Calculating e^x by math.h and by own means For this program, the user needs to enter an exponent and the program will calculate \$e\$ (Euler's number) to the power of the exponent the user inputs.

This is done by two ways: 


By the math.h library
By a for loop (Series)


Then the program calculates the difference of these two functions (which we now call \$d\$) and calculates \$n\$.

\$n = \frac{d}{0,001}\$

#include &lt;math.h&gt;
#include &lt;stdio.h&gt;




main()
{
int a, exponent, h;
float math, iteraties, c, stdio, z, verschil;

    char stop='n';
    while (stop!='j')
    {

      printf("vul exponent in:\t\t");
      scanf("%d",&amp;exponent);
      syntax: math= exp(exponent);
      c=1.0;
      h=0;
      stdio=1.0;
      a=0;
      for (iteraties=1.0; a&lt;1; iteraties++)
      {
        h=h+1;
        Syntax: z= pow(exponent,h);

        c=c*iteraties;
        stdio=stdio+(z/c);
        verschil=math-stdio;
        if(verschil&lt;0.01)
        {
            a=2;
        }
      }
      printf("\nexp (%d) uit math.h is:\t\t%f",exponent, math);
      printf("\nexp (%d) m.b.v. de reeks is:\t%f",exponent,stdio);
      printf("\nhet verschil is: \t\t%f", verschil);
      printf("\naantal iteraties is:\t\t%.f", iteraties);


      printf("\n\nstoppen n/j\t\t\t");
      getchar();
      scanf("%c",&amp;stop);
    }

}
What's the point of the syntax: and Syntax: labels?
You don't use them, so remove them.
And if you must use two labels, make them less similar to each other.

When the condition in the while loop is true from the start,
consider rewriting it as a do { ... } while loop.

You don't need the a variable.
You can easily rewrite the program without it.

The coding style is very odd:


Excessive vertical spacing (many empty lines) at many places
Too compact: it's recommended to put spaces around operators


Other tips:


Instead of h=h+1 you can do ++h, as you did in the for loop
Instead of x = x * y you can use x *= y
Instead of x = x + y you can use x += y


With the above suggestions (and some more) applied, the code becomes slightly simpler:

#include &lt;math.h&gt;
#include &lt;stdio.h&gt;

main()
{
    int exponent, h;
    float math, iteraties, c, stdio, z, verschil;

    char stop = 'n';
    do
    {
        printf("vul exponent in:\t\t");
        scanf("%d", &amp;exponent);
        math = exp(exponent);
        c = 1.0;
        h = 0;
        stdio = 1.0;
        for (iteraties = 1.0; ; iteraties++)
        {
            ++h;
            z = pow(exponent, h);

            c *= iteraties;
            stdio += z / c;
            verschil = math - stdio;
            if (verschil &lt; 0.01)
            {
                iteraties++;
                break;
            }
        }
        printf("\nexp (%d) uit math.h is:\t\t%f",exponent, math);
        printf("\nexp (%d) m.b.v. de reeks is:\t%f",exponent,stdio);
        printf("\nhet verschil is: \t\t%f", verschil);
        printf("\naantal iteraties is:\t\t%.f", iteraties);

        printf("\n\nstoppen n/j\t\t\t");
        getchar();
        scanf("%c", &amp;stop);
    } while (stop != 'j');
}