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1 Methods The problem is from the given function f(x), determine the root x ∗ such that f(x ∗ ) = 0 (1) Numerically, this equation may be approximated using a stopping tolerance . f(x ∗ ) < (2) where is close enough to zero. 1.1 Bisection Method The simplest method in root finding uses an interval [a, b] in the domain of the function where the bounds of the interval satisfy: f(a)f(b) < 0 (3) or in other words, f(a) and f(b) have opposing signs. The successive process finds the midpoint c bisecting the interval: c = a + b 2 (4) and tests if on that iteration, f(c) < which implies that c is a root of f. On the other hand, if f(c) ≮ , then the bounds are reassigned. If f(c) has the similar sign with f(a), the next iteration uses the interval [c, b] and if f(c) has the similar sign with f(b), the next iteration uses the interval [a, c]. Note that this method requires that the function is continuous on the interval [a, b]. 1.2 Newton’s Method This method requires continuity as well since the derivative of the function is needed. Additionally, it also requires an initial guess x0 and repetitively determines the root using the formula xi+1 = xi − f(xi) f 0(xi) (5) and tests if on that iteration, f(xi+1) < which implies that xi+1 is a root of f. 1.3 Secant Method Regarded as a quasi-Newton technique, this method is related to the previous one since f 0 (xi) ≈ f(xi) − f(xi−1) xi − xi−1 (6) and the difference between the two methods is that the secant method requires two initial guesses. Substitute this to the previous equation, xi+1 ≈ xi − f(xi) f(xi)−f(xi−1) xi−xi−1 (7) which can be rewritten as xi+1 ≈ xi − f(xi) xi − xi−1 f(xi) − f(xi−1) (8) If on that iteration, f(xi+1) < , then xi+1 is a root of f.
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