- PDF 無料ダウンロード

Similar documents

() 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)

36 3 D f(z) D z f(z) z Taylor z D C f(z) z C C f (z) C f(z) f (z) f(z) D C D D z C C 3.: f(z) 3. f (z) f 2 (z) D D D D D f (z) f 2 (z) D D f (z) f 2 (

2 N(ε 1 ) N(ε 2 ) ε 1 ε 2 α ε ε 2 1 n N(ɛ) N ɛ ɛ- (1.1.3) n > N(ɛ) a n α < ɛ n N(ɛ) a n

z z x = y = /x lim y = + x + lim y = x (x a ) a (x a+) lim z z f(z) = A, lim z z g(z) = B () lim z z {f(z) ± g(z)} = A ± B (2) lim {f(z) g(z)} = AB z

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

高等学校学習指導要領

高等学校学習指導要領

I.2 z x, y i z = x + iy. x, y z (real part), (imaginary part), x = Re(z), y = Im(z). () i. (2) 2 z = x + iy, z 2 = x 2 + iy 2,, z ± z 2 = (x ± x 2 ) +

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

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

1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

y 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 =

(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

Microsoft Word - 触ってみよう、Maximaに2.doc

notekiso1_09.dvi

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

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a

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

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

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 +

(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z

arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =

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 :

.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 =,

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

24.15章.微分方程式

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)

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 {

II (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1

2S III IV K A4 12:00-13:30 Cafe David 1 2 TA 1 appointment Cafe David K2-2S04-00 : C

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

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

°ÌÁê¿ô³ØII

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.

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

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

III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2 lim. (x,y) (1,0) x 2 + y 2 lim (x,y) (0,0) lim (x,y) (0,0) lim (x,y) (0,0) 5x 2 y x 2 + y 2. xy x2 + y

example2_time.eps

f(x) = e x2 25 d f(x) 0 x d2 dx f(x) 0 x dx2 f(x) (1 + ax 2 ) 2 lim x 0 x 4 a 3 2 a g(x) = 1 + ax 2 f(x) g(x) 1/2 f(x)dx n n A f(x) = Ax (x R

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

( ) a, b c a 2 + b 2 = c : 2 2 = p q, p, q 2q 2 = p 2. p 2 p q 2 p, q (QED)

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

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

( ) x y f(x, y) = ax

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,,

高等学校学習指導要領解説　数学編

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

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

Z: Q: R: C: 3. Green Cauchy

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y

120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2

(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

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

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

2000年度『数学展望 I』講義録

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

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

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,

.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 ft t 2 ft = t 2 ft+ t = t+ t d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0

( ) 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) (

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 (

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r

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

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

ψ(, v = u + v = (5.1 u = ψ, v = ψ (5.2 ψ 2 P P F ig.23 ds d d n P flow v : d/ds = (d/ds, d/ds 9 n=(d/ds, d/ds ds 2 = d 2 d v n P ψ( ψ

x (0, 6, N x 2 (4 + 2(4 + 3 < 6 2 3, a 2 + a 2+ > 0. x (0, 6 si x > 0. (2 cos [0, 6] (0, 6 (cos si < 0. ( (2 (3 cos 0, cos 3 < 0. cos 0 cos cos

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

Transcription:

1 1 1 1 1 1

2 f z 2 C 1, C 2 f 2 C 1, C 2 f(c 2 ) C 2 f(c 1 ) z C 1 f f(z)

xy uv ( u v ) = ( a b c d ) ( x y ) + ( p q ) (p + b, q + d) 1 (p + a, q + c) 1 (p, q)

1 1 (b, d) (a, c) 2 3 2 3 a = d, c = b ( a c ) c a

w = { z = x + iy u + iv v = { u = ax cy + p cx + ay + q w = ax cy + p + i(cx + ay + q) = a(x + iy) + ic(x + iy) + (p + iq) = Az + B (A = a + ic, B = p + iq) A arg A B

f(x) f(x lim 0 ) = f (x x x 0 ) 0 x x 0 f(x) x 0 ( ) f(x) = f(x 0 ) + f (x 0 )(x x 0 ) + ɛ lim ɛ/(x x x x 0 ) = 0 0 ɛ x x 0 f(x) = f(x 0 ) + f (x 0 )(x x 0 )

f(z) = f(z 0 ) + f (z 0 )(z z 0 ) + ɛ ɛ z 0 f(z) = f(z 0 ) + f (z 0 )(z z 0 ) = f (z 0 )z + ( f(z 0 ) + f (z 0 )z 0 ) f (z 0 ) 0

f u, v u x = v y u y = x v a, b, c, d u 2 u x 2 + 2 u y 2 = 0 x z

w = z 2 = r 2 (cos 2θ + i sin 2θ)

w = e z = e x (cos y + i sin y) π 2 i w = z 0 dz (z 5 1) 2/5

Riemann, Georg Friedrich Bernhard (1826.9.17-1866.7.20)

{ w < 1} (

f u. u(x, y) [0, 1] u(0) = a, u(1) = b u (x) = 0 u(x) = (b a)x + a

a b 1

(x, y) u(x, y)

4 4 z A z D G z B z C

G G G { w = 1} 1 1 G 3 ζ 1, ζ 2, ζ 3 { w = 1} 3 ξ 1, ξ 2, ξ 3 1

2 4 (G; z A, z B, z C, z D ), (G ; z A, z B, z C, z D ) G G z A, z B, z C, z D z A, z B, z C, z D z D z A z D f z A z C z C z B z B G G

4 (G; z A, z B, z C, z D ) (G ; z A, z B, z C, z D ) 4 (G; z A, z B, z C, z D ) 4 (G ; ξ A, ξ B, ξ C, ξ D ) z dz w = 0 (z ξ A )(z ξ B )(z ξ C )(z ξ D ) (G ; z A, z B, z C, z D )

z A z D z B z C z A z D z B z C

4 A D A D B C B C z A z D z D z A z C z B z C z B

2

A D A D B C BC AB = B C A B B C z A z D z D z A z C z B z C z B 4

4 BC AB 4 z A z B z C z D 4 z A z B z C z D

Γ D D ρ(z) Γ Γ γ ρ(z) 1 + γ ρ(z) dz 1 D D (ρ(z))2 dxdy Γ

a γ ρ(x, y) b = γ ρ(z) dz 1 D ρ2 dxdy ρ o (x, y) = 1 b D ρ2 0 dxdy = ab b 2

[ ] 2 4 (G; z A, z B, z C, z D ), (G; z A, z B, z C, z D ) z A z B z C z D z A z B z C z D

2 2 2

2

2

?

2πi r R e z log r2 R log r log R e z = e x (cos y + i sin y)

f r R r R f f

R r = R r r R r R 2

r R 2 {r < z < R} 2 2 log R r 2π 2

2 2

2 1 4 4 2

2π 2π 2 2 4 2

3 3 3 2

3 3 2 2 1 2 2

2 3 2

e z 2π log r log R