Sn(x) = (x-1)^2n+1/(2n+1)!(n+1) - C 语言

Question

I tried to use this code to solve the question in the picture but I can't solve it, I'd be happy to help.

how can I fix the code that works for the question in the picture? thanks.

#include

void main() {
    int n, i;
    double sum, x, last;
    do {
        printf("Enter an positive number: ");
        scanf("%d", & n);
    } while (n<0);
    printf("Please enter x: ");
    scanf("%lf", & x);
    sum = last = x;
    for (i = 1; i<= n; i++) {
        last = last * (-1) * ((x * x) / ((2 * i - 1) * (2 * i)));
        sum = sum + last;
    }
    printf("S = %lf\n", sum);
}

Answer 1

As mentioned in the comments, you are using the wrong quotient. Also, a closed expression for the value of the series is available.

From the last term of the formula you can read off the general form of the terms

    a[k] = (x-1)^(2k+1) / ( (2k+1)! * (k+1) ),  k = 0,..., n

The quotient of two such adjacent terms is

    a[k]/a[k-1] = (x-1)^2 / ( (2k+1)*2(k+1) )

so that the modified code computing your sum is

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

void main() {
    int n, i;
    double sum, x,x1s last;
    do {
        printf("Enter an positive number: ");
        scanf("%d", & n);
    } while (n<0);
    printf("Please enter x: ");
    scanf("%lf", & x);
    sum = last = x-1;
    x1s = last*last;
    for (i = 1; i<= n; i++) {
        last = last * (x1s / ((2*i+1) * 2 * (i+1)));
        sum = sum + last;
    }
    printf("S = %20.15lf,  S_infty = %20.15lf\n", sum, 2*(cosh(x-1)-1)/(x-1));
}