strictmath

StrictMath類log1p()方法 (StrictMath Class log1p() method)

• log1p() method is available in java.lang package.

  log1p()方法在java.lang包中可用。

• log1p() method is used to return (the logarithm of the sum of the given argument and 1 like log(1+d) in the method.

  log1p()方法用于返回(給定參數和1之和的對數，如log(1 + d)在該方法中。

• log1p() method is a static method so it is accessible with the class name and if we try to access the method with the class object then also we will not get an error.

  log1p()方法是靜態方法，因此可以使用類名進行訪問，如果嘗試使用類對象訪問該方法，則也不會收到錯誤。

• We need to remember one thing if we pass smaller values for the given argument so the final calculated result of log1p(d) is nearer to the exact result of ln(1+d) than the double floating-point calculation of log(1.0+d).

  如果為給定參數傳遞較小的值，則需要記住一件事，即與log(1.0+)的雙浮點計算相比， log1p(d)的最終計算結果更接近ln(1 + d)的精確結果。 d)。

• log1p() method does not throw any exception.

  log1p()方法不會引發任何異常。

Syntax:

句法：

    public static double log1p(double d);

Parameter(s):

參數：

• double d – represents the double type argument.

  double d –表示double類型的參數。

Return value:

返回值：

The return type of this method is double – it returns the logarithm (1+d) of the given argument.

此方法的返回類型為double-返回給定參數的對數(1 + d)。

Note:

注意：

• If we pass NaN, method returns NaN.

  如果傳遞NaN，則方法返回NaN。

• If we a value which is less than -1, method returns NaN.

  如果我們的值小于-1，則方法返回NaN。

• If we pass a positive infinity, method returns the same (i.e. positive infinity).

  如果我們傳遞一個正無窮大，則方法將返回相同的值(即正無窮大)。

• If we pass a negative infinity, method returns NaN.

  如果我們傳遞一個負無窮大，則方法返回NaN。

• If we pass 0 (negative or positive), method returns the same with the same sign.

  如果傳遞0(負數或正數)，則方法將返回相同的符號。

Example:

例：

// Java program to demonstrate the example
// of log1p(double d) method of StrictMath class.
public class Log1p {
public static void main(String[] args) {
// variable declarations
double d1 = 7.0 / 0.0;
double d2 = -7.0 / 0.0;
double d3 = 0.0;
double d4 = -0.0;
double d5 = 6054.2;
// Display previous value of d1,d2,d3,d4 and d5
System.out.println("d1: " + d1);
System.out.println("d2: " + d2);
System.out.println("d3: " + d3);
System.out.println("d4: " + d4);
System.out.println("d5: " + d5);
// Here , we will get (Infinity) because we are 
// passing parameter whose value is (Infinity)
System.out.println("StrictMath.log1p(d1): " + StrictMath.log1p(d1));
// Here , we will get (NaN) because we are 
// passing parameter whose value is (-Infinity)
System.out.println("StrictMath.log1p(d2): " + StrictMath.log1p(d2));
// Here , we will get (0.0) because we are 
// passing parameter whose value is (0.0)
System.out.println("StrictMath.log1p(d3): " + StrictMath.log1p(d3));
// Here , we will get (-0.0) because we are 
// passing parameter whose value is (-0.0)
System.out.println("StrictMath.log1p(d4): " + StrictMath.log1p(d4));
// Here , we will get (log [1 + d5]) and we are 
// passing parameter whose value is (6054.2)
System.out.println("StrictMath.log1p(d5): " + StrictMath.log1p(d5));
}
}

Output

輸出量

d1: Infinity
d2: -Infinity
d3: 0.0
d4: -0.0
d5: 6054.2
StrictMath.log1p(d1): Infinity
StrictMath.log1p(d2): NaN
StrictMath.log1p(d3): 0.0
StrictMath.log1p(d4): -0.0
StrictMath.log1p(d5): 8.708672685994957

翻譯自: https://www.includehelp.com/java/strictmath-log1p-method-with-example.aspx

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