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. 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.

Size: px

Start display at page:

Download "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. 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."

よしじろうももき
6 years ago
Views:

2 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. 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.

4 i ( ) 2 (1) ANSI (2) 2 (3) Web (4) C 2 Web 2 Windows 7 CPad for Boland C++ Compiler Ver ( ) C BASIC

5 ii ( ) 2 ( ) ( ) C C C (1) C ANSI (2) C BASIC C C C (3) BASIC GNUPLOT (4) C Sun Solaris 2.6 gcc (GNU C version 2.8.1) ( 5 5 )

6 iii C 2 ( ) ( 1 ) ( ) (1) (2) (3) BASIC (4) (5) ( ) N88BASIC

7 iv

8 v LU

9 vi

10 1 f(x) = O 1 O 2 O 3 O 1 1 m 3 m O 3 O 1 3 O 2 2 O O 2 x [m] π πx3 = 4 π(3 x)3 3 x 3 3x 2 + 9x 8 = 0 (1.1) x 3 f(x) = 0

11 f(x) [a, b] f(a) f(b) a b c f(c) = 0 (a < c < b) (1.1) 2 (1.1) f(x) f(x) = x 3 3x 2 + 9x 8 f(x) x x 1.2 f(1) = 1 < 0 f(2) = 6 > 0 [1, 2] f(x) = f(1.5) f(1.5) = > 0 f(1) < 0 f(1.5) > 0 [1, 1.5] [1, 2] 1.2

12 [1, 1.5] f(1.25) f(1.25) = 0.51 > 0 [1, 1.25] f(x) f(1.125) = 0.24 < 0, 4 : [1.125, 1.25] f(1.1875) = 0.13 > 0, 5 : [1.125, ] f( ) = 0.05 < 0, 6 : [ , ] f( ) = 0.03 > 0, 7 : [ , ] f( ) = 0.01 < 0, 8 : [ , ] f( ) = 0.01 > 0, 9 : [ , ] f( ) = > 0, 10 : [ , ] f( ) = < 0, 11 : [ , ] f( ) = < 0, 12 : [ , ] f( ) = < 0, 13 : [ , ] f( ) = < 0, 14 : [ , ] f( ) = > 0, 15 : [ , ] 8 x = 1.16 x = x = 1.16 x mm 15 x = x = O 2 1 m 16 cm 6 mm 2 f(x) = 0 f(x) f(x) a 1, b 1, (a 1 < b 1 ) [a 1, b 1 ] a 1 b 1 2 c 1 f(c 1 ) f(c 1 ) = 0 c 1 ( ) f(c 1 ) [a 1, c 1 ] [c 1, b 1 ] ( )

13 f(a 1 ) f(c 1 ) [a 1, c 1 ] 2 [a 1, c 1 ] ( 1.3(a))f(a 1 ) f(c 1 ) f(c 1 ) f(b 1 ) [c 1, b 1 ] ( 1.3(b)) 2 [c 1, b 1 ] 2 [a 2, b 2 ] [a 2, b 2 ] 3 [a 3, b 3 ] ( 1.4) 2 1.1

14 /* ************************************************* */ 2 /* 2 nibun.c */ 3 /* ************************************************* */ 4 # include < stdio.h> 5 /* ** ** */ 6 # define FNF (x) (x*x*x - 3* x*x + 9* x - 8) 7 int main ( void ) 8 { double a, b, c; 9 int k; 10 char z, zz; 11 while ( 1 ) { 12 printf ("f(a)*f(b)<0 a, b "); 13 printf (" \n\n"); 14 printf (" 1 [ a,b] a="); 15 scanf ("%lf%c",&a,& zz ); 16 printf (" 1 [ a,b] b="); 17 scanf ("%lf%c",&b,& zz ); 18 printf ("\ n (y/n)"); 19 scanf ("%c%c",&z,& zz ); 20 if(z == n ) continue ; 21 if ((z == y )&&( a < b )&&( FNF (a ) * FNF (b ) < 0)) 22 break ; 23 else { 24 printf ("\na >b f(a)*f(b ) >=0 \ n"); 25 printf (" \n"); 26 printf (" \n"); 27 scanf ("%c",&z ); continue ; 28 } 29 } 30 k = 0; 31 printf (" A B B-A\n"); 32 /* ** * * */ 33 while ( b - a >= ) { 34 k = k + 1; 35 printf ("%4d %8.5 lf %8.5 lf %8.5 lf\n",k,a,b,b-a); 36 /* a,b a,b */ 37 c = ( a + b ) / 2.0; 38 if ( FNF (a ) * FNF (c ) > 0 ) a = c; 39 else b = c; 40 if (( k % 10) == 0) { 41 printf ("\ n \n"); 42 printf (" (y/n ): "); 43 scanf ("%c%c",&z,& zz ); 44 if(z == n ) { 45 printf ("\ n \n" ); break ; 46 } else if( z == y ) 47 printf (" A B B-A\n"); 48 else { z = n ; break ; }

