The Math.expm1() static method returns e raised to the power of a number, subtracted by 1. That is
πΌπππ.ππ‘πππ·(π‘)=exβ1\mathtt{\operatorname{Math.expm1}(x)} = \mathrm{e}^x - 1
Syntax
Math.expm1(x)
Parameters
x- : A number.
Return value
A number representing ex - 1, where e is the base of the natural logarithm.
Description
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 limxβ0exβ1x=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, instead, you calculate Math.exp1m(1.1111111111e-15), 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).
Examples
Using Math.expm1()
Math.expm1(-Infinity); // -1
Math.expm1(-1); // -0.6321205588285577
Math.expm1(-0); // -0
Math.expm1(0); // 0
Math.expm1(1); // 1.718281828459045
Math.expm1(Infinity); // Infinity