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Java Math.sqrt function rounding down to zero [duplicate]

This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
How to divide 1 by 25(1/25) and get result .04 in java [duplicate]
(4 answers)
Closed 1 year ago.
I am trying to write a function to check a T score value and populate half of a 5x5 array.
public void calcTScores()
{
double temp = 0;
double tempSp = 0;
int n = 24;
this.tScores = new String [5][5];
for (int i = 0; i< this.Headers.length; i++){
for (int j = 0; j<this.Headers.length; j++)
{
if(i < j)
{
tempSp += (n-1)*this.SD[i] * this.SD[i] + (n-1)*this.SD[j] * this.SD[j];
tempSp = tempSp/(n+n-2);
tempSp = Math.sqrt(tempSp);
temp = tempSp * Math.sqrt(0.0833);
System.out.println(Math.sqrt(1/12));
temp = ((this.Mean[i] - this.Mean[j])/temp);
if(temp > 2.25 || temp< -2.25)
{
this.tScores[i][j] = "Y";
}
else
{
this.tScores[i][j] = "N";
}
temp = 0;
tempSp = 0;
}
}
}
}
Any idea why Math.sqrt(0.0833) and Math.sqrt(1/12) would evaluate to different values?
The T score when I add the 1/24 and 1/24 value and take the sqrt keeps evaluating to zero but when I plug in the actual decimal it gives me the answer I would expect
Any ideas why this is occuring?

1/12==0 as per integer division

There's nothing wrong with Math.sqrt. You're passing it the number zero.
Math.sqrt(1/12)
1/12 is a division operation on two integers (namely, 1 and 12), so it's going to produce an integer result. 12 goes into 1 zero times, so the integer result is zero. Thus, your expression is
Math.sqrt(0)
Consider
Math.sqrt(1.0/12.0)

Math.sqrt(1/12d)
Cast 1/12 to a double by casting one operand to a double.

Problem while truncating double to two decimal places Java [duplicate]

This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 2 years ago.
While I'm using truncating in this method, it is somewhere wrong as in picture
public static List<Double> interp(DoubleUnaryOperator f, double l, double u, int n) {
double d = (u - l)/n;
List<Double> result = new ArrayList<>(n);
double net = f.applyAsDouble(l);
result.add(net);
int i = 0;
while(i+1 < n){
result.add(Math.floor(f.applyAsDouble(l + d)*100)/100);
l += d;
i++;
}
return result;
}
Who know how to truncate correclty? P.S. its important for the number be in two decimal places.

Use this :
new DecimalFormat("#.##").format(dblVar);
df.setRoundingMode(RoundingMode.DOWN);
s = df.format(d);
and import the class java.text.DecimalFormat.

Java pow without rounding up the final value [duplicate]

This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Closed 3 years ago.
I am trying to write a formula for one of my functions where I need to raise my X to power of Y. The values I am working with are really small and will get rounded up as soon as I use pow function of Math, BigInteger and BigDecimal.
For example if I use the following values it will return 1000 whereas it should return 1006.931669!
T0 = 1000, TN = 1, k = 1, N = 1000
double X = (finalTemp/initialTemp);
double A = Math.pow(X, (k/n));
double tk = initialTemp * A;

They are being divided in integer arithmetics. So dividing integer a by integer b you get how many times b fits into a, If the types of the operands are double, then "real" division is performed.
double X = (finalTemp/(double)initialTemp);
double A = Math.pow(X, (k/(double)n));
double tk = initialTemp * A;

the output is correct according to calculator
public class Main
{
public static void main(String[] args) {
double finalTemp = 1.0;
double initialTemp = 1000.0;
double k = 1.0;
double n = 1000.0;
double X = (finalTemp/initialTemp);
double A = Math.pow(X, (k/n));
double tk = initialTemp * A;
System.out.println(tk);
}
}
output is 993.1160484209338

Best way to check if a value is an integer? [duplicate]

This question already has answers here:
How to test if a double is an integer
(18 answers)
Closed 8 years ago.
There are several ways to decide if a value is an integer or not.
All of the ways i know are using the divide operation.
Which method is the fastest and:
Is there a method to do this without doing floating point operations?
EDIT: For clarification, my program is only dealing with integer and double values. Here is some of my code:
for (int i = 6;; i++) {
for (int j = i - 1; j > 0; j--) {
double number1 = i;
double number2 = j;
double d = number1 / number2;
int help = (int)d;
if( (d - help) == 0.0){
System.out.println("Whole number found");
}
}
}

Note that int / int will also be an int due to the rules of integral division. You need to promote one of them to floating point or use the idiom 1.0 * a / b if you want to retain a remainder.
If you have a floating point number, f, then
java.lang.Math.floor(f) - f == 0 is probably the best way.
It avoids an intermediate cast to int which can overflow.

This will do what you want!
String str = "1";
try {
Integer.parseInt(str);
} catch (NumberFormatException e) {
System.err.println("Not a number");
}

For loop, dividing one [duplicate]

This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Closed 7 years ago.
I have tried this in Javascript and have gotten my answers, but the answer I need must be more exact. I am trying to divide 1 by every number between 2 and 1000, and simply print them.
public static void main(String [] args) {
for (int i=2;i<=1000;i++){
double g = (1/i);
System.out.println(g); // print "1/1,2,3,4.....1000"
}
}
I haven't done Java in a while, so I forget my correct variable names.

Since both 1 and i are integers, integer division is being used. Either 1 or i need to be double in the 1/i section of your code so that integer division is not used. You can do something like 1.0/i or 1/((double) i) to ensure that float division is used instead.

replace 1 by 1.0D that will result into double

try this
public static void main ( String args[] ) {
for (int i=2;i<=1000;i++){
double g = (1.0/i);
System.out.println("1/"+ig);
}
output:
0.5
0.3333333333333333
0.25
0.2
0.16666666666666666
0.14285714285714285
0.125
.
.
.
.

I would do something like this (note you can have as many digits of precision as you like) utilizing BigDecimal.
public static void main(String[] args) {
java.math.BigDecimal ONE = java.math.BigDecimal.ONE;
// 50 digits of precision.
java.math.MathContext mc = new java.math.MathContext(50);
for (int i = 2; i <= 1000; i++) {
java.math.BigDecimal divisor = new java.math.BigDecimal(i);
java.math.BigDecimal result = ONE.divide(divisor, mc);
result.round(mc);
System.out.printf("%d/%d = %s\n", 1, i,
result.toString());
}
}