創建一個方法計算sqrt

Question

我有這個f(x,num)=x*x-num。應該用#define來聲明。我應該使用mysqrt，它將計算一個數字num的平方根。主要的，我應該讀取一個數字num來「出現 - 顯示」變量num，mysqrt（num）和sqrt（num）。在函數mysqrt（num）中，我們給出x和oldx的初始值。 DO-而（| X-oldx |> 1E-15）。在我們將x保存到oldx之後。我們用方法x = x-f（x）/ f'（x）計算新的x。

我做錯了什麼？

#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

這被稱爲牛頓方法來計算sqrt。你發現它到處都有記錄。