Julien De Wisscher Julien De Wisscher - 5 months ago 35
Java Question

Sine approximation error in Java

I'm a bit annoyed with a method I wrote to approximate sine function in Java. Here it is, it's based on Taylor's series.

static double PI = 3.14159265358979323846;
static double eps = 0.0000000000000000001;

static void sin(double x) {
x = x % (2 * PI);
double term = 1.0;
double res = 0.0;

for (int i = 1; term > eps; i++) {
term = term * (x / i);
if (i % 4 == 1) res += term;
if (i % 4 == 3) res -= term;
}
System.out.println(sum);
}


For little values, I got very good approximation of sine, but for large values (e.g pow(10,22)), results seems very very wrong.

Here are the results :

sin(pow(10,22)) // 0.8740280612007599
Math.sin(pow(10,22)) // -0.8522008497671888


Does someone have an idea ? Thank you !

Best regards,

Answer

Be reassured that the Java sin function will be off too.

You problem is that the Taylor expansion for sin has a small radius of convergence and convergence is slow even if you're within that radius.

There are floating point considerations too: a floating point double gives you about 15 significant figures of accuracy.

So for large arguments for sin, the accuracy will deteriorate significantly especially given that sin is a periodic function:

sin(x + 2 * pi * n) = sin(x) for any integer n.