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.
Best Answer
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: $1.262$.