Basis of Bisection Method

Theorem An equation f(x)=0, where f(x) is a real continuous function,

has at least one root between xl and xu if f(xl) f(xu) < 0.
f(x)

x
x
xu

Figure 1 At least one root exists between the two points if the function is
real, continuous, and changes sign.

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 Basis of Bisection Method
f(x)

x x
xu

Figure 2 If function f x  does not change sign between two

points, roots of the equation f x   0 may still exist between the two
points.
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 Basis of Bisection Method
f(x)
f(x)

x xu
x x
x xu

Figure 3 If the function f x  does not change sign between two

points, there may not be any roots for the equation f x   0 between
the two points.

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 Basis of Bisection Method
f(x)

xu x
x

Figure 4 If the function f x  changes sign between two points,

more than one root for the equation f x   0 may exist between the two
points.

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 Algorithm for Bisection Method

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 Step 1
Choose x and xu as two guesses for the root such that
f(x) f(xu) < 0, or in other words, f(x) changes sign
between x and xu. This was demonstrated in Figure 1.
f(x)

x
x
xu

Figure 1

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 Step 2
Estimate the root, xm of the equation f (x) = 0 as the mid
point between x and xu as
f(x)

x  xu
xm =
2
x xm
x
xu

Figure 5 Estimate of xm

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 Step 3
Now check the following

a) If f xl  f xm   0 , then the root lies between x and

xm; then x = x ; xu = xm.

b) If f xl  f xm   0 , then the root lies between xm and

xu; then x = xm; xu = xu.

c) If f xl  f xm   0 ; then the root is xm. Stop the

algorithm if this is true.

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 Step 4
Find the new estimate of the root
x  xu
xm =
2
Find the absolute relative approximate error
x new  xm
old

a  m
new
100
x m

where
xmold  previous estimate of root
xmnew  current estimate of root

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 Step 5
Compare the absolute relative approximate error a with
the pre-specified error tolerance s .
Go to Step 2 using new
Yes upper and lower
Is a s ? guesses.

No Stop the algorithm

Note one should also check whether the number of

iterations is more than the maximum number of iterations
allowed. If so, one needs to terminate the algorithm and
notify the user about it.
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 Example 1
You are working for ‘DOWN THE TOILET COMPANY’ that
makes floats for ABC commodes. The floating ball has a
specific gravity of 0.6 and has a radius of 5.5 cm. You
are asked to find the depth to which the ball is
submerged when floating in water.

Figure 6 Diagram of the floating ball

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 Example 1 Cont.
The equation that gives the depth x to which the ball is
submerged under water is given by
x3  0.165x 2  3.993 104  0

a) Use the bisection method of finding roots of equations to

find the depth x to which the ball is submerged under
water. Conduct three iterations to estimate the root of
the above equation.
b) Find the absolute relative approximate error at the end
of each iteration, and the number of significant digits at
least correct at the end of each iteration.

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 Example 1 Cont.
From the physics of the problem, the ball would be
submerged between x = 0 and x = 2R,
where R = radius of the ball,
that is
0  x  2R
0  x  20.055
0  x  0.11

Figure 6 Diagram of the floating ball

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 Example 1 Cont.
Solution

To aid in the understanding

of how this method works to
find the root of an equation,
the graph of f(x) is shown to
the right,
where
f x   x3  0.165x 2  3.993 10- 4

Figure 7 Graph of the function f(x)

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 Example 1 Cont.
Let us assume
x  0.00
xu  0.11
Check if the function changes sign between x and xu .
f xl   f 0  0  0.1650  3.993 104  3.993 104
3 2

f xu   f 0.11  0.11  0.1650.11  3.993 104  2.662 104

3 2

Hence
  
f xl  f xu   f 0 f 0.11  3.993 104  2.662 104  0

So there is at least on root between x and xu, that is between 0 and 0.11

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 Example 1 Cont.

