3 ms·
Agreed, this is risky business. The intermediate values still need to fit into floats and are still losing precision. From the article: g(1e-9) returns 1
by DougBTX 3y ago
Agreed, this is risky business. The intermediate values still need to fit into floats and are still losing precision.
From the article:
g(1e-9) returns 1.0000000005,
g(1e-12) returns 1.0000000000005,
g(1e-15) returns 1.0000000000000004
but... g(1e-16) throws ZeroDivisionError: float division by zero.