Suppose we have a polynomial f(x) = 4 – x3.

We want to find the value of x that makes f(x) = 0. (This value, call it t, is the cube root of 4.
In fact we can use the bisection method to find the nth root of a number.)

If we know that the function f(x) is definitely 0 somewhere inside an interval [l, r], and that f(x) is strictly increasing (or decreasing) over that interval, then the bisection method is a fast way of finding the root of f(x) to within a good level of accuracy.

Our job then is to describe the bisection method succinctly: We assume that f(x) is strictly increasing over the interval chosen (call it [l, r]). Suppose also that somewhere inside the interval, f(x) is zero. Then clearly f(l) < 0 and f(r) > 0.

We can find the point t where f(t) = 0 by progressively splitting the interval, choosing smaller and smaller intervals until we find t, or the interval is small enough that we consider our answer found. In particular, take a point z that is the mid‐point of the interval [l, r]. If f(z) > 0 then clearly the root of f(x) is to be found in the interval [l, z]. Similarly if f(z) < 0 then …

What this means is that we can take the new interval [l, z] – replacing r with z – and ask the same question of its mid‐point! We can keep doing this until we find z such that f(z) = 0 or until the interval [l, z] (or [z, r] for that matter) is small.
In this assignment you will implement this bisection method for root finding.

Write a function bisect(), which calls your polynomial function f() repeatedly until it finds the root to within 3 decimal places. Check your bisect() function using f(x) = x2 – 2, whose root I am sure we all know!

Use your program to compute the 7th root of 126, the cube root of 43 and the 5th root of 729. Be careful to choose the right initial interval!
Submit the solution source code for your program.

Note: You should write separate functions for bisect() and the polynomial it iterates on.

This article has been dead for over six months. Start a new discussion instead.