C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1

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1 C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1

2 2 7x + 4 = 0 (2) 1 1 x + x + 5 = 0 2 sin x x = 0 e x + x = 0 x = cos x (3) x log x? 0.1% () 2.2 p12 3 x 3 3x 2 + 9x 8 = 0 (4) 1 [ ] 1/3 [ 2 1 ( x 1 = ) ] 1/ x 2 = 1 + (1 + [ ] 1/3 2 3i) (1 [ 1 3i) 2 x 2 = 1 + (1 [ ] 1/3 2 3i) (1 + [ 1 3i) 2 ( 1 + ) ] 1/3 33 ( 1 + ) ] 1/ x Mathematica 3 A.1 x (8) (5) (6) (7) 2

3 f(x) = 0 (9) x (9) (9) (9) f(x) x f(x) x 2 (, ) y = f(x) x f(x) = 0 () (4) f(x) = x 3 3x 2 + 9x 8 x f(x) 9 x 150 y y=f(x) 解は x 軸との交点数値計算により求める x -25 1: f(x) = x 3 3x 2 + 9x 8 x (4) 3 3

4 3 (bisection method) 3.1 [a, b] f(x) f(a)f(b) 0 (10) f(α) = 0 α [a, b] f(a)f(b) 0 2 a, b(a b) [a, b] 2 c = (a + b)/2 f(c) f(c)f(a) 0 b c f(c)f(a) 0 a c [a, b] b a ε 1/2 1 1/2 10 (1/2) 10 = 1/1024 (4) 2 f(a) f(b) (10) [a, b] 1/2 b a ε (11) ε b a 4

5 初期値 c a 25 a c a 初期値 b x 2 x x 3 x 1-25 b c b c b c x : f(x) = x 3 3x 2 + 9x 8 a = 1, b = 11 c = x 0 = 5 x 1, x 2, x 3, x = (x ) 3.2 f(x) C double f(double x){ double y; y=x*x*x-3*x*x+9*x-8; } return y; #define EPS 1.0e-10 EPS C 5

6 3 始め a, b を入力 f(a) f(b) > 0 yes b - a < 0 yes a と b を入れ替える b - a no yes c=(a+b)/2 yes f(c) f(a) < 0 no 解 c を表示 b c a c 終り 3: 4 (Newton s method) f(x) α x 0 f(x) (x 0, f(x 0 )) x x 1 x x 2 x 3, x 4, ( 4) (x 0, x 2, x 3, x 4, ) x 0 α f(x) (x i, f(x i )) x x i+1 x i+1 (x i, f(x i )) 6

7 f (x i ) y f(x i ) = f (x i )(x x i ) (12) 4 y = 0 x x i+1 x i+1 x i+1 = x i f(x i) f (x i ) (13) x i x i+1 x i+1 x i x i ε (14) 4 ε (4) x 3 x : f(x) = x 3 3x 2 + 9x 8 x 0 = 5 x x 1 x 0 4 f(x i+1 ) ε 7

8 A A (13) α x i+1 α f(α) = 0 f(x i )/f (x i ) α α x i+1 = α x i + f(x i) f (x i ) = α x i + f(α) [ f (α) + 1 f(α)f ] (α) f 2 (x i α) + O ( (α x i ) 2) (α) = O ( (α x i ) 2) i + 1 i (13) 10 6 (15) tan 1 (x 1) + x 4 = 0 (16) x 0 = x 0 = [a, b] f(a) < 0, f(b) >

9 f (x) > 0, f (x) > 0 x 0 = b f (x) > 0, f (x) < 0 x 0 = b f(a) > 0, f(b) < 0 f(x) x 5 x 3 x 2 x 4 x x 1 x x 1 x 5 x x 3 1 x 4 2 x : 6:

10 始め i = -1 x 0 を入力 i i+1 x i+1 = x i - f(x i ) / f (x i ) 解の精度検査 (x i+1 - x i ) / x i yes no 解 x i+1 を表示 yes 反復回数検査 i imax no 収束しないと表示 終り 7: = 1/

11 8: ( ) 5.2 /?? 11

12 1: ( 8) ( 5) [a, b] (robust ) b a 6 [ 1] x 3 3x 2 + 9x 8 = 0 x + e x + sin x = 0 [ 2] z 3 3z 2 + 9z 8 = 0 z = i z = i z =

