Collatzの問題 （数学/数理科学セレクト１）
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- しまな さかいざわ
- 5 years ago
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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
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Chap9.dvi
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no35.dvi
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(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
2016
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
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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%
, 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 ( 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](/thumbs/93/113428934.jpg ", 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
13 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
A(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
(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
![(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](/thumbs/67/56612994.jpg "(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 ]
V 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)
ORIGINAL 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
No2 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
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
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
ii 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. ( ),..
2.2 ( 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
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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.
2014 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
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
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)
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)
II 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.
II 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
19 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