Figure 8 Graph demonstrating sign change between initial limits

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 Example 1 Cont.
Iteration 1 x  xu 0  0.11
The estimate of the root is xm    0.055
2 2

f xm   f 0.055  0.055  0.1650.055  3.993 104  6.655 105

3 2

  
f xl  f xm   f 0 f 0.055  3.993 104 6.655 105  0

Hence the root is bracketed between xm and xu, that is, between 0.055
and 0.11. So, the lower and upper limits of the new bracket are
xl  0.055, xu  0.11
At this point, the absolute relative approximate error a cannot be
calculated as we do not have a previous approximation.

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 Example 1 Cont.

Figure 9 Estimate of the root for Iteration 1

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 Example 1 Cont.
Iteration 2 x  xu 0.055  0.11
The estimate of the root is xm    0.0825
2 2
f xm   f 0.0825  0.0825  0.1650.0825  3.993  10 4  1.622  10 4
3 2

f xl  f xm   f 0.055 f (0.0825)   1.622  10 4 6.655  10 5   0

Hence the root is bracketed between x and xm, that is, between 0.055
and 0.0825. So, the lower and upper limits of the new bracket are
xl  0.055, xu  0.0825

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 Example 1 Cont.

Figure 10 Estimate of the root for Iteration 2

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 Example 1 Cont.
The absolute relative approximate error a at the end of Iteration 2 is

xmnew  xmold
a  new
100
xm
0.0825  0.055
 100
0.0825
 33.333%
None of the significant digits are at least correct in the estimate root of
xm = 0.0825 because the absolute relative approximate error is greater
than 5%.

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 Example 1 Cont.
Iteration 3 x  xu 0.055  0.0825
The estimate of the root is xm    0.06875
2 2
f xm   f 0.06875  0.06875  0.1650.06875  3.993 104  5.563 105
3 2

  
f xl  f xm   f 0.055 f 0.06875  6.655 105  5.563 105  0

Hence the root is bracketed between x and xm, that is, between 0.055
and 0.06875. So, the lower and upper limits of the new bracket are
xl  0.055, xu  0.06875

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 Example 1 Cont.

Figure 11 Estimate of the root for Iteration 3

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 Example 1 Cont.
The absolute relative approximate error a at the end of Iteration 3 is

xmnew  xmold
a  new
100
xm
0.06875  0.0825
 100
0.06875
 20%
Still none of the significant digits are at least correct in the estimated
root of the equation as the absolute relative approximate error is
greater than 5%.
Seven more iterations were conducted and these iterations are shown in
Table 1.
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 Table 1 Cont.
Table 1 Root of f(x)=0 as function of number of iterations for
bisection method.
Iteration x xu xm a % f(xm)

1 0.00000 0.11 0.055 ---------- 6.655×10−5

2 0.055 0.11 0.0825 33.33 −1.622×10−4
3 0.055 0.0825 0.06875 20.00 −5.563×10−5
4 0.055 0.06875 0.06188 11.11 4.484×10−6
5 0.06188 0.06875 0.06531 5.263 −2.593×10−5
6 0.06188 0.06531 0.06359 2.702 −1.0804×10−5
7 0.06188 0.06359 0.06273 1.370 −3.176×10−6
8 0.06188 0.06273 0.0623 0.6897 6.497×10−7
9 0.0623 0.06273 0.06252 0.3436 −1.265×10−6
10 0.0623 0.06252 0.06241 0.1721 −3.0768×10−7

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 Table 1 Cont.
Hence the number of significant digits at least correct is given by the
largest value or m for which
a  0.5 10 2 m
0.1721  0.5 10 2 m
0.3442  10 2 m
log 0.3442   2  m
m  2  log 0.3442   2.463
So
m2
The number of significant digits at least correct in the estimated root
of 0.06241 at the end of the 10th iteration is 2.

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 Advantages
 Always convergent
 The root bracket gets halved with each
iteration - guaranteed.

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 Drawbacks
 Slow convergence
 If one of the initial guesses is close to
the root, the convergence is slower

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 Drawbacks (continued)
 If a function f(x) is such that it just
touches the x-axis it will be unable to find
the lower and upper guesses.
f(x)

f x   x 2

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 Drawbacks (continued)
 Function changes sign but root does not
exist

1
f(x)
f x  
x
x

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