13 [ 3] { (x 3) 2 + y 2 3 = 0 sin x + e y 1 1 = 0 (x = , y = ) (x = , y = ) [ 1] [ 1] [ 3] [ 2] ( ) AM 8:45 A4 1 5E 2 13

14 A 4 A.1 A f(x) = 0 x = α f(α) = 0 i x i x f(x i + x) = f(x i ) + f (x i ) x f (x i ) x 2 + x (17) f(x i ) + f (x i ) x f(x i + x) = 0 α = x i + x x (17) x f(x i) f (x i ) (18) α = x i + x x i+1 = x i f(x i) f (x i ) (19)

15 A f(z) = 0 z w(z) = 0 (20) z i+1 = z i w(z i) w (z i ) (21) A.2 (2 ) A.2.1 x 2 3x + 2 = 0 (22) 3x + 2y + z = 10 x + y + z = 6 x + 2y + z = 11 (23) x 3 sin x x n n n 1 1 n m 15

16 m?? A.2.2 { (x 3) 2 + y 2 3 = 0 sin x + e y 1 1 = 0 (24) 2??? 2 9 A B 16

17 g(x, y) = A B f(x, y) = 0 9: A B f(x, y) = (x 3) 2 + y 2 3 (25) g(x, y) = sin x + e y 1 1 (26) f(x, y) = 0 g(x, y) = 0 (x, y) (α x, α y ) f(α x, α y ) = 0 g(α x, α y ) = 0 i (x i, y i ) ( x, y) f(x i + x, y i + y) = f(x i, y i ) + f f x + x y y + O( 2 ) ( x, y) f(x i, y i ) + f f x + x y y (27) 17

18 g(x, y) 2 = f(x i + x, y i + y) = f(x i, y i ) + f f x + y x y (28) g(x i + x, y i + y) = g(x i, y i ) + g g x + y x y (29) f(x i + x, y i + y) = 0 g(x i + x, y i + y) = 0 x y x y (28) (29) ( f x g x f y g y ) ( ) ( ) x f(x i, y i ) = y g(x i, y i ) ( x, y) α x x i + x (30) { xi+1 = x i + x y i+1 = y i + y (31) A A.3 ( )

19 N f 1 (x 1, x 2, x 3,, x N ) = 0 f 2 (x 1, x 2, x 3,, x N ) = 0 f 3 (x 1, x 2, x 3,, x N ) = 0. f N (x 1, x 2, x 3,, x N ) = 0 (32) X = (x 1, x 2, x 3,, x N ) (33) i f i (X) f i (X + X) = f i (X) + f i x 1 x 1 + f i x 2 x 2 + f i x 3 x f i x N x N + O( X 2 ) (34) i = 1, 2, 3,, N f i (X+ X) = 0 X = ( x 1, x 2, x 3,, x N ) 2 f 1 f 1 f 1 f x 1 x 2 x x N x 1 f f 2 f 2 f 2 f x 1 x 2 x (X) x N x 2 f f 3 f 3 f 3 f x 1 x 2 x x N x 2 (X) 3 = f 3 (X).... f N x 3... x N f N (X) f N x 1 f N x 2 f N x N N 1 X = ( x 1, x 2, x 3,, x N ) X new X new = X old + X x new 1 = x old 1 + x 1 x new 2 = x old 2 + x 2 x new 3 = x old 3 + x 3 x new N. = x old N + x N (35) (36) 19

( ) C 2 f(x) = 0 (1) x (1) x 3 3x 2 + 9x 8 = 0 (2) 1

( ) C 2 f(x) = 0 (1) x (1) x 3 3x 2 + 9x 8 = 0 (2) 1 2004 8 25 1 1.1 ( ) 4 2 1.2 4 C 2 f(x) = 0 (1) x (1) x 3 3x 2 + 9x 8 = 0 (2) 1 [ ] 1/3 [ 2 1 ( x 1 = 1 2 1 + + 1 + ) ] 1/3 33 33 2 x 2 = 1 + (1 + [ ] 1/3 2 3i) 1 + 1 33 2 (1 [ 1 3i) 2 x 2 = 1 + (1 [ ]

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