I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17 6 4 7 1 7 8 7 15 * 7 0 * 7 7 9 * 59 F60 69 C70 79 B80 89 A90 100 S A4 A4
I013 00- Cafe David Cafe David 1:00 13:30 1 (1) C: () R: (3) Q: (4) Z: (5) N: (6) : (1) α: () β: (3) γ, Γ: (4) δ, : (5) ϵ: (6) ζ: (7) η: (8) θ, Θ: (9) ι: (10) κ: (11) λ, Λ: (1) µ: (13) ν: (14) ξ, Ξ: (15) o: (16) π, Π: (17) ρ: (18) σ, Σ: (19) τ: (0) υ, Υ: (1) ϕ, Φ: () χ: (3) ψ, Ψ: (4) ω, Ω: (1) x X x X x X () X {x X x } N = {n Z n > 0} (3) X Y X Y (4) f : X Y f X Y ( (5) A := B A B e := lim 1 + 1 n. n n) (6) 1: ( ) 1 {a n } : M n a n M
I013 01-1 : April, 013 Version : 1.1 1 (4/15) 1,, 3,... 1, 1 3, 3,..., etc. 5 = 1 3 + 1 15 p q = 1 a 1 + 1 a + + 1 a n 5C 1 1C 0 808 DCCCVIII 7C 0 18C 187 (Dedekind, 1831 1916) 1891 (Peano, 1858 193) 19C (Kronecker, 183 1891). 17C 150 a 0, b, c ax + bx + c = 0 x = b ± b 4ac 6C a x. (a, b, c) = (1, 0, ) 1 x + 0 x + ( ) = 0 x = 0 x = ±
I013 01- (a, b, c) x. (a, b, c) = (1, 0, 3) x + 3 = 0 x x 0 3 x x. x = 0 x = ± ( ) = ( 3) = 3 3 x + 3 = 0 x = ± 3. (a, b, c) = (1, 1, 1) x + x + 1 = 0. x = 1 ± 3 ( ) 1 ± 3 + 1 ± 3 + 1 0 16C 3 4 1541 1501 1576 3 4 10 40 5 + 15 5 15 17C 3 (imaginary number) 18C 1799 1777 1855 19C
I013 01-3 5 13 1 1 1-1. z = i, w = 3 + i z (1) z + w () z w (3) z w (4) w 1-. z = a + biw = c + di a, b, c, d (1) z w = z w. () w 0 z = z w w (3) z + w z + w 1-3.
I013 0-1 : May 13, 013 Version : 1.1 5/0 7 http://www.math.nagoya-u.ac.jp/~kawahira/courses/kansuron.pdf http://www.math.nagoya-u.ac.jp/~kawahira/courses/kansuron_in1.pdf A4 (4/) 1. x = x = ± x = 1 x = ± 1 1 i i. a b a + bi i (complex number) C C := {a + bi a, b R}. b = 0 a + 0i a C R C 3. C z = a + bi, w = c + di C (C0) a + bi = 0 a = b = 0 (C1) z ± w := (a ± c) + (b ± d)i (C) zw := (ac bd) + (ad + bc)i (C3) w 0 z ac + bd ad + bc := w c + + d c + d i 4. xy R R := {(a, b) a, b R}. (a, b) a + bi xy R C (a, b) a + bi i (0, 1) (C0) (C1) (C0) (a, b) = 0 a = b = 0 (C1) (a, b) ± (c, d) = (a ± c, b ± d)
I013 0- (C) (a, b) (c, d) := (ac bd, ad + bc) ( ) ac + bd ad + bc (C3) (a, b) / (c, d) := c, + d c + d (C) (C3) (C) (0, 1) (0, 1) = ( 1, 0) i = 1 i (C0) (C3) C xy xy R (a, b) (a, 0) (0, 1) (0, b) (0, 0) x y (a, b) x a (a, b) y b C a + bi a R i bi Ri 0 C R Ri a + bi a a + bi b. z = a + bi a Re z b Im z z z = a + b (a, b) (a, b) x z = a + bi arg z
