📜  Java中的 StrictMath exp() 方法

📅  最后修改于: 2022-05-13 01:54:38.969000             🧑  作者: Mango

Java中的 StrictMath exp() 方法

Java.lang.StrictMath 类的所有方法:Set 1, Set 2
Java.lang.StrictMath.exp()是Java中的一个内置方法,用于返回一个欧拉数,该数被提升到指定双精度值的幂。它产生了三个特殊的结果:

  • 当给定参数为正无穷大时,结果为正无穷大。
  • 当参数为负无穷大时,结果为正零。
  • 当给定参数为 NaN 时,结果为 NaN。

句法 :

public static double exp(double num)

参数:此方法接受一个参数num ,它是 double 类型,它是 e 提升到的指数。
返回值:该方法返回值e^num ,其中 e 是自然对数的底。
例子:

Input: num = 7
Output: 1096.6331584284585

Input: num = (1.0 / 0.0)
Output: Infinity

下面的程序说明了Java.lang.StrictMath.exp() 方法:
方案一:

java
// Java program to illustrate the
// java.lang.StrictMath.exp()
import java.lang.*;
 
public class Geeks {
 
    public static void main(String[] args)
    {
 
        double num1 = 0.0, num2 = (1.0 / 0.0);
        double num3 = 4;
 
        // Returns Euler's number e raised to the given power
        double expValue = StrictMath.exp(num1);
        System.out.println("The exp value of "+
                         num1+" = " + expValue);
 
        expValue = StrictMath.exp(num2);
        System.out.println("The exp value of "+
                         num2+" = " + expValue);
 
        expValue = StrictMath.exp(num3);
        System.out.println("The exp value of "+
                         num3+" = " + expValue);    }
}


java
// Java program to illustrate the
// java.lang.StrictMath.exp()
import java.lang.*;
 
public class Geeks {
 
    public static void main(String[] args)
    {
 
        double num1 = -0.0, num2 = (1.0 / 0.0);
        double num3 = 14;
 
        // Returns Euler's number e raised to the given power
        double expValue = StrictMath.exp(num1);
        System.out.println("The exp value of "+
                         num1+" = " + expValue);
 
        expValue = StrictMath.exp(num2);
        System.out.println("The exp value of "+
                         num2+" = " + expValue);
 
        expValue = StrictMath.exp(num3);
        System.out.println("The exp value of "+
                         num3+" = " + expValue);
    }
}


输出:
The exp value of 0.0 = 1.0
The exp value of Infinity = Infinity
The exp value of 4.0 = 54.598150033144236

方案二:

Java

// Java program to illustrate the
// java.lang.StrictMath.exp()
import java.lang.*;
 
public class Geeks {
 
    public static void main(String[] args)
    {
 
        double num1 = -0.0, num2 = (1.0 / 0.0);
        double num3 = 14;
 
        // Returns Euler's number e raised to the given power
        double expValue = StrictMath.exp(num1);
        System.out.println("The exp value of "+
                         num1+" = " + expValue);
 
        expValue = StrictMath.exp(num2);
        System.out.println("The exp value of "+
                         num2+" = " + expValue);
 
        expValue = StrictMath.exp(num3);
        System.out.println("The exp value of "+
                         num3+" = " + expValue);
    }
}
输出:
The exp value of -0.0 = 1.0
The exp value of Infinity = Infinity
The exp value of 14.0 = 1202604.2841647768