Series of Taylor for cos(x) - in C

Question

Series of Taylor for cos(x) - in C

Asked 5 years, 8 months ago

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I’ve been trying to calculate this sum, but I’m having trouble making the factorial work... someone can help me identify my mistake?

.

Where x is the angle in radians and N the number of terms of series minus 1.

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


int main()
{
    double x, z = 0, fat = 1;
    int n, i, j, fat2=0;
    scanf("%lf%d", &x, &n);

    j = 1;

    for(i = 0; i < n; i++ )
    {
        //printf("\n%d :: %lf :: %d ", fat2, fat, i); // DEBUG

        z = z + ( ( pow(-1 , i)*pow(x, (2*i)) ) / (fat) ); 

        fat2 = 2*j;
        fat = fat * (fat2 - i);

        j++;
    }
        printf("\ncos(x) = %lf\n", z);  

    return 0;
}

Edit:

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


int main()
{
    double x, z = 0.0;
    int n, i, j;
    scanf("%lf%d", &x, &n);
    double fat[n], fat2[n];

    fat[1] = 1.0;
    for(j = 1; j < n; j++){
        fat2[j] = 2.0 *i;
        fat[j] = fat[j] * (fat2[j] - j); 
        printf("\n%lf\t%lf", fat[j], fat2[j]);

    }

    fat[0] = 1.0;
    for(i = 0; i < n; i++ )
    {
        //printf("\n%d :: %lf :: %d ", fat2, fat, i); // DEBUG

        z = z + ( ( pow(-1 , i)*pow(x, (2*i)) ) / (fat[i]) ); 

    }

        printf("\n\ncos(x) = %lf\n", z);    

    return 0;
}

Edit 2 [solved]:

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

/*long int fatorial(int n){
    long int fat = n * (n - 1);
    if(n == 0 || fat == 0)
        fat = 1.0;

    return fat;}*/

long fat(int n) {
    if(n < 2)
        return 1;
    else
        return n * fat(n-1);    //return (n < 2) ? 1 : n * fat(n-1); 
    }


int main()
{
    double x, z = 0.0;
    int n, i, j;
    scanf("%lf%d", &x, &n);

    for(i = 0; i < n; i++ )
    {
        z = z + ( ( pow(-1.0 , i)*pow(x, (2.0*i)) ) / (fat(2*i)) ); 
        //printf("\n\t%ld", fat(2*i)); //debug
    }

        printf("\n\ncos(x) = %lf\n", z );

    return 0;
}

what’s the matter ?

– zentrunix

2019/04/21 at 19:46
I need to return the sum value of the photo above, but when calculating the factor of (2 * n) wrong, because it has that concept of 0! is equal to 1... And then mess a little, I have tried to put an if for this condition but keeps giving wrong values to the following factors

– Luciano

2019/04/21 at 19:50
(1) create a separate function to calculate factorial(n); create a separate table (array), and initialize in this table the values 0!, 2!, etc... (2) in each iteration of the command for you take the next element of this factorial table

– zentrunix

2019/04/21 at 19:58
Do as @zentrunix told you, create a factorial function.

– FourZeroFive

2019/04/21 at 20:04
I created an array... However only returns 0.0...

– Luciano

2019/04/21 at 21:02
I don’t understand your calculation of fat[j], it seems to me that you are not calculating the factorial correctly. Use a function like: long fat(int n) { Return (n < 2) ? 1 : n * fat(n-1); } , and call with fat(2*i). After your program is correct but you think you can optimize then optimize.

– anonimo

2019/04/21 at 21:55
Oops, I get it now... N had thought that way... I tried to function the factorial the way I put in parentheses in Edit 2... but returned 0.000000... Anyway, everyone, thank you so much for your help!

– Luciano

2019/04/21 at 22:16

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by zentrunix • **5,511** points · Answer 1 · 2019-04-21T22:51:54+00:00

The OP has already solved its doubt, but for the record, here is an optimized version, which calculates the factorial without recursion and without repeating operations. You can still improve a little bit on the accounts, for example calculate "2*i" only once...

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

static long fat = 1; // fat(2*0)

int main()
{
  int n, i, j;
  double x, z = 0.0;

  scanf("%lf%d", &x, &n);

  for (i = 0; i < n; i++ )
  {
    printf("* i=%d fat(%d)=%d\n", i, 2*i, fat);
    z = z + ( ( pow(-1.0 , i)*pow(x, (2.0*i)) ) / fat );
    fat = fat * (2*i+1) * (2*i+2); // proximo fat
  }

  printf("\n\ncos(x) = %lf\n", z );

  return 0;
}

Testing:

[~/Projects/testes/so]
$./377309
3 5
* i=0 fat(0)=1
* i=1 fat(2)=2
* i=2 fat(4)=24
* i=3 fat(6)=720
* i=4 fat(8)=40320

cos(x) = -0.974777

[~/Projects/testes/so]
$

Checking here on this page it seems that the factor calculation is right.