Java Math.sin() doesn't work as expected

Question

Okay, my assignment is to produce an array values a sine wave with a custom frequency. This is my sine function:

Math.sin(frequency*(j/samplesPerSecond)*Math.PI*2)

j is the array index and samplesPerSecond equals 44100.

Now for an array of 100,000 values it returns 0.0 for the first 33,000, 1.571509...E-14 for the next 33,000 and 3.143018...E-14 for the rest.

I've tried System.out.println(Math.sin(Math.PI)); and it doesn't return 0 but it returns 1.224646...E-16

This is my first java assignment so I might be overlooking something really obvious here, but I'm really confused.

Note that `1.2246467991473532E-16` is extremely close to `0`. `Math.sin(Math.PI))` is just limited by the precision of `double`. — Sotirios Delimanolis, Dec 20 '13 at 16:12

score 4 · Accepted Answer · answered Dec 20 '13 at 16:06

4

Try changing to this:

Math.sin(1.0 * frequency*(1.0 * j / samplesPerSecond) * Math.PI * 2)

See if it works better.

I suspect you have issue with integer division.

In Java 2/5 = 0, you know, while 1.0 * 2 / 5
is what you would expect usually ( 0.4 ).

answered Dec 20 '13 at 16:06

peter.petrov

38,363
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Yeah. Thanks a bunch. We used to use C, so this is something you want to get used to in Java. – Marv Dec 20 '13 at 16:09
2

@Marv Isn't it the same in C? ;) http://stackoverflow.com/questions/3602827/what-is-the-behavior-of-integer-division-in-c – peter.petrov Dec 20 '13 at 16:10

score 2 · Answer 2 · answered Dec 20 '13 at 16:11

2

Most probably you are dealing with integer division. So if the numerator is less than denominator it will round to zero. Try converting/casting j or samplesPerSecond to float or double.

answered Dec 20 '13 at 16:11

AbhinavRanjan

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score 1 · Answer 3 · answered Dec 20 '13 at 16:12

j is the array index and samplesPerSecond equals 44100.

As an array index, I assume j is of type int. This is your primary issue, as you're performing integer division, which truncates the result to a whole number. Either cast j to a double, or multiply it by a double before dividing:

((double)j / samplesPerSecond)
(1.0 * j / samplesPerSecond)

score 0 · Answer 4 · answered Dec 15 '18 at 16:43

I've tried System.out.println(Math.sin(Math.PI)); and it doesn't return 0 but it returns 1.224646...E-16

It has been already stated that this has to do with the finite precision of double. I use something like this in those cases:

 double result = Math.sin(2*Math.PI);
 for(double d=-1;d<=1;d=d+0.05){
        if(Math.abs(i-result) <1E-10){
            result = d;
        }
    }
 System.out.println(result); //prints 0

Java Math.sin() doesn't work as expected

4 Answers4