function that applies the secant method to iteratively approximate the nth root of a constant given n, the constant, and two initial guesses--x0 and x1.

Question

What will be an ideal response?

Answer 1

```
double secant( double c, double n, double x0, double x1 )
{
int i;
double tmp, tmpn, tmpd;

for( i = 2; (x0-x1) > TOLERANCE && i <= MAX_ITERATIONS; i++)
{
tmpn = ( x0 * ( pow(x1, n) - c ) - x1 * ( pow(x0, n) - c ) );
tmpd = ( (pow(x1 ,n) - c) - (pow(x0, n) - c) );

tmp = tmpn / tmpd;

x0 = x1;
x1 = tmp;
}

//if no convergence after MAX_ITERATIONS iterations,
//display error message and return 0
if( x0 - x1 > TOLERANCE)
{
cerr << "Error in function 'secant'...no convergence" << endl;
return 0;
}

return x1;
}

```