我就废话不多说了，大家还是直接看代码吧~
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def sq2(x,e): e = e ＃误差范围 low= 0 high = max(x,1.0) ＃处理大于0小于1的数 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) |
补充：数值计算方法：二分法求解方程的根（伪代码 python c/c++）
数值计算方法：
二分法求解方程的根
伪代码
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fun (input x) return x^2+x-6newton (input a, input b, input e)//a是区间下界，b是区间上界，e是精确度 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++:
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#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：
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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) |
以上为个人经验，希望能给大家一个参考，也希望大家多多支持服务器之家。如有错误或未考虑完全的地方，望不吝赐教。
原文链接：https://blog.csdn.net/sharkandshark/article/details/83625839