I013 03-1 : May 7, 013 Version : 1.1 6/3 7 7/15, 0, 9 3 (5/13) 1.. 3. z i iz z 90 = 4. z = a+bi 0 r = z > 0arg z = θ a = r cos θ, b = r sin θ z z = r(cos θ + i sin θ) z (polar form) (polar representation) 1 1 = cos 0 + i sin 0 i = cos π + i sin π 1 = cos π + i sin π i = cos 3π + i sin 3π 1 + i = (cos π 4 + i sin π 4 ) 3 + i = (cos 5π 6 + i sin 5π 6 ) 5. 0 z w z = r(cos θ + i sin θ), w = r (cos θ + i sin θ ). (C) zw = rr (cos θ + i sin θ)(cos θ + i sin θ ) = rr { (cos θ cos θ sin θ sin θ ) + i(sin θ cos θ + cos θ sin θ ) } = rr { cos(θ + θ ) + i sin(θ + θ ) } 1 r = 1 z = cos θ + i sin θ z = e iθ e iθ = cos θ + i sin θ
I013 03- w = i r = 1, θ = π/ iz = r{cos(θ + π/) + i sin(θ + π/)} z π/ i π/ w = r (cos θ + i sin θ ) r θ 0, 1, z w w arg w 0, w, zw (5/7). z, w 0 (1) zw = z w arg zw = arg z + arg w () z = z arg z = arg z arg w w w w de Moivrez = r(cos θ + i sin θ) n z n = r n (cos nθ + i sin nθ). A = (1 + 3 i) 10 1 + 3 i = (cos π 3 + i sin π ) de Moivre 3 A = 10 (cos 10π 3 + i sin 10π 3 ) = 104(cos 4π 3 + i sin 4π 3 ) = 51 51 3i.
I013 04-1 : June 10, 013 Version : 1.1 4 (5/7) 1. (de Moivre) z = r(cos θ + i sin θ) m z m = r m (cos mθ + i sin mθ). B = ( 3 + i) 10 3 + i = (cos 5π 6 + i sin 5π 6 ) B = 10 (cos 50π 6 + i sin 50π 6 ) = 104(cos π 3 + i sin π 3 ) = 51 + 51 3i. 3. θ R e iθ = cos θ + i sin θ. 4. θ = π e iπ = 1 e iπ + 1 = 0 e, i, π, 1, 0 5. z = r(cos θ + i sin θ) z = re iθ z 6. z, w z w 7. {z n } n=0 = {z 0, z 1, z,...} {z n } α C z n α 0 (n ) lim n z n = α z n α (n ) 8. z n = (i/) n z n 0 = (i/) n = i n 1/ n = 1/ n 0. z n 0 (n ). 9. {z n } n=0 S 0 = z 0, S 1 = z 0 + z 1, S = z 0 + z 1 + z,... {S n = z 0 + z 1 + + z n } lim n S n z 0 + z 1 + + z n + n=0 z n n 0 {z n } (series) z n
I013 04-10. z n = (1/) n 1 + 1/ + 1/ + = 11. z n = n 1 + + 3 + S n 1. ( ) i n 1.. n=0 n=1 1 n 3. n=1 n i n 4. n 1 i n n 1. 4. 13. z C e z := 1 + z + z! + z3 3! + = n=0 z n n! ( ) z (exponential function) exp(z) 14. ( ) e 100 := 1 + 100 + 10000/! + 1000000/3! + 100000000/4! + 15. z, w e z e w = e z+w ) ) (1 + z + z + (1 + w + w + = 1 + (z + w) +!! 16. (6/10) 17. ( ) z = iθ (θ R) e iθ = ) ) (1 θ! + θ4 4! + (θ θ3 3! + θ5 5! i (z + w)! + θ sin θ = θ θ3 3! + θ5 θ cos θ = 1 5!! + θ4 4! e iθ = cos θ + i sin θ 18.. e iθ = cos θ + i sin θ 1 θ
I013 04-3 6 4 1 1-1. z 0 r(cos θ + i sin θ) m z m = r m (cos mθ + i sin mθ) -1. N z N = 1 ( ) mπi z = exp N (m = 0, 1,..., N 1) N = 6-3 A. z z > 1 1/z z z w w 1/z w = u + vi 1/z = u vi0 < z < 1 z = 1 1/z -4 S. D N := ( N n=0 )( z n N n! n=0 w n n! z, w N ) N n=0 (z + w) n N 0 (Hint: D N N + 1 zi w m i i!(m i)! (m = N + 1, N +,, N) N(N + 1)/ K = max{ z, w, 1} mc ik m zi w m i i!(m i)! n! m! (K)m m! (K)N (N+1)!
