The Math.expm1() static method returns e raised to the power of a number, subtracted by 1. That is

π™ΌπšŠπšπš‘.πšŽπš‘πš™πš–πŸ· ( 𝚑 ) = e x βˆ’ 1 \mathtt{\operatorname{Math.expm1}(x)} = \mathrm{e}^x - 1

Try it





A number.

Return value

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 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 β†’ 0 e x βˆ’ 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(-Infinity); // -1
Math.expm1(-1); // -0.6321205588285577
Math.expm1(-0); // -0
Math.expm1(0); // 0
Math.expm1(1); // 1.718281828459045
Math.expm1(Infinity); // Infinity


Browser compatibility

Desktop Mobile Server
Chrome Edge Firefox Internet Explorer Opera Safari WebView Android Chrome Android Firefox for Android Opera Android Safari on IOS Samsung Internet Deno Node.js
expm1 38 12 25 No 25 8 38 38 25 25 8 3.0 1.0 0.12.0

See also

Β© 2005–2023 MDN contributors.
Licensed under the Creative Commons Attribution-ShareAlike License v2.5 or later.