Problem with float values and how to avoid infinite values of function f(x)

Question

I just started to learn C from absolute zero.

I have a task to solve function y = f(x) by putting in x values - function is given in my code. Total x and y values are 15(from 0 to 14). First of all, I ask about x0 - start x value and then I ask about what STEP is going to be

(take a look at code)

1st for loop:

For x values, logic should look like this(START is variable for x0 - means first value):

1) START
2) START + STEP
3) START + 2*STEP
4) START + 3*STEP
.
.
.
15) START + 15*STEP

2nd for loop:

Here I am trying to put in my x values into f(x) function.

---

So here are 2 problems:

• 1st problem is that when we take different START(x) values and STEPS some of them making the answer as the infinite number like x=4 and the program just crashes.
• 2nd problem I don't know why, but neither float nor double doesn't show answer as a floating-point value. Like if we put x=2, so y = 39.5 and the program shows it like 39.00.

Tried to change float or double, counters and different variables. Actually, I think I just have a lack of experience

Here's wolfram alpha to check different values: https://www.wolframalpha.com/input/?i=(14*x%5E3%2B7*x%5E2-2%2B20)%2F(x%5E2-4*x),+x%3D5

Compiler: Geany

#include <stdio.h>
#include <math.h>
#define MAX 14




int main()
{

    int i,j;            // counter
    int START;          // x0
    int STEP;           // 
    int x[MAX];         // x values array - total 15
    double y[MAX];      // f(x) values - total 15

    printf("START NUMBER: ");
    scanf("%d",&START);
    printf("\n");

    printf("STEP: ");
    scanf("%d",&STEP);
    printf("\n");

    /* This cycle is for calculating x values until 15(from 0 to 14)
    * Using logic of a homework it should be like this:
    * First value 1) x0; 2) x0 + STEP; 3) x0 + 2*STEP; 4) x0 + 3*STEP and etc
         */
    for(i=0;i <= MAX; i++)
    {
        x[i] = START + (STEP*i);
        /* printf to check if it counts at all: */ printf("%d\n",x[i]);
    }

    printf("\n");
    /* This cycle is for calculating y(function) values using array of x's
    * function y =   (14x^3 + 7x^2 - x + 20)
    *               ------------------------
    *                       (x^2 - 4x)
    */
    for(j=0;j <= MAX; j++)
    {
        y[j] = ((14*(x[j])*(x[j])*(x[j]) + 7*(x[j])*(x[j]) - x[j] + 20)/(x[j]*(x[j]) - 4*x[j]));
        /*  printf to check if it counts at all:*/ printf("%.2lf\n",y[j]);
    }



    return 0;

    }

I just want to make for a second for cycle to avoid infinite values and put it in y array, like just 0 instead of crashing after an infinite value, I think it could be done by if cycle, but I don't know how to put it logically into for loop, so for could just continue to solve in peace and if works as сhecker.

And the second thing is about floating-point. I just don't actually understand why if I put into f(x), x=2 it gives me 39.00, not 39.50 like wolfram did.

Answer 1

Your issue with the program printing 39.00 instead of 39.50 comes from the fact that you are using integer operations.
See Can integer division in C/C++ run into loss of precision issues?.
The lazy way to solve this issue would be to replace int x[MAX]; by float x[MAX];
Also, in for(i=0;i <= MAX; i++) & in for(j=0;j <= MAX; j++) you run the loops 15 times instead of 14 and create a buffer overflow (because you access x[14] and y[14]).
You also have a nasty division by 0 in there.
I tried to improve your code:

#include <stdio.h>
#include <math.h>
#define MAX 14

int main()
{
    int i,j;
    int START, STEP; 
    double y[MAX], x[MAX];

    printf("START NUMBER: ");
    scanf("%d", &START);
    printf("STEP: ");
    scanf("%d", &STEP);

    printf("counted:\n");
    for(i = 0; i < MAX; i++)
    {
        x[i] = START + (STEP * i);
        printf("%.0lf\n", x[i]);
    }
    printf("Results: \n");
    for(j=0;j < MAX; j++)
    {
        if (pow(x[j], 2) == 4 * x[j]) //prevent division by 0, if(x==4) also works
            y[j] = NAN; //the solution doesn't exist
        else
            y[j] = (14 * pow(x[j], 3) + 7 * pow(x[j], 2) - x[j] + 20)
                / (pow(x[j], 2) - 4 * x[j]);
        printf("%.2lf\n", y[j]);
    }
    return 0;
}

Note that I use pow() but you can use your own function (and it would be faster).