# Math.expm1()

## BaselineWidely available

This feature is well established and works across many devices and browser versions. It’s been available across browsers since July 2015.

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

$𝙼𝚊𝚝𝚑.𝚎𝚡𝚙𝚖𝟷 ( 𝚡 ) = e x − 1 \mathtt{\operatorname{Math.expm1}(x)} = \mathrm{e}^x - 1$

## Syntax

js
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 $\mathrm\left\{e\right\}^x$, where x is a number very close to 0, you should get an answer very close to 1 + x because: $\lim_\left\{x \to 0\right\} \frac\left\{\mathrm\left\{e\right\}^x - 1\right\}\left\{x\right\} = 1$. 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.exp1m(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).

## Examples

### Using Math.expm1()

js
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


## Specifications

Specification
ECMAScript Language Specification
# sec-math.expm1

## Browser compatibility

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