Download ASSIGNMENT ON NUMERIC ANALYSIS FOR ENGINEERS

ASSIGNMENTON
NUMERICANALYSIS
FORENGINEERS
Submitted by:
Tarek Salah Uddin Mahmud
BSSE0508
Details
about the
methods of
finding
roots in a
nonlinear
equation
Methods To Find The Roots Of A Function:
The solution or the values of ‘x’ of a given function f(x) are called the
roots of the equation f(x)=0.
To solve this problem, there are some methods devided into two catagories:
1. Bracating methods
2. Open methods
Bracating Methods: These methods require the limits between which the root
lies.
Example: 1. Bisection Method
2. Method of False Position (Regular Falsi Method)
Open Methods: These methods require the initial estimation of the solution.
Example: 3. Newton-Raphson Method (Newton’s method)
4. Successive Approximation Method.
Let’s discuss the methods in a short brief.
BISECTION METHOD:
Bisection method
After a root of f
Bisection method
accomplishes this
sufficiently small.
is also known as the interval halving method. Because
(x) = 0 has been bracketed in the interval (a, b).
can be used to close in on it. The Bisection method
by successfully halving the interval until it becomes
Algorithm:
 First Step: Define the first interval (a, b) such that solution exists
between them.
Check f (a) * f (b) < 0.
 Second Step: Compute the first estimate of the numerical solution ‘c’
by
c = (a+b)/2
 Third Step: Determine whether the true solution is between ‘a’ and ‘c’
or between ‘c’ and ‘b’
If f (a) * f (c) < 0, the true solution is between a and c.
If f (a) * f (c) > 0, the true solution is between c and b.
If |f(c)| = ∈, then accept c as the root and stop. ∈ is the
error tolerance, ∈ > 0.
 Fourth Step: Choose the subinterval that contains the true solution (a
to c or c to b) as the new interval (a, b) and go back to ‘Second
Step’.
‘Second Step’ to ‘fourth Step’ are repeated until a
specified tolerance or error bound is attained.
Program:
A program to solve the corresponding problem is given below:
#include <stdio.h>
#define f(x) ((x)*(x)*(x)-4*(x)-8.95)
int main ()
{
double a=2, b=3, eps=1.0e-3,c;
if((f(a)*f(b))>=0) return 0;
while(1){
c = (a+b)/2.0;
if(f(c)>-eps && f(c)<eps){
printf("%lf\n",c);
break;
}
if((f(a))*(f(c))<0) b=c;
else a=c;
}
return 0;
}
METHOD OF FALSE POSITION:
It is very similar to Bisection method. But rather than bisecting the interval
(a, b), it locates the root by joining f (a1) and
f (b1) with a
straight line. The intersection of this line with the x-axis represents an
improved estimate of the root.
Algorithm:
 First Step: Define the first interval (a, b) such that solution exists
between them.
Check f (a) f (b) < 0.
 Second Step: Compute the first estimate of the numerical solution ‘c’
by
c = ((a*f(b))-(b*f(a)))/(f(b)-f(a))
 Third Step: Determine whether the true solution is between ‘a’ and ‘c’
or between ‘c’ and ‘b’
If f (a) f (c) < 0, the true solution is between a and c.
If f (a) f (c) > 0, the true solution is between c and b.
If |f(c)| = ∈, then accept c as the root and stop. ∈ is the
error tolerance, ∈ > 0.
 Fourth Step: Choose the subinterval that contains the true solution (a
to c or c to b) as the new interval (a, b) and go back to ‘Second
Step’.
‘Second Step’ to ‘fourth Step’ are repeated until a
specified tolerance or error bound is attained.
Program:
A program to solve the corresponding problem is given below:
#include <stdio.h>
#define f(x) ((x)*(x)*(x)-4*(x)-8.95)
int main ()
{
double a=2, b=3, eps=1.0e-3,c;
if((f(a)*f(b))>=0) return 0;
while(1){
c = ((a*f(b))-(b*f(a)))/(f(b)-f(a));
if(f(c)>-eps && f(c)<eps)
{
printf("%lf\n",c);
break;
}
if((f(a))*(f(c))<0) b=c;
else a=c;
}
return 0;
}
NEWTON-RAPHSON METHOD:
The Newton-Raphson method is the best-known method of finding roots of
a function f (x) as it is simple and fast.
This method is used only in problems where f '(x)can be readily computed.
The solution process starts by selecting point x1 as the first estimate of the
solution. The second estimate x2 is found by drawing the tangent line to f
(x) at the point (x1, f (x1)) and determining the intersection point of the
tangent line with the x-axis. Thus the process goes on. The slope,
f '(x1) = (f(x1)-0)/(x1-x2)
Algorithm:
 First Step: Select a point x1 as an initial guess of the solution.
 Second Step: For i =1,2,3... Compute Xi+1 by
Xi+1 = (Xi - ((f(Xi))/( f ' (Xi))))
This step runs until the error is smaller than a specified
value. That means,
|( Xi+1 - Xi )/ Xi | <= ∈
Program:
A program to solve the corresponding problem is given below:
#include <stdio.h>
#define f(x) ((x)*(x)*(x)*(x)-11*(x)+8)
#define df(x) (4*(x)*(x)*(x)-11)
int main ()
{
double eps=1.0e-3, x1=2.0,x2;
while(1)
{
x2 = x1 - (f(x1)/df(x1)) ;
if(f(x2)>-eps && f(x2)<eps)
{
printf("%lf\n",x2);
break;
}
x1 = x2;
}
return 0;
}
SUCCESSIVE APPROXIMATION METHOD:
If an equation f(x) = 0 is given whose roots are to be determined, it can
be written in the form
x = f(x)
Let x = Xi be an initial approximation to the desired root. Then, the first
approximation Xi+1 is given by
Xi+1 = g(Xi)
This iterative sequence of solution is called Successive Approximation Method.
Algorithm:
 First Step: Select a point x1 as an initial guess of the solution.
 Second Step: Compute Xi+1 by
Xi+1 = g(Xi)
This step runs until the error is smaller than a specified
value. That means,
|( Xi+1 - Xi )/ Xi | <= ∈
Program:
A program to solve the corresponding problem is given below:
#include <stdio.h>
#define f(x) ((x)*(x)*(x)*(x)-11*(x)+8)
#define g(x) ((x)*(x)*(x)*(x)+8)/11
int main ()
{
double eps=1.0e-6, x1=2,x2;
while(1)
{
x2 = g(x1);
if(f(x2)>-eps && f(x2)<eps){
printf("%lf\n",x2);
break;
}
x1 = x2;
}
return 0;
}