用C编程语言创建一个Sin公式

Question

Hello everyone I'm trying to implement a programm like the fomular sin The program will compile but when running it i am not getting the correct values back from my inputs. I'm still getting a negativ value. can someone help me please ? I had a look at the other posts but that didn't help me :(. my code is:

#include <stdio.h>
#include <stdlib.h>

int fac (int a) { // fac. => factorial and i is for the loop

    int i,fac;
        fac=1;

    for (i=1; i<=a; i++){
        fac=fac*i;
    }
    return fac;
}
int power_func(int x,int y) // x is exponent and y is the number that would be multiplied by itself.
     {
         int i;//i is for the loop
         int ret = 1;
         for(i=1;i<=x;i++)
         {
             ret *= y;
         }
         return ret;
     }
int main()
{
    int num,denom,i;//num. is numerator and denom. is denominator
    int sin,x,result=0;
    printf("Enter the number of x \n");
    scanf("%d",&x);
    for(i=0;i<x;i++)
    {

    num= power_func(2*i+1,x);
    denom=fac((2*i+1));
    sin=power_func(i,-1)*num/denom;
    result =result+sin;
    printf("%d \n",result);
    }

    return 0;
}

Answer 1

You have various misconceptions about your code. First, let's look at the formula you have provided:

sin(x) = sum((−1)^k * x^(2*k + 1) / (2*k + 1)!   for x ∈ R;   k = 0, ..., infinity

The sine function takes a real and returns a real. Therefore, you should use a floating-point type for x and sin(x). Use a double . Let's also write a function that emulates sin from <math.h> :

double my_sin(double x);

The above series is accurate when there are infinitely many terms. We can't calculate that many, of course, and it would be a waste of time, too, because the terms are getting ever smaller until they can no longer be represented by a double . So let's chose a maximum number of terms, say

enum {
    nTerms = 8
};

Factorials grow big fast. A regular 32-bit int can hold 12! = 479,001,600. A 64-bit int can hold 20! = 2,432,902,008,176,640,000. Since we are going to use these factorials in a double calculation, we can just as well use double here. That will even allow us to represent 22! = 1,124,000,727,777,607,680,000 accurately.

Your power function should also have a double base. The exponent is integer. (But please use the more natural order power(base, exp) .

Finally, (−1)^k is just an alternating sign. It is positive when k is even and odd otherwise.

Putting all this together:

double fact(int n)
{
    double result = 1.0;

    while (n > 0) {
        result *= n;
        n--;
    }

    return result;
}

double power(double a, int n)
{
    double result = 1.0;

    while (n > 0) {
        result *= a;
        n--;
    }

    return result;
}

enum {
    nTerms = 8
};

double my_sin(double x)
{
    double result = 0.0;
    double sign = 1.0;

    for(int k = 0; k < nTerms; k++)
    {
        double num = power(x, 2*k + 1);
        double denom = fact(2*k + 1);
        double term = sign * num / denom;

        result = result + term;
        sign = -sign;
    }

    return result;
}

If we write a driver program to print out some test values and compare them with the standard math library's implementation of sin :

int main(void)
{
    for (int i = 0; i < 15; i++) {
        double x = 0.1 * i;
        double m = my_sin(x);       // series approximation
        double s = sin(x);          // <math.h> implementation

        printf("%16g%16g%16g%16g\n", x, m, s, m - s);
    }

    return 0;
}

we can see that we're not doing so badly:

           x       my_sin(x)          sin(x)      difference
    --------    ------------    ------------    ------------
           0               0               0               0
         0.1       0.0998334       0.0998334     1.38778e-17
         0.2        0.198669        0.198669     2.77556e-17
         0.3         0.29552         0.29552               0
         0.4        0.389418        0.389418    -5.55112e-17
         0.5        0.479426        0.479426               0
         0.6        0.564642        0.564642               0
         0.7        0.644218        0.644218               0
         0.8        0.717356        0.717356               0
         0.9        0.783327        0.783327    -4.44089e-16
           1        0.841471        0.841471    -2.77556e-15
         1.1        0.891207        0.891207    -1.43219e-14
         1.2        0.932039        0.932039    -6.20615e-14
         1.3        0.963558        0.963558    -2.42029e-13
         1.4         0.98545         0.98545    -8.52318e-13

(But it gets worse the farther we go from zero. Try other values for nTerms .)

---

I've said in a comment above that you don't need to calculate factorials and powers and that's true. If you look at the terms of the series, you will see that:

s[n] = -1 * s[n - 1] * x^2 / (2*n * (2*n +1))

s[0] = x
s[1] = x^3 / (1 * 2 * 3)         =                 x * x^2 / (2 * 3)
s[2] = x^5 / (1 * 2 * 3 * 4 * 5) = x^3 / (1 * 2 * 3) * x^2 / (4 * 5)
s[3] = ...

Here's a function that implements that.It calculates terms until adding them to the sum doesn't change it, because they are too small:

double sin_r(double x)
{
    double sum = x;
    double a = x;
    int n;

    for (n = 1; ; n++) {
        double was = sum;

        a = -a * x*x / (2*n) / (2*n + 1);
        sum += a;

        if (was == sum) break;
    }

    return sum;
}

The addition still loses some precision by summing the first terms first, but it has the benefit that it doesn't have to calculate factorials and powers. You don't even need <math.h> .