编写 C function 来评估系列 // sin(x) = x-(x3 /3!)+(x5 /5!)-(x7 /7!)+... // 最多 10 个术语

Question

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

int series(float,float);

int main()

{

    float x,n,series_value;
    printf("Enter the value of x: ");
    scanf("%f",&x);
    printf("\nEnter the value of n: ");
    scanf("%f",&n);
    series_value=series(x,n);
    printf("\nValue of series sin (%.2f) is: %f\n",x,series_value);
    return 0;
}

int series(float x,float n)

{

    int i,sum=0,sign=-1;
    int j,fact=1,p=1;
    for (i=1; i<=(2*n)-1; i+=2)
    {
        for (j=1; j<=i; j++)
        {
        p=p*x;
        fact=fact*j;
        }
        sign=-1*sign;
        sum=sum + sign*p/fact;       
    }
    return (sum);
}

Output: Enter the value of x: 5

Enter the value of n: 10 (lldb) and this message Thread 1: EXC_ARITHMETIC (code=EXC_I386_DIV, subcode=0x0) :[Thread 1 Queue. com.apple.main-thread (serial) ] 1

Why is this message coming? and what is wrong in the program as answer is not coming right

Answer 1

There is a few problems with your code. As @PaulHankin said, when fact overflows and becoms zero, you will have a division by zero, and "weird things" happen.

Your factorial and power calculation is also wrong. You are recalculating it in each iteration of the outer loop without reseting fact and p first:

fact = 1; // You need to reset fact and p to its start value here
p = 1;
for (j=1; j<=i; j++)
  {
    p=p*x;
    fact=fact*j;
  }

Your third problem is that for your function calculate the correct value for sin , which is not an integer value, you need to use float , or even better double , when calculating sum . So sum must be declared float , and the division p/fact must use float division. By also declaring p and fact as float, you will solve both the overflow issue, and use the correct division. Naturally your function must also return a float

float series(float x,float n)
{
    int i,sign=-1;
    int j, 
    float sum = 0;
    float fact = 1;
    float p = 1;
    for (i=1; i<=(2*n)-1; i+=2)
    {
        fact = 1;
        p = 1;
        for (j=1; j<=i; j++)
        {
           p=p*x;
           fact=fact*j;
        }
        sign=-1*sign;
        sum=sum + sign*p/fact;       
    }
    return (sum);
}

This code still has a minor problem. By having an inner loop, it is slower than necessary. Since this probably is homework, I am not getting rid of that loop for you, just giving you a hint: You don't have to recalculate fact from scratch on each iteration of the outer loop, just try to find out how fact changes from one iteration to the next. The same goes for p .

Answer 2

//Series of Sinx

#include<stdio.h>

#include<math.h>

#define ACCURACY 0.0001

int factorial(int n);

int main()
{
    float x,sum,term;
    int i,power;

    printf("Enter value of X: ");
    scanf("%f",&x);

    i=1;
    power=3;
    sum=x;
    term=x;

    while(term>=ACCURACY)
    {
        term = pow(x,power) / factorial(power);

        if(i%2==1)
        {
            sum -= term;
        }
        else
        {
            sum += term;
        }

        power+=2;
        i++;
    }
    printf("sin(%f) = %.6f\n",x,sum);
    
    return 0;
}

int factorial(int n){
    int i=n,fact=1;
    for(i=1;i<=n;i++)
    {
        fact=fact*i;
    }
    return fact;
}

Answer 3

plenty bugs. To do not caclulate the fact values all the time they are in the lookup table

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

double series(double,int);

long long fact[] = { 1, 2, 6, 24,
                     120, 720, 5040, 40320,
                     362880, 3628800, 39916800, 479001600,
                     6227020800, 87178291200, };

double mypow(double x, unsigned p)
{
    double result = x;
    while(p && --p)
        result *= x;
    return result;
}

int main()
{
    for(double x = 0; x <= M_PI + M_PI / 60; x += M_PI / 30)
        printf("Value of series sin (%.2f) is: %f\n",x,series(x, 5));
    fflush(stdout);
}


double series(double x,int n)
{
    double sum = x;
    int i,sign=1;
    for (i=3; i<=(2*n)-1; i+=2)
    {
        sign=-1*sign;
        sum += sign*(mypow(x, i)/fact[i -1]);
    }
    return (sum);
}

https://godbolt.org/z/U6dULN

Answer 4

maybe its due to floating-point exception as u have declared that the function should return int type value int series(float,float);//hear so u can try editing the return type of this function as float Note:-also u need to change at function definition and the datatype of int i,sum=0,sign=-1; int j,fact=1,p=1; to float as it is returning the value (sum) which should also be float