In the method of bisection, we approximate a solution of the equation f(x) = 0, where f(x) is a continuous function, by means of a successive reduction of intervals that contain the solution.
Let f(x) be a continuous function, and let 
and 
be in the
domain of f(x), where 
, such that 
,
, and 
and 
have opposite signs. Then, as
we have seen, there is some 
between and such that
. There may be several such values, but let us assume that
is unique in this respect. Without any previous knowledge of
f(x), we can make no conclusions about the location of with
respect to and , other than the fact that lies in the
interval 
. If we choose some point 
between and
, then this solution must lie either in the interval
, or in the interval 
. We can determine which
interval by considering the value of 
. If 
, then
our solution is 
, and no additional work must be done to
find . However, if 
, then our solution will
lie in the interval whose endpoints yield opposite signs when evaluated
by f(x). Thus lies in if and have
opposite signs, and lies in if and
have opposite signs. Since is the unique solution to f(x) = 0
on , then only one of the two intervals will be such that
f(x) yields opposite signs when evaluated at the endpoints. We have
now found a subinterval of which contains --let us
denote it as 
. We then proceed in the same manner with this
subinterval, , to enclose on a smaller subinterval,
. We continue this until we find a subinterval 
,
containing , such that every point in the subinterval lies within
a given distance of . To maximize the effectiveness of the choice
of 
for each i, 
, since we have no previous
knowledge of f(x), we take to be the midpoint of the interval
,
thus bisecting into two smaller subintervals.
Example 1 Let us employ the method of bisection to approximate the positive solution of the equation
Here we can actually solve our equation to obtain the solutions
, the value that we seek to approximate being
, to ten decimal places. For this problem our
function 
, which is continuous over the set of all real
numbers. To apply the bisection method, we need initial upper and lower
bounds for the solution. Note that f(1) = -1 and f(2) = 2, which
implies that we can take 
and 
as our initial bounds.
The following table illustrates the rate of convergence of the bisection
method for the positive solution of the equation 
with the
initial bounds on the solution being and .
Choosing 
as our final approximation, we see
that it took more than thirty steps in order to obtain an approximation
to the solution that is accurate to nine decimal places. 
Let us now consider the issue of how many times we must apply the
method of bisection until we have reduced our interval to
one in which our approximation lies within a given distance 
of the solution .
After n - 1 applications of the bisection method, we reduce the
interval to the 
subinterval . Note
that in bisecting the 
subinterval 
,
we obtain some subinterval whose width is 1/2 of the width
of the previous subinterval. Thus the intervals are being reduced at a
linear rate of 1/2, with
By induction, we see that this implies that the subinterval
has width
The midpoint, 
, of the subinterval is a distance of
away from either endpoint of the subinterval, and so
it must lie at most units away from the solution .
Therefore
The implications here are that, since 
, in
order to gain an additional three decimal places of accuracy using the
bisection method, one must make about ten additional iterations. Thus
we see that the bisection method converges slowly, at a linear rate,
although it does converge. For n to be such that is within
of ,
we need
Thus if we take
then (1) will be satisfied and we will find an approximation
that is within of if
However, it is not necessary to know n beforehand in order to be able to determine when to terminate a bisection procedure. It is sufficient to continue the bisection method until n is found such that
Since must lie between 
and 
, it must also lie within
of either or .
Example 2 In Example 1 we employed the bisection method
to approximate the positive solution of to nine decimal
places of accuracy. Thus we found such that
Here our original interval was [1,2], implying and
, and 
. Therefore, by
(2), we know that will satisfy (3) for
any value of n such that
Since n must be an integer, we see that we need no more than 31
iterations to give us an approximation that is within
of the solution. Note that the magnitude of the
error in the 
approximation is given by
whereas the magnitude of the error in the 
approximation
is
Thus, for this example, the estimate that we would need at most 31
iterations in order to approximate the solution to within
turned out to be exact.
Example 3 In this example we shall use the bisection method to approximate the values of x for which
Since a significant amount of work is involved for each decimal place of accuracy, we shall limit our approximation to three places of accuracy.
Note that the natural logarithm function, 
, is increasing over
all of its domain 
. Thus, for x > e, the natural logarithm
function satisfies 
. Since 
for all x, this implies that (4) can hold only for values
of x in the interval (0,e]. The function 
is decreasing
in this interval, and since is increasing over this interval,
we see that there is at most one value of x for which (4)
holds. Let us denote 
. Then any solution of
(4) is also a solution of f(x) = 0. Note that
, whereas 
. Thus we see
that f(x) = 0 has a unique solution on [1,e].
Therefore, taking , 
, and 
for
each n, we obtain the following values
To three decimal places, 
. Therefore the
approximate value of x that satisfies (4) is x = 1.303.
The method of bisection will always produce an approximation of a root of a continuous function f(x) provided one can find values a and b, in the domain of f(x), such that f(a) < 0 and f(b) > 0. However, as we have seen, the major drawback of this method is that it converges rather slowly. Because of this, the method of bisection is commonly utilized for the purpose of reducing the size of the interval on which a root of a given function is known to exist, and then an alternative method of root approximation is utilized to bring the approximation to within the desired amount of accuracy.
