Java Math.expm1(double x) method returns ex – 1. Here, e is a Euler’s number, whose approximate value is 2.718281828459045.
public class JavaExample
{
public static void main(String[] args)
{
double x = 1;
// returns e (2.718281828459045) to power of 1 minus 1
System.out.println(Math.expm1(x));
}
}
Output:
1.718281828459045
Note: An interesting point to note is that for the values of x near 0, the sum of expm1(x) + 1 gives more accurate value of ex than exp(x) method.
Syntax of Math.expm1() method
Math.expm1(0); //returns 0.0
expm1() Description
public static double expm1(double x): Returns ex – 1.
expm1() Parameters
- x: A double value that is the exponent in computation of ex – 1.
expm1() Return Value
- Returns
eraised to the power of given argumentx, minus 1. - If the argument is NaN (Not a number), then it returns NaN.
- If the argument is positive infinity, then it returns positive infinity.
- If the argument is negative infinity, then it returns -1.0.
- If the argument is zero, then it returns zero with the same sign.
Example 1
public class JavaExample
{
public static void main(String[] args)
{
double x = 1, x2 = 10;
System.out.println(Math.expm1(x));
System.out.println(Math.expm1(x2));
}
}
Output:
Example 2
public class JavaExample
{
public static void main(String[] args)
{
double x = 2, x2 = -2, x3 = 0;
System.out.println(Math.expm1(x));
System.out.println(Math.expm1(x2));
System.out.println(Math.expm1(x3));
}
}
Output:
Example 3
public class JavaExample
{
public static void main(String[] args)
{
double x = Double.NaN;
double x2 = Double.MAX_VALUE;
double x3 = Double.MIN_VALUE;
System.out.println(Math.expm1(x));
System.out.println(Math.expm1(x2));
System.out.println(Math.expm1(x3));
}
}
Output:
