【发布时间】:2020-10-05 01:21:14
【问题描述】:
我正在尝试使用 Newton-Raphson 方法计算数字的平方根。这部分相对简单。但是,当我想确保最终答案返回的值始终为小数点后 100 位时,我似乎无法让它工作。我在 VS Code 和 Jupyter Notebooks 上试过,似乎只能显示到小数点后 17 位或 18 位。
我的代码是:
import decimal as Dec
Dec.getcontext().prec = 100
Number = 5
bStart = 1 # Start value is 1
zStart = 1 # Start value is 1 (x0)
aCount = 0 # Iteration count
yNewNum = 0 # New value of estimate (x1)
xRoot = 0 # Used to transfer values between yNewNum and zStart
while aCount < 101: # While the difference between zStart and yNewNum is greater than 0.0000001 (7 decimal places)
yNewNum = (zStart + (Number / zStart)) / 2 # x1 = (x0 + (S/x0)) / 2
zStart = yNewNum # Replace the value x0 with the value of x1
yNewNum = xRoot # Replace the value of x1 with the transfer value
xRoot = zStart # Replace the transfer value with x0
aCount += 1 # aCount iterates by 1
print()
print(aCount, ":", Dec.Decimal(zStart))
print(len(str(zStart)))
print("Newton-Raphson method = ", Dec.Decimal(zStart))
print("Length:", len(str(zStart)))
最后几次迭代的输出在两个平台上都是这样的,当起始值为 15 时:
98 : 3.87298334620741702138957407441921532154083251953125
长度:17
99 : 3.87298334620741702138957407441921532154083251953125
长度:17
100 : 3.87298334620741702138957407441921532154083251953125
长度:17
101:3.87298334620741702138957407441921532154083251953125
长度:17
Newton-Raphson 方法 =
3.87298334620741702138957407441921532154083251953125长度:17
关于如何获得小数位以显示 100 个小数位的任何建议?请注意,我必须使用 Newton-Raphson 方法,因为它是必需的。
【问题讨论】:
标签: python python-3.x