Python用二分法求平方根的案例
我就廢話不多說瞭,大傢還是直接看代碼吧~
def sq2(x,e): e = e #誤差范圍 low= 0 high = max(x,1.0) #處理大於0小於1的數 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)
補充:數值計算方法:二分法求解方程的根(偽代碼 python c/c++)
數值計算方法:
二分法求解方程的根
偽代碼
fun (input x) return x^2+x-6 newton (input a, input b, input e) //a是區間下界,b是區間上界,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 - 6 def 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)
以上為個人經驗,希望能給大傢一個參考,也希望大傢多多支持WalkonNet。如有錯誤或未考慮完全的地方,望不吝賜教。