Math.expm1()
Math.expm1()
静态方法返回 e 的数字次方减去 1。那是
¥The Math.expm1()
static method returns e raised to the power of a number, subtracted by 1. That is
Try it
语法
参数
返回值
¥Return value
代表 ex 的数字 - 1,其中 e 是 自然对数的底。
¥A number representing ex - 1, where e is the base of the natural logarithm.
描述
¥Description
对于非常小的 x 值,加 1 会降低或消除精度。JS 中使用的双浮点可为你提供大约 15 位的精度。1 + 1e-15 = 1.000000000000001,但 1 + 1e-16 = 1.000000000000000,因此在该算术中恰好是 1.0,因为超过 15 的数字会四舍五入。
¥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.
当你计算
时,其中 x 是一个非常接近 0 的数字,你应该得到一个非常接近 1 + x 的答案,因为:
Math.exp(1.1111111111e-15) - 1
,你应该得到接近 1.1111111111e-15
的答案。相反,由于 Math.exp
结果中的最高有效数字是个位 1
,所以最终的结果是 1.1102230246251565e-15
,只有 3 位正确。如果你改为计算 Math.exp1m(1.1111111111e-15)
,你将得到更准确的答案 1.1111111111000007e-15
,其精度为 11 位正确数字。
¥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 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.
因为 expm1()
是 Math
的静态方法,所以你始终将其用作 Math.expm1()
,而不是用作你创建的 Math
对象的方法(Math
不是构造函数)。
¥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).
示例
使用 Math.expm1()
规范
Specification |
---|
ECMAScript Language Specification # sec-math.expm1 |
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