The Math.expm1() function returns ex - 1, where x is the argument, and e is the base of natural logarithms.

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A number.

Return value

A number representing ex - 1, where e is the base of natural logarithms and x is the argument.


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 e x \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 0 e x 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, 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).


Using Math.expm1()

Math.expm1(-1); // -0.6321205588285577
Math.expm1(0);  // 0
Math.expm1(1);  // 1.718281828459045


ECMAScript Language Specification
# sec-math.expm1

Browser compatibility

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See also