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Newton method solution in Speedy category for Super Root by Sim0000
from math import log
def super_root(number):
x = 10
while abs(x**x - number) >= 0.001:
t = x**x
x = x - (t - number) / (t + t * log(x))
return x
# Newton method
# f(x) = x**x - a
# f'(x) = (x**x) * (log(x) + 1)
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
def check_result(function, number):
result = function(number)
if not isinstance(result, (int, float)):
print("The result should be a float or an integer.")
return False
p = result ** result
if number - 0.001 < p < number + 0.001:
return True
return False
assert check_result(super_root, 4), "Square"
assert check_result(super_root, 9), "Cube"
assert check_result(super_root, 81), "Eighty one"
July 2, 2014
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