I013 05-1 : June 17, 013 Version : 1.1 5 (6/10) 1. ( ) i n 1.. n=0 n=1 1 n 3. n=1 n i n 4. n 1 1. 4. 1. /( i)4. log + πi/4 1.0 0.5 1.0 0.8 0.6 0.4 i n n 10 5 10 5 5 10 1. 0.5 0.5 1.0 0.5 3. 0. 1.0 0.8 0.6 0.4 0. 0. 0.4 0. 4. 5 10.. 1 + 1/ + 1/3 + 0 (1 + 1/ + + 1/n) log(n + 1) 1 n 1 1/nα α > 1 1+1/ +1/3 + = π /6 3. z 0 + z 1 + z + z 0 + z 1 + z + 4. 4. 5.. e z = 1 + z + z! + z z = 0 1 + 0 + 0 + = 1 z 0 z n = z n /n! z n+1 = zn+1 /(n + 1)! z n z n /n = z n + 1. z < N n N z n+1 z n = z n + 1 < N N + 1.
I013 05- r = N/(N + 1) < 1 k z n+k < r k z N z 0 + z 1 + + z N+k z 0 + z 1 + + z N 1 + z N + + z N+k < z 0 + z 1 + + z N 1 + z N (1 + r + r + + r k ) 1 < z 0 + z 1 + + z N 1 + z N 1 r < z 0 + z 1 + + z N 1 + (N + 1) z N. k 6. z, w e z e w = e z+w ) ) (1 + z + z + (1 + w + w (z + w) + = 1 + (z + w) + +!!! 7. 0 z = re iθ w = r e iθ zw = re iθ r e iθ = rr e i(θ+θ ) 30 ( 30 ) (1989/04) T. (00/01) http://www.math.nagoya-u.ac.jp/~kawahira/courses/kansuron.pdf
I013 06-1 : June 4, 013 Version : 1.1 3 6 (6/17) 1. x + bx + c = 0 x = b ± b 4c. 3 x 3 + ax + bx + c = 0 x = (a, b, c ) n a, b, c R 3. (Cardano) x = A + B a 3, x = ωa + ω B a 3, x = ω A + ωb a 3.. A = 3 q + q + p 3, B = 3 q q + p 3, p = 1 ) (b a, q = 1 3 3 (c + a3 7 ab 3 ω = 1 + 3i ). = e πi/3, 4. (Bombelli) x 3 = 15x+4 x = 4 a = 0, b = 15, c = 4 p = 5, q = A = 3 + 11 = 3 + 11i, B = 3 11i x = A + B ( ± i) 3 = ± 11i A = + i, B = i x = A + B = 4 3 5. 3 α 0 z 3 = α z 6. α = 1 z 3 1 = (z 1)(z + z + 1) = 0 z = 1, ω, ω z = e 0i, e πi/3, e 4πi/3 α = re θi A = 3 re θi/3 A 3 = α z 3 = α = A 3 (z/a) 3 = 1 z/a = 1, ω, ω z = A, ωa, ω A.
