Mathematical expression in Objective C [closed] - objective-c

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Closed 11 years ago.
Please see the formulæ below.
It's the solution to a circle chord problem where A is the arc length and C is the chord length.
I want to find the Radius r.
The formula below uses factorials and powers etc which I don't know how to express in Objective C.
Can somebody help by converting the formula into Objective C language?
a = A² – C²
b = -2A^4/4!
c = 2A^6/6!
d = -2A^8/8!
//Let:
f = c/a-b²/3a²
g = d/a+2b³/27a³-bc/3a²
x = cuberoot(-g/2 + sqrt(g²/4+f³/27)) – cuberoot(g/2 + sqrt(g²/4+f³/27)) - b/3a
r = sqrt(x)

Answer 1

This should work:
double factorial(int x)
{
double result = 1;
for (x = x; x > 0; x--)
{
result *= x;
}
return result;
}
...
int A = 20;
int C = 25;
double a = A*A - C*C;
double b = -2.0*pow(A, 4.0)/factorial(4);
double c = 2.0*pow(A, 6.0)/factorial(6);
double d = -2.0*pow(A, 8.0)/factorial(8);
//Let:
double f = c/a-(b*b)/(3.0*(a*a));
double g = d/a + 2.0*pow(b, 3.0)/(27.0*pow(a, 3.0))-(b*c)/(3.0*a*a);
double x = pow(-g/2.0 + sqrt(g*g/4.0+pow(f, 3.0)/27.0), 1.0/3.0) - pow(g/2.0 + sqrt(g*g/4.0+pow(f, 3.0)/27.0), 1.0/3.0) - b/(3.0*a);
double r = sqrt(x);

Answer 2

The header is <math.h>. You square by multiplying a factor by itself; you cube by multiplying a factor by itself twice. The factorials are constants; you precalculate them, or let the computer do so:
enum { factorial_3 = 1 * 2 * 3 }; // Not needed, but an example
enum { factorial_3 = 3 * factorial_2 }; // Alternative technique
The function sqrt() gives you the square root. If it is available, then cbrt() from C99 gives you a cube root. When it isn't available, you can use pow() instead. If the notation A^4, A^6 and A^8 is equivalent to A4, A6, A8, then those are easily handled as multiplications.
You can assemble the bits and pieces using this information.

Answer 3

For exponents, you can use math.h(which you'll need to import)'s pow(float base, float exponent) function. Then for factorials, you'll have to create one yourself. The following function should work (I didn't try it out yet).
long factorial(int number)
{
long numb = 1;
for (int i=1;i<=number;i++)
{
numb*=i;
}
return numb;
}
However, be aware that it's very easy to get too huge numbers with factorials, so use long long (or even malloc if necessary) if you need to get the factorial of a large number.