Question

我有f(x,num)=x*x-num。它应该用#define声明。我应该使用mysqrt来计算数字num的平方根。主要是，我应该读一个num到＃34;出现 - 显示＆＃34;变量num，mysqrt（num）和sqrt（num）。在函数mysqrt（num）中，我们在x和oldx上给出了首字母。 DO-而（| X-oldx |＆GT; 1E-15）。我们将x内部保存到oldx后。我们用方法x = x-f（x）/ f＆＃39;（x）计算新的x。

我做错了什么？

#include <stdio.h>
#include <stdlib.h> 
#define f(x) ((x) * (x) - (num))
#define fdx(x) (2 * (x))
double mysqrt(double num);
main() {
  double num;
  printf("the num is ", &num, &sqrt);

system("Pause");

} //end main
double mysqrt(double num) { //method
  double x = 3; // i dont know if i did good i gave 3 a random number from my mind
  double oldx = 1; // the same and here i did i gave a random number
  double result;
  double residual;
  unsigned double iterations = 0;
  result = sqrt(num);
  x = num / 2; /* initial guess */
  printf("sqrt(%g)=%.16g\n", num, result);
  do {
    printf("x%u=%.16g\n", iterations, x);
    iterations++;
    oldx = x;
    // x=x-f(x)/f'(x)
    x = x - f(x) / fdx(x);
   residual = x - oldx;
  } while (fabs(residual) > 1e-15);
  printf("x%u=%.16g residual=%.16g\n", iterations, x, residual);
  return x;
}

Answer 1

这称为牛顿计算sqrt的方法。你发现它随处可见。

创建方法计算sqrt

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