Math.expm1() static method returns e raised to the power of a number, subtracted by 1. That is
A number representing ex - 1, where e is the base of the natural logarithm.
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
where x is a number very close to 0, you should get an answer very close to 1 + x, because
. 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.
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).
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
|ECMAScript Language Specification |
BCD tables only load in the browser