How to calculate an arbitrary power/root?

问题

I have a application which needs to raise a number to a fractional power. The target platform is an FPGA and I can get estimates on an FPU size for it, but I need an algorithm for raising a number to a fractional power just for a feasibility study. I'm assuming floating point as a worst case, I expect in practice we will be able to use short cuts, but for now I want to show that worst case can be implemented on our part.

Thought I'd ask here and see if there were any common methods that I could checkout. I know there are software methods of doing this, I want just a reasonably efficient algorithm to start with. I'll worry about the FPGA implementation.

回答1:

Is your range of inputs arbitrary, or known within a certain range?

in either case, x^m = exp(m log x), so if you can create functions to evaluate exp(x) and log(x) and have a multiply, you're probably all set.

You'll have to figure out how you want to handle nonpositive values of x.

(hint for log(x): if this is IEEE-754 floating point, shift the significand if necessary until you end up with a range of numbers between 2^k and 2^k+1 for some value K. This lets you deal with a 2:1 range which isn't too hard to approximate with polynomials. Then you just have a small number of possibilities to handle for the exponent and the shift number.

Corresponding hint for exp(x): write x = k+b where 0 <= b < 1 and k is an integer. Then exp(x) = exp(k)*exp(b); b has limited range and k has a limited number of discrete possibilities.)

(hint #2: the numbers probably work out better for x^m = g(m f(x)) where f(x) = log₂x and g(x) = 2^x.)

回答2:

As Jason S said, this is done using the identity x^m = exp(m log x). In practice though, you will have to deal with truncation errors. I think that the way this is usually done is

1. Use this identity: x^m = (2ⁿ * x / 2ⁿ)^m = 2^nm * (x/2ⁿ)^m and find an integer n such that 1 <= x/2ⁿ < 2.
2. Calculate t = log2(x / 2ⁿ). This can be done either using a sufficiently high degree Taylor expansion, or with good ol' Newton-Raphson. You have to make sure that the maximum error over the interval [1, 2[ is not too large for you.
3. Calculate u = nm + tm.
4. Our goal is now to calculate 2^u. Use the fact that 2^u = 2^v * 2^u-v and find an integer v such that 0 <= u-v < 1.
5. Calculate w = 2^u-v, again using our friend Taylor or Newton-Raphson. The interval of interest is 0 <= u-v < 1.
6. Your answer is now 2^v * w.

来源：https://stackoverflow.com/questions/1375953/how-to-calculate-an-arbitrary-power-root