Collatzの問題 (数学/数理科学セレクト1)
|
|
- しまな さかいざわ
- 5 years ago
- Views:
Transcription
1 / AICHI UNIVERSITY OF EDUCATION
2 A { z = x + iy <x<0.700, <y<0.600 }. z F (z) F n (z). 4.2
3 B { z = x + iy <x<1.500, <y<1.200 }. z F (z) F n (z). 4.2 i
4 ii
5 m n (1) (2) (3) m (4) (5) iii
6 f(x) =y F iv
7 1 1.1 Syracuse Hasse Ulam 3n +1 Lothar Collatz( )
8 y y y 1 y 2 y n 1 y n y
9 ( : x f f f(x) 3x +1 2 k 2 k+1 k 3x +1 y = 2 k f y = f(x) f f(x) 3x +1=2 k y y f. x =53 3x + 1 = = 2 5 f(53) = 5 3
10 f(x) f 2 (x) = f(f(x)) f(f(x)) f(x) f. f 3 (x) = f(f(f(x))) { n }} { f n (x) = f(f n 1 (x)) = f(f( (f(x)) )) f n (x), n =2, 3, f n (x) f(x) n f(x) n = n {}}{ f(x) f(x) f(x) f n (x) (n =2, 3, ) f(x) f 2p +1 n f n (2p +1)=1, f f 4
11 x f(4x +1)=f(x) 3(4x +1)+1=12x +4=2 2 (3x +1). 1.2 f(x) =y x, y f(x) =y x x, y k 3x +1=2 k y x = 2k y 1 3 3m +1,3m +2,3m +3 (m =0, 1, 2, ) y =3m + r (r =0, 1, 2) x = 2k (3m + r) 1 3 =2 k m + 2k r 1 3 y =3m + r 3, r =0 5
12 , p =0, 1, 2, p 2 +4 p 1 +4 p = 4p = 22p p 2 +4 p 1 + 4p 3 = 4p p 3 = 22p y 3 1, r =1 x k 4 y 1 3, 4 2 y 1 3, 4 3 y 1 3,, 4 n y 1 3,. y 3 2, r =2 2y 1 3, 4 2y 1 3, 4 2 2y 1 3,, 4 n 2y 1 3,. 3m +1, 3m +2, 3m +3 (m =0, 1, 2, ) y 6n +1, 6n +3,6n +5(n =0, 1, 2, ) 6
13 x, y f(x) =y x y =6n p+2 (6n +1) 1 3 =4 p (8n +1)+4 p (p =0, 1, 2, ). y =6n +3. y =6n p+1 (6n +5) 1 3 =4 p (4n +3)+4 p (p =0, 1, 2, ). 1.3, n y 1,y 2,,y n 1,y n = y 0 y n = y 0 y 1 y 2 y n 1 y n n 7
14 Computer x f 3x +1 2 k 2 k+1 k y = 3x +1 2 k f y = f(x) 3x +1=2 k y y f f(x) 1 ; 1 ( 2 )
15 y y = y 0 y 1 y 2 y n 1 y n y ( ) 2 p1 y 1 =3y 0 +1 (p 1 1) 2 p2 y 2 =3y 1 +1 (p 2 1). 2 pi y i =3y i 1 +1 (p i 1). 2 pn y n =3y n 1 +1 (p n 1) ( ) y y n ( ) 2 p1+p2+ +pn y n = 2 p1+p2+ +pn p1+p2+ +pn i 2 p1+p2+ +pn i n 2 2 p1 +3 n 1 +3 n y n y n = y n ( ) (2 p1+p2+ +pn 3 n )y =2 p1+p2+ +pn p1+p2+ +pn i 2 p1+p2+ +pn i n 2 2 p1 +3 n 1 9
16 y, p 1 1,p 2 1,,p n 1 ( ) y =1,p 1 = p 2 = = p n =2 y = 1,p 1 = p 2 = = p n =1. (1) ( ) n =2 p 1 =1,p 2 =2,y= 5 ( ) ( )( 5) = (2) n =7, p 1 = p 2 = p 3 = p 5 = p 6 =1,p 4 =2,p 7 =4,y= 17 p 1 + p 2 + p 3 + p 4 + p 5 + p 6 + p 7 = = = ( 139)( 17) = 2363 ( ). 10
17 p 3 q = ±1 p, q ( ) n = q p 3 q =1 p, q 2 p 3 q = 1 2 p 3 q =1 p>n= q 2 p 1 =2, p 2 = p 3 = = p n 1 =1,p n = p n p 1 + p p n = p 2 p 3 n =1 y =2 p1+p2+ +pn p1+p2+ +pn i 2 p1+p2+ +pn i n 2 2 p1 +3 n 1 11
18 ( ) (2 p1+p2+ +pn 3 n )y =2 p1+p2+ +pn p1+p2+ +pn i 2 p1+p2+ +pn i n 2 2 p1 +3 n 1. 2 p 3 q =1 p, q. 2 p 3 q = 1 p =3,q=2 Catalan x a y b =1 a =2,b=3,x=3,y=2. n =2, 3, 4 ( ) ( ) y, p 1,p 2,,p n 12
