In this article, as part of my output practice, I will note the method of java.lang.Math, which is an API used in mathematical calculations. If there is something wrong with the explanation, it would be helpful if you could point it out. Reference: Official java.lang.Math
abs Find the absolute value of the integer taken as an argument.
int num1 = Math.abs(-1); //num1 = 1
max, min Returns the larger (smaller) of the two arguments.
int num2 = Math.max(1,2); //num2 = 2
int num3 = Math.min(1,2); //num3 = 1
random Returns a positively signed double value greater than or equal to 0.0 and less than 1.0. If you want to make it an integer value, multiply it by an arbitrary multiple before casting.
//0~Take a random value in the range of 1000
int num4 = (int)(Math.random()*1000);
log, log10
log
returns the natural logarithm value and log10
returns the common logarithm value.
Math.E
: The base $ e $ of the natural logarithm. It is defined as a constant. Math.PI
(Pi $ π $) is also defined. double num5 = Math.log(Math.E); //num5 = 1.0
double num6 = Math.log10(100); //num6 = 2.0
sin, cos, tan Trigonometric method. Enter radians as arguments. ~~ If you want to input the argument of the radian method, you can convert it by multiplying by $ \ frac {π} {180} $. ~~
Math.toRadians (double angdeg)
method to convert it to radians. double num7 = Math.sin(Math.PI/4); //1/Approximate value of √2
double num8 = Math.cos(Math.PI/4); //1/Approximate value of √2
double num9 = Math.tan(Math.toRadians(45));//1(0.9999999999999999)
pow Returns a double value that is the power of the first argument to the second argument. Cast if treated as an integer value.
double num10 = Math.pow(10, 4); //num10 = 10000.0
exp, expm1 The former returns the power value of the base $ e $ of the natural logarithm, and the latter returns the number obtained by subtracting 1 from the former. Let's use it in physics and statistics ...?
double num11 = Math.exp(2); // e^2
double num12 = Math.expm1(2);// e^2 -1
sqrt, cbrt Returns the square root and cube root.
double num13 = Math.sqrt(9); //num13 = 3.0
double num14 = Math.cbrt(64); //num14 = 4.0
... I will update it when I find something that I think I will use again.
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