FORTRAN 90
Lecturer : Rafel Hekmat Hameed
University of Babylon
Subject : Fortran 90
College of Engineering
Year : Second B.Sc.
Mechanical Engineering Dep.
S
SO
OL
LU
UTTIIO
ON
N O
OF
F N
NO
ON
N-L
LIIN
NE
EA
AR
R E
EQ
QU
UA
ATTIIO
ON
N
False Position Method
This method attempts to solve an equation of the form f(x)=0. (This is
very common in most numerical analysis applications.) Any equation can be
written in this form.
The algorithm requires a function f(x) and two points a and b for which
f(x) is positive for one of the values and negative for the other. We can write
this condition as f(a)f(b)<0.
If the function f(x) is continuous on the interval [a,b] with f(a)f(b)<0,
the algorithm will eventually converge to a solution.
The idea for the False position method is to connect the points (a,f(a)) and
(b,f(b)) with a straight line.
Since linear equations are the simplest equations to solve for find the regulafalsi point (xrfp) which is the solution to the linear equation connecting the
endpoints.
Look at the sign of f(xrfp):
If sign (f(xrfp)) = 0 then end algorithm
else If sign(f(xrfp)) = sign(f(a)) then set a = xrfp
else set b = xrfp
ϭ
Regula Falsi (Method of False Position) is a modification of the
Bisection Method.. Instead of halving the interval on which there exists a root#_>ADF;;==@
r of f , we use the root of the line joining the points(a1,f(a1)) and (b1,f(b1)) as
out approximation to .
By similar triangles we have that,
and so,
If f(c1)=0,, then we have found a solution and may stop looking.
If f(a1) and f(c1) have opposite signs, then a root must lie on [a1,c1], and so we
may shorten our interval by b1-c1. We set b1=c1 and leave a as is. The points
a1 is stationary.
If f(a1) and f(c1) have the same sign, then a root must lie on [c1,b1], and so we
may shorten our interval by c1-a1, and assign a1=c1.
Ϯ
program false_position
implicit none
real::a,b,c,x,f,error=1e-5
f(x)= x**3 - 2
10 read(*,*)a,b
If (f (a) * f (b).gt.0) goto 10
20 c=b-(f(b)*(b-a))/(f(b)-f(a))
If (f(c)==0)t hen
print*,"the root =",c
elseif (f(c)*f(a)<0) then
b=c
else
a=c
;
endif
if (abs(b-a).lt.error) then
print*,"the root=",c
else
;
goto 20
endif
end
ϯ