Roots of nonlinear equation

2 3 4 1 0 r r r r _______________(2.1)

Figure (2.1)

Let r1 lie a long the x-axis. Equation (2.1) can be written as two scalar

equations, corresponding to the x, and y components of the r vectors. Thus,

r2cos( 2) + r3cos( 3) + r4cos( 4) r1=0 ___________(2.2a)

r2sin( 2) + r3 sin( 3) + r4 sin( 4) =0 ___________(2.2b)

combining equation (2.2a) and (2.2b) , letting 2= and 4= + thus.

R1cos( ) R2cos( ) + R3 cos( - ) =0 ____________(2.3)

Where

R1=r1/r2 , R2=r1/r4 , R3=(r1

2+r2

2+r3

2+r4

2) / (2r2r4) _______(2.4)

Roots of nonlinear equation Chapter 3

Nizar Salim 2 lecture 1

Consider the particular four bar linkage specified by

r1=10 , r2=6 , r3=8 and r4=4

which is illustrated in figure (2.1). Thus, equation (2.3) becomes:

(5/3)cos( ) (5/2)cos( ) + (11/6) cos( - ) =0 ____________(2.5)

This problem will be used in this chapter to illustrate methods of solving for

the roots of nonlinear equations.

Solving for the zeros of an equation, a process known as root finding, is

one of the oldest problems in mathematics.

There are two distinct phases in finding the roots of a nonlinear equation:

1) bounding the solution and 2) refining the solution.

Bounding the solution involves finding a rough estimate of the solution

that can be used as the initial approximation or the starting point, in a

systematic procedure that refines the solution to a specified tolerance in an

efficient manner if possible, the root should be bracketed between two points

at which the value of the nonlinear function has opposite signs.

Several possible bounding procedures, one of there is (Graphing the

function).

Graphing the function involves plotting the nonlinear function over the

range of interest. As an example of graphing a function to bound a root,

consider the for bar linkage problem consider an input of =40 deg. The

graph of equation (2.5) with =40 deg is presented in figure (2.2)

Roots of nonlinear equation Chapter 3

Nizar Salim 3 lecture 1

Figure (2.2)

The graph shows that there are two roots of equation (2.5) when =40 deg:

one root between =30 deg and =40 deg and one root between =350(or

-10) deg and 360 (or 0)deg

Refining the solution involves determining the solution to a specified

tolerance by an efficient systematic procedure. Several methods for refining

the solution are:

1. Trial and error.

2. Closed domain (bracketing) methods.

3. Open domain methods.

Trial and error methods simply guess the root, x= evaluate f(x), and

compare to zero. If f(x) is close enough to zero quit. If not, guess another ,

and continue until f(x) is close enough to zero. This approach is totally

unacceptable.

Closed domain (bracketing) methods are methods that start with two values

of (x) which bracket the root x= , and systematically reduce the interval

Roots of nonlinear equation Chapter 3

Nizar Salim 4 lecture 1

while keeping the root trapped with in the interval. Two such methods are

presented in section 2.3

Nonlinear equation can behave in various ways in the vicinity of a root.

Figure (2.3) illustrates several distinct type of behavior of nonlinear

equations in the vicinity of a root. Figure 2.3a illustrates a case of a single

real root, which is called a simple root. Figure 2.3b illustrates a case where

no real roots exist. Complex roots may exist in such a case. Situations with

two and three simple roots are illustrated in figure 2.3c and 2.3d

respectively. Situations with two and three multiple roots are illustrated in

figure 2.3e and 2.3f respectively. A situation with one simple root and two

multiple roots is illustrated in figure 2.3g. lastly, figure 2.3h illustrates the

general case where any number of simple or multiple roots can exist.

Roots of nonlinear equation Chapter 3

Nizar Salim 5 lecture 1

Figure (2.3)

Roots of nonlinear equation Chapter 3

Nizar Salim 6 lecture 1

Two of the simplest methods for finding the roots of a nonlinear

equation are:

1. Interval halving (Bisection).

2. False position.

One of the simplest methods for finding a root of a nonlinear

equation is interval halving (also known as bisection). In this method, two

estimates of the root, x=a to the left of the root and x=b to the right of the

root, which bracket the root, must first be obtained, as illustrated in figure

2.4

Figure 2.4

Which illustrates the two possibilities which f(x)>0 and f(x)<0.

The root, x= , obviously lies between a and b, that is in the interval (a,b).

The interval between a and b can be halved by averaging a and b. thus

c=(a+b)/2 .there are now two intervals: (a,c) and (c,b). The interval

containing the root, x= depends on the value of f(c). if f(a)*f(c) <0, which

is the case in figure (2.4a) the root is in the interval (a,c). thus set b=c and

continue. If f(a)*f(c)>0, which is the case in figure (2.4b), the root is in the

interval (c,b) thus , set a=c and continue. If f(a)*f(c)=0, c is the root.

The algorithm is as follows:

Roots of nonlinear equation Chapter 3

Nizar Salim 7 lecture 1

c=(a+b)/2 _____________(2.6)

if f(a)*f(c)<0 a=a and b=c ___(2.7a)

if f(a)*f(c)>0 a=c and b=b ___(2.7b)

a=30deg , b=40deg

f( a)=f(30)=(5/3)cos(40) (5/2)cos(30) + (11/6) cos(40- 30) = -0.03979719 ____(2.9a)

f( b)=f(40)=(5/3)cos(40) (5/2)cos(40) + (11/6) cos(40- 40) = 0.19496296 ____(2.9b)

from equation 2.6

c=( a+ b)/2 = (30+40)/2 =35deg _____________(2.10)

substituting c=35deg into equation (2.8) yields

f( c)=f(35)=(5/3)cos(40) (5/2)cos(35) + (11/6) cos(40- 35) = 0.06599926 ___(2.11)

since f( a)* f( c)<0, b= c for the next iteration and a remains the same.

The solution is presented in table 2.3

Table 2.1

the results in the table are rounded in the sixth digit after the decimal place.

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