Wondering what is the right way to represent x^(1/3)? Here is my code and it returns right value 2 for 8^(1/3). Wondering if any other better methods?
int a = 8;
System.out.println(Math.pow(8, 1/3.0)); // returns 2
regards, Lin
Asked
Active
Viewed 111 times
-2
Lin Ma
- 9,739
- 32
- 105
- 175
-
2Better in what sense? – azurefrog Mar 15 '17 at 22:47
-
@azurefrog, more elegant and efficient way. – Lin Ma Mar 15 '17 at 22:48
-
@azurefrog, how would you write the same logic, btw? – Lin Ma Mar 15 '17 at 22:48
-
2What does "elegant" mean? Efficient in what dimension(s)? – Lew Bloch Mar 15 '17 at 22:48
-
2For specifically a cube root, I'd probably use `Math::cbrt(double a)`, but for general exponentiation, `Math::pow(double a, double b)` is what I'd use. – azurefrog Mar 15 '17 at 22:50
-
1But then, I'm usually working on engineering problems, so 2.9999 == 3 as far as I care. If you're working on something algebraic, floating numbers would be bad. – azurefrog Mar 15 '17 at 22:53
-
1You might get better answers on http://codereview.stackexchange.com, which is where this question belongs. – Dawood ibn Kareem Mar 15 '17 at 23:26
-
@LewBloch, for elegant I mean more formal. How do you calculate n^(1/m) btw? – Lin Ma Mar 16 '17 at 00:29
-
@DavidWallace, thanks but since it is just one line of code so I just ask here. How do you calculate n^(1/m) btw? – Lin Ma Mar 16 '17 at 00:30
-
@azurefrog, thanks and it seems you are also using `Math::pow(double a, double b)`, but how it is related to your comments "floating numbers would be bad"? I think return float/double is more general case for `Math.pow`? – Lin Ma Mar 16 '17 at 00:31
-
1What you started with is about as formal as it gets, and probably fairly efficient because it uses a library method. I doubt that you can improve on it. – Lew Bloch Mar 16 '17 at 04:48
-
@LewBloch, thanks for the confirmation. – Lin Ma Mar 16 '17 at 06:12
1 Answers
6
There is a Math.cbrt method for x1/3.
System.out.println(Math.cbrt(8));
See also:
- Time complexity of the Math.cbrt(a) method in java.math library
- nth root implementation (If you want to compute x1/n)
-
2
-
Thanks kennytm, if for any root, not necessarily to be cube root, is there a general function (e.g. for n ^ (1/m))? – Lin Ma Mar 16 '17 at 00:28
-
1@LinMa No there isn't. Use `Math.pow(x, 1/n)` as you have written, or the implementation by the asker of the second linked question if you want to deal with odd roots of negative numbers as well. – kennytm Mar 16 '17 at 01:00
-