I013 06-7. x 3 + ax + bx + c = 0 y = x + a/3 y 3 + 3py + q = 0 p, q y x y = u + v y y 3 + 3py + q = (u 3 + v 3 + q) + 3(uv + p)(u + v) 0 u 3 + v 3 = q uv = p u, v u 3 v 3 = p 3 u 3 v 3 t qt p 3 = 0 u α 3 v β 3 uv = p u 3 = q + q + p 3 =: α, v 3 = q q + p 3 =: β 3 (u, v) = (A, B) y = u + v (u, v) = (ωa, ω B), (ω A, ωb) 3 y = A + B, y = ωa + ω B, y = ω A + ωb 8. (A, B) q + p 3 0 α, β 3 A = 3 α, B = 3 β q + p 3 < 0 α = q + q + p 3 i = re θi, β = q q + p 3 i = re θi A = 3 re θi/3 B = 3 re θi/3 AB = p R 9. a, b, c 3 WikiPedia http://ja.wikipedia.org/wiki/ 3 a, b, c http://hooktail.sub.jp/algebra/cubicequation/ (1987/01) / (011/11) (001/11/1)
I013 07-1 4 : July 1, 013 Version : 1.1 7 (6/4) 1. 3 16C 156 1545 4. 4 x 4 + ax 3 + bx + cx + d = 0 a, b, c, d R n 3. Step x 4 + ax 3 + bx + cx + d = 0 y = x + a/4 y 4 + py + qy + r = 0 (1) p, q, r a, b, c y x 4. Step q = 0 (1) y 4 + py + r = 0 t = y q 0 y 4 + py + qy + r = ( y + p + u ) u( y q ) u u 0 u 3 u(p + u) 4ru = q q 0 u 0 3 u 5. Step u (1) {( y + p + u ) ( u y q ) }{( y + p + u u ) + ( u y q ) } = 0 u { } 0 y 6. 5 5 7. a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 a n 0, a n 1,..., a 0 C 8. a n x n + + a 1 x + a 0 = a n (x α 1 ) (x α n ) α 1,..., α n n n
I013 08-1 : July 8, 013 Version : 1.1 7/15 7/9 7/9 8 (7/1) 1. a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 a n 0, a n 1,..., a 0 C. n = 3, a 3 = 1 ( ) f(z) = z 3 + a z + a 1 z + a 0 = 0. 3. z, w C 4. R, M > 0 z w z + w z + w. z R = f(z) M > 0. R M = R 3 / z R f(z) = 0 5. f(z) = z 3 1 + ( a z + b + c ) ( z 3 1 a z z 3 z + b + c z z R a z 3 ). z 3 a z + b + c z z a 3 R + b + c a, b, c R R R 3 a R + b + c 1/ f(z) z 3 (1 1/) = z 3 / R 3 /. R R 3 M := R 3 / > 0 6. α C f(z) f(α) ( 0). z + b + c z z C α f(z) 7. f(z) = 0 f(α) = 0 f(α) > 0 8. f(0) < R 3 / R > 0 K = { z R} z f(z) 0 α K z R f(z) f(α) z R f(z) R 3 / > f(0) 0 K f(0) f(α) z C f(z) f(α)
I013 08-9. f(z) = 0 f(α) > 0 f(z) f(α) f(z) f(α) = (z α) 3 + a(z α) + b(z α) a b a, a 1, α f(z) (z α)3 a(z α) b(z α) = 1 + + + f(α) f(α) f(α) f(α) f(α) 0 w = z α A = 1/f(α)B = a/f(α)c = b/f(α) f(α + w) f(α) = 1 + Aw 3 + Bw + Cw f(α+w) f(α) < 1 w f(α) 10. C 0 Cw < 0 w w C = re iθ w = ρe i(π θ) Cw = rρe πi < rρ e πi = 1 ρ η(w) := (Aw + Bw)/C f(α + w) f(α) = 1 + Cw(1 + η(w)) η(w) w Aw + B / C w ( A w + B)/ C = ρ(ρ A + B )/ C A, B, C ρ η(w) 1/ ρ 0 < rρ < 1 w = ρe i(π θ) f(α + w) f(α) = 1 + Cw(1 + η(w)) = 1 rρ rρη(w) 1 rρ + rρη(w). 1 rρ = 1 rρ > 0 rρη(w) = rρ η(w) rρ/ f(α + w) f(α) 1 rρ + rρ/ = 1 rρ/ < 1 f(α) 11. C = 0, B 0 Bw < 0 w w 1. C = B = 0, A 0 Aw 3 < 0 w w
