document.write('<div>正数n的平方根可以通过计算一系列近似值来获得，每个近似值都比前一个更加接近准确值，直到找出满足精度要求的那个数位置。具体而言，可以找出第一个近似值是1，接下来的近似值则可以通过下面的公式来获得：ai+1=(ai+n/ai)/2。实现代码如下：</div><div>&nbsp; &nbsp; #获取n的平方根, e为精度要求</div><div>&nbsp; &nbsp; def squareRoot(n,e):</div><div>&nbsp; &nbsp; new_one=n</div><div>&nbsp; &nbsp; last_one=1.0 #第一个近似值为1</div><div>&nbsp; &nbsp; while new_one-last_one＞e:#直到满足精度要求为止</div><div>&nbsp; &nbsp; new_one=(new_one+last_one)/2# 求下一个近似值</div><div>&nbsp; &nbsp; last_one=n/new_one</div><div>&nbsp; &nbsp; return new_one</div><div>&nbsp; &nbsp;&nbsp;</div><div>&nbsp;</div><div>&nbsp; &nbsp; if __name__==&quot;__main__&quot;:</div><div>&nbsp; &nbsp; n=50</div><div>&nbsp; &nbsp; e=0.000001</div><div>&nbsp; &nbsp; print str(n)+&quot;的平方根为&quot;+str(squareRoot(n,e))</div><div>&nbsp; &nbsp; n=4</div><div>&nbsp; &nbsp; print str(n)+&quot;的平方根为&quot;+str(squareRoot(n,e))</div><div>&nbsp; &nbsp; 程序的运行结果为：</div><div>&nbsp; &nbsp; 50的平方根为7.071068</div><p>&nbsp; &nbsp; 4的平方根为2.000000&nbsp;</p>');