f(x) at two points
xn and xn-1 by
f(xn) - f(xn-1)
f'(x) = --------------
xn - xn-1
Geometrically, Newton's method uses the tangent line and the secant
method approximates the tangent line by a secant line.
This gives the iteration scheme
f(xn)(xn - xn-1)
xn+1 = xn - --------------
f(xn) - f(xn-1)
The secant method has superlinear convergence (approximately 1.6,
compared to 2 for Newton's method).
When the secant method converges it usually does so in 5 to 7
iterations if the initial guesses are good enough.