What is the correct way to obtain (-1)^n?

Question

Many algorithms require to compute (-1)^n (both integer), usually as a factor in a series. That is, a factor that is -1 for odd n and 1 fo

Answer 1

Well if we are performing the calculation in a series, why not handle the calculation in a positive loop and a negative loop, skipping the evaluation completely?

The Taylor series expansion to approximate the natural log of (1+x) is a perfect example of this type of problem. Each term has (-1)^(n+1), or (1)^(n-1). There is no need to calculate this factor. You can "slice" the problem by either executing 1 loop for every two terms, or two loops, one for the odd terms and one for the even terms.

Of course, since the calculation, by its nature, is one over the domain of real numbers, you will be using a floating point processor to evaluate the individual terms anyway. Once you have decided to do that, you should just use the library implementation for the natural logarithm. But if for some reason, you decide not to, it will certainly be faster, but not by much, not to waste cycles calculating the value of -1 to the nth power.

Perhaps each can even be done in separate threads. Maybe the problem can be vectorized, even.