15 } 50 } 51 if(z!= n ) { 52 printf ("\n %3 d \ n",k); 53 printf ("\ n = %10.6 lf\n", (a+b )/2.0); 54 } 55 return 0; 56 } 1.2 (1.1) x 3 3x 2 + 9x 8 = 0 (1.1 ) f(x) [1, 2] f (x) = 3x 2 6x + 9 = 3(x 1) 2 + 6, f (x) = 6x 6 (1, 2) f (x) > 0 f (x) > 0 y = f(x) [1, 2] y = f(x) (2, 6) x x 1 x 1 1 (x 1, f(x 1 )) x x 2 (x 2, f(x 2 )) x x 3 x 1, x 2, x 3,, x n,

17 L = , [ U b ] = b = t [1, 0.4, , ] U b Ux = b u = , z = , y = , x = LU (2.4) (2.5) (1) 2x 4y + 6z = 1 (2) 2x + 8y + 2z 3w = 2 x + 7y 8z = 0 4x + 6y 2z w = 1 x + y 2z = 3 2x 4y 2z w = 3 x 5y + 2z + w = 2 (3) 2x + 7y z + 5u 3w = 6 x 4y + 2z u + 6w = 1 3x + y 9z 2u + w = 2 10x 2y 5z + 8u 7w = 4 4x + 3y + 12z 4u 2w =

18 (1) x + 3y 2z = 2 3x 2y + z = 0 2x + y 3z = 1 (2) 3y + 2z + u = 7 3x + 2y 3z = 1 x + 2y 3z + 2u = 3 3x + 4y + z + 2u = ( 2.2) 2.5 (1) (2) (3) (1) x + y z = 0 (2) 2x 2y z = 0 (3) 5x + 2y 6z = 0 x 3y 2z = 0 4x + y 5z = 0 6x + 2y + 3z = 0 x + 2y 2z = 0 2x 2y + 2z = 0 x + 6y 6z = { (1) 2x + y 4z + 5u = 1 (2) 2x 6y + 4z = 0 3x + y 2z + u = 3 x + 3y 2z = 0 3x 9y + 6z = 0 (3) x + y z = 2 (4) 2x 2y z = 12 5x + 2y 6z = 5 x 3y 2z = 19 (5) (7) 4x + y 5z = 3 x + 2y 2z = 2 2x 2y + 2z = 3 x + 6y 6z = 1 (6) 6x + 2y + 3z = 16 x + y + z = 1 2x + y 2z = 3 x y + 2z = 0 3x y + 2z = 2 x + y 2z = 2 x 1 + 2x 2 x 4 + x 6 7x 7 = 4 x 1 + 3x 2 x 3 + 3x 4 + 2x 5 + 3x 7 = 3 3x 1 + 7x 2 + x 3 + 2x 4 + 4x 5 + 3x 6 6x 7 = 13 x 1 + 2x 2 + x 4 + 4x 5 + 3x 6 11x 7 = P Q R 3 A B C A 1 kg P 1 kg Q 1 kg R 2 kg B 1 kg P 2 kg Q 3 kg R 2 kg C 1 kg P 1 kg Q 2 kg R 3 kg A B C 1 kg

