x x 0 (1-1-8) {ϕ m (x)} m ϕ m+1 (x) = o(ϕ m (x)) x x 0 (1-1-9) asymptotic sequences {ϕ m (x)} a m ϕ m (x) k m=0 k y(x) a m ϕ m (x) ϕ k (x) x x 0 (1-1-10) m=0 a m ϕ m (x) y(x) asymptotic series m=0 y(x) a m ϕ m (x) x x 0 (1-1-11) m=0 {(x x 0 ) m } a m (x x 0 ) m y(x) a m = y(m) (x 0 ) m=0 m! k y(x) m=0 a m (x x 0 ) m = y(k+1) (x 0 + θ(x x 0 )) (x x 0 ) k+1 (x x 0 ) k (k + 1)! (0 <θ<1) x x 0 (1-1-12) a m (x x 0 ) m y(x) m=0
13 4 1 y(x) a m (x x 0 ) m x x 0 (1-1-13) m=0 x + {x m } y(x) a 0 + a 1 x + a 2 x + = a 2 m x m (1-1-14) m=0 leading term leading behavior controlling factor f (x) a m ϕ m (x), g(x) b m ϕ m (x) x x 0 (1-1-15) m=0 m=0 cf(x) + dg(x) (ca m + db m )ϕ m (x) c, d x x 0 (1-1-16) m=0 f (x) a n (x x 0 ) n, g(x) b n (x x 0 ) n n=0 f (x)g(x) h n (x x 0 ) n, n h n = a m b n m x x 0 (1-1-17) n=0 m=0 n=0 x 0 x + {x m } y(x) a 0 + a 1 x + a 2 x + = a 2 m x m (1-1-18) x ( y(x) a 0 a 1 x m=0 ) dt a 2 x + a 3 2x 2 + a 4 3x 3 + = m=2 a m x m+1 (m 1) (1-1-19) f (x) g(x) x x 0 (1-1-6 ),
14 1-1 5 x x f (t)dt g(t) dt x x 0 (1-1-20) f (x) f (x) f (x) f (x) y = tan x y = x y 1 y = tan x y = x y = sin x º 30º x
15 6 1 y = sin x sin x x, tan x x x 0 (1-1-21) 1-1 x = sin x x, x 0 x 0.52(π/6) 30 4% tan x x, x 0 x = 0.52(π/6) 30 sin x x = x3 6 + O(x5 ) x 0 (1-1-22) tan x x = x3 3 + O(x5 ) x 0 (1-1-23) O(x 3 ) 1/6 1/3 d 2 y dx + 2 ax2 y = 0, a > 0 (1-1-24) y = [ ( ) ( )] a a x AJ 1/4 2 x2 + BY 1/4 2 x2 A, B (1-1-25) x x + (1-1-24) y(x) = exp(q(x)) (1-1-26) [5] 5 exp(q(x)) d 2 [ ] 2 Q(x) dq(x) + + ax 2 = 0 (1-1-27) dx 2 dx (1-1-24) (1-1-27) 1 2
16 1-1 7 d 2 Q dx 2 ( ) 2 dq (1-1-28) dx 1 2 (1-1-27) 1 (1-1-28) dq(x) ±i ax dx x + (1-1-29) Q(x) ± i a 2 x2 x + (1-1-30) Q(x) = ± i a 2 x2 + R(x), x 2 R(x) x + (1-1-31) (1-1-27) ±i ( a + d2 R dx + ±i ax + dr ) 2 2 dx + ax2 = 0 (1-1-32) [5] 2 (1-1-31) 2x R (x), 2 R (x) x + (1-1-33) R(x) (1-1-32) (1-1-33) (1-1-32) i a 2i ax dr dx x + (1-1-34) R(x) 1 log x x + (1-1-35) 2 R(x) (1-1-33) x + [ y(x) exp ± i a 2 x2 1 ] 2 log x = 1 [ exp ± i ] a x 2 x2 x + (1-1-24) y(x) 1 [ ] ax 2 ax 2 exp A sin + B cos A, B x + (1-1-36) x 2 2
17 118 4 d 2 r dt = (4-0-1) 2 r = ( x(t),y(t) ) r 0 (x 0,y 0 ) u 0 y y 0 = (x x 0 )tanθ (x x 0) 2 2v 2 0 cos2 θ (4-0-2) r 0 (x 0,y 0 ) v 0 θ (4-0-2)
18 m 3 2 h = [m/s 2 ] 1000 [m] 140 [m/s] 500 [km/h] ICBM 2 (4-0-2) (4-0-1) 6400 km d 2 r dt = GM e 2 r (4-0-3) 2 GM r e r Brook Taylor 1
19 120 4 ( ) ( ) x 2 2 dy = 4y 2 (y 1) (4-1-1) dx [6] y = u 2 (1 + x 2 ) (4-1-2) (4-1-1) ( ) 2 du (1 + x 2 ) 2xu du dx dx + u2 1 = 0 (4-1-3) (4-1-3) 1 d 2 [ u (1 + x 2 ) du ] dx 2 dx xu = 0 (4-1-4) (4-1-4) A d2 u dx 2 = 0 du dx = c c (4-1-5a) (4-1-5b) ( B 1 + x 2) du xu = 0 (4-1-6) dx A (4-1-5b) (4-1-3) u c 2 ( 1 + x 2) 2cxu + u 2 1 = 0 (4-1-7) (4-1-2) (4-1-7) y y = (1 + x 2 ) ( cx ± 1 c 2 ) 2 (4-1-8) 4-1 B (4-1-6) (4-1-3) ( ) 1 + x 2 u 2 = 0 (4-1-9a) (4-1-2) (4-1-9a) y = 1 (4-1-9b)
20 y y= x (4-1-9b) (4-1-8) c 1-7 II ( F x,y, dy ) = 0 (4-2-1) dx p dy dx *1 (4-2-1) x,y,p (4-2-2) 1 2 (x,y) z(x,y) p = z(x,y), q = z(x,y) x y, r = 2 z(x,y) x 2, s = 2 z(x,y) x y, t = 2 z(x,y) y 2
