[Math] How to calculate non-integer exponents

approximationarithmeticexponentiationnumerical methods

I can calculate the result of $x^y$ provided that $y \in\mathbb{N}, x \neq 0$ using a simple recursive function:

$$
f(x,y) =
\begin {cases}
1 & y = 0 \\
(x)f(x, y-1) & y > 0
\end {cases}
$$

or, perhaps more simply stated, by multiplying $x$ by itself $y$ times.

Unfortunately, I am unsure how I can numerically approximate $x^y$ for non-integer rationals.

For example, what method can I use to approximate 33.3?

If possible, I would like to be able to do this using only elementary arithmetic operations, i.e. addition, subtraction, multiplication, and division.

Best Answer

I'll consider the problem of computing $x^\frac1{q}, \; q > 0$; as I've already mentioned in the comments, one can decompose any positive rational number as $m+\dfrac{p}{q}$, where $m,p$ are nonnegative integers, $q$ is a positive integer, and $p < q$. Thus for computing $x^{m+\frac{p}{q}}$, one could use binary exponentiation on $x^m$ and $\left(x^\frac1{q}\right)^p$ and multiply the results accordingly.

A.N. Khovanskiĭ, in his book on continued fractions, displays a continued fraction representation for the binomial function:

$$(1+z)^\alpha=1+\cfrac{2\alpha z}{2+(1-\alpha)z-\cfrac{(1-\alpha^2)z^2}{3(z+2)-\cfrac{(4-\alpha^2)z^2}{5(z+2)-\cfrac{(9-\alpha^2)z^2}{7(z+2)-\cdots}}}}$$

which converges for $|\arg(z+1)| < \pi$.

Letting $z=x-1$ and $\alpha=\dfrac1{q}$, one can then evaluate this continued fraction (with, say, Lentz-Thompson-Barnett) to generate a "seed" that can be subsequently polished with Newton-Raphson, Halley, or any of a number of iterations with high-order convergence. You'll have to experiment with how accurate a seed you need to start up the iteration, by picking a not-too-small tolerance when evaluating the continued fraction.

Here's some Mathematica code demonstrating what I've been saying earlier, for computing $\sqrt[3]{55}$:

With[{q = 3, t = 55, prec = 30},
 y = N[2 + (1 - 1/q) (t - 1), prec];
 c = y; d = 0; k = 1;
 While[True,
  u = (k^2 - q^-2) (t - 1)^2; v = (2 k + 1) (t + 1);
  c = v - u/c; d = 1/(v - u d);
  h = c*d; y *= h;
  If[Abs[h - 1] <= 10^-4, Break[]];
  k++];
 FixedPoint[
  Function[x, x ((1 + q) t - x^q (1 - q))/(x^q (1 + q) - (1 - q) t)], 
  1 + 2 (t - 1)/q/y]]

Here, I've arbitrarily chosen to stop when the continued fraction has already converged to $\approx 4$ digits, and then polished the result with Halley's method. The result here is good to $\approx 28$ digits. Again, you'll have to experiment on the accuracy versus expense of evaluating the "seed", as well as picking the appropriate iteration method for polishing the seed.

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