( ) C 2 f(x) = 0 (1) x (1) x 3 3x 2 + 9x 8 = 0 (2) 1
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- らむ あいきょう
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1 ( ) C 2 f(x) = 0 (1) x (1) x 3 3x 2 + 9x 8 = 0 (2) 1
2 [ ] 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 1 2 x (6) (2) 3 4 (,,, ) y = f(x) x f(x) = 0 ( ) (2) f(x) = x 3 3x 2 + 9x 8 x f(x) 11 x (3) (4) (5) 1 Mathematica
3 解は 軸との交点数値計算により求める 1: f(x) = x 3 3x 2 + 9x 8 x 3 (bisection method) 3.1 [a, b] f(x) f(a)f(b) < 0 (7) 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 (2) 2 f(a) f(b) (7) [a, b] 1/2 [a, b] (robust ) 3
4 b a ( 8) 2: f(x) = x 3 3x 2 + 9x 8 a = 1, b = 11 c = x 0 = 5 x 1, x 2, x 3, x = (x )
5 a b f(a) f(b) > 0 yes b - a < 0 yes a b b - a > ε no yes c=(a+b)/2 yes f(c) f(a) < 0 no c b c a c 3: 4 (Newton s method) 4.1 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 )) f (x i ) y f(x i ) = f (x i )(x x i ) (8) 5
6 4 y = 0 x x i+1 x i+1 x i+1 = x i f(x i) f (x i ) (9) x i x i+1 x i+1 x i x i < ε (10) ε (2) 4 (9) α x i+1 α f(α) = 0 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 (9) 10 6 ( 8) ( 9) 8 8 (11) 6
7 x 3 x : f(x) = x 3 3x 2 + 9x 8 x 0 = 5 x x 1 x
8 i=-1 x 0 i i+1 x i+1 =x i -f(x i ) / f'(x i ) (x i+1 - x i x i ε yes no x i+1 yes i imax no "!#%$ &(' 5: c [a, b] 2 (a, f(a)) (b, f(b)) x c x c c = af(a) bf(b)) f(b) f(a) (12) α [a, b] c α b a ε f(c) < ε (2) 6 8
9 x 0 x 1 x 2 6: f(x) = x 3 3x 2 + 9x 8 a = 0.5, b = 5.5 x x 0, x 1, x 2, x 3 x = (x i 1, f(x i 1 )) (x i, f(x i )) ( ) x x i+1 f(x i ) x x i x i 1 x i+1 = x i f(x i ) f(x i ) f(x i 1 ) (13) (9) x 0, x 1 (13) x 2, x 3, i x i α 1 f(x i ) f (x i ) 2 f(x i ) (2) 7 9
10 x 4 x 3 x 1 x 0 7: f(x) = x 3 3x 2 + 9x 8 x 2 = 5.5, x 1 = 5.0 x x 0, x 1, x 2, x 3 x = = 1/
11 8: ( ) tan 1 (x 1) + x 4 = 0 (14) x 0 = x 0 =
12 [a, b] f(a) < 0, f(b) > 0 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 x 2 x 3 4 x x x 0 9: 12
13 4 2 x 1 x 5 x x 3 1 x 4 2 x : 8 ( )
14 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 (15) f(x i ) + f (x i ) x f(x i + x) = 0 α = x i + x x (15) x f(x i) f (x i ) (16) α = x i + x x i+1 = x i f(x i) f (x i ) (17) w(z) = 0 (18) z i+1 = z i w(z i) w (z i ) (19) FORTRAN C C FORTRAN C p.471 #include <complex.h> 14
15 float complex double complex long double complex double complex (+,-,*,/) 1 1: complex.h csin() csinf() csinl() ccos() ccosf() ccosl() ctan() ctanf() ctanl() casin() casinf() casinl() cacos() cacosf() cacosl() catan() catanf() catanl() csinh() csinhf() csinhl() ccosh() ccoshf() ccoshl() ctanh() ctanhf() ctanhl() casinh() casinhf() casinhl() cacosh() cacoshf() cacoshl() catanh() catanhf() catanhl() cexp() cexpf() cexpl() clog() clogf() clogl() cabs() cabsf() cabsl() csqrt() csqrtf() csqrtl() cpow() cpowf() cpowl() creal() crealf() creall() cimag() cimagf() cimagl() carg() cargf() cargl() conj() conjf() conjl() cproj() cprojf() cprojl() 15
16 8.2 (2 ) x 2 3x + 2 = 0 (20) 3x + 2y + z = 10 x + y + z = 6 x + 2y + z = 11 (21) x 3 sin x x n n n 1 1 n m m?? { (x 3) 2 + y 2 3 = 0 sin x + e y 1 1 = 0 (22) 2??? 16
17 2 11 A B 11: A B f(x, y) = (x 3) 2 + y 2 3 (23) g(x, y) = sin x + e y 1 1 (24) 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 (25) 17
18 g(x, y) 2 = f(x i + x, y i + y) = f(x i, y i ) + f f x + y x y (26) g(x i + x, y i + y) = g(x i, y i ) + g g x + y x y (27) f(x i + x, y i + y) = 0 g(x i + x, y i + y) = 0 x y x y (26) (27) ( 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 α y x i + x { xi+1 = x i + x y i+1 = y i + y (28) (29) ( )
19 N f 1 (x 1 + x 2 + x x N ) = 0 f 2 (x 1 + x 2 + x x N ) = 0 f 3 (x 1 + x 2 + x x N ) = 0. f N (x 1 + x 2 + x x N ) = 0 (30) X = (x 1, x 2, x 3, + x N ) (31) 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 ) (32) 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 (33) (34) 19
20 9 [ 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 = [ 3] { (x 3) 2 + y 2 3 = 0 sin x + e y 1 1 = 0 (x = , y = ) (x = , y = ) 10 [ 1] x + e x + sin x = 0 20
21 9 9 ( ) PM 6:00 A4 1 5E ( ) 21
C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1
2006 7 18 1 2 C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1 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 + 5 + log x? 0.1% () 2.2 p12 3 x 3 3x 2 + 9x 8 = 0 (4) 1 [ ] 1/3 [ 2 1 ( x 1
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sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V
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C による数値計算法入門 ( 第 2 版 ) 新装版 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/009383 このサンプルページの内容は, 新装版 1 刷発行時のものです. i 2 22 2 13 ( ) 2 (1) ANSI (2) 2 (3) Web http://www.morikita.co.jp/books/mid/009383
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応用数学特論.dvi
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方程式を解く 知的情報処理 ３ 非線形方程式を解く 一変数 代数方程式を解くことは昔から重要な問題であった 算木にもたくさんある 数学競技会(例: 30題を40 50日で解く)で出された ３次 ４次代数方程式が一般的に解けた Scipione del Ferro (465-56), Niccoló Fontana Tartaglia(499547), Girolamo Cardano (50-576)
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