MAB3053 Roots Bracketingbisection Week3 Lec5

7/26/2019 MAB3053 Roots Bracketingbisection Week3 Lec5

1/21

Chap5/1

CHAPTER 5 : ROOTS OF EQUATIONS

Bracketing Methods

LESSON PLAN

To calculate roots of equation using

Bracketing Methods:

1. Bisection Method

2. False Position Method

y

x

- when a function is zero

- when a function crosses x-axis

roots


2/21

Chap5/2

2 TYPES OF EQUATIONS

0... 011

1

fyfyfyf

n

i

n

i

AlgebraicEquations

- A functiony=f(x) is algebraic if it can be expressed in

the form

- Example: Quadratic equation 2x2 x 1= 0

TranscendentalEquations

- Non-algebraic equations

- Trigonometric, exponential, logarithmic are examples of

transcendental equations

- Example

01ln

2

x


3/21

Chap5/3

METHODS TO FIND ROOTS

-1

0

1

2

3

4

5

6

7

0.

0

0.4

0.

8

1.2

1.

6

2.

0

2.4

2.

8

3.2

3.

6

4.

0

x

y

plot the data points & connect them in a smooth curve

locate the points at which the curve crosses thex-axis

it provides rough estimate of theroots

useful visualisation tool

y= x23x+2

Try this in MATLAB:

>> f=inline(x^2-3*x+2)

>> fplot(f,[0,3])

roots


4/21

To find the roots of

Method 1:

You have previously learned that:

Method 2:

How

about


5/21

BRACKETING METHODS

2 types:

1.Bisection Method

2. False-Position Method

Exploit the fact that a

function typically changes

sign in the vicinity of the

root f(xL)f(xU) < 0

It is called bracketing

because two initial guesses

that bracket the root on

either side are required

Note:


6/21

Chap5/6

BISECTION METHOD

It is also known as Binary ChoppingorInterval Halving

In general, if there are two points xL(lower bound) and

xU(upper bound) such that f(xL)f(xU) < 0, then there is at least

one rootbetweenxLandxU

Next, the interval is

successivelybisected into

half

After each bisection, the

upper and lower bounds

are updated.

Interval is

bisected

into half


7/21

Chap5/7

BISECTION METHOD: 3-Step

ALGORITHM

STEP 1: Choose upperxUand lowerxLguesses such that

f(xL) f(xU)< 0

STEP 2: Estimate the root, xr

by dividing the interval [xL

xU

]

into half

STEP 3: Evaluate the next sub-interval

Iff(xL) f(xr) < 0, root lies in the lower interval, setxU=xr, return to step 2

Iff(xL) f(xr) > 0, root lies in the upper interval, setxL=xr, return to step 2

Iff(xL) f(xr) = 0, root equals toxr, terminate computation

2

ULr

xxx

Midpoint:


8/21

Chap5/8


f(xL) f(xU)< 0

root


9/21

Chap5/9

STEP 2: Estimate the root,xrby dividing the

interval [xLxU] into half

2

ULr

xxx

Midpoint:

Estimated root


10/21


If f(xL) f(xr) < 0, root lies in the lower interval, set

xU=xr, return to step 2

Case 1 Case 2

New iteration:xU=xr


11/21

Chap5/11

If f(xL) f(xr) > 0, root lies in the upper interval, set

xL=xr, return to step 2

Case 1 Case 2

New iteration:xL=xr


12/21

If f(xL) f(xr) = 0, root equals to xr, terminate

computation

Case 1 Case 2

Root =xr


13/21

Chap5/13

BISECTION METHOD

STOPPING CRITERION

2 ULnewr

xxx

s100

new

r

old

r

new

ra

x

xx

When to stop the iteration if it is hard to get f(xL) f(xr) = 0 ?

Apply the convergence (stopping) criterionas follows:

If error |a| < stopping error |s| , then root= xrnew

where xrnew is the root for the current iteration,

s

LU

LUa

xx

xx

100or

only valid for

Bisection method (p121)


14/21

Chap5/14

CLASS ACTIVITY

Use analytical method to determine the radius, r of a

cylindrical can so that it holds 400 cm3 of liquid. Given

r=h/3,wherehis the height.

Now, use 3 iterations of the bisection method to determine

the radius of the can. Use initial guesses of rL=3 cm and

rU=4 cm. Also, calculate the estimated error,

aand the true

percent relative error, tafter each iteration.


15/21

Chap5/15

BISECTION METHOD

PROS AND CONS

Pros

Easy and simple

Can always find a

single root

Convergence is

guaranteed

Cons

Slow to converge

Must know xL

and xU

that bound

the root

Cannot detect multiple roots

No account is taken of the

magnitudes of f(xL) and f(xU). Iff(xL) is closer to zero, it is likely

that the root is closer toxLthan to

xU use False-Position Method


16/21

Chap5/16

FALSE-POSITION METHOD

Iff(xL) is closer to zero, it is likely

that the rootxris closer toxL

The curve is replaced by a straight

line that crosses the x-axis & givesfalse position atxr

Based on similar triangle

principle, we can approximate the

solution by using linearinterpolationbetween the lower &

upper bounds to find the root,xr

Gives better estimate than

Bisection method

)()(

))((

:forrearrange

0)()()(

uL

uLuur

r

u

ru

Lu

Lu

xfxf

xxxfxx

x

xf

xx

xfxf

xx

Lower bound

Upper bound


17/21


f(xL) f(xU)< 0

STEP 2: Use similar triangle to interpolate the new estimate

of the root


Iff(xL) f(xr) < 0, root lies in the lower interval, setxU=xr, return to step 2

Iff(xL) f(xr) > 0, root lies in the upper interval, setxL=xr, return to step 2

Iff(xL) f(xr) = 0, root equals toxr, terminate computation Chap5/17

FALSE-POSITION METHODALGORITHM

uL

uLuur

xfxf

xxxfxx

Refer to Box 5.1

(p125)


18/21

Stopping criterion : Check for convergence criterion

whereby error, a< s. If not, go back toSTEP 2.

Why this method ?

converge faster than Bisection methodalways converges to a single root

Chap5/18

FALSE-POSITION METHOD

STOPPING CRITERION

s%100

new

r

old

r

new

ra

x

xx


19/21

Chap5/19

CLASS ACTIVITY

Use 3 iterations of false-position method to find a root of

equationx3 + 4x2 10 =0. Employ initial guesses of

xL=1 and xU=2. Calculate the approximate error, aand the

true error, tafter each iteration. True value = 1.36523.

uL

uLuur

xfxf

xxxfxx


20/21

False-position

Faster than Bisection method

Always converge to a single

root Step 1:f(xL)f(xu) < 0

Step 2:

Step 3:

Chap5/20

SUMMARY: BRACKETING METHODS

uLuLu

ur

xfxf

xxxfxx

Bisection

Easy and simple

Always converge to a

single root Step 1:f(xL)f(xu) < 0

Step 2:

Step 3:

2 UL

r

xxx

rUrL xxxfxf set,0If

rLrL xxxfxf set,0If

rUrL xxxfxf set,0If

rLrL xxxfxf set,0If


21/21

Chap5/21

Thank You

MAB3053 Roots Bracketingbisection Week3 Lec5

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