keith keith - 4 years ago 117
C++ Question

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

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 Source

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

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