numerical methods - marmara Üniversitesi · 2019-03-07 · lecture 4: roots of equations -...

Numerical Methods

Mustafa YILMAZSchool of Mechanical Engineering

University of Marmara

Lecture 4 –Roots of Equations Bracketing Methods

MATH259 Numerical AnalysisLecture 4: Roots of Equations - Bracketting 2

Root Concept

• The root of an equation is the value of x that f(x) = 0.

• Roots are called the zeros of equation.

• There are many functions for which the root cannot be determined so easily.


Methods for Determining Roots

• Noncomputer methods.

• Computer methods.


Manual Methods (Graphical Method)

• Graphical method consists to plot the function and determines where it crosses the x axis. The x value for which f(x) = 0 (the root).

• This process is repeated until a guess is obtained that results in an f(x) than is close to zero.


Manual Methods (Trial and Error)

• Graphical method and trail and error are inefficient and inadequate for the requirements of engineering practice.

• Trial and error consists of guessing a value of x and evaluating whether f(x) is zero. If not , another guess is made, f(x) is again evaluated.


Example: the parachutist problem

Given m = 68.1 kg, v = 40 m/s, t = 10 s, g = 9.8 m/s2, find the corresponding c

Graphical Methods

40138667401168891 146843010168 −−=−−=−−= −−− )(.)().(.)()( .)./()/( cctmc ec

ec

vec

gmcf

)( )/( tmcec

gmv −−= 1

Make a plot of the function and observe where it crosses the x axis

Solution:

c f(c)4 34.115

8 17.653

12 6.067

16 -2.269

20 -8.401

Root is between 12 and 16


A root may occur in an interval

• f (xmin) and f (xmax) have the same sign.• (a) there will be no roots.• (c) there will be an even numbers of roots within

the interval.

Where : xmin: lower bound and xmax: upper bound


• f (xmin) and f (xmax) have different sign, • (b) and (d) there will be an odd numbers of roots

within the interval.

A root may occur in an interval

Where : xminl: lower bound and xmax: upper bound


Cases Where the General Rule does not Hold

• For this case, the end points are of the opposite signs,

• there are an even number of axis intersection for the interval.

• multiple roots that occur when the function is tangential to the x axis.


Cases Where the General Rule does not Hold

• (b) Discontinuous function where end points of opposite sign bracket an even number of roots.


Estimate Properties of Roots By Graphical Methods

From (a) and (c):

if both f(xl) and f(xu) have the same sign, there must be 0 or even number of roots

From (b) and (d): if f(xl) and f(xu) have different signs, there must be 1 or odd number of roots

Exceptions:

multiple roots

f(x) = (x-2)2(x-4)

discontinuous f(x)


Example: Interest Rate

Suppose you want to buy an electronic appliance from a shop and you can either pay an amount of 12,000 or have a monthly payment of 1,065 for 12 months. What is the corresponding interest rate?

1)1()1(−+

+= n

n

xxxPA

A is the monthly paymentP is the loan amountx is the interest rate per period of timen is the loan period

To find the yearly interest rate, x, you have to find the zero of

1)1()1(000,12065,1 12

12

121212

−++

−x

xx

We know the payment formulae is:


Finding Roots Graphically

• Not accurate

• However, graphical view can provide useful info about a function.– Multiple roots?– Continuous?– etc.


The following root finding methods will be introduced:

A. Bracketing MethodsA.1. Bisection MethodA.2. Regula Falsi

B. Open MethodsB.1. Fixed Point IterationB.2. Newton Raphson's MethodB.3. Secant Method


Bracketing Methods

• Bracketing methods consider the fact that a function typically changes sign in the vicinity of the root.

• In this process two initial guesses for the root are required.

• These two initial guesses are called xmin(lower, xl) and xmax (upper, xu).


Bracketing Methods

• These guesses must “bracket” or be on either side of, the root.

• The particular methods employ different strategies to systematically reduce the width of the bracket, and find the correct answer.


Bracketing Methods

By Mean Value Theorem,

we know that if a function f(x) is

continuous in the interval [a, b]

and f(a)f(b) < 0, then the

equation f(x) = 0 has at least one

real root in the interval (a, b).


Usually• f(a)f(b) > 0 implies zero or even

number of roots – [figure (a) and (c)]

• f(a)f(b) < 0 implies odd number of roots– [figure (b) and (d)]


Exceptional Cases• Multiple roots

– Roots that overlap at one point.– e.g.: f(x) = (x-1)(x-1)(x-2) has a

multiple root at x=1.

• Functions that discontinue within the interval


Algorithm for bracketing methods

Step 1: Choose two points xl and xu such that f(xl)f(xu) < 0

Step 2: Estimate the root xr (note: xl < xr < xu)

Step 3: Determine which subinterval the root lies:if f(xl)f(xr) < 0 // the root lies in the lower subinterval

set xu to xr and goto step 2

if f(xl)f(xr) > 0 // the root lies in the upper subintervalset xl to xr and goto step 2

if f(xl)f(xr) = 0xr is the root than stop


How to select xr in step 2?

