I was given a task by the University to evaluate the following exponential formulas using C:
1. (a + b)2 = a2 + 2ab + b2
2. (a - b)2 = a2 - 2ab + b2
3. (a + b)3 = a3 + 3a2b + 3ab2 + b3
4. (a - b)3 = a3 - 3a2b + 3ab2 - b3
5. a2 - b2 = (a - b) (a + b)
6. a3 - b3 = (a - b)3 + 3 a b (a - b)
7. a3 - b3 = (a - b) (a2 + a b + b2)
8. a3 + b3 = (a + b) (a2 - a b + b2)
9. a3 + b3 = (a + b)3 - 3 a b (a + b)
And here's my code, which takes two int values from the user and then calculates and prints them using the above algebraic expressions:
//Lab 1 Task Putting the values in the math formulas and solving them
#include <stdio.h>
#include <math.h>
int main (void)
{
int a;
int b;
printf("\nEnter The First Integer : ");
scanf("%i" , &a);
printf("\nEnter The Second Integer : ");
scanf("%i" , &b);
double f1 = (a * a) + 2 * (a * b) + (b * b);
double f2 = (a * a) - 2 * (a * b) + (b * b);
double f3 = pow(a,3) + 3 * pow(a,2) * b + 3 * a * pow(b,3) + pow(b,3);
double f4 = pow(a , 3) - 3 * (pow(a , 2) * b) + 3 * (pow((a * b) , 2) - pow(b , 3));
double f5 = (a - b) * (a + b);
double f6 = pow((a - b) , 3) + 3 * ((a * b) * (a - b));
double f7 = (a - b) * (pow(a , 2) + (a * b) + pow(b , 2));
double f8 = (a + b) * (pow(a , 2) - (a * b) + pow(b , 2));
double f9 = pow((a + b) , 3) - 3 * ((a * b) * (a + b));
printf("\n1. (%i + %i) ^ 2 = %.0lf\n\n" , a , b , f1);
printf("2. (%i - %i) ^ 2 = %.0lf\n\n" , a , b , f2);
printf("3. (%i + %i) ^ 3 = %.0lf\n\n" , a , b , f3);
printf("4. (%i - %i) ^ 3 = %.0lf\n\n" , a , b , f4);
printf("5. %i ^ 2 - %i ^ 2 = %.0lf\n\n" , a , b , f5);
printf("6. %i ^ 3 - %i ^ 3 = %.0lf\n\n" , a , b , f6);
printf("7. %i ^ 3 - %i ^ 3 = %.0lf\n\n" , a , b , f7);
printf("8. %i ^ 3 + %i ^ 3 = %.0lf\n\n" , a , b , f8);
printf("9. %i ^ 3 + %i ^ 3 = %.0lf\n\n" , a , b , f9);
}
It works fine except for the 3rd and 4th where it gives much-more-than-expected output. What am I doing wrong?
Here's the output being received with the expected and faulty output mentioned:
~/FinalLabProjects/ $ ./formulas <--- I run the program
Enter The First Integer : 7
Enter The Second Integer : 5
1. (7 + 5) ^ 2 = 144 <----fine
2. (7 - 5) ^ 2 = 4 <----fine
3. (7 + 5) ^ 3 = 3828 <----expected 1728
4. (7 - 5) ^ 3 = 2908 <----expected 8
5. 7 ^ 2 - 5 ^ 2 = 24 <----fine
6. 7 ^ 3 - 5 ^ 3 = 218 <----fine
7. 7 ^ 3 - 5 ^ 3 = 218 <----fine
8. 7 ^ 3 + 5 ^ 3 = 468 <----fine
9. 7 ^ 3 + 5 ^ 3 = 468 <----fine
You wrote:
double f3 = pow(a,3) + 3 * pow(a,2) * b + 3 * a * pow(b,3) + pow(b,3);
which equals
f3 = a^3 + 3ba^2 + 3ab^3 + b^3
But it has to be:
f3 = a^3 + 3ba^2 + 3ab^2 + b^3
so change this line to:
double f3 = pow(a,3) + 3 * pow(a,2) * b + 3 * a * pow(b,2) + pow(b,3);
And also
double f4 = pow(a , 3) - 3 * (pow(a , 2) * b) + 3 * (pow((a * b) , 2) - pow(b,3));
equals:
f4 = a^3 - 3(ba^2) + 3[(ab)^2 - b^3]
This is wrong. Instead:
f4 = a^3 - 3ba^2 + 3ab^2 - b^3
so:
double f4 = pow(a , 3) - 3 * (pow(a , 2) * b) + 3 * a * pow(b , 2) - pow(b,3);
Lastly, both bases and exponents are int numbers. In fact, pow() function is of form:
double pow( double base, double exponent );
You may want to use round() with pow() function. Or better write your own function to compute your expressions because you will work with relatively small integers. (I guess)