19 ( ) Computer Tomas Oliveira e Silva, 1999 n y n = y 0 y 1 y 2 y n 1 y n y n = y 0 > 1, y 1 > 1,,y n 1 > ( ) 2 p1+p2+ +pn y 1 y 2 y n = (3y 0 + 1)(3y 1 +1) (3y n 1 +1) = 3 n y 0 y 1 y n 1 ( y 0 )( y 1 ) (1 + 2 p1+p2+ +pn y n =3 n y 0 ( y 0 )( y 1 ) (1 + 2 p1+p2+ +pn =3 n ( y 0 )( y 1 ) (1 + a (y n =)y 0, y 1,, y n 1 >a y n 1 ) 1 3y n 1 ) 1 3y n 1 )
20 10 10 a = n < 2 p1+p2+ +pn =3 n ( y 0 )( y 1 ) (1 + < 3 n ( a )n 1 3y n 1 ) II n ( ) x 1+x 8 y=k(x) y=3x/
21 k(x) k(x) =log 2 (1 + x) < 3x 2 III 3 n < 2 p1+p2+ +pn < 3 n ( a )n log 2 (1 + x) < 3x 2 (x >0) log 2 3 < p 1 + p p n n < log a log 2 3= a < p 1 + p p n n a =10 10 < 2.0. log 2 3= > = < p 1 + p p n n < p 1 + p p n n log = (p 1 + p p n ) n n ( ) log 3 < log = log 3 log < <
22 (p 1 + p p n ) n n > n (p 1 + p p n ) n < 10 9 n n>10 9 /( ) = x e =2.718 log(1 + x) x log(1 + x) <x 3log2= =2.079 > 2 x log 2 (1 + x) < 3x 2 16
23 HP( ) 10 = BASIC C++,Mathematica Maple. 17
24 ., Windows BASIC Discoversoft ActiveBasic, BASIC.., BASIC, JIS Full BASIC Windows, Full BASIC (ISO) BASIC. 18
25 BASIC [CollatzTest.BAS] #structcode STARTF=1 DO PRINT CollatzTest PRINT INPUT i( 0)=, i k=0 Collatz c=i WHILE c >1 IF c MOD 2 = 0 THEN c=c/2 ELSE c=3*c+1 END IF PRINT ;c; k=k+1 WEND PRINT ;k;! PRINT INPUT <Yes(Return)/No>=,S$ IF (S$ <> AND S$ <> Y AND S$ <> y ) THEN STARTF=0 LOOP WHILE STARTF=1 END 1 ( 19
26 20
27 11 1 ( BASIC [OddCollatzTest.BAS] #structcode STARTF=1 DO PRINT OddCollatzTest PRINT INPUT i( 0)=,i k=0 Collatz c=i WHILE c>1 c=3*c+1 r=c Mod 2 WHILE r=0 c=c/2 r=c Mod 2 WEND PRINT ;c; k=k+1 WEND PRINT ;k;! PRINT INPUT <Yes(Return)/No>=,S$ IF (S$ <> AND S$ <> Y AND S$ <> y ) THEN STARTF=0 LOOP WHILE STARTF=1 END 21
28 2.2 1 n 1 n n k 1 j k b n = j 1 + j j n ( = 1 n ) j n n k b n log n k=1 q(n) = log n n n k=1 j k b n q(n)
29 , , C BASIC
30 1 n k 1 i k a n = i 1 + i i 2n 1 ( = 1 n ) i n n 2k 1 a n log n log 2 2 p(n) =2.406 log n log 2 2 q(n) 3p(n) 6.56 k=1 n n k=1 i 2k 1 a n p(n)
31 n ( 10 9 ) b n q(n) = log n = log n log a n p(n) =2.406 log n log 2 2 3a n b n 6.56 (10 5 n 10 9 ) n 25
32 Pierre de Fermat, x n + y n = z n n 3 0 x, y, z Andrew Wiles. n =2 0. Christian Goldbach, L. Euler, J. Richstein (Erdös-Strauss) n 4 n = 1 a + 1 b + 1 c a, b, c. p 4 2+3p = 1 2+3p p + 1 (1 + p)(2 + 3p) n
33 3 3.1 f N = {1, 2, 3, } Z = {0, ±1, ±2, } OZ = {±1, ±3, ±5, }. x 2-0 x 2 k a (k Z, a, b OZ ) b k k x 2-, e(x) - 27