1. Bisection MethodGuess without considering the characteristics of f(x)

in (xl , xu)

2. False Position Method (Regula Falsi)Use "average slope" to predict the root


The Bisection Method (Bolzano’s Method)

• Given a bracketed root, repeatedly halve the interval while continuing to bracketing the root.

• Bisection always converges if the original interval contained a root.


The Bisection Method

• Given an initial bracket [xl , xu], an approximation to the root ( xm )is the midpoint of the interval.

• An alternative for the midpoint.

𝑥𝑥𝑟𝑟 =12

(𝑥𝑥𝑢𝑢 + 𝑥𝑥𝑙𝑙)

𝑥𝑥𝑟𝑟 = 𝑥𝑥𝑙𝑙 +(𝑥𝑥𝑢𝑢 − 𝑥𝑥𝑙𝑙)

2


The Bisection Method

• if f(xm)*f(a) < 0, the root lies in the lower subinterval. Set b = xm , a = a.

•• if f(xm)*f(a) > 0, the root lies in the upper

subinterval. Set a = xm , b = b.•• The new value of the midpoint xm is

calculated, and the process is repeated until the

value of the root is found (or an approximate value).


Use the bisection method to locate the

lowest root of the following polynomial

f(x) = -2 + 7x - 5x2 + 6x3

initial guesses a = 0 and b = 1, εs = 10 %.

Example: The Bisection Method


Iteration I:

f(xl)=f(0) = -2, f(xu)=f(1) = 6, f(0)*f(1) < 0

xm = (a + b) / 2= (0 + 1) / 2 = 0.5

f(xm)=f(0.5) = 1 ,

f(a)*f(xm) = f(0)*f(0.5) = (-2) (1) < 0 ,

a = 0 and b= xm=0.5, 𝜀𝜀𝑎𝑎 =xm,i − xm,i−1

xm,i∗ 100%



Iteration II:

xm = (a + b) / 2= (0 + 0.5) / 2 = 0.25

f(0.25) = - 0.46875

f(a)*f(xm) = (-2) (- 0.46875) > 0 ,

a= xm=0.25 & b = 0.5, εa = 100 %.



Iteration III:

xm = (a + b) / 2= (0.25 + 0.5) / 2 = 0.375

f(0.375) = 0.2383

f(a)*f(xm) = (- 0.46875) (0.2383) < 0 ,

a= 0.25 & b = xm = 0.375, εa = 33%.



Iteration IV:

xm =(a + b)/2=(0.25 + 0.375) / 2=0.3125

f(0.3125) = - 1.6802

f(a)*f(xm) = (- 0.46875) (- 1.6802) > 0 ,

b = 0.375 & a = xm = 0.3125, εa = 20%.



Iteration V:

xm = (a + b) / 2= (0.3125 + 0.375) / 2 = 0.34375

f(0.34375) = 0.0591428

f(a)*f(xm) = (- 1.6802) (0.0591428) < 0 ,

a = 0.3125 & b= xm = 0.34375

εa = 9.09% < εs = 10 % The process stops.




Apply Bisection to x – x1/3 – 2 = 0.

Use the MATLAB function

demoBisect( xleft, xright, n)

where

xleft, xright : left and right brackets of the root.

n: optional number of iterations.

Default: n = 15


Bisection Method – Example

𝑓𝑓(𝑥𝑥) =667.38𝑥𝑥

(1 − 𝑒𝑒−0.146843𝑥𝑥) − 40

Find the root of f(x) = 0 with an

approximated error below 0.5%.

(True root: α=14.7802)


Example (continue)

n xl xr xu f(xl) f(xr) f(xu) f(xl)f(xr) εa (%)0 12 16 6.067 -2.2691 12 14 16 6.067 1.569 -2.269 > 02 14 15 16 1.569 -0.425 -2.269 < 0 6.7

3 14 14.5 15 1.569 0.552 -0.425 > 0 3.5

4 14.5 14.75 15 0.552 0.0590 -0.425 > 0 1.7

5 14.75 14.875 15 0.0590 -0.184 -0.425 < 0 0.8

6 14.75 14.8125 14.875 0.0590 -0.0629 -0.184 < 0 0.4

%100newr

oldr

newr

a xxx −

=ε

%219.0%1007802.14

8125.147802.14=

−=tε𝜀𝜀𝑡𝑡 =

xt − xr

xt∗ 100%


Error Bounds

The true root, xr , must lie between xl and xu.

xl xuxr

xl(1) xu

(1)

After the 1st iteration, the solution, xr(1), should be within an

accuracy of)1()1(

21

lu xx −

xr(1)

Let xr(n) denotes xr in the nth iteration


Error Bounds

xl(2) xu

(1)xu(2)

Suppose the root lies in the lower subinterval.

xr(2)

)1()1(2

)2()2(

21

21

lulu xxxx −=−

After the 2nd iteration, the solution, xr(2), should be

within an accuracy of


Error Bounds

)1()1()1()1(2

)()(

21...

