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1 ) :00 12:00 e iπ 1. i ). e π. MacTutor History of Mathematics archive www-history.mcs.standrews.ac.uk/history/ FAQ kawaguch))math.sci.osaka-u.ac.jp

2

3 Part 1 e πi = 1 e = π = i i 2 = 1 1/70

4 e πi + 1 = 0 π e i 1 0 2/70

5 1 = 100%) e e e π e e = /70

6 Benjamin Pierce 19 e πi = 1 We cannot understand it, and we don t know what it means, but we have proved it, and therefore we know it must be the truth. 4/70

7 David Mumford e πi = 1 i ) cf. Mumford, Calculus Reform For the Millions, Notices AMS, May 1997 Mumford 5/70

8 Part =100%) = 1 2 t 6/70

9 = = 4 8 t = 2 t = t 7/70

10 2 1 = = = = = = = = = = 1024 t 2 t /70

11 /70

12 2 80 = = /70

13 g = kg ) ) kg 1 60kg = 33 11/70

14 1 6,000 12/70

15 Coffee Break log 2 80 = log 10 2 = log = 80 log 10 2 = = = /70

16 2-2 14/70

17 = = 2 5 = = /70

18 5 16/70

19 5 17/70

20 18/70

21 5 80 = = /70

22 /70

23 TV MIT Happy Ending 21/70

24 Coffee Break 0.1mm = mm 44 km 38 km 22/70

25 2-4 1 =100%) 1 1 = 2 t = 2 t 2 t t /70

26 Part 3 e 1 =100%) 1, = ) 2 24/70

27 1 = ) ) ) ) 12 2 = ) /70

28 1 2 1 = = ) 26/70

29 1 =100%) 1, = ) 3 27/70

30 ) ) = ) ) ) /70

31 1 = ) 2 = ) ) ) ) /70

32 1 3 1 = = ) 30/70

33 1 =100%) 1, 1 n n = 2, 3 1 = 1 + n) 1 n n = 1/n = 1 =100%) 1 + n) 1 n n 31/70

34 a n := n) n n a 1 = 2 a 6 = a 2 = 2.25 a 7 = a 3 = a 8 = a 4 = a 9 = a 5 = a 10 = a 1 < a 2 < a 3 < a 4 < a 5 < a n < 3 n = 1, 2,...) 32/70

35 3-1 e a n := ) n n n = 1) a 1 < a 2 < a 3 < a 4 < a 5 < 2) a n < 3 n = 1, 2,...) 1), 2) cf 1) 1 33/70

36 n a n = 1 + n) 1 n e e = e e 1 =100%) 1 1 = e 34/70

37 3-2 t = e = t t = 1 n n 1 = 1 + n) 1 n e { t = 1 + n) 1 n } t e t 35/70

38 3-3 1 =100%) 1 1 = e = t = e t t cf Part 1 1 = 2 t = 2 t a a ) 1 t = e at t 36/70

39 Part 4 i 2 1 i 2 = 1 complex number z = a + b i a, b i ) i, 1 + i, 2 37/70

40 Coffee Break 2 x 2 + 2x + 2 = 0 x 2 + 2x + 2 = 0 = x + 1) = 0 = x + 1) 2 = 1 = x + 1 = ±i, = x = 1 + i, 1 i 38/70

41 4-1 i 2 1 a + bi) + c + di) = a + c) + b + d)i a + bi) c + di) = a c) + b d)i 2 + 7i) i) = i 39/70

42 i 2 = 1 a + bi)c + di) = ac + adi + bci + bdi 2 = ac + adi + bci + bd 1) = ac bd) + ad + bc)i 1+2i) 3+4i) = i+2 3i+2 4i 2 = 3 8) )i = i 40/70

43 Coffee Break ) 3 x 3 = 15x i imaginary number) i real number) real) 41/70

44 = n = 42/70

45 1, 2 negative number) negative false positive number) = /70

46 4-3 z = x + yi i, 1 + 2i, 3 i iy 1 + 2i i 0 3 i x 44/70

47 z = x + iy r θ x = r cos θ, θ r x y = r sin θ z = r cos θ + ir sin θ z = x + iy z r z θ z y 45/70

48 Coffee Break r = 2, z = 1 + i 1 + i r 1 θ 1 θ = π 4 z = 1 + i = 2 cos π 4 + i 2 sin π 4 46/70

49 4-4 z 1, z 2 z 1 = r 1 cos θ 1 + ir 1 sin θ 1 z 2 = r 2 cos θ 2 + ir 2 sin θ 2 z 1 z 2 z 1 z 2 47/70

