Math.log1p() static method returns the natural logarithm (base e) of
1 + x, where
x is the argument. That is:
A number greater than or equal to -1.
For very small values of x, adding 1 can reduce or eliminate precision. The double floats used in JS give you about 15 digits of precision. 1 + 1e-15 = 1.000000000000001, but 1 + 1e-16 = 1.000000000000000 and therefore exactly 1.0 in that arithmetic, because digits past 15 are rounded off.
When you calculate log(1 + x) where x is a small positive number, you should get an answer very close to x, because
. If you calculate
Math.log(1 + 1.1111111111e-15), you should get an answer close to
1.1111111111e-15. Instead, you will end up taking the logarithm of
1.00000000000000111022 (the roundoff is in binary, so sometimes it gets ugly), and get the answer 1.11022…e-15, with only 3 correct digits. If, instead, you calculate
Math.log1p(1.1111111111e-15), you will get a much more accurate answer
1.1111111110999995e-15, with 15 correct digits of precision (actually 16 in this case).
If the value of
x is less than -1, the return value is always
log1p() is a static method of
Math, you always use it as
Math.log1p(), rather than as a method of a
Math object you created (
Math is not a constructor).
Math.log1p(-2); // NaN Math.log1p(-1); // -Infinity Math.log1p(-0); // -0 Math.log1p(0); // 0 Math.log1p(1); // 0.6931471805599453 Math.log1p(Infinity); // Infinity
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