Excuse me, but I’m going to toot Stata’s horn.
I got an email from Nicholas Cox (an Editor of the Stata Journal) yesterday. He said he was writing something for the Stata Journal and wanted the details on how we calculated a^b. He was focusing on examples such as (-8)^(1/3), where Stata produces a missing value rather than -2, and he wanted to know if our calculation of that was exp((1/3)*ln(-8)). He didn’t say where he was going, but I answered his question.
I have rather a lot to say about this.
Nick’s supposition was correct, in this particular case, and for most values of a and b, Stata calculates a^b as exp(b*ln(a)). In the case of a=-8 and b=1/3, ln(-8)==., and thus (-8)^(1/3)==.. Read more…