It’s exactly the same principle as for x^n == x*(x^(n-1)): Insert your recursive function for x^(n/2) and (…)^2, but make sure you don’t enter an infinite recursion for n == 2 (as 2 is even, too): if (n % 2 == 0 && n > 2) return power(power(x, n / 2), 2); } Alternatively, you … Read more
None. As you can read in the ES7 spec, both Math.pow and the ** exponentation operator cast their arguments/operands to numbers and use the very same algorithm to determine the result. Addendum: this changed with the introduction of the BigInt type in ES2020, whose values are only supported by operators (including **) but not the … Read more
Oh, it’s much “worse” than that: In [2]: numpy.power(10,-1) Out[2]: 0 But this is a hint to what’s going on: 10 is an integer, and numpy.power doesn’t coerce the numbers to floats. But this works: In [3]: numpy.power(10.,-1) Out[3]: 0.10000000000000001 In [4]: numpy.power(10.,-100) Out[4]: 1e-100 Note, however, that the power operator, **, does convert to … Read more
1e5 is a number expressed using scientific notation, specifically E notation, and it means 1 multiplied by 10 to the 5th power (the ‘e’ meaning ‘exponent’). So 1e5 equals 1*100000 and is equal to 100000. The three notations are interchangeable meaning the same.
Executing it in the browser console says SyntaxError: Unexpected token **. Because that’s the spec. Designed that way to avoid confusion about whether it’s the square of the negation of one (i.e. (-1) ** 2), or the negation of the square of one (i.e. -(1 ** 2)). This design was the result of extensive discussion … Read more
math.sqrt is the C implementation of square root and is therefore different from using the ** operator which implements Python’s built-in pow function. Thus, using math.sqrt actually gives a different answer than using the ** operator and there is indeed a computational reason to prefer numpy or math module implementation over the built-in. Specifically the … Read more
A pretty fast one might be something like this: int IntPow(int x, uint pow) { int ret = 1; while ( pow != 0 ) { if ( (pow & 1) == 1 ) ret *= x; x *= x; pow >>= 1; } return ret; } Note that this does not allow negative powers. … Read more
There isn’t an operator but you can use the pow function like this: return pow(num, power) If you want to, you could also make an operator call the pow function like this: infix operator ** { associativity left precedence 170 } func ** (num: Double, power: Double) -> Double{ return pow(num, power) } 2.0**2.0 //4.0
There are actually three exponentiation operators: (^), (^^) and (**). ^ is non-negative integral exponentiation, ^^ is integer exponentiation, and ** is floating-point exponentiation: (^) :: (Num a, Integral b) => a -> b -> a (^^) :: (Fractional a, Integral b) => a -> b -> a (**) :: Floating a => a -> … Read more
Rust provides exponentiation via methods pow and checked_pow. The latter guards against overflows. Thus, to raise 2 to the power of 10, do: let base: i32 = 2; // an explicit type is required assert_eq!(base.pow(10), 1024); The caret operator ^ is not used for exponentiation, it’s the bitwise XOR operator.