34 GQ = { 2 k 2m +1 m, n Z, k =0, 1, 2, } 2n OQ = { 2m +1 m, n Z } 2n GQ 1 3 OQ - 1 x 3 y( OQ f - k x =2 k y ( GQ, y OQ) f(x) =y x ( OQ ) k =1, 2, 3x +1=2 k y (y OQ) f(x) =y f(x) GQ x OQ f(x) f GQ OQ f(x) 28
35 - k x = 2k r (r, s OZ ) s f(x) = r ( OQ s x = r s OQ 3r + s - k t 3r + s =2 k t 3 x +1=3 r s +1= 3r + s s = 2k t s f(x) = 3x +1 2 k = t s ( OQ 3 5 f( 3 5 )= 7 5 f( 7 13 )= 5 5 f GQ OQ f OQ OQ f(x) OQ a(x) f OQ OQ OQ a(x) =2x = 6x +1 3 (x OQ) f(a(x)) = f(x) (x OQ ) a(a(x)) = 4x +1 (x OQ) f(4x +1)=f(x) (x OQ ) 29
36 f f(x) =y x, y f(x) =y x x 2 k y 1 3 (k =1, 2, 3, ) f OQ OQ f OQ { 1 2 k 3 k =1, 2, 3, f f(x) =x. } k, x = x 0,,x k = f k (x) OQ, 30
37 {x 0,x 1,x 2, } x f {x k } {x k } k=0. p k e(3x k 1 +1), 3x k 1 +1=2 p k x k p k, {p 1,p 2,p 4, } x f 2- {p k } k=1 f {x k } k=0 n x 0 = x n x 1 = x n+1, 2 p1+p2+ +pn p1+p2+ +pn i 2 p1+p2+ +pn i n 2 2 p1 +3 n 1 C(p 1,p 2,,p n 1 ) 1.3 ( ),. x = x 0,x k = f k (x) OQ (k =1, 2, 3, ) x f 2- {p k } k=1, n 2 p1+p2+ +pn x n = C(p 1,p 2,,p n 1 )+3 n x 31
38 ,. x = x 0,x k = f k (x) OQ (k =1, 2, 3, ) x f 2- {p k } k=1, f {x k } k=0 n, (2 p1+p2+ +pn 3 n )x = C(p 1,p 2,,p n 1 ) x = C(p 1,p 2,,p n 1 ) 2 p1+p2+ +pn 3 n OQ,, f 2- f. x = x 0, x k = f k (x), y = y 0, y k = f k (y) OQ (k = 1, 2, 3, ) x y f 2-, x y., k e(3x k 1 +1)=e(3y k 1 +1) x = y 32
39 x = x 0,x k = f k (x) OQ (k =1, 2, 3, ) f {x k } k=0 n, x f 2- {p k } k=1 n t N k =1, 2,,n p tn+k = p k p 1,p 2,,p n (1) f OQ OQ 1 2 p 3 (p =1, 2, 3, ) (f ) 2- {p k = p} 1 (2) f 2- {p, q} 2 2 p +3 2 p+q 9 (p, q =1, 2, 3, ) 7 f 2- {2, 1} = 7 1 = 7 f 2- {1, 3} =
40 3.3 N n 2 n N<2 n+1 { N =2 n + M 0 M<2 n, M, N N =Σ n k=0 a k2 k ( a k 0 1) r =2 k a b (0 k Z, a,b OZ) 34
41 -. r =Σ k=0 a k2 k ( a k 0 1). 1 = Σ k=0 2k 0 = 1+( 1) = 1 + Σ k=02 k 2 = Σ k=1 2k, 2=2 1 3 = 1+Σ k=22 k, 3=1+2 4 = Σ k=2 2k, 4=2 2 5 = 1+2+Σ k=32 k, 5= p 1 = Σ p 1 k=0 2k 2 p = Σ k=p2 k p 1 = =Σ k=02 2k = 2+Σ k=1 22k = 1+Σ k=1 22k 1 = Σ k=02 2k+1 = =(1+2)Σ k=02 4k =Σ k=02 4k +Σ k=02 4k+1 = =1+Σ k=0 24k+2 +Σ k=0 24k+3 = = p =Σ k=02 pk 35
42 = 1+2 p +2 2p +2 3p +2 4p + f x = 1 5 = 1 2 p 3, p =3 -. x = 1-1, 0,, 5 2x +1-1, 0,, 3x +1-1, 0,. 2x +1 - x - 1, 1. 3x +1 - x - 2x +1 -, x -. x = 1 f =1+22 +( )Σ k=02 12k =
43 f x = 1 13 = 1 2 p 3, p =4, x = 1 -, 2x +1-13, 3x +1-1, 0,. 3x +1 - x - 2x +1 -, x -. f x = = 1 29, x = = =1+( )Σ k=02 28k =
44 1 61 =1+( )Σ k=0 260k =
45 4 4.1, 1 90 =radian =radius 39
46 t x t 180 = x π 180 x πx cos(180 x) = cosπx cos(90 x) = cos πx 2 sin(90 x) = sin πx 2 x : g(x) = x 2 cos2 πx 2 = x 4 +(3x +1)sin2 πx 2 5x +2 4 cos πx. g(x) x x x 2 3x +1 g(x) g(x) g(x) 10 <x<22. 40
