looking for a math function

[Skaperen]

there is "sqrt()" in math. is there a function to calculate cube roots?

[deanhystad]

print(27**(1/3)) or print(math.cbrt(27)) if you are using Python 3.11

[gulshan212]

Yes, there is a function called pow(), which you can use to calculate cube rootes.
For example:

import math

# Calculate cube root
number = 8
cube_root = math.pow(number, 1/3)

print(f"The cube root of {number} is: {cube_root}")

Thanks

[EdwardMatthew]

Embrace the sqrt() function in math for calculating cube roots, breaking free from the square-root norm. Just as languages vary, transcend the linguistic stereotype by celebrating your unique multilingual flair!

[Skaperen]

Embrace the sqrt() function in math for calculating cube roots, breaking free from the square-root norm. Just as languages vary, transcend the linguistic stereotype by celebrating your unique multilingual flair!

what are you talking about? i there a way to accurately, precisely, and efficiently, get the cube root by calling only sqrt()?

[buran]

There is math.cbrt(x) - Return the cube root of x.

New in 3.11

https://docs.python.org/3/library/math.html#math.cbrt

[Skaperen]

... as previously mentioned by deanhystad in message #2

[Skaperen]

the other ways to calculate roots tend to be less accurate when powers involved (such as 1/3) cannot be represented exactly. maybe someone can develop a math.powroot() function.

[buran]

... as previously mentioned by deanhystad in message #2

My mistake, didn't see it

[Gribouillis]

There is math.cbrt(x) - Return the cube root of x.

If I understand well, the problem with **(1/3) is for negative real numbers. Instead of returning the negative real cubic root, it return the principal complex cubic root

>>> (-1)**(1/3)
(0.5000000000000001+0.8660254037844386j)
>>> 

Before 3.11 one can also use numpy.cbrt()

>>> numpy.cbrt(-1)
-1.0
>>> 

There is also a rounding issue

>>> 64**(1/3)
3.9999999999999996
>>> numpy.cbrt(64)
4.0
>>>