Java - Math log1p(double) method

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Description

The Java Math log1p(double x) returns the natural logarithm of the sum of the argument and 1. Note that for small values x, the result of log1p(x) is much closer to the true result of ln(1 + x) than the floating-point evaluation of log(1.0+x).Special cases −

• If the argument is NaN or less than -1, then the result is NaN.

• If the argument is positive infinity, then the result is positive infinity.

• If the argument is negative one, then the result is negative infinity.

• If the argument is zero, then the result is a zero with the same sign as the argument.

Declaration

Following is the declaration for java.lang.Math.log1p() method

public static double log1p(double x)

Parameters

x − a value

Return Value

This method returns the value ln(x + 1), the natural log of x + 1

Exception

NA

Example 1

The following example shows the usage of Math log1p() method.

package com.tutorialspoint;
public class MathDemo {
   public static void main(String[] args) {

      // get a double number
      double x = 10.7;

      // print the log1p of the number
      System.out.println("Math.log1p(" + x + ")=" + Math.log1p(x));
   }
}

Output

Let us compile and run the above program, this will produce the following result −

Math.log1p(10.7)=2.4595888418037104

Example 2

The following example shows the usage of Math log1p() method of zero value.

package com.tutorialspoint;
public class MathDemo {
   public static void main(String[] args) {

      // get a double number
      double x = 0.0;

      // print the log1p of the number
      System.out.println("Math.log1p(" + x + ")=" + Math.log1p(x));
   }
}

Output

Let us compile and run the above program, this will produce the following result −

Math.log1p(0.0)=0.0

Example 3

The following example shows the usage of Math log1p() method of a negative number.

package com.tutorialspoint;
public class MathDemo {
   public static void main(String[] args) {

      // get a double number
      double x = -10.7;

      // print the log1p of the number
      System.out.println("Math.log1p(" + x + ")=" + Math.log1p(x));
   }
}

Output

Let us compile and run the above program, this will produce the following result −

Math.log1p(-10.7)=NaN