47 70 60 y=g(x) g 2 (x) = g(g(x)) g 3 (x) = g(g(g(x))) g n (x) = g(g n 1 (x)) = g(g( (g(x)) )) x g n (x) g(x), h(x) h(x) = x 2 = 1 4 cos2 πx 2 + 3x x 2x +1 4 cos πx. sin 2 πx 2 41
48 h(x) x x 2 x 3x +1 2 h(x). h(x) 10 <x< y=h(x) F., F (x) = 4 πx ( ) 1 2 cos2 f(2k +1) π 2 (x 2k 1) 2 1 (2k +1) 2. k=0 42
49 F (x), 4 π 2 πx cos2 2 n k=0 ( ) f(2k +1) 1 (x 2k 1) 2 1 (2k +1) 2 n. F (x) 10 <x< y=f(x) cos πx cos πx =1+ q=1 ( 1)q π 2q x 2q (2q)! =1 π2 x 2 2! + π4 x 4 4! π6 x 6 6! π8 x 8 8! +. F (x) x, 43
50 F (x) = 2 (2+ π 2 ( 1) q π2q x 2q ) { ( p (2q)! q=1 p=2 m=0 { ( = 2 f(2m +1) ) ( π 2 4 (2m +1) 3 x + 6 m=0 m=0 ( f(2m +1) + 8 (2m +1) 5 f(2m +1) ) π2 (2m +1) 3 m=0 m=0 ( f(2m +1) + 10 (2m +1) 6 3π2 2!. m=0 m=0 } f(2m +1) ) (2m +1) p+1 x p 1 f(2m +1) ) (2m +1) 4 x 2 x 3 f(2m +1) ) (2m +1) 4 x 4 +, F (0) = 8 f(2n +1) π 2 (2n +1) 3 n=0 = 8 ( ) 6k +1 π 2 (4 n (8k +1)+4 n ) 3. k=0 n=0 + 8 ( π 2 k=0 n=0 (=1.04 ) > 1 ) 6k +5 (4 n (4k +3)+4 n ) 3 0 F 0 F z F (1) = 0 1 F 1 F z } F (x) x z F (z) 44
51 A B C E F (z) z F n (z). lim n F n (z) =1 z lim n F n (z) =z 0 z z 0 = 0.03 F z 0 F z n u + iv = F n (z), u> v>70 z C { z = x + iy 4 <x<4, 1.5 <y<4.5 } z F n (z) E { z = x + iy <x<3.0200, <y< } z F n (z) x cos πx cos πx =1+ q=1 ( 1)q π 2q x 2q (2q)! =1 π2 x 2 2! + π4 x 4 4! π6 x 6 6! π8 x 8 8! + x z cos πz 45
52 C D 46
53 E cos πz =1+ q=1 ( 1)q π 2q z 2q (2q)! =1 π2 z 2 2! + π4 z 4 4! π6 z 6 6! z π8 z 8 8! + 47
54 g(z) = z 2 = z 4 cos2 πz 2 +(3z +1)sin2 πz 2 5z +2 4 cos πz z x z = x g(z) =g(x) D g(z) { z = x + iy 4 <x<4, 3 <y<3 } z g 3 (z). g 3 (z) 1 < 0.1 z u + iv = g 3 (z) u>10 20 v>70 z F (z) z lim F n (z) =1 n? (Paul Erdös, ) 48
55 ( Mathematics is not yet ready for such problems. ) : Richard K. Guy(Ed.), : 1983 Jeffrey C. Lagarias : The 3x + 1 Problem and its Generalizations, Amer. Math. Monthly 92(1985),
56 f 3,8,27,28 f GQ OQ 28 f OQ OQ 29,30,33 F 42,44 f 31 f 2-31 f(x) a(x) Catalan 12 f(x) 4 27, y 2,9 3,21,24 1,22 1,19, n +1 1 N ,33 7 2,7,8,14 GQ 28 GQ 28 OQ 28 OQ 28 OZ 27 Z ,35 34,36, ,30,33 BASIC 18 22,
57 / c T. Urata 2002 / (0566) /2315
58 /. AICHI UNIVERSITY OF EDUCATION
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(,,
More informationp-sylow :
p-sylow :15114075 30 2 20 1 2 1.1................................... 2 1.2.................................. 2 1.3.................................. 3 2 3 2.1................................... 3 2.2................................