21

21

lunn

ln

un

ln

u xxxxxx −==−=− −−

In general, after the nth iteration, the solution, xr(n),

should be within an accuracy of

If we want to achieve an absolute error of no more than Exr

)(log

21

)1()1(

2

)1()1(

rx

lu

lun

Exxn

Exx

−≥⇒

≤− α


Implementation Issues

• The condition f(xl)*f(xr) = 0 (in step 3) is difficult to achieve due to errors.

• We should repeat until xr is close enough to the root, but we don't know what the root is!

• Therefore, we have to estimate the error as

and repeat until εa < εs (acceptable error)

ε𝑎𝑎 =𝑥𝑥𝑟𝑟𝑛𝑛𝑛𝑛𝑛𝑛 − 𝑥𝑥𝑟𝑟𝑜𝑜𝑙𝑙𝑜𝑜

𝑥𝑥𝑟𝑟𝑛𝑛𝑛𝑛𝑛𝑛100%


Pseudocode of Bisection Algorithm


A.2. Regula Falsi Method

• Also known as the false-position method, or linear interpolation

method.

• Unlike the bisection method which divides the search interval by

half, regula falsi interpolates f(xu) and f(xl) by a straight line and

the intersection of this line with the x-axis Is used as the new

search position.

• The slope of the line connecting f(xu) and f(xl) represents the

"average slope" (i.e., the value of f'(x)) of the points in [xl, xu ].


=>−

=− rl

l

ru

u

xxxf

xxxf )()(

𝑥𝑥𝑟𝑟 = 𝑥𝑥𝑢𝑢 −𝑓𝑓(𝑥𝑥𝑢𝑢)(𝑥𝑥𝑙𝑙 − 𝑥𝑥𝑢𝑢)𝑓𝑓(𝑥𝑥𝑙𝑙) − 𝑓𝑓(𝑥𝑥𝑢𝑢)

Similar to the bisection method except that the new division point xr is not the middle point of the interval but this:

f(xl) is much closer to 0 than f(xu) the root should be closer to xl than to xu

Using similar triangles, the intersection of the straight line with the x axis can be estimated as


40138667 1468430 −−= − )(.)( . cec

cf

Example Linear Interpolation Method

1st iteration:xl = 12 ⇒ f(xl) = 6.067xu = 16 ⇒ f(xu) = -2.2688xr = 16 – (-2.268(12 – 16))/(6.067 – (-2.2688)) =14.9113εt = 0.89%

2nd iteration:f(xl)f(xr) = -1.5426 < 0xl = 12, xu = 14.9113;xr = 14.9113 – (-0.2543(12 – 14.9113))/(6.0669-(- 0.2543) = 14.7942εt = 0.09% < εt = 0.2% (Bisection @ 6th iteration)

(True root: α=14.7802)


Example (continue)

n xl xr xu f(xl) f(xr) f(xu) f(xl)f(xr) εt (%)0 12 16 6.067 -2.2691 12 14.9113 16 6.067 -0.2543 -2.269 > 0 0.89

2 12 14.7942 14.9113 6.067 -0.2543 < 0 0.09

%100newr

oldr

newr

a xxx −

=ε

𝜀𝜀𝑡𝑡 =14.7802 − 14.7942

14.7802100% = 0.09%

𝜀𝜀𝑡𝑡 =xt − xr

xt∗ 100%


• The error for false

position decreases

much faster than for

bisection.

•The false-position

method is more

efficient.


False-position vs Bisection

• False position in general performs better than bisection method.

• Exceptional Cases:– (Usually) When the deviation of f'(x) is high and

the end points of the interval are selected poorly.– For example,

3.1,0with1)( 10

==−=

ul xxxxf


Iteration xl xu xr εa (%) εt (%)1 0 1.3 0.65 352 0.65 1.3 0.975 33.3 253 0.975 1.3 1.1375 14.3 13.84 0.975 1.1375 1.05625 7.7 5.65 0.975 1.05625 1.015625 4.0 1.6

Iteration xl xu xr εa (%) εt (%)1 0 1.3 0.09430 90.62 0.09430 1.3 0.18176 48.1 81.83 0.18176 1.3 0.26287 30.9 73.74 0.26287 1.3 0.33811 22.3 66.25 0.33811 1.3 0.40788 17.1 59.2

Bisection Method (Converge quicker)

False-position Method


3.1,0with1)( 10

==−=

ul xxxxf

The slow convergence of the false-position method.


Summary

• Bracketing Methods– f(x) has the be continuous in the interval [xl, xu]

and f(xl)f(xu) < 0

– Always converge

– Usually slower than open methods

• Bisection Method– Slow but guarantee the best worst-case convergent rate.

• False-position method– In general performs better than bisection method (with some

exceptions).


Incremental Search and Determining Initial Guesses

• For every interval xi – xi+1, apply bisection or false-position method to find the roots in the interval

• Only roots in x4-x5 will be found


Home Work of Chapter 5

From the text book:Problem 5.5, 5.6, 5.13.

Additional:None.

numerical methods - marmara Üniversitesi · 2019-03-07 · lecture 4: roots of equations -...

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