CODEWITHSUNDEEP
ctrigonometrytaylor-series
ProgrammingEnthusiast
ProgrammingEnthusiast

Reputation: 480

This code of taylor series doesn't work for n= 1 or anything other than 0. Why?

First of all, let me tell you, I am learning programming.

Today, I tried to find the approximate value of cosine by using the taylor series. When I put n=0, my code gives me correct result of 1. But when I put n=1 or anything else, my code does not give correct result.

I am unable to understand where the problem is. Can anyone help?

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(int argc, char *argv[])
{
    float xnot = atof(argv[1]);
    float n = atof(argv[2]);
    float cosine = cos(xnot*(3.14159265/180));
    float result;
    printf("%.4f\n", cosine);
    float min;


    float d, c, b;
    c = 1;
    d = 2 * n; 
    for(b = 1; b <= d; b++){
        c = c * b; /*value of the factorial is in c*/
    }
    c = c;
    float power;
    power = pow((-1), n);
    xnot = pow(xnot, 2*n);


    for(min = 0; min <= n; min++)
    {

        result += ((power * xnot) / c);
    }
    printf("%.4f", result);
}

Upvotes: 0

Views: 139

Answers (3)

4386427
4386427

Reputation: 44246

You need to redo all calculations inside the for-loop. Keeping as much as your original code as possible, it could be something like:

int n = atoi(argv[2]);  // n is an integer
...
...
float result = 1;  // The first term (n=0) gives 1

for(int i = 1; i <= n; i++)   // Start the loop from 1
{
    float d, c, b;
    c = 1;
    d = 2 * i;     // Use i instead of n
    for(b = 1; b <= d; b++){
        c = c * b; /*value of the factorial is in c*/
    }
    float power;
    power = pow((-1), i);    // Use i instead of n
    xnot = pow(xnot, 2*i);   // Use i instead of n

    result += ((power * xnot) / c);
}

The code can be optimized - both for performance and precision - but as already stated I tried to keep it close to your original code.

Upvotes: 1

David C. Rankin
David C. Rankin

Reputation: 84521

When computing either sine or cosine with Taylor series, you also need to take the angular quadrants into consideration to minimize error growth. The following is a short example:

#define TSLIM 20    /* Series Limit (no. of terms) */
...
/** cos with taylor series expansion to n = TSLIM
 *  (no function reliance, quadrants handled)
 */
double cosenfq (const double deg)
{
    double fp = deg - (int64_t)deg, /* save fractional part of deg */
        qdeg = (int64_t)deg % 360,  /* get equivalent 0-359 deg angle */
        rad, cose_deg = 1.0;        /* radians, cose_deg */
    int pos_quad = 1,               /* positive quadrant flag 1,4  */
        sign = -1;                  /* taylor series term sign */

    qdeg += fp;                     /* add fractional part back to angle */

    /* get equivalent 0-90 degree angle, set pos_quad flag */
    if (90 < qdeg && qdeg <= 180) {         /* in 2nd quadrant */
        qdeg = 180 - qdeg;
        pos_quad = 0;
    }
    else if (180 < qdeg && qdeg <= 270) {   /* in 3rd quadrant */
        qdeg = qdeg - 180;
        pos_quad = 0;
    }
    else if (270 < qdeg && qdeg <= 360)     /* in 4th quadrant */
        qdeg = 360 - qdeg;

    rad = qdeg * M_PI / 180.0;      /* convert to radians */

    /* compute Taylor-Series expansion for sine for TSLIM / 2 terms */
    for (int n = 2; n < TSLIM; n += 2, sign *= -1) {
        double p = rad;
        uint64_t f = n;

        for (int i = 1; i < n; i++)     /* pow */
            p *= rad;

        for (int i = 1; i < n; i++)     /* nfact */
            f *= i;

        cose_deg += sign * p / f;       /* Taylor-series term */
    }

    return pos_quad ? cose_deg : -cose_deg;
}

With a 20 term limit, max error is approximately 1.2E-15 (compared to math.h cos())

Upvotes: 0

Bob Jarvis - Слава Україні
Bob Jarvis - Слава Україні

Reputation: 50007

When implementing the Taylor series you have to recompute the value of the terms for each value of 'n'. Here it looks like you've computed the value of -1^n (as xnot) for the maximum value of n and then you're just multiplying by that value for each iteration. That's wrong. Same for the values of x^2n / (2n)! - you have to recompute this for each value of n as you increment it, then sum up the values.

Best of luck.

Upvotes: 2

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