solution of nonlinear equations topic: bisection method

Solution of Nonlinear Equations

Topic: Bisection method

Numerical AnalysisNumerical Analysis

Bisection Method• The method is known as the Bolzano method

and can be called interval halving technique.• The method is simple and straight-forward.• Given a bracketed root, the method repeatedly

halves the interval while continuing to bracket the root and it will converge on the solution.

Step 1

Choose xl and xu as two guesses for the root such that f(xl) f(xu) < 0, or in other words, f(x) changes sign between xl and xu.

x

f(x)

xu x

Step 2

x

f(x)

xu x

Estimate the root, xm of the equation f (x) = 0 as the mid-point between xl and xu as

xx

m = x u

2

Step 3Now check the following

If f(xl) f(xm) < 0, then the root lies between xl and xm; then xl = xl ; xu = xm.

If f(xl ) f(xm) > 0, then the root lies between xm and xu; then xl = xm; xu = xu.

If f(xl) f(xm) = 0; then the root is xm. Stop the algorithm if this is true.

x

f(x)

xu x

xm

Step 4

xx

m = xu

2

100

newm

oldm

new

a x

xxm

root of estimatecurrent newmx

root of estimate previousoldmx

New estimate

Absolute Relative Approximate Error

Step 5

Check if absolute relative approximate error is lessthan pre-specified tolerance or if maximum numberof iterations is reached.

Yes

No

Stop

Using the new upper and lower

guesses from Step 3, go to Step 2.

Example• You are working for ‘DOWN THE TOILET COMPANY’ that

makes 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.

SolutionThe equation that gives the depth ‘x’ to which the ball is

submerged under water is given by

423 10x99331650 -.+x.-xxf

Use the Bisection method of finding roots of equations to find the depth ‘x’ to which the ball is submerged under water. Conduct three iterations to estimate the root of the above equation.

Graph of function f(x)

423 10x99331650 -.+x.-xxf

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

xx

5

4

4

10x655.6055.0

10x662.211.0

10x993.30

f

f

f

11.0

055.0

ux

x

Iteration #2

%33.33

0825.02

11.0055.0

11.0,055.0

a

m

u

x

xx

0825.0,055.0

10x62216.10825.0

10x662.211.0

10x655.6055.0

4

4

5

uxx

f

f

f

Iteration #3

06875.02

0825.0055.0

0825.0,055.0

m

u

x

xx

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 + x2)/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 a function f(x) is shown here.

Bisection Method

DemoBisect: the example program does a simple bisection for a cubic 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 and see 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

The graph for the function 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; endend

Bisection Method k a xmid b f(xmid) 1 -2.00000000 -1.50000000 -1.00000000 5.313e-001 2 -2.00000000 -1.75000000 -1.50000000 -6.272e+000 3 -1.75000000 -1.62500000 -1.50000000 -2.184e+000 4 -1.62500000 -1.56250000 -1.50000000 -6.748e-001 5 -1.56250000 -1.53125000 -1.50000000 -3.612e-002 6 -1.53125000 -1.51562500 -1.50000000 2.562e-001 7 -1.53125000 -1.52343750 -1.51562500 1.122e-001 8 -1.53125000 -1.52734375 -1.52343750 3.860e-002 9 -1.53125000 -1.52929688 -1.52734375 1.379e-003 10 -1.53125000 -1.53027344 -1.52929688 -1.734e-002 11 -1.53027344 -1.52978516 -1.52929688 -7.971e-003 12 -1.52978516 -1.52954102 -1.52929688 -3.294e-003 13 -1.52954102 -1.52941895 -1.52929688 -9.572e-004 14 -1.52941895 -1.52935791 -1.52929688 2.108e-004 15 -1.52941895 -1.52938843 -1.52935791 -3.732e-004 16 -1.52938843 -1.52937317 -1.52935791 -8.117e-005 17 -1.52937317 -1.52936554 -1.52935791 6.482e-005 18 -1.52937317 -1.52936935 -1.52936554 -8.174e-006 19 -1.52936935 -1.52936745 -1.52936554 2.832e-005 20 -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,

the convergence is slower

Drawbacks (continued)• If a function f(x) is such that it just touches the x-axis it

will be unable to find the lower and upper guesses.

2xxf

Drawbacks (continued) Function changes sign but root does not exist

x

xf1