50 z 1 z 2 = r 1 cos θ 1 + ir 1 sin θ 1 )r 2 cos θ 2 + ir 2 sin θ 2 ) = r 1 r 2 cos θ 1 cos θ 2 sin θ 1 sin θ 2 ) + ir 1 r 2 sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) = r 1 r 2 cosθ 1 + θ 2 ) + ir 1 r 2 sinθ 1 + θ 2 ) z 1 z 2 r 1 r 2 θ 1 + θ 2 z 1 z 2 = z 1 z 2 z 1 z 2 = z 1 + z 2 48/70

51 1 θ = 1 θ + θ = 2θ cos θ + i sin θ) 2 = cos2θ) + i sin2θ). n cos θ + i sin θ) n = cosnθ) + i sinnθ) de Moivre 49/70

52 4-5 i i 2 = 1 z = x + yi x, y z z = r cos θ + ir sin θ r z θ z cos θ + i sin θ) n = cosnθ) + i sinnθ) de Moivre 50/70

53 Part 5 e iπ = 1 e iπ e iπ Part 2 a a 1 t = e at i i 1 t e it 51/70

54 e iπ = 1 i 1 π 1) i t 52/70

55 5-1 1 i 1 1 = e i e i e i = cos 1 + i sin 1 53/70

56 i i = i 1 + i 1 = 1 + i i 1 54/70

57 1 2 i i 2 = i i 2 55/70

58 1 = i ) i ) 2 ) = i i ) i ) i 2 i 2 56/70

59 1 2 1 = 1 + i 2) ) i i ) 2 57/70

60 Coffee Break i 1 1 = 1 + i 1 i = 1 + i ) 2 = i 1 3/4 = 7500 i 58/70

61 1 n 1 = 1 + i n) n n 1 + i n) n e i e i = cos 1 + i sin 1 59/70

62 1 + i 1 n n 1 + i 1 n cos 1 n + i sin 1 n cos 1 n + i sin 1 n i n 1 1 n 1 n 1 60/70

63 cos 1 n + i sin 1 n 1, 1 n radian de Moivre cos 1 n + i sin 1 n ) n = cos ) 1 n n = cos 1 + i sin 1 + i sin ) 1 n n 61/70

64 n 1 + i 1 n cos 1 n + i sin 1 n = 1 + i 1 n ) n cos 1 n + i sin 1 n ) n = cos 1 + i sin 1 n 1 + n) i n cos 1 + i sin 1 e i = cos 1 + i sin 1 62/70

65 5-2 t i 1 t t = e it e it 1 1 n n t = 1 + n) i nt e it 63/70

66 1 + i n) nt n 1 + i 1 n cos 1 n + i sin 1 n cos 1 n + i sin 1 n ) nt = cos ) 1 n nt = cos t + i sin t. + i sin ) 1 n nt 64/70

67 n 1 + i n cos 1 n + i sin 1 n = 1 + i n) nt cos 1 n + i sin 1 n) nt = cos t + i sin t n e it = cos t + i sin t 65/70

68 5-3 π t e it = cos t + i sin t t t = π e iπ = cos π + i sin π = 1 66/70

69 5-4 i 1 t t ) = e it = cos t + i sin t π = e πi = 1 1 2π = e 2πi = cos2π) + i sin2π) = /70

70 e = ) n n n i 1 t t ) = e it = cos t + i sin t t π n t 68/70

71 1 + 1 ) n n e := lim n e it := lim n 1 + i n ) nt t = lim 1 + i ) nt = cos t + i sin t n n e it = cos t + i sin t t t = π t = e iπ = 1 69/70

72 Leonhard Euler ) MacTutor History of Mathematics archive www-history.mcs.st-andrews.ac.uk/history/ 70/70

73 1. e 1= 100%) 1 1 n n) n ) a 1 < a 2 < a 3 < ) 2) a n < 3 n = 1, 2,...) a n = ) n n, a n = n ) n 1 = 1 + n C 1 n + 1 nc 2 n nc n n n = 1 + n 1 nn 1) 1 + n 2 n = ! 1 1 n nn 1) ) n! n n n! n ) 1 2 n ) 1 n 1 ) n ), a n < a n+1 a n < a n+1 ) ) ) ) 1 2 ) 1 n 1 ) 2! n n! n n n < ) ) 1 2 ) 1 n 1 ) 2! n + 1 n! n + 1 n + 1 n ) 1 2 ) 1 n 1 ) 1 n ) n + 1)! n + 1 n + 1 n + 1 n + 1 a 2 < a ) < ! 2 2! k = 1, 2,, n ) 1 2 ) 1 k ) k! n n n < 1 k! 1 1 ) ) 1 2 ) 3 3! ) 1 2 ) 1 k ) n + 1 n + 1 n + 1 ) 1) 2) k = 1, 2,, n ) 1 2 k! n n ) 1 k ) < 1 n k!

