Chapter 2 Solutions of Equations in One Variable
2.1 The Bisection Method
In this section, we consider one of the most basic problems of numerical
approximation, the root-finding problem. This process involves finding a root, or
solution, of an equation of the form f  x   0 , for a given function f . A root of
this equation is also called a zero of the function f .
The first technique, based on the Intermediate Value Theorem, is called the
Bisection, or Binary-search, method.
Intermediate Value Theorem (IVT): If f  C  a, b and K is any number
between f  a  and f  b  , then there exists a number c in
 a, b 
for which
f c  K .
Suppose f is a continuous function defined on the interval  a, b , with f  a 
and f  b  of opposite sign. By the IVT, there exists a number p in
 a, b 
with
f  p   0 . Although the procedure will work when there is more than one root in the
interval
 a, b  , we assume for simplicity that the root in this interval is unique. The
method calls for a repeated halving of subintervals of  a, b and, at each step,
locating the half containing p .
To begin, set a1  a and b1  b , and let p1 be the midpoint of  a, b ; that is
p1  a1 
b1  a1 a1  b1

.
2
2
If f  p1   0 , then p  p1 , and we are done. If f  p1   0 , then f  p1  has the
same sign as either f  a  or f  b  . When f  p1  and f  a1  have the same sign,
p   p1 , b1  , and we set a2  p1 and b2  b1 . When f  p1  and f  a1  have
opposite sign, p   a1 , p1  , and we set a2  a1 and b2  p1 . We then reapply the
1
process to the interval  a2 , b2  . This produces the method described in Algorithm 2.1.
Algorithm 2.1 Bisection
PURPOSE: To find a solution to F(x)=0 given the continuous function f on the
interval [A,B], where F(A) and F(B) have opposite signs
INPUT: endpoints: A,B, tolerance: TOL>0.0, maximum number of iterations N
OUTPUT:
approximation solution P or message that the algorithm fails
Step 1 Set I=1
FA = F(A); FB=F(B)
Step 2 While I  N do Steps3-6
Step 3 Set P=A+(B-A)/2; FP=F(P).
Step 4 IF FP=0 or (B-A)/2<TOL THEN
OUTPUT (P) (Procedure completed successfully)
STOP
END IF
Step 5 Set I=I+1
Step 6 IF FA FP  0 THEN
Set
A=P; FA=FP
ELSE
Set B=P; FB=FP
END IF
Step 7 OUTPUT (Method failed after N iteration)
STOP
We can select a tolerance (TOL>0) and generate p1 , p2 , , pi until one of the
following conditions (stopping criteria) is met:
(bi  ai )
 TOL ,
2
( pi  pi 1 )  TOL ,
( pi  pi 1 )
pi
 TOL ,
2
f  pi   TOL
This produces the method (Algorithm 2.1) described in the following SUBROUTINE.
SUBROUTINE BISECTION(N,A,B,TOL,P)
C=====================================================
C ALGORITHM 2.1 Bisection Method
C PURPOSE
C
To find a solution to F(x)=0 given the continuous function f
C
on the interval [A,B], where F(A) and F(B) have opposite signs
C
INPUT
C
endpoints: A,B
C
tolerance: TOL>0.0
C
maximum number of iterations N
C OUTPUT
C
approximation solution P or message that the algorithm fails
C------------------------------------------------------------------------------------------C
INTEGER N
REAL A,B,TOL,P
REAL FA,FB,FP
EXTERNAL F
INTRINSIC ABS
C
FA = F(A)
FB = F(B)
IF (FA*FB.GT.0.0) THEN
WRITE(*,*) 'FA and FB are the same signs'
STOP
END IF
DO 10 I = 1,N
P = A+(B-A)/2.0
FP = F(P)
C *** Print the Bisection Algorithm gives the values ***
9
WRITE(10,9) I,A,B,P,FP
FORMAT(1X,I5,F14.9,F14.9,F14.9,F14.9)
IF ((ABS(FP).LE.1.0E-20).OR.((B-A)/2.0.LT.TOL)) THEN
C *** Procedure completed successfully ***
RETURN
3
END IF
IF (FA*FP.GT.0.0) THEN
A= P
FA = FP
ELSE
B=P
FB = FP
END IF
10 CONTINUE
WRITE(*,*) 'Method failed after N iterations'
RETURN
END
Example 1 The equation f  x   x3  4 x2  10 has a root in [1,2] since f 1  5
and f  2  14 .
C *******************************************************************
C
Example 1 (Chapter 2.1) Using the Bisection Method to find a solution
C
accurate to within 0.0002 for f(x)=x^3+x^2-10 on [1,2]
C
C*******************************************************************
*
C
REAL A,B,TOL
INTEGER N
A = 1.0
B = 2.0
TOL = 0.0002
N = 30
CALL BISECTION(N,A,B,TOL,P)
WRITE(*,9) P
9
FORMAT(1X,'Approximation solution P:',F14.9)
STOP
END
REAL FUNCTION F(X)
C============================================================
C PURPOSE
C
Find the value of function f(x)=x^3+4x^2-10
C----------------------------------------------------------------------------------------------------C
F = X*X*X+4.0*X*X-10
4
RETURN
END
The Bisection Algorithm gives the values in Table 2.1:
Table 2.1
-----------------------------------------------------------------i
bi
ai
f  pi 
pi
-----------------------------------------------------------------1 1.000000000 2.000000000 1.500000000 2.375000000
2 1.000000000 1.500000000 1.250000000 -1.796875000
3 1.250000000 1.500000000 1.375000000 0.162109375
4 1.250000000 1.375000000 1.312500000 -0.848388672
5 1.312500000 1.375000000 1.343750000 -0.350982666
6 1.343750000 1.375000000 1.359375000 -0.096408844
7 1.359375000 1.375000000 1.367187500 0.032355785
8 1.359375000 1.367187500 1.363281250 -0.032149971
9 1.363281250 1.367187500 1.365234375 0.000072025
10 1.363281250 1.365234375 1.364257812 -0.016046692
11 1.364257812 1.365234375 1.364746094 -0.007989263
12 1.364746094 1.365234375 1.364990234 -0.003959102
13 1.364990234 1.365234375 1.365112305 -0.001943659
-----------------------------------------------------------------Approximation solution P: 1.365112305
The Bisection method, though conceptually clear, has significant drawbacks. It is
slow to converge (that is, N may become quite large before pi  pi 1 is sufficiently
small), and a good intermediate approximation can be inadvertently discarded.
However, the method has the important property that it always converges to a solution,
and for that reason it is often used as starter for the more efficient methods we will
present later in this chapter.
Theorem 2.1: Suppose that f  C  a, b and f  a   f  b   0 . The Bisection
method generates a sequence
 pi i 1
ba
,
2i
For each n  1, we have
ba
bn  an  n 1
2
pi  p 
Proof

