Numerical Calculation
Part 1: Equations &Roots
(Bisection's Method)
Dr.Entesar Ganash
Bisection's Method
A general idea: This is another numerical method , which could be used to solve a
general equation f(x)=0. The method depends on the intermediate value theorem for
continuous functions. If a function f(x) is continuous on the interval [a, b] and sign of
f(a) ≠ sign of f(b), then: there is a value c ∈ [a..b] such that: f(c) = 0 i.e., there is a root
c in the interval [a, b]
f(x)
In [a,b]
x
From: http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/bisection.html
•It is a successive approximation method that makes an interval, which has a root of
the function f(x) narrow.
• It will bisect the interval in two parts and test which one has a function root.
• It will continue bisect the interval until the resulting interval becomes really small.
Then the root is approximately equal to any value in the final interval.
• The efficient thing of Bisection's method is number of bisections, which is needed
for getting a good accuracy can be calculated before beginning as will be seen below.
A Structure Plan:
A structure plan to execute Bisection's method:
1. Estimate a required accuracy (Epsilon).
2.
try to find two initial values of x ( and ) such that
that
means both functions have different signs. So, at least one root must lie some
where between them.
3.
Calculate maximum bisection N from inequality
4.
Repeat N times (a) & (b) steps
a) Find the mid point, call it
It is given by
b) Test whether
. .then let
5. Print root
N
log( xL  xR / Eps)
log( 2)
, which is the estimated root of the interval [
, also Find
.
if this condition is right
otherwise let
and then end a program .
, ].
Example
Write a program using Bisection’s method to solve this equation
starting with two guesses repetitions.
Solving
PROGRAM BISECTION
IMPLICIT NONE
Integer::k,N
! a counter & a maximum bisection
Real :: eps ,XL,XR !required accuracy (epsilon)& two initial guesses
Real :: xm
! a mid point
Eps=1E-6
xL=1.0 ; xR=2.0
N=log(ABS(XL-XR)/eps)/log(2.0)+1 !N must exceed formula value
DO K=1,N
XM=(XL+XR)/2.0
If(F(XL)*F(XM) <0.0)THEN
XR=XM
Else
XL=XM
END IF
END DO
Write(*,*)'Number of bisection:‘ N
Write(*,*)'The root is approximately =',XM
Contains
!-------------------------!Function Subprogram f is to solve f(x)=0
!-------------------------REAL FUNCTION f(x)
Implicit none
Real::f,x
f=x**3+x-3
End FUNCTION f
!-------------------------END PROGRAM BISECTION
continue
Solving
1
The result:
Exercise
Write a program compute the position( x & y co-ordinate) and the velocity (magnitude
&direction) of projectile, given t, the time since launch, u, the launch velocity, a, the initial angle
of launch(in degrees) and g, the acceleration due gravity.
The horizontal & vertical displacement are given by the formulae
x  ut cos a, y  ut sin a  gt 2 / 2
The velocity has magnitude V such that
V  Vx2  Vy2
Where its horizontal & vertical components
Vx  u cos a, V y  u sin a  gt
And V makes an angle  with the ground such that
tan  
Vy
Vx
References
•Hahn, B.D., 1994, Fortran 90 For Scientists and Engineers, Elsevier.
•http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/bisection.html
•http://mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/bracketing
%20methods/bisection/bisection.html
•Univ.,