Numerical trade offs between 1/sqrt(x) and std::exp(-0.5 * std::log(x))

Question

I have come across some old code that computes

double y = 1 / std::sqrt(x);

Using:

constexpr double base16 = 16.0;
double log_base16 = std::log(base16);
double y = std::pow(base16, -0.5 * std::log(x) / log_base16);

Which is essentially:

double y = std::exp(-0.5 * std::log(x));

Is there any justifications with regard to numerical benefits (such as accuracy or more likely to avoid underflow/overflow) between the methods? The original author may have thought so.

Answer 1

The original code is to be considered to be very naughty indeed, particularly in modern C++ standards and IEEE754 floating point:

std::sqrt is required by the IEEE standard be exact. [sic.]

Furthermore, std::pow has no such requirements.

Therefore I'd be tempted to rewrite it as 1 / std::sqrt(x) , testing of course.

Reference: http://en.cppreference.com/w/cpp/numeric/math/sqrt

Answer 2

I can't see any particularly good reasons for implementing sqrt() in terms of pow() and log(). It could be that there was a bug in the implementation of sqrt() and this was used as a workaround.

In general, I would expect this implementation to be slower and less precise.