Question

C语言编程 牛顿迭代法求sinx-x/2在x=pi附近的一个实根，精度小于10（-4）

Answer 1

#include <stdio.h>
#include <math.h>

#define PI 3.141592653

int main()
{
double dXPrev, dX;

dX = PI;
do {
dXPrev = dX;
dX = dXPrev - (sin(dXPrev) - dXPrev / 2) / (cos(dXPrev) - 0.5);
} while (fabs(dX - dXPrev) >= 1e-4);

printf("Result: %f\n", dX);

return 0;
}

Answer 2

(n))^2-sinx(n)-1]/[18x(n)-cosx(n)].
取x(0)=0.5,
x(1)=0.405129911,
x(2)=0.392101462,
x(3)=0.391847004,
x(4)=0.391846907,
3次迭代已经得到四位近似值x=0.3918.

二分法: f(x)=9x^2-sinx-1.
f(0)=-1,f(1)=7.15853,f(0.5)=0.770577.
f(0.25)=-0.68490,
f[(0.25+0.5)/2]=f(0.375)=-0.10065,
f[(0.375+0.5)/2]=f(0.4375)=0.29898,
f[(0.375+0.4375)/2]=f(0.40625)=0.09018,
......
要13次左右才能得到四位近似值.