我就廢話不多說了，大家還是直接看代碼吧~

def sq2(x,e): e = e ＃誤差范圍 low= 0 high = max(x,1.0) ＃處理大于0小于1的數(shù) guess = (low + high) / 2.0 ctr = 1 while abs(guess**2 - x) > e and ctr<= 1000: if guess**2 < x: low = guess else: high = guess guess = (low + high) / 2.0 ctr += 1 print(guess)

補(bǔ)充：數(shù)值計(jì)算方法：二分法求解方程的根（偽代碼 python c/c++）

數(shù)值計(jì)算方法：

偽代碼

fun (input x) return x^2+x-6newton (input a, input b, input e)//a是區(qū)間下界，b是區(qū)間上界，e是精確度 x <- (a + b) / 2 if abs(b - 1) < e: return x else: if fun(a) * fun(b) < 0: return newton(a, x, e) else: return newton(x, b, e)c/c++:

#include <iostream>#include <cmath>using namespace std; double fun (double x);double newton (double a, double b,double e); int main(){ cout << newton(-5,0,0.5e-5); return 0;} double fun(double x){ return pow(x,2)+x-6;} double newton (double a, double b, double e){ double x; x = (a + b)/2; cout << x << endl; if ( abs(b-a) < e) return x; else if (fun(a)*fun(x) < 0) return newton(a,x,e); else return newton(x,b,e);}python：

def fun(x): return x ** 2 + x - 6def newton(a,b,e): x = (a + b)/2.0 if abs(b-a) < e: return x else: if fun(a) * fun(x) < 0: return newton(a, x, e) else: return newton(x, b, e)print newton(-5, 0, 5e-5)

以上為個(gè)人經(jīng)驗(yàn)，希望能給大家一個(gè)參考，也希望大家多多支持好吧啦網(wǎng)。如有錯(cuò)誤或未考慮完全的地方，望不吝賜教。