MATH 340
LAB 4 Assignment
Root-finding: Secant Method
To find the root of a function of the form f(x) = 0, the secant iterative method is:
xn+1 = xn −
f(xn)(xn − xn−1)
f(xn) − f(xn−1)
, n ≥ 1
Note that this is a two-step method, therefore we need two initial guesses for our
solution: x0 and x1.
A variation: Regula Falsi
Since the Secant Method does not guarantee the bracketing of the solution (as it
was for Bisection Method), it may migrate away from it. Hence, we present here a
variation which takes care of bracketing at each step: the Regula Falsi. This could
be seen as a mashup of the Bisection Method and the Secant Method. In fact, it
consists in taking one step of Secant Method to find the new intermediate value,
called c in the Bisection Method, and just continue with the Bisection Method
algorithm. The pseudo-code is given in your textbook.
To define the error, once again we can use as an estimate the difference in
absolute value of two consecutive approximations errn = |xn+1 − xn|
Problem 1:
Do Problem 28 of your homework. Write a one sentence conclusion comparing the
number of steps needed with bisection, Newton, secant and regula falsi.
Problem 2:
Do Problem 29 of your homework. Write a one sentence conclusion comparing the
number of steps needed with bisection, Newton, secant and regula falsi.
1