More information1. A0 A B A0 A : A1,...,A5 B : B1,...,B
1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A B f : A B 4 (i) f (ii) f (iii) C 2 g, h: C A f g = f h g = h (iv) C 2 g, h: B C g f = h f g = h 4 (1) (i) (iii) (2) (iii) (i) (3) (ii) (iv) (4)
More information1 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
More information17 ( ) 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
More information2016 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 16 2 1 () X O 3 (O1) X O, O (O2) O O (O3) O O O X (X, O) O X X (O1), (O2), (O3) (O2) (O3) n (O2) U 1,..., U n O U k O k=1 (O3) U λ O( λ Λ) λ Λ U λ O 0 X 0 (O2) n =
More information4 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
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
More information36 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 (
3 3. D f(z) D D D D D D D D f(z) D f (z) f (z) f(z) D (i) (ii) (iii) f(z) = ( ) n z n = z + z 2 z 3 + n= z < z < z > f (z) = e t(+z) dt Re z> Re z> [ ] f (z) = e t(+z) = (Rez> ) +z +z t= z < f(z) Taylor
More informationAppendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1) (2) ( ) BASIC BAS
Appendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1 (2 ( BASIC BASIC download TUTORIAL.PDF http://hp.vector.co.jp/authors/va008683/
More informationコンピュータ概論
4.1 For Check Point 1. For 2. 4.1.1 For (For) For = To Step (Next) 4.1.1 Next 4.1.1 4.1.2 1 i 10 For Next Cells(i,1) Cells(1, 1) Cells(2, 1) Cells(10, 1) 4.1.2 50 1. 2 1 10 3. 0 360 10 sin() 4.1.2 For
More information(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n
. 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n
More information熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
More information18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (
8 ) ) [ ] [ ) 8 5 5 II III A B ),,, 5, 6 II III A B ) ),,, 7, 8 II III A B ) [ ]),,, 5, 7 II III A B ) [ ] ) ) 7, 8, 9 II A B 9 ) ) 5, 7, 9 II B 9 ) A, ) B 6, ) l ) P, ) l A C ) ) C l l ) π < θ < π sin
More informationfunction2.pdf
2... 1 2009, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 38 : 5) i) [], : 84 85 86 87 88 89 1000 ) 13 22 33 56 92 147 140 120 100 80 60 40 20 1 2 3 4 5 7.1 7 7.1 1. *1 e = 2.7182 ) fx) e x, x R : 7.1)
More information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
More information1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +
( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n
More information³ÎΨÏÀ
2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p
More information() 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 ()
More information, 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,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 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,
More information, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
More informationx () 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 ),
More informationx, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
More information.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
More information1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 5 3. 4. 5. A0 (1) A, B A B f K K A ϕ 1, ϕ 2 f ϕ 1 = f ϕ 2 ϕ 1 = ϕ 2 (2) N A 1, A 2, A 3,... N A n X N n X N, A n N n=1 1 A1 d (d 2) A (, k A k = O), A O. f
More informationA S- hara/lectures/lectures-j.html r A = A 5 : 5 = max{ A, } A A A A B A, B A A A %
A S- http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html r A S- 3.4.5. 9 phone: 9-8-444, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office
More information入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
More informationExcel ではじめる数値解析 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 刷発行時のものです.
Excel ではじめる数値解析 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. http://www.morikita.co.jp/books/mid/009631 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. Excel URL http://www.morikita.co.jp/books/mid/009631 i Microsoft Windows
More information「産業上利用することができる発明」の審査の運用指針(案)
1 1.... 2 1.1... 2 2.... 4 2.1... 4 3.... 6 4.... 6 1 1 29 1 29 1 1 1. 2 1 1.1 (1) (2) (3) 1 (4) 2 4 1 2 2 3 4 31 12 5 7 2.2 (5) ( a ) ( b ) 1 3 2 ( c ) (6) 2. 2.1 2.1 (1) 4 ( i ) ( ii ) ( iii ) ( iv)
More informationuntitled
yoshi@image.med.osaka-u.ac.jp http://www.image.med.osaka-u.ac.jp/member/yoshi/ II Excel, Mathematica Mathematica Osaka Electro-Communication University (2007 Apr) 09849-31503-64015-30704-18799-390 http://www.image.med.osaka-u.ac.jp/member/yoshi/
More information,2,4
2005 12 2006 1,2,4 iii 1 Hilbert 14 1 1.............................................. 1 2............................................... 2 3............................................... 3 4.............................................
More information1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +
6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +
More information離散数理工学 第 2回 数え上げの基礎:漸化式の立て方
2 okamotoy@uec.ac.jp 2014 10 21 2014 10 29 10:48 ( ) (2) 2014 10 21 1 / 44 ( ) 1 (10/7) ( ) (10/14) 2 (10/21) 3 ( ) (10/28) 4 ( ) (11/4) 5 (11/11) 6 (11/18) 7 (11/25) ( ) (2) 2014 10 21 2 / 44 ( ) 8 (12/2)
More information1990 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,
More information(, Goo Ishikawa, Go-o Ishikawa) ( ) 1
(, Goo Ishikawa, Go-o Ishikawa) ( ) 1 ( ) ( ) ( ) G7( ) ( ) ( ) () ( ) BD = 1 DC CE EA AF FB 0 0 BD DC CE EA AF FB =1 ( ) 2 (geometry) ( ) ( ) 3 (?) (Topology) ( ) DNA ( ) 4 ( ) ( ) 5 ( ) H. 1 : 1+ 5 2
More informationf(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a
3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (
More information1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
More information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)
More information1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.
