math.Expm1 Function() in Golang With Examples
Go language provides inbuilt support for basic constants and mathematical functions to perform operations on the numbers with the help of the math package. You are allowed to find base-e exponential of the specified number minus 1, i.e., e**a -1 with the help of Expm1() function provided by the math package. So, you need to add a math package in your program with the help of the import keyword to access Expm1() function.
Syntax:
func Expm1(a float64) float64
- If you pass +Inf in this function, then this function will return +Inf.
- If you pass -Inf in this function, then this function will return -1.
- In this function, very large values overflow to -1 or +Inf.
- If you pass NaN in this function, then this function will return NaN.
Example 1:
// Golang program to illustrate the Exmp1 function package main import ( "fmt" "math" ) // Main function func main() { // Using Expm1() function res_1 := math.Expm1(4) res_2 := math.Expm1(-1) res_3 := math.Expm1(math.Inf(1)) res_4 := math.Expm1(math.NaN()) // Displaying the result fmt.Printf( "Result 1: %.1f" , res_1) fmt.Printf( "\nResult 2: %.1f" , res_2) fmt.Printf( "\nResult 3: %.1f" , res_3) fmt.Printf( "\nResult 4: %.1f" , res_4) } |
Output:
Result 1: 53.6 Result 2: -0.6 Result 3: +Inf Result 4: NaN
Example 2:
// Golang program to illustrate the Exmp1 function package main import ( "fmt" "math" ) // Main function func main() { // Using Expm1() function nvalue_1 := math.Expm1(7.3) nvalue_2 := math.Expm1(-3) res := nvalue_1 + nvalue_2 fmt.Printf( "%.1f + %.1f = %.1f" , nvalue_1, nvalue_2, res) } |
Output:
1479.3 + -1.0 = 1478.3
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