Implementing Taylor Series for sine and cosine in C


I've been following the guide my prof gave us, but I just can't find where I went wrong. I've also been going through some other questions about implementing the Taylor Series in C.

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Just assume that RaiseTo(raise a number to the power of x) is there.

double factorial (int n)
{
    int fact = 1,
    flag;

   for (flag = 1; flag <= n; flag++)
   {
        fact *= flag;
   }

   return flag;
}

double sine (double rad)
{

int flag_2,
    plusOrMinus2 = 0; //1 for plus, 0 for minus 
double sin, 
    val2 = rad,
    radRaisedToX2,
    terms;

terms = NUMBER_OF_TERMS; //10 terms

    for (flag_2 = 1; flag_2 <= 2 * terms; flag_2 += 2)
    {
        radRaisedToX2 = RaiseTo(rad, flag_2);   

        if (plusOrMinus2 == 0)
        {
            val2 -=  radRaisedToX2/factorial(flag_2);
            plusOrMinus2++; //Add the next number
        }

        else
        {
            val2 += radRaisedToX2/factorial(flag_2);
            plusOrMinus2--; //Subtract the next number
        }
    }

    sin = val2;
    return sin;
 }

int main()
{
    int degree;
    scanf("%d", &degree);
    double rad, cosx, sinx;
    rad = degree * PI / 180.00;
    //cosx = cosine (rad);
    sinx = sine (rad);
    printf("%lf \n%lf", rad, sinx);
}

So during the loop, I get the rad^x, divide it by the factorial of the odd number series starting from 1, then add or subtract it depending on what's needed, but when I run the program, I get outputs way above one, and we all know that the limits of sin(x) are 1 and -1, I'd really like to know where I went wrong so I could improve, sorry if it's a pretty bad question.


Answers:


Anything over 12! is larger than can fit into a 32-bit int, so such values will overflow and therefore won't return what you expect.

Instead of computing the full factorial each time, take a look at each term in the sequence relative to the previous one. For any given term, the next one is -((x*x)/(flag_2*(flag_2-1)) times the previous one. So start with a term of x, then multiply by that factor for each successive term.

There's also a trick to calculating the result to the precision of a double without knowing how many terms you need. I'll leave that as an exercise to the reader.