1 1 n 0, 1, 2,, n 1 1.1 n 2 a, b a n b n a, b n a b (mod n) 1 1. n = 10 1567 237 (mod 10) 2. n = 9 1567 1826578 (mod 9) n II Z n := {0, 1, 2,, n 1} 1.2 a b a = bq + r (0 r < b) q, r q a b r 2 1. a = 456,
More informationChap9.dvi
.,. f(),, f(),,.,. () lim 2 +3 2 9 (2) lim 3 3 2 9 (4) lim ( ) 2 3 +3 (5) lim 2 9 (6) lim + (7) lim (8) lim (9) lim (0) lim 2 3 + 3 9 2 2 +3 () lim sin 2 sin 2 (2) lim +3 () lim 2 2 9 = 5 5 = 3 (2) lim
More informationf(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
22 A 3,4 No.3 () (2) (3) (4), (5) (6) (7) (8) () n x = (x,, x n ), = (,, n ), x = ( (x i i ) 2 ) /2 f(x) R n f(x) = f() + i α i (x ) i + o( x ) α,, α n g(x) = o( x )) lim x g(x) x = y = f() + i α i(x )
More informationr 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B
1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n
More information2000年度『数学展望 I』講義録
2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53
More informationZ[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)
3 3 22 Z[i] Z[i] π 4, (x) π 4,3 (x) x (x ) 2 log x π m,a (x) x ϕ(m) log x. ( ). π(x) x (a, m) = π m,a (x) x modm a π m,a (x) ϕ(m) π(x) ϕ(m) x log x ϕ(m) m f(x) g(x) (x α) lim f(x)/g(x) = x α mod m (a,
More information5 Armitage x 1,, x n y i = 10x i + 3 y i = log x i {x i } {y i } 1.2 n i i x ij i j y ij, z ij i j 2 1 y = a x + b ( cm) x ij (i j )
5 Armitage. x,, x n y i = 0x i + 3 y i = log x i x i y i.2 n i i x ij i j y ij, z ij i j 2 y = a x + b 2 2. ( cm) x ij (i j ) (i) x, x 2 σ 2 x,, σ 2 x,2 σ x,, σ x,2 t t x * (ii) (i) m y ij = x ij /00 y
More information1. A0 A B A0 A : A1,...,A5 B : B1,...,B
1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A, B Z Z m, n Z m n m, n A m, n B m=n (1) A, B (2) A B = A B = Z/ π : Z Z/ (3) A B Z/ (4) Z/ A, B (5) f : Z Z f(n) = n f = g π g : Z/ Z A, B (6)
More information1 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
More information離散数理工学 第 2回 数え上げの基礎:漸化式の立て方
2 okamotoy@uec.ac.jp 2015 10 20 2015 10 18 15:29 ( ) (2) 2015 10 20 1 / 45 ( ) 1 (10/6) ( ) (10/13) 2 (10/20) 3 ( ) (10/27) (11/3) 4 ( ) (11/10) 5 (11/17) 6 (11/24) 7 (12/1) 8 (12/8) ( ) (2) 2015 10 20
More informationno35.dvi
p.16 1 sin x, cos x, tan x a x a, a>0, a 1 log a x a III 2 II 2 III III [3, p.36] [6] 2 [3, p.16] sin x sin x lim =1 ( ) [3, p.42] x 0 x ( ) sin x e [3, p.42] III [3, p.42] 3 3.1 5 8 *1 [5, pp.48 49] sin
More informationII Time-stamp: <05/09/30 17:14:06 waki> ii
II waki@cc.hirosaki-u.ac.jp 18 1 30 II Time-stamp: ii 1 1 1.1.................................................. 1 1.2................................................... 3 1.3..................................................
More information() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
More information2010 II / y = e x y = log x = log e x 2. ( e x ) = e x 3. ( ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0
2010 II 6 10.11.15/ 10.11.11 1 1 5.6 1.1 1. y = e x y = log x = log e x 2. e x ) = e x 3. ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0 log a 1 a 1 log a a a r+s log a M + log a N 1 0 a 1 a r
More information44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)
(1) I 44 II 45 III 47 IV 52 44 4 I (1) ( ) 1945 8 9 (10 15 ) ( 17 ) ( 3 1 ) (2) 45 II 1 (3) 511 ( 451 1 ) ( ) 365 1 2 512 1 2 365 1 2 363 2 ( ) 3 ( ) ( 451 2 ( 314 1 ) ( 339 1 4 ) 337 2 3 ) 363 (4) 46
More informationi ii i iii iv 1 3 3 10 14 17 17 18 22 23 28 29 31 36 37 39 40 43 48 59 70 75 75 77 90 95 102 107 109 110 118 125 128 130 132 134 48 43 43 51 52 61 61 64 62 124 70 58 3 10 17 29 78 82 85 102 95 109 iii
More informationhttp://know-star.com/ 3 1 7 1.1................................. 7 1.2................................ 8 1.3 x n.................................. 8 1.4 e x.................................. 10 1.5 sin
More information(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
More information( )
18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................
