Java.lang.StrictMath.expm1() Method
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Description
The java.lang.StrictMath.expm1() method returns ex -1. For values of x near 0, the exact sum of expm1(x) + 1 is much closer to the true result of ex than exp(x).It include some cases −
- If the argument is NaN, the result is NaN.
- If the argument is positive infinity, then the result is positive infinity.
- If the argument is negative infinity, then the result is -1.0.
- 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.StrictMath.expm1() method
public static double expm1(double x)
Parameters
x − This is the exponent to raise e to in the computation of ex -1.
Return Value
This method returns the value ex - 1.
Exception
NA
Example
The following example shows the usage of java.lang.StrictMath.expm1() method.
package com.tutorialspoint; import java.lang.*; public class StrictMathDemo { public static void main(String[] args) { double d1 = 0.0 , d2 = -(1.0/0.0), d3 = (1.0/0.0); /* returns (Euler's number e raised to the power of a double value)-1 i.e ex - 1 */ double eulerValue = StrictMath.expm1(d1); System.out.println("Value = " + eulerValue); eulerValue = StrictMath.expm1(d2); System.out.println("Value = " + eulerValue); eulerValue = StrictMath.expm1(d3); System.out.println("Value = " + eulerValue); } }
Let us compile and run the above program, this will produce the following result −
Value = 0.0 Value = -1.0 Value = Infinity
java_lang_strictmath.htm
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