Luca Citi
http://users.neurostat.mit.edu/lciti
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Faster / more precise logarithm of complimentary error function?
Not much faster but certainly more accurate when erfc(x)⩬1: log1p(erf(-x)) So switching between log1p(erf(-x)) and log(erfc(x)...
Faster / more precise logarithm of complimentary error function?
Not much faster but certainly more accurate when erfc(x)⩬1: log1p(erf(-x)) So switching between log1p(erf(-x)) and log(erfc(x)...
12개월 전 | 0