Java StrictMath expm1()方法及示例

java.lang.StrictMath.expm1()是Java中的一个内置方法，用于返回给定 num 值的 指数e^num-1 。 该方法产生了四种不同的情况。

语法:

public static double expm1(double num)

参数： 该方法接受一个双倍类型的参数num，指的是要进行指数运算的值。
返回值： 该方法将返回e num – 1 的运算结果。
示例:

Input: num = (1.0/0.0)Output: InfinityInput: 32.2Output: 9.644557735961714E13

下面的程序说明了java.lang.StrictMath.expm1()方法：
程序1

// Java program to illustrate the// java.lang.StrictMath.expm1()import java.lang.*; public class Geeks { public static void main(String[] args) {     double num1 = 0.0, num2 = -(1.0/0.0);        double num3 = (1.0/0.0), num4 = 32.2;         /*It  returns e^num - 1 */    double eValue = StrictMath.expm1(num1);    System.out.println("The expm1 Value of "+                              num1+" = "+eValue);         eValue = StrictMath.expm1(num2);    System.out.println("The expm1 Value of "+                              num2+" = "+eValue);         eValue = StrictMath.expm1(num3);    System.out.println("The expm1 Value of "+                              num3+" = "+eValue);         eValue = StrictMath.expm1(num4);    System.out.println("The expm1 Value of "+                              num4+" = "+eValue);}}

程序2

// Java program to illustrate the// java.lang.StrictMath.expm1()import java.lang.*; public class Geeks { public static void main(String[] args) {     double num1 = 2.0 , num2 = -51.8;        double num3 = 61.0, num4 = -32.2;         /*It  returns e^num - 1 */    double eValue = StrictMath.expm1(num1);    System.out.println("The expm1 Value of "+                              num1+" = "+eValue);         eValue = StrictMath.expm1(num2);    System.out.println("The expm1 Value of "+                              num2+" = "+eValue);         eValue = StrictMath.expm1(num3);    System.out.println("The expm1 Value of "+                              num3+" = "+eValue);         eValue = StrictMath.expm1(num4);    System.out.println("The expm1 Value of "+                              num4+" = "+eValue);}}