74 2 ) ) a n = n ) n n 2! + 1 3! n! n! = n n 1) n 1 ) ! + 1 3! n! n 1 = n 1 < 3 ), ), a n < 3 2) ), 2) {a n }, {a n }. e. e = log log x = 100%) 1 2 a 2 1 a t 1 + a ) t a ) t = T = log 2 log ) 1 + a S = 72 a /1 = /6 = 12 2 T 72 S a) x 0 b) 0 < a 10 x x2 2 log1 + x) x a ) < S log T < 0.72 log < S T < < log 2 <

75 3 3. e i lim 1 + i n = cos 1 + i sin 1 n n) e i e i = cos 1 + i sin ) x 0 1 x2 2 cos x 1 2) x 0 x x3 6 sin x x x 0 sin x x x sin x) = 1 cos x 0 x sin x x = 0 0 x 0 x sin x 0 1) cos x 1 x 0 1 x2 x2 2 cos x fx) = cos x f x) = sin x + x 0. fx) x 0 f0) = 0 x 0 fx) 0 2) sin x x x x3 x3 6 sin x gx) = sin x x + 6 1) x 0 g x) = cos x 1 + x2 2 0 gx) x 0 g0) = 0 x 0 gx) 0 z = 1 + i n, w = cos 1 n + i sin 1 n n w = cos 1 n + i sin 1 n z = 1 + i n 1 Figure 1. z w 1 n w z 1 1) z w 1 n 2) z n w n ) z = n 2 4n n n n 2 ) n n n 6 = 1 n n 1) 3.11)2) z w = 1 cos 1 ) 1 + i n n sin 1 ) = n n n, w = 1 max{ z, w } = 2 + w n 1 ) n 2 ) n n 1 n 1 cos 1 ) n n sin 1 ) 2 n 10 1 n 2. z n w n = z w z n 1 + z n 2 w + + w n n z n w n = z w)z n 1 + z n 2 w + 2

76 4 z w 1) z n 1 + z n 2 w + + w n 1 z n 1 + z n 2 w + + w n 1 z n 1 + z n 2 w + + w n 1 ) n 1 n max{ z, w } n 1 = n 1 + 1n n2 ) n 1 = 1 + 1n 2 ) n n) n < ) 1) z n w n = z w z n 1 + z n 2 w + + w n 1 ) n n 2 n + 1n n 3.3. n z w z n w n = cos 1 + i sin 1 n z w 1 n 3.21) z n 2 w n = cos 1 + i sin 1 1 n 3.22) 3.4. lim 1 + i n = cos 1 + i sin 1. n n) w n = cos 1 n n) + i sin 1 n ) ) 1 1 = cos n n + i sin n n = cos 1 + i sin 1 de Moivre 3.22) z n cos 1 + i sin 1) n n 4. e e 4.1. e = 1 + k=1 1 k! 1 *) n ) n = ! n 1 1 ) ) 1 2 ) 1 n 1 ) n n! n n n ! n! 1 + e 1 + k=1 k=1 1 k! 1 k!

77 5 n m 1 *) ) n = ) ) 1 2 n 2! n m! n n ) 1 2 ) 1 n 1 ) n! n n n ) ) 1 2 2! n m! n n m n e ! m! = 1 + m m m 1 e 1 + k! e = e k=1 1 k! e = a a, b ) b e = b b 1 e 1 k! = 1 k! b! ab!) b k=1 b! b k=1 k=1 k=b+1 b! k! = k=b+1 k=1 b! k! 1 k! ) 1 m 1 ) n ) 1 m 1 ) n ab!) b = ab 1)!) 1 k b b! k! = bb 1) b k + 1) k b + 1 b + 1 > 2 k=b+1 0 < b! k! = 1 b + 1)b + 2) k < 1 2 k b k=b+1 b! k! < k=b k b = 1 2 i = 1 b! 0 1 k! e i=1

29

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