BISECTION METHOD
Sunitha A C
Assistant Professor
Department of Computer Science and Applications
St. Mary’s College Thrissur
Bisection Method, Sunitha A C, St.Mary’s College
INTRODUCTION
 One of the simplest and most reliable iterative method
for the solution of non linear equations.
 Also known as binary chopping or half-interval method.
 Based on the repeated application of the intermediate
value theorem.
 The Bisection Method is given an initial interval [a,b]
that contains a root (We can choose a and b such that
f(a) and f(b) are of opposite sign)
 The Bisection Method will cut the interval into 2 halves
and check which half interval contains a root of the
function
 The Bisection Method will keep cut the interval in
halves until the resulting interval is extremely small.
The root is then approximately equal to any value
in the final interval.
Bisection Method, Sunitha A C, St.Mary’s College
 Graphical Representation
• Suppose the interval [a,b] is as follows:
Bisection Method, Sunitha A C, St.Mary’s College
We cut the interval [a,b] in the middle: m = (a+b)/2
Bisection Method, Sunitha A C, St.Mary’s College
Bisection Method, Sunitha A C, St.Mary’s College
Because sign of f(m) ≠ sign of f(a) , we proceed with the
search in the new interval [a,b]:
Bisection Method, Sunitha A C, St.Mary’s College
 Example execution:
Bisection Method, Sunitha A C, St.Mary’s College
• We will use a simple function to illustrate the execution of
the Bisection Method
• Function used:
f(x) = x3 – 5x+1
Bisection Method, Sunitha A C, St.Mary’s College
We will use the starting interval [0,1] since:
The interval [0,1] contains a root because: sign of f(0) ≠ sign
of f(1)
• f(0) = 03āˆ’5Ɨ0+1 = 1
• f(1) = 13āˆ’5Ɨ1+1 = -3
 Iteration 1
Let š‘„1 = 0 š‘Žš‘›š‘‘ š‘„2 = 1
then š‘„0=
š‘„1+ š‘„2
2
=
0+1
2
= 0.5
š‘“ 0.5 = āˆ’1.375 š‘Žš‘›š‘‘ š‘“(š‘„1)š‘“(š‘„0)Ė‚0
Thus, the root lies in the interval (0,0.5)
Bisection Method, Sunitha A C, St.Mary’s College
Bisection Method, Sunitha A C, St.Mary’s College
 Iteration 2
Let š‘„1 = 0 š‘Žš‘›š‘‘ š‘„2 = 0.5
then š‘„0 =
š‘„1+ š‘„2
2
=
0+0.5
2
= 0.25
š‘“ 0.25 = āˆ’0.234375 š‘Žš‘›š‘‘ š‘“(š‘„1))š‘“ š‘„0 Ė‚0
Thus, the root lies in the interval (0,0.25)
Bisection Method, Sunitha A C, St.Mary’s College
Iteration š’™ šŸ š’™ šŸ š’™ šŸŽ š’‡(š’™ šŸŽ)
1 0 1 0.5 Ė‚
2 0 0.5 0.25 Ė‚
3 0 0.25 0.125 ˃
4 0.125 0.25 0.1875 ˃
5 0.1875 0.25 0.21875 Ė‚
Hence,the root lies in (0.1875,0.21875).The approximate
root is taken as the midpoint of this interval,that is
0.203125.
Sequence of intervals is as follows:
Simple and easy to implement
One function evaluation per iteration
The size of the interval containing the zero is
reduced after each iteration
No knowledge of the derivative is needed
Slow to converge
Good intermediate approximations may be
discarded
DISADVANTAGES
ADVANTAGES
Bisection Method, Sunitha A C, St.Mary’s College
REFERENCE
1. Numerical Methods for Scientific and Engineering
Computation,M.K.Jain,S.R.K.Iyengar,R.K.Jain,New Age
International Publishers
2. Numerical Methods,E.Balagurusamy,Tata McGraw-Hill
Bisection Method, Sunitha A C, St.Mary’s College

BISECTION METHOD

  • 1.
