C 程序使用 taylor put sin(1) 得到 0.841

Question

Im stuck maybe someone here can help me plzzz:) I did wrong program that calc from radian to degri I need to build a program like you calc Sin(x) in radian mode in calculator I put radian x like 1 in radian and it need to give me in radian mode the calc of sin(1) Like if i put sin(1) it need to give me 0.8414 and i cant use sin() and all this only standart and need to use taylor to calc. help plz:):') my wrong code:

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
double my_sin(double x);
/*my_sin func calc sin(x) from the main*/
double my_sin(double x){
double min_num = 0.000001;
double radian = x*(3.14/180);
double sol = radian;
int i;

int sign = 1;
int n;
int aseret = 1;

//Calc the solution if i <= 0.000001 it stop calc.
for(i=3;i <= min_num; i = i+2){
    n = n * radian * radian;        
    sign = -(sign);                 //change the sign every round.
    aseret = aseret * i * (i-1);    //!3 = 3*2*1.
    sol = sol + (pow(x,i)/aseret) *sign;    // calc the solution.

}
return sol;
}
/*main func*/
int main()
{
double x = 0;
double result;
printf("please enter the number to check the SIN of it:");
scanf("%lf",&x);
result = my_sin(x);
printf("SIN(%f) = %f\n",x,result);
return 0;
}

Answer 1

Your main problem is that you are comparing the index i with the minimum value of the term. One consequence is that the loop stops immediately. It is the term x^i/i(i-1) which must be compared with the minimum value. As the series is alternating, the error at the end is lower than this minimum value.

It is also useles to perform the conversion to radian, as it seems you are entering a radian value.

Moreover, it is useless and inefficient to use pow() function to calculate x^i . Better to alculate this term iteratively.

Output:

please enter the number to check the SIN of it: 1
SIN(1.000000) = 0.841471
error = -1.59828e-010

Code:

#include <stdio.h>
#include <math.h>

double my_sin(double x){
    double min_num = 0.000001;

    int sign = 1;
    double term = x;
    double sol = term;
    double x2 = x*x;
    int i = 1;

    //Calc the solution if term <= 0.000001 it stop calc.
    do {
        i += 2;
        term *= x2 / (i * (i-1));        
        sign = -sign;                 //change the sign every round.
        sol += term * sign;    
    } while (term > min_num);
    return sol;
}

int main() {
    double x;
    printf("please enter the number to check the SIN of it: ");
    scanf("%lf", &x);
    double result = my_sin(x);
    printf("SIN(%f) = %f\n", x, result);
    double result_exact = sin(x);
    double delta = result - result_exact;
    printf ("error = %g\n", delta);
    return 0;
}

Answer 2

• You convert x to radians when it is already in radians.
• Your loop condition is wrong, the loop terminates immediately and compares an integer with a double.
• The variable n is uninitialised and the result of the expression it is used in is unused in any case.
• Unnecessary headers included.

Change as per <<<<<< comments:

#include <stdio.h>
// #include <string.h>      // <<<<<<<<<<<<<< REDUNDANT
// #include <ctype.h>       // <<<<<<<<<<<<<< REDUNDANT
#include <math.h>

#define LIMIT 10        // <<<<<<<<<<<<<< ADD

double my_sin( double x );


/*my_sin func calc sin(x) from the main*/
double my_sin( double x )
{
    // double min_num = 0.000001;       // <<<<<<<<<<<<<< REMOVE   
    // double radian = x*(3.14/180);    // <<<<<<<<<<<<<< REMOVE
    double sol = x ;                    // <<<<<<<<<<<<<< CHANGE
    // int i;                           // <<<<<<<<<<<<<< REMOVE

    int sign = 1;
    // int n = 0 ;                      // <<<<<<<<<<<<<< REMOVE - NOT USED
    int aseret = 1;

    //Calc the solution if i <= LIMIT it stop calc.    // <<<<<<<<<<<<<< CHANGE
    for( int i = 3; i <= LIMIT; i += 2 )               // <<<<<<<<<<<<<< CHANGE
    {
        // n = n * x * x;                              // <<<<<<<<<<<<<< REMOVE - RESULT NOT USED
        sign = -(sign);                 //change the sign every round.
        aseret = aseret * i * (i - 1);    //!3 = 3*2*1.
        sol = sol + (pow( x, i ) / aseret) * sign;    // calc the solution.

    }
    return sol;
}