21 122 4 F(x,y,p) = 0 (4-2-3) F(x,y,p) x,y 1 p (x 0,y 0 ) (x,y) p F(x 0,y 0, p) = 0 (4-2-4) p 0 (4-2-3) (x 0,y 0 ) (x,y) = (x 0,y 0 ) p = p 0 p = f (x,y) (4-2-5) f (x,y) (x 0,y 0 ) 1 f (x,y) (x 0,y 0 ) (5-3-2a) dy = f (x,y) (4-2-6) dx (x,y) = (x 0,y 0 ) (4-2-4) p 0 f (x,y) p (x 0,y 0 ) p 0 m ] ] ] F(x 0,y 0, p) = 2 F(x 0,y 0, p) = = m 1 F(x 0,y 0, p) = 0 p p=p 0 p 2 p=p 0 p m 1 p=p 0 m ] F(x 0,y 0, p) p m 0 (4-2-7) p=p 0 (4-2-3) x = x 0 y = y 0 p = p 0 ] ] F(x,y,p) F(x,y,p) + p 0 x x=x 0,y=y 0 p=p 0 y x x=x 0,y=y 0 p=p 0 ( p)m m ] F(x,y,p) m! p m x 0 (4-2-8) x=x 0,y=y 0 p=p 0 x = x x 0, y = y y 0, p = p p 0 (4-2-9) ] ] F(x,y,p) F(x,y,p) + p 0 x x=x 0,y=y 0 p=p 0 y x=x 0,y=y 0 p=p 0 0 (4-2-10)
22 (4-2-8) p A( x) 1/m A x 0 (4-2-11a) y p 0 x + B( x) 1+1/m B x 0 (4-2-11b) x = x 0 y = y 0 y y 0 + p 0 (x x 0 ) + B(x x 0 ) 1+1/m x 0 (4-2-12) (x 0,y 0 ) (x,y) = (x 0,y 0 ) m 2 (x,y) (x 0,y 0 ) p (4-2-4) 2 (4-2-3) F(x,y,p) = x x 0 + y y 0 C(p p 0 ) 2 = 0 (4-2-13) C (x 0,y 0, p 0 ) (4-2-8) (1 + p 0 ) x C( p) 2 x 0 (4-2-14) p ± (1 + p 0 ) x/c x 0 (4-2-15a) y p 0 x ± p0 C ( x)3/2 x 0 (4-2-15b) (4-2-15b) y y 0 + p 0 (x x 0 ) ± p0 3 C (x x 0) 3/2 x 0 (4-2-15c) (x,y) (x 0,y 0 ) (x 0,y 0, p 0 ) 4-2 (4-2-12) (x 0,y 0 ) cusps
23 [5] B A. separation of variables P(y) dy = Q(x) (5-1-1) dx y x P(t) dt = Q(t) dt + C C (5-1-2) P 1 (y)q 1 (x) dy dx = P 2(y)Q 2 (x) (5-1-3)
24 y P 1 (t) x P 2 (t) dt = Q 2 (t) dt + C C (5-1-4) Q 1 (t) P 2 (y)q 1 (x) = xy dy dx = (x + 1)(y2 1) (5-1-5) x + 1 dx = 2y dy (5-1-6) x y 2 1 x + log x = log y C y = ± 1 ± x exp(x C) C (5-1-7) (5-1-6) = 0 x = 0 y = ±1 (5-1-8) (5-1-5) y = ±1 (5-1-7) C + x = 0 B. P(x,y) dy = Q(x,y) (5-1-9) dx P(x,y), Q(x,y) x y n P(x,y) = x n P(1,v), Q(x,y) = x n Q(1,v) (5-1-10) v = y x (5-1-11) y v (5-1-11) (5-1-9) xp(1,v) dv = [Q(1,v) vp(1,v)] dx (5-1-12) P(1,v) dx dv = Q(1,v) vp(1,v) x (5-1-13)
25 142 5 (x 2 xy) dy dx = (y2 + xy) (5-1-14) (5-1-11) x(1 v) dv = [ (v 2 + v) v(1 v) ] dx 1 v dx dv = 2v2 x 1 ( log v + 1 ) = log x + C C 2 v (5-1-15) log xy + x y = C (5-1-16) C. Bernoulli equations 1695 James Bernoulli dy dx = a(x)y(x) + b(x)y(x)n n 1 (5-1-17) z(x) = y(x) 1 n (5-1-18) dz = (1 n)[a(x)z(x) + b(x)] (5-1-19) dx y(x) 1 n = Ce F(x) + (1 n)e F(x) x e F(t) b(t) dt, F(x) (1 n) x a(t) dt D. Riccati equations C (5-1-20) 1712 dy dx = a(x)y(x)2 + b(x)y(x) + c(x) (5-1-21)
26 dz y(x) = a(x)z(x) dx (5-1-22) 2 d 2 [ z a ] dx (x) dz 2 a(x) + b(x) + a(x)c(x)z(x) = 0 (5-1-23) dx (5-1-23) (5-1-21) (5-1-21) y 1 (x) y(x) = y 1 (x) + z(x) (5-1-24) (n = 2) dz dx = [ 2a(x)y 1 (x) + b(x) ] z(x) + a(x)z(x) 2 (5-1-25) E. exact equations P(x,y) + Q(x,y) dy dx = 0, f (x,y) = C, P(x,y) y = Q(x,y) x P(x,y) = f (x,y), Q(x,y) = f (x,y) x y (5-1-26) C (5-1-27) ( F x,y, dy ) [ ] n [ ] n 1 [ ] 2 dy dy dy + q n 1(x,y) + + q 2 (x,y) + q 1 (x,y) dy dx dx dx dx dx + q 0 (x,y) = 0 (5-1-28) q k (x,y)(k = 0, 1,...,n 1) (x,y) A. 1 (5-1-28) dy/dx n
27 144 5 [ ][ ] [ ] dy dy dy dx r 1(x,y) dx r 2(x,y) dx r n(x,y) = 0 (5-1-29) dy dx r k(x,y) = 0 (5-1-30) f k (x,y,c) = 0 C (5-1-31) (5-1-29) f 1 (x,y,c) f 2 (x,y,c) f n (x,y,c) = 0 (5-1-32) 1 ( ) 2 dy 2 dy dx dx + 1 y2 = 0 (5-1-33) ( )( ) dy dy dx + y 1 dx y 1 = 0 (5-1-34) (Ce x y + 1)(Ce x y 1) = 0 C (5-1-35) B. 1 (5-1-28) (x,y) k ( y ) F(x,y,p) x k G x, p = 0 (5-1-36) ( y ) p = f x (5-1-37) (5-1-28) (5-1-9, 10) C. 2 (x,y) x,y 1
