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 ex \mathrm{e}^x , where x is a number very close to 0, you should get an answer very close to 1 + x because: lim x â 0 ex â 1 x = 1 \lim_{x \to 0} \frac{\mathrm{e}^x - 1}{x} = 1 . If you calculate Math.exp(1.1111111111e-15) - 1
, you should get an answer close to 1.1111111111e-15
. Instead, due to the highest significant figure in the result of Math.exp
being the units digit 1
, the final value ends up being 1.1102230246251565e-15
, with only 3 correct digits. If you calculate Math.expm1(1.1111111111e-15)
instead, you will get a much more accurate answer, 1.1111111111000007e-15
, with 11 correct digits of precision.
Because expm1()
is a static method of Math
, you always use it as Math.expm1()
, rather than as a method of a Math
object you created (Math
is not a constructor).
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