關於牛頓法的計算，xn+1 初始值xn-(f(xn)/f (xn))，以及誤差判斷,每疊代一次，牛頓法結果的有效數字將增加一倍。-Newton on the calculation method, xn+1 initial value xn- (f (xn)/f ' (xn)), and an error judgment time, Newton' s method results significant figures will double every iteration.