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---
title: Math.expm1()
short-title: expm1()
slug: Web/JavaScript/Reference/Global_Objects/Math/expm1
page-type: javascript-static-method
browser-compat: javascript.builtins.Math.expm1
sidebar: jsref
---
The **`Math.expm1()`** static method returns [e](/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/E) raised to the power of a number, subtracted by 1. That is
<!-- prettier-ignore-start -->
<math display="block">
<semantics><mrow><mrow><mo lspace="0em" rspace="0.16666666666666666em">πΌπππ.ππ‘πππ·</mo><mo stretchy="false">(</mo><mi>π‘</mi><mo stretchy="false">)</mo></mrow><mo>=</mo><msup><mi mathvariant="normal">e</mi><mi>x</mi></msup><mo>β</mo><mn>1</mn></mrow><annotation encoding="TeX">\mathtt{\operatorname{Math.expm1}(x)} = \mathrm{e}^x - 1</annotation></semantics>
</math>
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{{InteractiveExample("JavaScript Demo: Math.expm1()")}}
```js interactive-example
console.log(Math.expm1(0));
// Expected output: 0
console.log(Math.expm1(1));
// Expected output: 1.718281828459045
console.log(Math.expm1(-1));
// Expected output: -0.6321205588285577
console.log(Math.expm1(2));
// Expected output: 6.38905609893065
```
## Syntax
```js-nolint
Math.expm1(x)
```
### Parameters
- `x`
- : A number.
### Return value
A number representing e<sup>x</sup> - 1, where e is [the base of the natural logarithm](/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/E).
## 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.
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When you calculate <math><semantics><msup><mi mathvariant="normal">e</mi><mi>x</mi></msup><annotation encoding="TeX">\mathrm{e}^x</annotation></semantics></math>, where x is a number very close to 0, you should get an answer very close to 1 + x because: <math><semantics><mrow><munder><mo lspace="0em" rspace="0em">lim</mo><mrow><mi>x</mi><mo stretchy="false">β</mo><mn>0</mn></mrow></munder><mfrac><mrow><msup><mi mathvariant="normal">e</mi><mi>x</mi></msup><mo>β</mo><mn>1</mn></mrow><mi>x</mi></mfrac><mo>=</mo><mn>1</mn></mrow><annotation encoding="TeX">\lim_{x \to 0} \frac{\mathrm{e}^x - 1}{x} = 1</annotation></semantics></math>. 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.expm1(1.1111111111e-15)` instead, you will get a much more accurate answer, `1.1111111111000007e-15`, with 11 correct digits of precision.
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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
{{Specifications}}
## Browser compatibility
{{Compat}}
## See also
- [Polyfill of `Math.expm1` in `core-js`](https://github.com/zloirock/core-js#ecmascript-math)
- {{jsxref("Math.E")}}
- {{jsxref("Math.exp()")}}
- {{jsxref("Math.log()")}}
- {{jsxref("Math.log10()")}}
- {{jsxref("Math.log1p()")}}
- {{jsxref("Math.log2()")}}
- {{jsxref("Math.pow()")}}
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