How to calculate power of a number with decimal exponent programatically?

Question

I am trying to code and algorithm that can allow me to calculate power of a function with decimal exponent. The language that I am using to code in doesn't has any predefined power functions.

I already have a simple formula coded which allows me to calculate power of a function with non-decimal exponents.

I don't necessarily need a generalized solution. I only need to find value of a number raised to the power 2.4.

I tried to break down my problem as follows:

x ^ (2.4) = (x ^ 2) * (x ^ 0.4)
= (x ^ 2) * (x ^ (2/5))
= (x ^ 2) * ((x ^ 2) ^ 1/5)

Finding square of x can be done, so my problem breaks down to calculating the "5th root" of a number.

I also thought of breaking my problem into a logarithmic equation but didn't reached anywhere with that.

I am thinking of writing a code implementing the long-divison method of calculating roots but that seems highly inefficient and non-elegant.

Can anyone suggest me a more simpler and efficient way of solving my problem? If not, then has anyone tried coding the long-divison method of calculating roots?

Thanx in advance!!

Note: this is a template that i would be using many many times in my execution so efficiency is all the more important for me.

Answer 1

Suppose you want to compute ${3.21}^{1/5}$. If you have a logarithm table (say base 10), you only need the logarithms of numbers between 0.1 and 1 stored (alternatively between 1 and 10), as many as is relevant for your precision.

Then because

$$\log (3.21^{1/5}) = \frac{1}{5}\left(\log(10) + \log(0.321)\right)= \frac{1}{5}\left(1+\log(0.321)\right)$$

Now you look up $\log(0.321)$ in your table, which will look something like this

$$\begin{array}{c|c} \text{Input} & \text{Output} \\ \hline \ldots & \ldots \\ \color{red}{0.321} & -0.493 \\ \ldots & \ldots \end{array}$$

do the above computation

$$\frac{1}{5}(1+\log(0.321)) = \frac{1}{5}(1-0.493) = 0.101$$

and look up the result in the "answer column" of your table to revert the $\log$ operation. Since the answer is positive, and we worked with a table containing logarithms of numbers between 0 and 1, we'll need to look up the opposite first

$$\begin{array}{c|c} \text{Input} & \text{Output} \\ \hline \ldots & \ldots \\ 0.792 & \color{red}{-0.101} \\ \ldots & \ldots \end{array}$$

and now take the inverse of that number to obtain the result: $1.262$.

Answer 2

Just a Travis commented, say that you need to compute $A=x^b$ whatever $b$ could be. So taking the logarithms of both sides gives $$\log(A)=b\log(x)$$ Taking the exponentials of both sides gives $$e^{\log(A)}=A=e^{b\log(x)}$$ and you are done for any value of $b$ (integer, rational, irrational, ...).

Answer 3

To compute fifth roots without logarithms, you can use Newton's method. Suppose we want $a^{1/5}$. Define $f(x) = x^5 - a$; we want to find a root of $f$.

So let $x_0$ be a reasonable starting point (see Where to start), and define for $n \ge 0$ $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^5-a}{5x_n^4} = \frac{4x_n^5 - a}{5x_n^4}$$

Now iterate this equation until $|x_{n+1}-x_n|$ is small enough (see When to stop).

Where to start
In this instance, $a$ is strictly positive, and $f$ is well-behaved $-$ monotonic increasing, and convex on $(0,\infty)$. So you can start just about anywhere in $(0,\infty)$. But the algorithm runs faster if you start with a reasonable estimate. A simple method, good enough in this case, would be to try $x_0=1,2,4,8,\ldots$ until $x_0^5 \ge a$.

When to stop
Because of the convexity of $f$, an iteration that starts with $x_0^5 \ge a$ would satisfy $x_n \ge x_{n+1} \ge a^{1/5}$ for all $n$, if all calculations were done to infinite precision. You can't do that, obviously; instead, you can stop as soon as $x_{n+1} \ge x_n$. This means that you have reached the limits of your floating-point precision.