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Java Math Class static double log1p(double d) with example

Java Math Class static double log1p(double d) method: Here, we are going to learn about the static double log1p(double d) method of Math Class with its syntax and example.
Submitted by Preeti Jain, on September 04, 2019

Math Class static double log1p(double d)

• This method is available in java.lang package.
• This method is used to return (the logarithm of the sum of the given argument and 1 like log(1+d)) in the method.
• This is a static method so it is accessible with the class name too.
• We need to remember one thing if we pass smaller values for the given argument so the final calculated result of log1p(d) is nearer to the exact result of ln(1+d) than the double floating-point calculation of log(1.0+d)
• The return type of this method is double, it returns the logarithm (1+d) of the given argument.
• In this method, we pass only one parameter as an argument of double type.
• This method does not throw any exception.

Syntax:

    public static double log1p(double d){
    }

Parameter(s): It accepts a double value, whose logarithm of the sum of the given argument and 1 like log(1+d)

Return value:

The return type of this method is double, it returns the logarithm (1+d) of the given argument.

Note:

• If we pass "NaN", it returns "NaN".
• If we pass value less than -1, it returns "NaN".
• If we pass a positive infinity, it returns the same value (positive infinity).
• If we pass a negative infinity, it returns the "NaN".
• If we pass zero (-0 or 0), it returns the same value.

Java program to demonstrate example of log1p(double d) method

// Java program to demonstrate the example of 
// log1p(double d) method of Math Class.

public class Log1pMethod {
    public static void main(String[] args) {
        // Here we are declaring few variables
        double d1 = 7.0 / 0.0;
        double d2 = -7.0 / 0.0;
        double d3 = 0.0;
        double d4 = -0.0;
        double d5 = 6054.2;

        // displaying the values
        System.out.println("d1 :" + d1);
        System.out.println("d2 :" + d2);
        System.out.println("d3 :" + d3);
        System.out.println("d4 :" + d4);
        System.out.println("d5 :" + d5);

        // Here , we will get (Infinity) because we are passing 
        // parameter whose value is (Infinity)
        System.out.println("Math.log1p(d1): " + Math.log1p(d1));

        // Here , we will get (NaN) because we are passing 
        // parameter whose value is (-Infinity)
        System.out.println("Math.log1p(d2): " + Math.log1p(d2));

        // Here , we will get (0.0) because we are passing 
        // parameter whose value is (0.0)
        System.out.println("Math.log1p(d3):" + Math.log1p(d3));

        // Here , we will get (-0.0) because we are passing 
        // parameter whose value is (-0.0)
        System.out.println("Math.log1p(d4):" + Math.log1p(d4));

        // Here , we will get (log [1 + d5]) and we are 
        // passing parameter whose value is (6054.2)
        System.out.println("Math.log1p(d5):" + Math.log1p(d5));
    }
}

Output

E:\Programs>javac Log1pMethod.java

E:\Programs>java Log1pMethod
d1 :Infinity
d2 :-Infinity
d3 :0.0
d4 :-0.0
d5 :6054.2
Math.log1p(d1): Infinity
Math.log1p(d2): NaN
Math.log1p(d3):0.0
Math.log1p(d4):-0.0
Math.log1p(d5):8.708672685994957