数值分析课后第二章习题解答.pdf

-5 1 10 1x – ln x=2 [24] x 2x e – 1 = 0 [01] 3 2 3x + 4x – 10 = 0 [12] 1x 18 = 3.14618 2-1 a=2;b=4;k=0; f=inline(x-log(x)-2); ya=f(a); while (b-a)0.00001 x0=.5*(a+b);y0=f(x0); if ya*y00 b=x0; else a=x0;ya=y0; end k=k+1; end format long disp([x0,k]) 2 x17 = 0.567146 3 x 17 = 1.365226 1 −4 2 1 – x – sin x = 0 [01] ×10 2 f (x) = 1 – x – sin x f (0) = 1f (1)= – sin 1 f(0)f (1) 0 [01] f ′(x) −1−cos x f(x) [01][01] 1−0 1 −4 ≤ ×10 2n+1 2 n 4 2 10 14 3 x e + 10 x – 2 =0 1[01] 1 x 2x (2 −e n ) x = 0 n+1 0 10 111 x 11 = 0.0901 25 x = 0.0905 5 x0=0;k=0;er=1; fi=inline((2-exp(x))/10); while er0.0001 6 x1=fi(x0); er=abs(x1-x0); x0=x1;k=k+1; end disp([x0,k]) 4 x 2 + 2 +L+ 2 lim x 2 n n n→∞ x1 2 xn+1 2 +xn ( n = 12 ) x 2k x 2 ( k+ 1) 1 k x 2 +x 2 +2 2