Sine function using Taylor expansion (C Programming) - c

Question

Here is the question..
This is what I've done so far,
#include <stdio.h>
#include <math.h>
long int factorial(int m)
{
if (m==0 || m==1) return (1);
else return (m*factorial(m-1));
}
double power(double x,int n)
{
double val=1;
int i;
for (i=1;i<=n;i++)
{
val*=x;
}
return val;
}
double sine(double x)
{
int n;
double val=0;
for (n=0;n<8;n++)
{
double p = power(-1,n);
double px = power(x,2*n+1);
long fac = factorial(2*n+1);
val += p * px / fac;
}
return val;
}
int main()
{
double x;
printf("Enter angles in degrees: ");
scanf("%lf",&x);
printf("\nValue of sine of %.2f is %.2lf\n",x,sine(x * M_PI / 180));
printf("\nValue of sine of %.2f from library function is %.2lf\n",x,sin(x * M_PI / 180));
return 0;
}
The problem is that the program works perfectly fine from 0 to 180 degrees, but beyond that it gives error.. Also when I increase the value of n in for (n=0;n<8;n++) beyond 8, i get significant error.. There is nothing wrong with the algorithm, I've tested it in my calculator, and the program seems to be fine as well.. I think the problem is due to the range of the data type.. what should i correct to get rid of this error?
Thanks..

Answer 1

You are correct that the error is due to the range of the data type. In sine(), you are calculating the factorial of 15, which is a huge number and does not fit in 32 bits (which is presumably what long int is implemented as on your system). To fix this, you could either:
Redefine factorial to return a double.
Rework your code to combine power and factorial into one loop, which alternately multiplies by x, and divides by i. This will be messier-looking but will avoid the possibility of overflowing a double (granted, I don't think that's a problem for your use case).

Answer 2

15! is indeed beyond range that a 32bit integer can hold. I'd use doubles throughout if I were you.
The taylor series for sin(x) converges more slowly for large values of x. For x outside -π,π. I'd add/subtract multiples of 2*π to get as small an x as possible.

Answer 3

You need range reduction. Note that a Taylor series is best near zero and that in the negative range it is the (negative) mirror image of it's positive range. So, in short: reduce the range (by the modula of 2 PI) to wrap it it the range where you have the highest accuracy. The range beyond 1/2 PI is getting less accurate, so you also want to use the formula: sin(1/2 PI + x) = sin(1/2 PI - x). For negative vales use the formula: sin(-x) = -sin(x). Now you only need to evaluate the interval 0 - 1/2 PI while spanning the whole range. Of course for VERY large values accuracy of the modula of 2 PI will suffer.

Answer 4

You may be having a problem with 15!.
I would print out the values for p, px, fac, and the value for the term for each iteration, and check them out.

Answer 5

You're only including 8 terms in an infinite series. If you think about it for a second in terms of a polynomial, you should see that you don't have a good enough fit for the entire curve.
The fact is that you only need to write the function for 0 <= x <=\pi; all other values will follow using these relationships:
sin(-x) = -sin(x)
and
sin(x+\pi;) = -sin(x)
and
sin(x+2n\pi) = sin(x)
I'd recommend that you normalize your input angle using these to make your function work for all angles as written.
There's a lot of inefficiency built into your code (e.g. you keep recalculating factorials that would easily fit in a table lookup; you use power() to oscillate between -1 and +1). But first make it work correctly, then make it faster.