牛顿法是一种常用的求方程数值解,具体方法如下
若在区间III中,f(x)f(x)f(x)连续可导,且有唯一零点x0x_{0}x0,则任取x1∈Ix_{1}\in Ix1∈I,定义数列xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{'}(x_{n})}xn+1=xn−f′(xn)f(xn)经过多次迭代后xnx_{n}xn会趋向于x0x_{0}x0
输入是xxx,求t=x→t2−x=0t=\sqrt{x}\rightarrow t^{2}-x=0t=x→t2−x=0,那么假设有函数f(t)=t2−xf(t)=t^{2}-xf(t)=t2−x,求函数零点。tn+1=tn−f(tn)f′(tn)=tn2+x2tnt_{n+1}=t_{n}-\frac{f(t_{n})}{f^{'}(t_{n})}=\frac{t_{n}}{2}+\frac{x}{2t_{n}}tn+1=tn−f′(tn)f(tn)=2tn+2tnx
class Solution:
def mySqrt(self, x: int) -> int:
t=x
while abs(t*t-x)>=1e-6:
t=t/2+x/(2*t)
return t