28 ( ) ( ) dy dy y = f x + g dx dx x p = f (p) + [ xf (p) + g (p) ] dp dx p x dx dp f (p) p f (p) x (5-1-38) (5-1-39) g (p) p f (p) = 0 (5-1-40) y = x dy dx 1 dy/dx x ( x + 1 p 2 ) dp dx = 0 ( ) ( ) A. d n y = f (x) (5-1-41) dxn y = x x 0 dx 1 x1 x 0 dx 2 xn 1 n 1 (x x 0 ) k f (x n )dx n + C k x 0 k! k=0 C 0, C 1,...,C n 1 (5-1-42) y = 1 (n 1)! x n 1 (x t) n 1 (x x 0 ) k f (t)dt + C k x 0 k! k=0 (5-1-43) B. y k
29 190 c- 129 p , 97, , , , 136, , , 96 89, , , 100,
30 , , , , , 99 91, P
31 2010 Printed in Japan ISBN
dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp
+ P (x)y = Q(x) (1) = P (x)y + Q(x) P (x), Q(x) y Q(x) 0 homogeneous = P (x)y 1 y = P (x) log y = P (x) + C y = C exp{ P (x) } = C e R P (x) 5.1 + P (x)y = 0 (2) y = C exp{ P (x) } = Ce R P (x) (3) αy
DVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
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)
() 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 ()
1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (
1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
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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.
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(,,
,. 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,
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 ),
0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9
1-1. 1, 2, 3, 4, 5, 6, 7,, 100,, 1000, n, m m m n n 0 n, m m n 1-2. 0 m n m n 0 2 = 1.41421356 π = 3.141516 1-3. 1 0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c),
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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
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
Gmech08.dvi
145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
mugensho.dvi
1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim
Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
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
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5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h 0 f(a + h, b) f(a, b) h............................................................... ( ) f(x, y) (a, b) x A (a, b) x
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CALCULUS II (Hiroshi SUZUKI ) 16 1 1 1.1 1.1 f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b) lim f(x, y) = lim f(x, y) = lim f(x, y) = c. x a, y b
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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
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(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 )
f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f
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m d y dt = F m d y = mg dt V y = dy dt d y dt = d dy dt dt = dv y dt dv y dt = g dv y dt = g dt dt dv y = g dt V y ( t) = gt + C V y ( ) = V y ( ) = C = V y t ( ) = gt V y ( t) = dy dt = gt dy = g t dt
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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
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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 =
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II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
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