why python stops calculating more decimal places after 14 digits?-CodePudding

I am trying to implement newtons algorithm to get square root of a number and well while it works , I can get accuracy only upto 14 places after decimal. I tried to use getcontext().prec but it didnt work. Here's the code

number = float(input("number less than 1000 : "))
li =[]
for i in range(32):

    if i*i <= number:
        li.append(i)

firstroot = (li[-1])
li.clear()

wants = int(input("how many digits: "))

count = 0

while count < wants:
    root = 0.5*(firstroot   (number/firstroot))
    firstroot = root
    count  = 1

answer = str(firstroot)
print(answer[0:wants 1])

even if I enter 50 in "wants" input I'll only get upto 14 decimal places. How to solve this?

CodePudding user response：

That's because firstroot is simply a float number, which has a fixed maximum precision. If you need more digits, you must either implement the calculation yourself or use a package like mpmath which already has done that for you.

In your special case, as you already have hand-implemented the square root from basic operations, you could alternatively use the standard library decimal and crank up its precision.

CodePudding user response：

You can use the decimal module for more precision in your variables (instead of the default float type).

import decimal

## Your code:
number = float(input("number less than 1000 : "))
li =[]
for i in range(32):

    if i*i <= number:
        li.append(i)

firstroot = (li[-1])
li.clear()

wants = int(input("how many digits: "))

##
decimal.getcontext().prec = wants
count = 0
wants = decimal.Decimal(wants)
number = decimal.Decimal(number)
firstroot = decimal.Decimal(firstroot)

half = decimal.Decimal(0.5)
## Your code:
while count < wants:
    root = half*(firstroot   (number/firstroot))
    firstroot = root
    count  = 1

print(firstroot)
# or a string:
str_firstroot = str(firstroot)