I013 08-3 3 7 1 1 3-1. (1) α, β 0 0, α, β α αβ + β = 0 Hint: α β ±π/3 () α, β, γ α, β, γ α + β + γ αβ βγ γα = 0 3-. ABC BCCAAB m : n A B C ABC A B C Hint: 3-1 3-3 A. f(z) = z n + a n 1 z n 1 + + a 1 z + a 0 (a n 1,..., a 0 C) n f(z) = 0 α C z f(z) z C f(z) f(α). 3-4 S. sin nβ ( cos α + n 1 ) β cos α + cos(α + β) + cos(α + β) + + cos(α + (n 1)β) = sin α + sin(α + β) + sin(α + β) + + sin(α + (n 1)β) = sin β sin nβ ( sin α + n 1 ) β Hint z n 1 + + z + 1(= 1 zn 1 z ) sin β 7 9
I013 09-1 : July 15, 013 Version : 1.1 9 (7/8) 1. a n+1 = a n + 1; a 1 = 1 x = x + 1 x = 1 a n+1 + 1 = (a n + 1); a 1 + 1 = b n := a n +1 b n+1 = b n ; b 1 = b n = n a n = b n 1 = n 1. x = x + 1 3. b n+1 = b n ; b 1 = b 0 = 1, b 1 = 1/, b = 1/4,... {b n } 0 a n = b n 1 a n 1 a n+1 = a n + 1-1 0 1 3 7 15 b n+1 = b n 0 1 4 8 16 4. 3 b n R 0, ±1, ±, x 1 x g(x) = x R g(x) = x (dynamical system) x 0 R x 0 g g(x 0 ) = x 0 g g(g(x 0 )) = x 0 g x 0 (orbit) b n b 1 = 5. a n 6. R f(x) = x + 1 a n+1 = f(a n ) a n+1 = f(f( f(a 1 ) )) f n a n+1 = f n (a 1 ) 7. p y = f(x) p f(p) f(f(p)) f(f(f(p)))... y = f(x) xy (p, p) (f(p), f(p))
I013 09- Step 0 y = f(x) y = x Step 1 (p, p) y = f(x) (p, f(p)) Step (p, f(p)) y = x (f(p), f(p)) Step 1 y = x p (graphical analysis) web diagram y f(p) y = x p y = f(x) O p f(p) x 8. g 6 4 5 4 3 1 1 4 1 1 3 4 5 x 0 = 1/4 6 x 0 = 1/4 9. g x 0 0n { g n + (x 0 > 0) (x) (x 0 < 0) g n (x 0 ) 0 x 0 = 0 g(x 0 ) = x 0 0
I013 10-1 / : July, 013 Version : 1.1 10 (7/15) 1. f(x) f(x 0 ) = x 0 f. g(x) = x x = x x = 0g 0 3. f(x) = x + 1 x + 1 = x x = 1f 1 4. a n a n+1 = f(a n ) = a n + 1 x = x + 1 a n+1 + 1 = (a n + 1) 1 5. f g x R f : x x + 1 X R g : X X X = T (x) = x + 1 x = T 1 (X) = X 1 T f T 1 (X) = T f(x 1) = T ((X 1) + 1) = X = g(x) g = T f T 1 f f R R R T T f T f f R g R g R g T f g g = T f T 1 g T = T f X = T (x) f g (conjugate) 6. f(x) = ax + b (a, b, c, d R, ad bc 0) cx + d 7. c = 0 ad bc = ad 0 d 0 f(x) = (ax + b)/d f(x) = x + 1 1 8. R 1 xy C : x + (y 1/) = 1/4 N= (0, 1) x R X= (x, 0) 1
