Finding Log1p() of the Given Number in Golang
Last Updated :
01 Apr, 2020
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 the natural logarithm of 1 plus its specified argument with the help of the Log1p() 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 the Log1p() function.
Syntax:
func Log1p(a float64) float64
- It is much more accurate than log(a+1) when the value of a is near zero.
- If you pass +Inf in this function, then this function will return +Inf.
- If you pass +0 or -0 in this function, then this function will return +0 or -0.
- If you pass -1 in this function, then this function will return -Inf.
- If the value of a < -1, then this function will return NaN.
- If you pass NaN in this function, then this function will return NaN.
Example 1:
package main
import (
"fmt"
"math"
)
func main() {
res_1 := math.Log1p(0)
res_2 := math.Log1p(1)
res_3 := math.Log1p(math.Inf(1))
res_4 := math.Log1p(math.NaN())
res_5 := math.Log1p(36)
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)
fmt.Printf("\nResult 5: %.1f", res_5)
}
|
Output:
Result 1: 0.0
Result 2: 0.7
Result 3: +Inf
Result 4: NaN
Result 5: 3.6
Example 2:
package main
import (
"fmt"
"math"
)
func main() {
nvalue_1 := math.Log1p(100)
nvalue_2 := math.Log1p(26)
res := nvalue_1 + nvalue_2
fmt.Printf("%.5f + %.5f = %.5f",
nvalue_1, nvalue_2, res)
}
|
Output:
4.61512 + 3.29584 = 7.91096
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