Java Math expm1() 方法与示例
Java.lang.Math.expm1()返回e x -1。请注意,对于接近 0 的 x 值,expm1(x) + 1 的精确总和比 exp(x) 更接近 e x的真实结果。
- 如果参数为NaN,则结果为NaN。
- 如果参数是正无穷大,那么结果是正无穷大。
- 如果参数为负无穷大,则结果为-1.0。
- 如果参数为零,则结果为零,其符号与参数相同。
句法:
public static double expm1(double x)
Parameter:
x-the exponent part which raises to e.
回报:
该方法返回值e x -1 ,其中 e 是自然对数的底。
示例:显示Java.lang.Math.expm1()函数的工作
// Java program to demonstrate working
// of java.lang.Math.expm1() method
import java.lang.Math;
class Gfg {
// driver code
public static void main(String args[])
{
double x = 3;
// when both are not infinity
double result = Math.expm1(x);
System.out.println(result);
double positiveInfinity = Double.POSITIVE_INFINITY;
double negativeInfinity = Double.NEGATIVE_INFINITY;
double nan = Double.NaN;
// when x is NAN
result = Math.expm1(nan);
System.out.println(result);
// when argument is +INF
result = Math.expm1(positiveInfinity);
System.out.println(result);
// when argument is -INF
result = Math.expm1(negativeInfinity);
System.out.println(result);
x = -0;
result = Math.expm1(x);
// same sign as 0
System.out.println(result);
}
}
输出:
19.085536923187668
NaN
Infinity
-1.0
0.0