## Friday, March 21, 2014

### Powers

bn is familiar when n is a whole integer (positive or negative). Here's an algorithm to compute it.
let rec pow (i, u) =
if i < 0 then 1.0 / (pow (-i, u))
else if i = 0 then 1.0 else u * pow ((i-1), u)

It is also familiar to us when c/d = n that is, for n a rational number (then the solution is found by taking the dth root of bc). Less so though I think when x = n is some arbitrary real. That is, generally speaking how is bx computed?

Logarithms and exponentiation. Since b = e log b then bx = (elog b)x = e x (log b).
let powf (b, x) = exp (x * log b)

This only works when b is real and positive. The case negative b leads into the theory of complex powers.

“Sometimes,' said Pooh, 'the smallest things take up the most room in your heart.”