19 120 f(x, y) dy dx = f(x, y), y(x 0) = y 0 y(x) 1 f(x, y) y(x) x 0, x 1, x 2,, x n y(x 0 ), y(x 1 ), y(x 2 ),, y(x n ) dy dx = f(x, y), y(x 0) = y 0 (7.1) x 1 = x 0 + h y 1 y(x 0 + h) y(x 0 + h) = y(x 0 ) + y (x 0 )h + 1 2! y (x 0 )h 2 + (7.2) (7.1) y (x 0 ) = f(x 0, y 0 ) f(x 0, y 0 ) y(x 1 ) y 1 (7.2) 2 y 1 = y 0 + f(x 0, y 0 )h x 1 h x 2 y(x 2 ) = y(x 1 + h) y 2 x 1 y 1 y 2 = y 1 + f(x 1, y 1 )h x x n = x 0 + nh y(x n ) y n (Euler)

20 dy dx = f(x, y), y(x 0) = y 0 h x n = x 0 + nh y(x n ) y n y n = y n 1 + f(x n 1, y n 1 )h, (n = 1, 2, ) (7.2) h 1 h x n = x 0 + nh y(x n ) y n 7.1 y = y 12x + 3, y(0) = 1 h = 0.1 y 1, y 2, y 3, y 4, y 5, y 6, y 7, y 8, y 9, y 10 0 x 1

21 122 7 x j y j f(x, y) hf(x, y) x j y j f(x, y) hf(x, y) y = 12x 8e x + 9 x = y(0.1) = , y(0.5) = , y(1.0) = (7.2) h O(h n ) n h k h 0 k/h n c( ) k = O(h n ) ( c = 0 ) O(h n ) + O(h n ) = O(h n ), ho(h n ) = O(h n+1 ) 0 < n < m O(h n ) + O(h m ) = O(h n ) (7.3) y(x) C n+1 y(x) y(x 0 + h) = y(x 0 ) + y (x 0 )h + 1 2! y (x 0 )h n! y(n) (x 0 )h n + O(h n+1 ) f(x, y) y f(a, b + h) = f(a, b) + O(h) (7.4) y(x 0 +h) h n y(x 0 +h) h n y(x 0 ) = y 0 y (x 0 ) = y 0 y (x 0 ) = y 0

22 h 2 y(x 0 + h) = y 0 + y 0h + 1 2! y 0 h 2 + O(h 3 ) (7.5) (7.5) y 0 = f(x 0, y 0 ) y 0 y 0 f(x, y) (7.5) h 2 f(x, y) y = y(x 0 + h) y(x 0 ) = hy (x 0 + θh), (0 < θ < 1) y (x 0 + θh) θ = 0 θ = 1 θ = 0 y hy (x 0 ) θ = 1 y hy (x 0 + h) y h 2 αhy (x 0 ) + βhy (x 0 + h) α β y h 2 αhy (x 0 ) + βhy (x 0 + h) = αhy (x 0 ) + βh{y (x 0 ) + y (x 0 )h + O(h 2 )} = (α + β)hy 0 + βh 2 y 0 + O(h 3 ) (7.5) α + β = 1, β = 1 2, α = 1 2, β = 1 2 y = 1 2 hy (x 0 ) hy (x 0 + h) + O(h 3 ) k 1 = hy (x 0 ) = hf(x 0, y 0 ) (7.6) (7.5) (7.4) hy (x 0 + h) = hf(x 0 + h, y(x 0 + h)) = hf(x 0 + h, y(x 0 ) + y (x 0 )h + O(h 2 )) = hf(x 0 + h, y 0 + k 1 + O(h 2 )) = hf(x 0 + h, y 0 + k 1 ) + O(h 3 ) k 2 = hf(x 0 + h, y 0 + k 1 ) (7.7)

23 124 7 y = 1 2 k k 2 + O(h 3 ) k = 1 2 (k 1 + k 2 ), y 1 = y 0 + k (7.8) y(x 0 + h) = y 1 + O(h 3 ) y 1 h 2 y 2 x 1 y ( ) y 1 (7.6) (7.7) (7.8) k 1 = hf(x 1, y 1 ), k 2 = hf(x 1 + h, y 1 + k 1 ), k = 1 2 (k 1 + k 2 ) y 2 = y 1 + k y 3 y 4 (Runge-Kutta) dy dx = f(x, y), y(x 0) = y 0 h x n = x 0 + nh y n y n+1 y n+1 = y n + k, (n = 0, 1, 2, ) k k 1 = hf(x n, y n ), k 2 = hf(x n + h, y n + k 1 ), k = 1 2 (k 1 + k 2 ) x j y j k j k x j y j x j y j

