Doubt in exercise with factorial and large numbers

Question

I need to do an exercise that asks for the Y value given by Y = x + x 2/2! + x 3/3! + x 4/4! + ... , considering the first 15 terms. I’m having problems because the factorial results from the 10 term already start to become giant. For this, I used long long int. I BELIEVE this is part of of(operations) that is wrong, because I tried to change the countPotency to <= 3, and tested the program by putting x = 2 and the result would have to be 6. Even so, gave an absurd number of great. What may be happening may also be the types of variables are wrong, but I don’t know what else to try to fix it.

#include <stdio.h>

int main()
{
    //VARIAVEIS
    unsigned long long int valorX, cont = 3, contPotencia = 1, fat = 2, potencia, x;
    float divisao, resultado;

    //INICIO
    printf("Insira X: ");
    scanf("%llu", &valorX);

    if (valorX < 1){
        printf("Número inválido, insira um maior que 0.");
    }
    else {
        printf("Y = %llu", valorX);
        x = valorX;
        do {
            x = valorX*x;
            printf(" + %llu/%llu ", x, fat);
            contPotencia++;

            fat = cont*fat;
            cont++;

            divisao = x/fat;

            resultado = resultado+divisao;

        } while (contPotencia < 15);
    }
    printf("\nY = %f", resultado);

}

Answer 1

Problem 1: You have to do the floating point math, otherwise the x/y split result when x < y will be 0.

Problem 2: The "result" variable needs to be initialized.

I did a test (well) fast, it seems to be working.

#include <stdio.h>

int main()
{
  //VARIAVEIS
  double valorX, cont = 3, contPotencia = 1, fat = 2, potencia, x;
  float divisao, resultado = 0;

  //INICIO
  printf("Insira X: ");
  scanf("%lf", &valorX);

  // if (x < 1){ // <------ erro
  if (valorX < 1) {
    printf("Número inválido, insira um maior que 0.");
  }
  else {
    printf("Y = %lf", valorX);
    x = valorX;
    do {
      x = valorX*x;
      printf(" + %lf/%lf ", x, fat);
      contPotencia++;

      fat = cont*fat;
      cont++;

      divisao = x/fat;
      printf("\n*");
      printf("\n* x=%lf fat=%lf divisao=%f", x, fat, divisao);

      resultado = resultado+divisao;
      printf("\n* resultado: %f", resultado);

    } while (contPotencia <= 3);
  }

  printf("\n*");
  printf("\nY = %f", resultado);
}

Testing:

$ 380126.exe                                           
Insira X: 2                                            
Y = 2.000000 + 4.000000/2.000000                       
*                                                      
* x=4.000000 fat=6.000000 divisao=0.666667             
* resultado: 0.666667 + 8.000000/6.000000              
*                                                      
* x=8.000000 fat=24.000000 divisao=0.333333            
* resultado: 1.000000 + 16.000000/24.000000            
*                                                      
* x=16.000000 fat=120.000000 divisao=0.133333          
* resultado: 1.133333                                  
*                                                      
Y = 1.133333                                           

$                                                      

Answer 2

I was able to answer the exercise and arrive at the expected result. The problem was in the very order of operations.

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

int main(){
    //VARIAVEIS
    unsigned long long int valorX, cont = 3, contPotencia = 1;
    float divisao, resultado, fat = 2, x;

     //INICIO
    printf("Insira X: ");
    scanf("%llu", &valorX);

    if (valorX < 1){
        printf("Número inválido, insira um maior que 0.");
    }
    else {
        printf("Y = %llu", valorX);
        x = valorX;
        resultado = x;
        do {
            x = valorX*x;
            printf(" + %.f/%.f", x, fat);

             divisao = x/fat;

            fat = cont*fat;

            contPotencia++;
            cont++;

            resultado = resultado+divisao;

        } while (contPotencia < 15);
    }
    printf("\nResultado = %f", resultado);

}