I013 10- NX C P(x) R x P(x) C N C N C N y C N P(x) 0.8 0.6 0.4 O X x R 0. x ± P(x) N N N (point at infinity) C = (C N) {N} C = R { } C ˆR C = ˆR x + x = x + =. 0 x x = x =, x = 0, x 0 =., ±, /, 0. x = (1) x + 1 x (3) 1 x () 3x + 1 (4) x 1 x 4 (extended reals) C C
I013 11-1 : July 9, 013 Version : 1.1 11 (7/) 1 (7/9) 1. ˆR ˆR. f(x) = x + 1 x f() = + 1 = 3 0 =. f(x) = 1 + 1/x 1 /x 1 + 1/ f( ) = 1 / = 1 + 0 1 0 = 1. 3. f(x) = ax + b a, b, c, d R, ad bc 0 cx + d 1 4. f(x) = ax + b cx + d ˆR 5. f(x) = x, x + 1, x + 1 x ˆR ˆRx x + 1 x + 1 x 6. f(x) = ax + b cx + d 7. f(x) = ax + b cx + d (1) g(x) = λx (x ˆR, λ > 1) () g(x) = x + 1 (x ˆR) (3) g(z) = e iθ z (z C, z = 1, θ R) 1
I013 11-8. f(x) = 3x + g(x) = 4X x + f(x) = x x = 1, 0 X = T (x) = x + 1 T ( 1) = 0, T () = x ( ) ( X + 1 3 X+1 T f T 1 X 1 (X) = T f = T + ) X 1 X+1 X 1 + = = 4X T f T = g f g T 9. a n+1 = 3a n + a n + = f(a n); a 1 = 1 a n = f n 1 (a 1 ) g n = T f n T 1 f n = T 1 g n T ( ) ( a n = T 1 g n 1 T (a 1 ) = T 1 g n 1 a1 + 1 = T 1 4 n 1 1 + 1 ) = = 4n 1 a 1 1 4 n 1 + 1. 10. f(x) = 3x + 1 g(x) = X + 1 x + 5 f(x) = x (x 1) = 0 x = 1 X = T (x) = 4 T (1) = x 1 ( ) X 4 T f T 1 (X) = T f = = X + 1 X T f T = g g R f g T 11. f(x) = x + 1 x + 1 a ˆR a f f(a) f f (a) 4 f f 3 (a) f f 4 (a) = a
I013 11-3 1. R ˆR C Ĉ C Ĉ = C { } ˆR 13. f(x) = x + 1 x + 1 f(x) = x x + 1 = 0 x = ±i 0 z = T (x) = x i x + i T (i) = 0T ( i) = T f T 1 (z) = T f ( ) i(z + 1) = = iz 1 z f g(z) = iz x ˆR T (x) = x i x + i = x + 1 x + 1 = 1 T (ˆR) = {z C z = 1} =: T ˆR f(x) = x + 1 T T x + 1 g(z) = iz g(z) = iz = e πi/ z 90 g 4 (z) = i 4 z = z f 4 (x) = T 1 g 4 T (x) = T 1 T (x) = x f 4 T (ˆR) T z = T (x)
I013 11-4. f(x) = ax + b cx + d x = f(x) x = ax + b cx + d (ad bc 0) (a, b, c, d) = (1, 0, 0, 1) ˆR c = 0 ad bc = ad 0 d 0 1 (*) (*) ˆR 3 x = x = c 0 f(x) = a d x + b d x = f(x) dx = ax + b b d a x = ax + b cx + d cx + (d a)x b = 0 a = d x = (*) 3 Case 1 ˆR Case ˆR Case 3 ˆR 3. ( ) Case 1 3 α, β X = T (x) = (x α)/(x β) f g = T f T 1 g(x) = λx (λ C {0, 1}) α, β β = f 1 x px+q T (x) = x α g = T f T 1 Case 1 f T ˆR ˆR g = T f T 1 ˆR ˆR λ 4 λ < 1 0 X = S(X) = 1/X G = S g S 1 = (S T ) f(s T ) 1 G(X ) = X /λ Case 3 α, β a ± bi T (ˆR) = T f ˆR ˆR g = T f T 1 T T f ˆR ˆR T T T g T g(z) = λz z = 1 g(z) = 1 λz = λ = 1 λ = e iθ (θ R) 3 R 4 R g
I013 11-5 ( ) ˆR Case α α X = T (x) = 1/(x α) g = T f T 1 g(x) = X + µ (µ R {0}) X = S(X) = X/µ G = S g S 1 = (S T ) f (S T ) 1 G(X ) = X + 1 α = f x x + µ (µ R {0}) g = S f S 1