    BISECTION METHOD Sunitha AC Assistant Professor Department of Computer Science and Applications St. Mary’s College Thrissur
  • 2.
    Bisection Method, SunithaA C, St.Mary’s College INTRODUCTION  One of the simplest and most reliable iterative method for the solution of non linear equations.  Also known as binary chopping or half-interval method.  Based on the repeated application of the intermediate value theorem.
  • 3.
     The BisectionMethod is given an initial interval [a,b] that contains a root (We can choose a and b such that f(a) and f(b) are of opposite sign)  The Bisection Method will cut the interval into 2 halves and check which half interval contains a root of the function  The Bisection Method will keep cut the interval in halves until the resulting interval is extremely small. The root is then approximately equal to any value in the final interval. Bisection Method, Sunitha A C, St.Mary’s College
  • 4.
     Graphical Representation •Suppose the interval [a,b] is as follows: Bisection Method, Sunitha A C, St.Mary’s College
  • 5.
    We cut theinterval [a,b] in the middle: m = (a+b)/2 Bisection Method, Sunitha A C, St.Mary’s College
  • 6.
    Bisection Method, SunithaA C, St.Mary’s College Because sign of f(m) ≠ sign of f(a) , we proceed with the search in the new interval [a,b]:
  • 7.
    Bisection Method, SunithaA C, St.Mary’s College
  • 8.
     Example execution: BisectionMethod, Sunitha A C, St.Mary’s College • We will use a simple function to illustrate the execution of the Bisection Method • Function used: f(x) = x3 – 5x+1
  • 9.
    Bisection Method, SunithaA C, St.Mary’s College We will use the starting interval [0,1] since: The interval [0,1] contains a root because: sign of f(0) ≠ sign of f(1) • f(0) = 03āˆ’5Ɨ0+1 = 1 • f(1) = 13āˆ’5Ɨ1+1 = -3
  • 10.
     Iteration 1 Letš‘„1 = 0 š‘Žš‘›š‘‘ š‘„2 = 1 then š‘„0= š‘„1+ š‘„2 2 = 0+1 2 = 0.5 š‘“ 0.5 = āˆ’1.375 š‘Žš‘›š‘‘ š‘“(š‘„1)š‘“(š‘„0)Ė‚0 Thus, the root lies in the interval (0,0.5) Bisection Method, Sunitha A C, St.Mary’s College
  • 11.
    Bisection Method, SunithaA C, St.Mary’s College  Iteration 2 Let š‘„1 = 0 š‘Žš‘›š‘‘ š‘„2 = 0.5 then š‘„0 = š‘„1+ š‘„2 2 = 0+0.5 2 = 0.25 š‘“ 0.25 = āˆ’0.234375 š‘Žš‘›š‘‘ š‘“(š‘„1))š‘“ š‘„0 Ė‚0 Thus, the root lies in the interval (0,0.25)
  • 12.
    Bisection Method, SunithaA C, St.Mary’s College Iteration š’™ šŸ š’™ šŸ š’™ šŸŽ š’‡(š’™ šŸŽ) 1 0 1 0.5 Ė‚ 2 0 0.5 0.25 Ė‚ 3 0 0.25 0.125 ˃ 4 0.125 0.25 0.1875 ˃ 5 0.1875 0.25 0.21875 Ė‚ Hence,the root lies in (0.1875,0.21875).The approximate root is taken as the midpoint of this interval,that is 0.203125. Sequence of intervals is as follows:
  • 13.
    Simple and easyto implement One function evaluation per iteration The size of the interval containing the zero is reduced after each iteration No knowledge of the derivative is needed Slow to converge Good intermediate approximations may be discarded DISADVANTAGES ADVANTAGES Bisection Method, Sunitha A C, St.Mary’s College
  • 14.
    REFERENCE 1. Numerical Methodsfor Scientific and Engineering Computation,M.K.Jain,S.R.K.Iyengar,R.K.Jain,New Age International Publishers 2. Numerical Methods,E.Balagurusamy,Tata McGraw-Hill Bisection Method, Sunitha A C, St.Mary’s College