Reputation: 45
I am trying to calculate the square root of a number using the Newton-Raphson method. This part is relatively straight forward. However, when I want to ensure that the value returned on the final answer is always to 100 decimal places, I can't seem to get it to work. I have tried on VS Code and Jupyter Notebooks, and can only seem to get it to display to 17 or 18 decimal places.
My code is:
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)))
The output of the last few iterations looks like this on both platforms, when the starting value is 15:
98 : 3.87298334620741702138957407441921532154083251953125
Length: 17
99 : 3.87298334620741702138957407441921532154083251953125
Length: 17
100 : 3.87298334620741702138957407441921532154083251953125
Length: 17
101 : 3.87298334620741702138957407441921532154083251953125
Length: 17
Newton-Raphson method =
3.87298334620741702138957407441921532154083251953125Length: 17
Any suggestions on how to get the decimal places to show the 100 decimal places? Note that I have to use the Newton-Raphson method as it is a requirement.
Upvotes: 3
Views: 56
Reputation: 333
At least one of the variable must be initialized with Decimal type in the floating point operations.
You have to use Number = Dec.Decimal(5)
Upvotes: 5