Lec 6 Bisection Method

7/27/2019 Lec 6 Bisection Method

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Solution of Nonlinear Equations

Topic: Bisection method

Dr. Nasir M Mirza

Numerical Methods

Email: [email protected]



Bisection Method

• The method is known as the Bolzano methodand can be called interval halving technique.

• The method is simple and straight-forward.

• Given a bracketed root, the method repeatedlyhalves the interval while continuing to bracketthe root and it will converge on the solution.







Step 3

Now check the following

If f(x l ) f(x m ) < 0, then the root

lies between x l and x m ;

then x l = x l ; x u = x m .

If f(x l ) f(x m ) > 0, then the root

lies between x m and x u ;

then x l = x m ; x u = x u .

If f(x l ) f(x m ) = 0; then the root

is x m. Stop the algorithm

if this is true.

x

f(x)

xu x

xm







Example

• You are working for ‘DOWN THE TOILET COMPANY’ thatmakes floats for ABC commodes. The ball has a specific

gravity of 0.6 and has a radius of 5.5 cm. You are asked to

find the distance to which the ball will get submerged when

floating in water.



Solution

The equation that gives the depth ‘x’ to which the ball issubmerged under water is given by

42310x99331650

-.+ x.- x x f

Use the Bisection method of finding

roots of equations to find the depth ‘x’ to which the ball is submergedunder water. Conduct threeiterations to estimate the root of the above equation.



Graph of function f(x)

423 10x99331650 -.+ x.- x x f



Checking if the bracket is valid

Choose the bracket

4

4

10x662.211.0

10x993.30.0

11.0

00.0

f

f

x

x

u



Iteration #1

055.02

11.00

11.0,0

m

u

x

x x

5

4

4

10x655.6055.0

10x662.211.0

10x993.30

f

f

f

11.0

055.0

u x

x



Iteration #2

%33.33

0825.02

11.0055.0

11.0,055.0

a

m

u

x

x x

0825.0,055.0

10x62216.10825.0

10x662.211.0

10x655.6055.0

4

4

5

u x x

f

f

f



Iteration #3

06875.02

0825.0055.0

0825.0,055.0

m

u

x

x x

5

4

5

10x5632.506875.0

10x62216.10825.0

10x655.6055.0

%20

f

f

f

a



Convergence

Table 1: Root of f(x)=0 as function of number of iterations for bisection method.

Iteration x xu xm a% f(xm)

1

2

3

4

5

6

7

8

9

10

0.00000

0.055

0.055

0.055

0.06188

0.06188

0.06188

0.06188

0.0623

0.0623

0.11

0.11

0.0825

0.06875

0.06875

0.06531

0.06359

0.06273

0.06273

0.06252

0.055

0.0825

0.06875

0.06188

0.06531

0.06359

0.06273

0.0623

0.06252

0.06241

----------

33.33

20.00

11.11

5.263

2.702

1.369

0.6896

0.3436

0.1721

6.655x10-5

-1.6222x10-4

-5.5632x10-5

4.4843x10-6

-2.5939x10-5

-1.0804x10-5

-3.1768x10-6

6.4973x10-7

-1.2646x10-6

-3.0767x10-7



Bisection Method

DO while 0.5*|x1 - x2| >= tolerance_value

Set xmid =(x1+ x

2)/2

IF f(xmid) of opposite sign of f(x1);

Set x2 = xmid;

ELSE

Set x1 = xxmid;

ENDIF

END loop

A basic loop that is used to find root between two points for afunction f(x) is shown here.



Bisection Method

DemoBisect: the example program does a simple bisection for acubic polynomial equation.



Bisection Method

• Example Problem:y(x) = a5 x5 + a4 x4 + a3 x3 + a2 x2 + a1 x + a0

• Let us consider constants to be following:

a0 = -2, a1 = -3, a2 = 4, a3 = 1, a4 = 0, a5 = 1

Then the function:

y(x) = x5 + x3 + 4x2 - 3x - 2

First we plot this function in a range from -3 to 3 andsee visually where roots are located.



