exp() returns Euler's number e raised to the power of a double value. If the argument is NaN, the result is NaN. If the argument is positive infinity, after that the result is constructive perpetuity. If the argument is downbeat infinity, then the product is positive zero. The process returns the base of the ordinary logarithms to the power of the argument. The scheme returns the support of the natural logarithms to the rule of the argument.
double exp(double d)
Here is the detail of parameters −
public class Test { public static void main(String args[]) { double x = 11.635; double y = 2.76; System.out.printf("The value of e is %.4f%n", Math.E); System.out.printf("exp(%.3f) is %.3f%n", x, Math.exp(x)); } }
This will produce the following result −
The value of e is 2.7183 exp(11.635) is 112983.831
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