The secant method of root-finding is an iterative method that is a variation on Newton's iteration formula. Newton's iterator is x

Under some conditions, the method will not converge, so various refinements have been suggested by different authors.

_{n+1}= x_{n}- f(x_{n})/f'(x_{n}) The secant method replaces f'(x_{n}) with a finite difference, so two starting values are required.**f'(x**_{n}) ≈ (f(x_{n}) - f(x_{n-1}))/(x_{n}- x_{n-1}) So the iterator becomes**x**_{n+1}**= x**_{n}**- f(x**_{n}**)*((x**_{n}**- x**_{n-1}**)/(f(x**_{n}**) - f(x**_{n-1}**)))**__Example__Suppose we have**f(x) = x^2 - 2**, for which we would like to find a root. (We know that one root is √2 ≈ 1.4142.) Further suppose that we want to start with x_{0}=1 and x_{1}=2.**x**_{2}**= x**_{1}**- (x**_{1}**^2 - 2)*(x**_{1}**- x**_{0}**)/((x**_{1}**^2 - 2) - (x**_{0}**^2 - x))**(the secant method iterator) x_{2}= x_{1}- (x_{1}^2 - 2)/(x_{1}+ x_{0}) (factor the denominator & cancel the common factor from the numerator) x_{2}= (2+x_{0}*x_{1})/(x_{0}+x_{1}) (we simplify the iterator first, so the steps below are not so difficult) Substituting our starting values, we get x_{2}= (2+1*2)/(1+2) = 4/3 (≈1.3333) x_{3}= (2 + (2)*(4/3))/(2 + 4/3) = 7/5 (= 1.4000) x_{3}= (2 + (4/3)*(7/5))/(4/3 + 7/5) = 58/41 (≈1.4146) x_{4}= (2 + (7/5)*(58/41))/(7/5 + 58/41) = 816/577 (≈1.4142)Under some conditions, the method will not converge, so various refinements have been suggested by different authors.