Page 238 - DMTH201_Basic Mathematics-1

Unit 8: Continuity

8.3 Bisection Method Notes

Let f(x) be a continuous function on the interval [a, b]. If d [f(a), f(b)], then there is a c [a, b]
such that f(c) = d.

By replacing f(x) by f(x) – d, we may assume that d = 0; it then suffices to obtain the following
version: Let f(x) be a continuous function on the interval [a, b]. If f(a) and f(b) have opposite signs,
then there is a c [a, b] such that f(c) = 0.

Here is an outline of its proof: Let’s assume that f(a) < 0, while f(b) > 0, the other case being
handled similarly. Set a = a and b = b.
0 0
Now consider the midpoint m = and evaluate f(m ). If f(m ) < 0, set a = m and b = b .
0 0 0 1 0 1 0
If f(m ) > 0, set a = a and b = m . (If f(m ) = 0, you’re already done.) What situation are we in
0 1 0 1 0 0
now? f(a ) and f(b ) still have opposite signs, but the length of the interval [a , b ] is only half of
1 1 1 1
the length of the original interval [a , b ]. Note also that a a and that b b .
0 0 0 1 0 1
You probably guess this by now: we will do this procedure again and again.
Here is the second step: Consider the midpoint m = and evaluate f(m ). If f(m ) < 0, set
1 1 1
a = m and b = b . If f(m ) > 0, set a = a and b = m . (If f(m ) = 0, you’re already done.) What
2 1 2 1 1 2 1 2 1 1
situation are we in now? f(a ) and f(b ) still have opposite signs, but the length of the interval [a ,
2 2 2
b ] is only a quarter of the length of the original interval [a , b ]. Note also that a a a and that
2 0 0 0 1 2
b b b .
0 1 2

The red line shows the interval [a , b ].
n n
Continuing in this fashion we construct by induction two sequences:
(a ) and (b )
n n = 1 n n = 1
with the following properties:
1. (a ) is an increasing sequence, (b ) is a decreasing sequence.
n n
2. a b for all n.
n n
3. f(a ) < 0 for all n, f(b ) > 0 for all n.
n n
–n
4. b – a = 2 (b – a) for all n.
n n

It follows from the first two properties that the sequences (a ) and (b ) converge; set
n n
=

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