24 /* ************************************************* */ 2 /* 2 rungekt2.c */ 3 /* ************************************************* */ 4 # include <stdio.h> 5 double fnf ( double x, double y) 6 { return ( y * x + 3.0); } 7 int main ( void ) 8 { int i; 9 double x, y, h, k1, k2, k; 10 char zz; 11 printf (" 2 \ n\n"); 12 printf ("dy/dx = y * x "); 13 printf ("\n\ n \n"); 14 scanf ("%c",&zz ); 15 printf (" X Y\n"); 16 x = 0.0; y = 1.0; h = 0.1; 17 printf (" %10.6 lf %10.6 lf\n",x,y); 18 for (i =1; i <=20; i ++) { 19 k1 = h * fnf (x,y); 20 k2 = h * fnf (x+h,y+k1 ); 21 k = ( k1 + k2 ) / 2.0; 22 x = x + h; 23 y = y + k; 24 printf (" %10.6 lf %10.6 lf\n",x,y); 25 } 26 return 0; 27 } 2 h 4 4 ( [17]) dy dx = f(x, y), y(x 0) = y 0 h x n = x 0 + nh y n y n+1 = y n + k, (n = 0, 1, 2, ) k

26 171 (2-1) (2-1) (1) (2) (3) cm 1.5 t = x 1 = b 1 a 21 x 1 + x 2 = b 2 x 1 = b 1 j a 31 x 1 + a 32 x 2 + x 3 = b 3,. x j = b j a jk x k. k=1 a n1 x 1 + a n2 x a n,n 1 x n 1 + x n = b n (j = 2, 3,, n) 2.2 (1) x = 1.8 y = 1 z = 1.1 (2) x = y = z = w = (3) x = y = z = u = w = (1) x = y = z = (2) x = 2 y = 1 z = 3 u = (2-1) (1) (2) 2.6 (2-2) (1) x = t 1 (2) x = t (t ) (3) (3) x = t s t x 1

27 (1) x = s t 13 0 (2) x = s 1 + t (3) x = s 1 (4) x = t 3 (5) (6) x = s 1 (7) x = 8 + r 2 + s 1 + t x = 11 3, y = 1 3, z = 5 79 kg, (1) = 40 (2) (3) = = (1) = (2) =

29 LU (1 ) ( ) () 166

30 179 [a 1, a 2,, a n ] 10 a 1 a a n [a ij ] a ij i j 11 t a a 12 t A A 75 A 1 A 22 A A 37 n a j a 1 a 2 a n 45 j=1 L k (x) n j=1 j k x x j x k x j 45 f[x 0, x 1 ] 1 49 f[x 0, x 1, x 2 ] 2 49 f[x 0, x 1,, x n ] n 50 C n n 51 ( ) n n 56 nc j 58 ( n j ) 58 T n (x) n 82 ζ 0, ζ 1, ζ 2,, ζ n T n+1 (x) 87 P n (x) n 90 H n (k) 104 O(h n ) h n ( ) 122 x x 149

31 C n DE h n LU n n n

32 x

x h = (b a)/n [x i, x i+1 ] = [a+i h, a+ (i + 1) h] A(x i ) A(x i ) = h 2 {f(x i) + f(x i+1 ) = h {f(a + i h) + f(a + (i + 1) h), (2) 2 a b n A(x i )

1 f(x) a b f(x)dx = n A(x i ) (1) ix [a, b] n i A(x i ) x i 1 f(x) [a, b] n h = (b a)/n y h = (b-a)/n y = f (x) h h a a+h a+2h a+(n-1)h b x 1: 1 x h = (b a)/n [x i, x i+1 ] = [a+i h, a+ (i + 1) h] A(x

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. 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.

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. 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.