Bisectional Method

There are 3 roots

(a) -2 < x < -1

(b) -1 < x < 0

(c) 0.5 < x <1.5

-2 -1 0 1 2

-40

-20

0

20

40

60

y ( x )

x

The graph for thefunction is shown here.



Computer Program for Bisection method

% A program in matlab

% bisection method to find the roots of x^5+x^3+4*x^2-3*x-2=0

xleft = -2.0 ; xright = -1.0 ; n = 100

% Input: xleft,xright = left and right brackets of the root

% n = (optional) number of iterations; default: n =15

% Output: x = estimate of the root

if nargin<3, n=15; end % Default number of iterations

a = xleft; b =xright; % Copy orig bracket to local variablesfa = a^5 + a^3 +4*a^2 - 3*a - 2; % Initial values

fb = b^5 + b^3 +4*b^2 - 3*b - 2; % Initial values

fprintf(' k a xmid b f(xmid)\n');

for k=1:n

xm = a + 0.5*(b-a); % Minimize roundoff in the midpoint

fm = xm^5 + xm^3 +4*xm^2 - 3*xm - 2; % f(x) at midpoint

fprintf('%3d %12.8f %12.8f %12.8f %12.3e\n',k,a,xm,b,fm);

if sign(fm) == sign(fa)

a = xm; fa = fm;

else

b = xm; fb = fm;

end

end



Bisection Method

k a xmid b f(xmid)

1 -2.00000000 -1.50000000 -1.00000000 5.313e-0012 -2.00000000 -1.75000000 -1.50000000 -6.272e+0003 -1.75000000 -1.62500000 -1.50000000 -2.184e+0004 -1.62500000 -1.56250000 -1.50000000 -6.748e-0015 -1.56250000 -1.53125000 -1.50000000 -3.612e-0026 -1.53125000 -1.51562500 -1.50000000 2.562e-0017 -1.53125000 -1.52343750 -1.51562500 1.122e-001

8 -1.53125000 -1.52734375 -1.52343750 3.860e-0029 -1.53125000 -1.52929688 -1.52734375 1.379e-00310 -1.53125000 -1.53027344 -1.52929688 -1.734e-00211 -1.53027344 -1.52978516 -1.52929688 -7.971e-00312 -1.52978516 -1.52954102 -1.52929688 -3.294e-00313 -1.52954102 -1.52941895 -1.52929688 -9.572e-00414 -1.52941895 -1.52935791 -1.52929688 2.108e-00415 -1.52941895 -1.52938843 -1.52935791 -3.732e-004

16 -1.52938843 -1.52937317 -1.52935791 -8.117e-00517 -1.52937317 -1.52936554 -1.52935791 6.482e-00518 -1.52937317 -1.52936935 -1.52936554 -8.174e-00619 -1.52936935 -1.52936745 -1.52936554 2.832e-00520 -1.52936935 -1.52936840 -1.52936745 1.008e-005

For range of x = [-2, -1]; we see x = -1.5293684 as root here.



Advantages & Disadvantages

Advatages: • Always convergent

• The root bracket gets halved with each iteration -guaranteed.

Disadvatanges:

• Slow convergence

• If one of the initial guesses is close to the root, theconvergence is slower



Drawbacks (continued)

• If a function f(x) is such that it just touches the x-axis it will beunable to find the lower and upper guesses.

2 x x f

-10 -5 0 5 10

0

20

40

60

80

100

y ( x )

x

y = x2



Drawbacks (continued)

Function changes sign but root does not exist

x

x f 1

-0.9 -0.6 -0.3 0.0 0.3 0.6 0.9

-100

-80

-60

-40

-20

0

20

40

60

80

100

y ( x )

x

y = 1/x

Documents

Lec 6 Bisection Method