Optimizing pow()¶

The implementation of $x^a$ can be optimized for constant $a$ .

It is obvious that for constant integer exponent you can just expand:

$x^3 = x^1 \cdot x^2 x^9 = x^1 \cdot x^8$

and since $13_{10} = 1101_2 = 2^0 + 2^2 + 2^3$ :

$x^{13} = x^1 \cdot x^4 \cdot x^8$

Likewise for fractional $a$ :

$x^\frac{3}{4} = x^\frac{1}{2} \cdot x^\frac{1}{4}$

So for $a = 2.4_{10} \simeq 10.011001101_2$ :

$x^{2.4} \simeq x^2 \cdot x^\frac{1}{4} \cdot x^\frac{1}{8} \cdot x^\frac{1}{64} \cdot x^\frac{1}{128} \cdot x^\frac{1}{512}$