approximating a zero p of f with
when i  1 .
and
5
p   an , bn  .
Since pn 
1
 an  bn  for all n  1, it follows that
2
1
ba
pn  p   bn  an   n .
2
2
Thus, the sequence
 1
is pn  p  O  n
2
 pn n1

 1 
converges to p with rate of convergence O  n  ; that
2 

.

It is important to realize that Theorem 2.1 gives only a bound for approximation
error and that this bound may be quite conservative. For example, this boundary
applied to the problem in Example 1 ensures only that
2 1
p  p9  9  2 103 ,
2
but the actual error is much smaller:
p  p9  1.365230013 1.365234375  4.4 106 .
Example 2 To determine the number of iterations necessary to solve
f  x   x3  4 x2  10 =0 with accuracy 0.001 using a1  1 and b1  2 requires
finding an integer i that satisfies
ba
pi  p  i  2 i  0.001 .
2
To determine i we will use logarithms. Although logarithms to any base would
suffice, we will use base-10 logarithms since the tolerance is given as a power of 10.
2i  0.001 implies that log10 2i  log10 103  3 , we have
i log10 2  3
and
i
3
 9.96.
log10 2
Hence, ten iterations will ensure an approximation accurate to within 0.001. It is
important to keep in in mine that the error analysis gives only a bound for the number
of iterations, and in many cases this bound is much larger than the actual number
required.
EXERCISE SET 2.1
Q4. Use the Bisection method to find solutions accurate to within 103 for
f  x   x4  2 x3  4 x2  4 x  4  0 on each interval (a) [-2,-1] (b) [-1,0] (c) [0,2]
(d) [2,3].
Q14. Let
f  x    x  1 ,
10
pn  1  1 n . Show that
p  1 and
6
f  pn   103
whenever n  1 but that p  pn  103 requires that n  1000 .
7