Bisection Method [Linear Convergence]
1)
Decide initial values for x1
and x2 and accuracy level E.
2)
Compute f1 = f(x1)
and f2 = f(x2).
3)
If (f1*f2) >
0, x1 and x2 do not bracket any root and go to step number
7; otherwise continue.
4)
Compute x0 = (x2-x1)/
x2 and compute f0 = f(x0).
5)
If (f1 = f0)
> 0
{
Set x1 = x0
Set f1 = f0
}
Otherwise
{
Set x2 = x0
}
6)
If absolute value of (x2-x1)/
x2 < E
{
root
= (x2-x1)/2 and write the value of root
go
to step number 7.
}
Otherwise
{
go
to step number 4.
}
7)
Stop.
Example:
Find the root of the equation x2-4x-10 = 0, where root lies between 5
and 6 using bisection method.
Soln: Here given, f(x) = x2-4x-10 and root lies between 5 and 6.
Step 1: Initial value for x1 = 5, x2 = 6 and accuracy
level (error criteria) E = 0.01(you can assume yourself as per your need).
Step 2: Computing f1 and
f2
f1 =
f(x1) = f(5) = 52-4*5-10 = 25 – 20 – 10 = -5
f2
= f(x2) = f(6) = 62-4*6-10 = 36 -24 -10 = 2
Step 3: Checking Existence of root
f1*f2
= -5*2 = -10
It shows that f1*f2
< 0. Thus required root of equation exists between 5 and 6.
Step 4: Iteration
I:
Computing
mid-point x0 = (x1+x2)/2 and compute f0
= f(x0).
x0 =
(5+6)/2 = 11/2 = 5.5
f0
= f(5.5) = (5.5)2 - 4*5.5 – 10 = -1.75
Step 5: Checking, where (left or
right) root lays form mid-point?
f1 * f0 = -5 *
-1.75 = 8.75
Here f1 * f0 > 0 so,
x1 = x0 = 5.5 and f1 = f0
= -1.75
Step 6: Checking error criteria, absolute
value of (x2-x1)/ x2 < E
(x2-x1)/x2 = (6-5.5)/6 =
0.0833
Or 0.0833 < 0.01
Error criteria not satisfied. So go to step
number 4.
Iteration II:
Computing
mid-point x0 = (x1+x2)/2 and compute f0
= f(x0).
…
Repeat the iterations until error
criteria satisfied.
If error criteria is satisfied then,
Compute root using,
root = (x2-x1)/2 and write the value of
root and stop.
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