More information1 1 [1] ( 2,625 [2] ( 2, ( ) /
[] (,65 [] (,3 ( ) 67 84 76 7 8 6 7 65 68 7 75 73 68 7 73 7 7 59 67 68 65 75 56 6 58 /=45 /=45 6 65 63 3 4 3/=36 4/=8 66 7 68 7 7/=38 /=5 7 75 73 8 9 8/=364 9/=864 76 8 78 /=45 /=99 8 85 83 /=9 /= ( )
More information2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta
009 IA 5 I, 3, 4, 5, 6, 7 6 3. () Arcsin ( (4) Arccos ) 3 () Arcsin( ) (3) Arccos (5) Arctan (6) Arctan ( 3 ) 3. n () tan x (nπ π/, nπ + π/) f n (x) f n (x) fn (x) Arctan x () sin x [nπ π/, nπ +π/] g n
More information1 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
More information1.2 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
1 1.1 R(x) = 0 y + P (x)y + Q(x)y = R(x)...(1) y + P (x)y + Q(x)y = 0...(2) 1 2 u(x) v(x) c 1 u(x)+ c 2 v(x) = 0 c 1 = c 2 = 0 c 1 = c 2 = 0 2 0 2 u(x) v(x) u(x) u (x) W (u, v)(x) = v(x) v (x) 0 1 1.2
More information70 : 20 : A B (20 ) (30 ) 50 1
70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................
More informationZ: Q: R: C: sin 6 5 ζ a, b
Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,
More information- II
- II- - -.................................................................................................... 3.3.............................................. 4 6...........................................
More information2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a
More informationwiles05.dvi
Andrew Wiles 1953, 20 Fermat.. Fermat 10,. 1 Wiles. 19 20., Fermat 1. (Fermat). p 3 x p + y p =1 xy 0 x, y 2., n- t n =1 ζ n Q Q(ζ n ). Q F,., F = Q( 5) 6=2 3 = (1 + 5)(1 5) 2. Kummer Q(ζ p ), p Fermat
More informationA, B, C. (1) A = A. (2) A = B B = A. (3) A = B, B = C A = C. A = B. (3)., f : A B g : B C. g f : A C, A = C. 7.1, A, B,. A = B, A, A A., A, A
91 7,.,, ( ).,,.,.,. 7.1 A B, A B, A = B. 1), 1,.,. 7.1 A, B, 3. (i) A B. (ii) f : A B. (iii) A B. (i) (ii)., 6.9, (ii) (iii).,,,. 1), Ā = B.. A, Ā, Ā,. 92 7 7.2 A, B, C. (1) A = A. (2) A = B B = A. (3)
More information() 3 3 2 5 3 6 4 2 5 4 2 (; ) () 8 2 4 0 0 2 ex. 3 n n =, 2,, 20 : 3 2 : 9 3 : 27 4 : 8 5 : 243 6 : 729 7 : 287 8 : 656 9 : 9683 0 : 59049 : 7747 2 : 5344 3 : 594323 4 : 4782969 5 : 4348907 6 : 4304672
More informationChap10.dvi
=0. f = 2 +3 { 2 +3 0 2 f = 1 =0 { sin 0 3 f = 1 =0 2 sin 1 0 4 f = 0 =0 { 1 0 5 f = 0 =0 f 3 2 lim = lim 0 0 0 =0 =0. f 0 = 0. 2 =0. 3 4 f 1 lim 0 0 = lim 0 sin 2 cos 1 = lim 0 2 sin = lim =0 0 2 =0.
More informationN88 BASIC 0.3 C: My Documents 0.6: 0.3: (R) (G) : enterreturn : (F) BA- SIC.bas 0.8: (V) 0.9: 0.5:
BASIC 20 4 10 0 N88 Basic 1 0.0 N88 Basic..................................... 1 0.1............................................... 3 1 4 2 5 3 6 4 7 5 10 6 13 7 14 0 N88 Basic 0.0 N88 Basic 0.1: N88Basic
More information数学の基礎訓練I
I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
More information) 2008 8 21 22 21 10:00 12:00 e iπ 1. i ). e π. T @ MacTutor History of Mathematics archive www-history.mcs.standrews.ac.uk/history/ FAQ kawaguch))math.sci.osaka-u.ac.jp 2008 8 21 Part 1 e πi = 1 e = 2.71828...
More informationIMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a
1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =
More information6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
More informationx (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
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
More informationI
I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
More information(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n
3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4
More information2016
2016 1 G x x G d G (x) 1 ( ) G d G (x) = 2 E(G). x V (G) 2 ( ) 1.1 1: n m on-off ( 1 ) off on 1: on-off ( on ) G v v N(v) on-off G S V (G) N(v) S { 3 G v S v S G G = 1 OK ( ) G 2 3.1 u S u u u 1 G u S
More informationuntitled
1 1 1. 2. 3. 2 2 1 (5/6) 4 =0.517... 5/6 (5/6) 4 1 (5/6) 4 1 (35/36) 24 =0.491... 0.5 2.7 3 1 n =rand() 0 1 = rand() () rand 6 0,1,2,3,4,5 1 1 6 6 *6 int() integer 1 6 = int(rand()*6)+1 1 4 3 500 260 52%
More information, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x
1 1.1 4n 2 x, x 1 2n f n (x) = 4n 2 ( 1 x), 1 x 1 n 2n n, 1 x n n 1 1 f n (x)dx = 1, n = 1, 2,.. 1 lim 1 lim 1 f n (x)dx = 1 lim f n(x) = ( lim f n (x))dx = f n (x)dx 1 ( lim f n (x))dx d dx ( lim f d
More information13 0 1 1 4 11 4 12 5 13 6 2 10 21 10 22 14 3 20 31 20 32 25 33 28 4 31 41 32 42 34 43 38 5 41 51 41 52 43 53 54 6 57 61 57 62 60 70 0 Gauss a, b, c x, y f(x, y) = ax 2 + bxy + cy 2 = x y a b/2 b/2 c x
More informationA(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
More information(2 X Poisso P (λ ϕ X (t = E[e itx ] = k= itk λk e k! e λ = (e it λ k e λ = e eitλ e λ = e λ(eit 1. k! k= 6.7 X N(, 1 ϕ X (t = e 1 2 t2 : Cauchy ϕ X (t
6 6.1 6.1 (1 Z ( X = e Z, Y = Im Z ( Z = X + iy, i = 1 (2 Z E[ e Z ] < E[ Im Z ] < Z E[Z] = E[e Z] + ie[im Z] 6.2 Z E[Z] E[ Z ] : E[ Z ] < e Z Z, Im Z Z E[Z] α = E[Z], Z = Z Z 1 {Z } E[Z] = α = α [ α ]
More informationV 0 = + r pv (H) + qv (T ) = + r ps (H) + qs (T ) = S 0 X n+ (T ) = n S n+ (T ) + ( + r)(x n n S n ) = ( + r)x n + n (d r)s n = ( + r)v n + V n+(h) V
I (..2) (0 < d < + r < u) X 0, X X = 0 S + ( + r)(x 0 0 S 0 ) () X 0 = 0, P (X 0) =, P (X > 0) > 0 0 H, T () X 0 = 0, X (H) = 0 us 0 ( + r) 0 S 0 = 0 S 0 (u r) X (T ) = 0 ds 0 ( + r) 0 S 0 = 0 S 0 (d r)
More informationORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2
More informationNo2 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
More informationt θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ
More informationii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.
(1 C205) 4 10 (2 C206) 4 11 (2 B202) 4 12 25(2013) http://www.math.is.tohoku.ac.jp/~obata,.,,,..,,. 1. 2. 3. 4. 5. 6. 7. 8. 1., 2007 ( ).,. 2. P. G., 1995. 3. J. C., 1988. 1... 2.,,. ii 3.,. 4. F. ( ),..
More information2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (
(. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2
More information<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
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.
More information2014 x n 1 : : :
2014 x n 1 : : 2015 1 30 : 5510113 1 x n 1 n x 2 1 = (x 1)(x+1) x 3 1 = (x 1)(x 2 +x+1) x 4 1 = (x 1)(x + 1)(x 2 + 1) x 5 1 = (x 1)(x 4 + x 3 + x 2 + x + 1) 1, 1,0 n = 105 2 1 n x n 1 Maple 1, 1,0 n 2
More informationt = 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
I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
More informationDVIOUT
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)
More informationII R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k
II 231017 1 1.1. R n k +1 v 0,, v k k v 1 v 0,, v k v 0 1.2. v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ kσ dimσ = k 1.3. k σ {v 0,...,v k } {v i0,...,v il } l σ τ < τ τ σ 1.4.
More informationII 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
More information19 Systematization of Problem Solving Strategy in High School Mathematics for Improving Metacognitive Ability
19 Systematization of Problem Solving Strategy in High School Mathematics for Improving Metacognitive Ability 1105402 2008 2 4 2,, i Abstract Systematization of Problem Solving Strategy in High School
More information