numerical methods for use in root discovery

8/8/2019 Numerical Methods for Use in Root Discovery

1/15

Numeri

DanCook

Abstract

Theneed

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functions

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tofind

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ng. In this

ng, andmor

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tion:

areoneoft

phenomeno

atemanyva

inedfrom

a

ptofaroot

itten in 155

La geomet

[7].

ny methods

ticaltoolsin

Analysis:

e roots of

s, engineeri

ng.

ollowingdia

sforusei

otof

afun

eport, roots

e specifically

sepositive,fi

hemostbasi

. Once a f

uesoverthe

functionis

it

datesbackt

7, the equal

ie in 1637

of root find

ludingtheu

a function i

g, and biol

ramasane

RootDis

tionis

of

vi

of a functio

circuitdesi

xedpointite

ideasinma

nction is d

rangeofth

sroot(s),

or

aboutthef

to sign wa

hich broug

ing have be

eofsigncha

s important

ogy. One s

ample:

overy

talimportan

n will be fo

n.Various r

ration,Newt

thematics.F

eveloped fo

function.

herethe

fu

ifteenthcent

s first intro

t about the

n develope

nge,derivati

in many di

uch exampl

cein

almos

nd that ha

ot finding

onRaphson,

nctionscan

r a physical

neofthem

ctionchang

ury. InRobe

uced. Shor

use of the

. These me

es,andtan

ferent area

is related

tall

aspects

its physical

ethods will

andsecant

beusedmod

l process, it

ost importan

ssign.

rtRecordes

tly after th

xy graph

thods rely o

entlinesto

including b

to circuit d

anddiscipli

basis in ele

be used incl

ethods.

elawidera

can be us

tvalue(s)th

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t Rene Des

to plot and

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ut not limit

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2/15

Whentheswitchisclosedattimet=0,resultingintheremovalofthebattery,thecircuitwillexperience

a transient response. During thisperiod thecurrent flowing in thecircuitwilldecreaseexponentially

withtimeaccordingtothefollowingequation:

2/2+/+(/=0WhereListheinductorvalue,Ristheresistorvalue,andCisthecapacitorvalue

Forthegivenproblemwewanttoimplementacircuitwhosetransientresponseallowsthechargeinthe

circuittodissipateto1percentof its initialvalue inatimeof.05seconds. GivenarethevaluesofL

andC,soonlytheresistorvalueRneedstobesolvedfor. TheRfunctionisgivenas:

()=(2)cos(1/(/2)2)/0

Tofindthecorrectresistorvaluetherootofthisequationmustbesolvedfor. Thebisection,modified

falsepositive,fixedpointiteration,NewtonRaphson,andsecantmethodswillbeusedtofindthisvalue.

NumericalAnalysis

BracketMethods

Thefirsttwomethodsdescribedhereareknowncollectivelyasbracketingmethods.Thesemethodsare

basedonthefactthat inthe localareaofarootthefunctionwillchangesignsonthe leftandrightof

theroot.

BisectionMethod

The first root findingmethod thatwillbediscussed is thebisectionmethod. With thismethod two

guessesareenteredsuchthattheroot isbetweenthem(i.e.thefunction ispositiveforonevalueand

negativefor

the

other

value).

Itthen

takes

the

midpoint

of

these

values

as

an

approximate

root.

Ifthe

signsof themidpointandupperguessareof thesame sign then it isknown that the root lies in the

lowersubinterval.Ifthesignsaredifferentthenitisknowthattherootliesintheuppersubintervaland

anewmidpointbetweentheupperguessandtheoldmidpoint istakenandtheprocessrepeatsuntil

therootisfound.Thiscanbeshownas[1]:

2

Whenf(xl)f(xr)0therootisintheupperinterval;setxl=xr

Whenf(xl)f(xr)=0

the

root

isxr


3/15

Figure1:

Bi

FalsePos

Thefalse

method

whereth

Thisappr

oneend

thefuncti

theuppe

ectionMethod

tionMethod

positionmet

onstructs a

approximat

oximationof

fthelinewi

onuntilitsi

andlowerg

.www.leda

tut

hodrelieso

line betwee

erootwillbe

theroot,Xr

llalwaysbe

terceptiseq

uess[1].This

orial.org

themagnit

f(xupper)

.Thisinterse

oot,thenre

ixedonone

ualtothero

processcon

desoftheu

nd f(xlower

ctionofthe

lacesthegu

oftheguess

ot. Thisisu

inuesuntila

pperandlo

and where

axiscanbe

essthathas

esandtheo

efulasther

rootisfoun

erguessest

this line cr

foundusing

thesamesi

therendpoi

otwillalwa

.

ofindaroot

ssed the x

1]:

gnasXroot.

twillmove

sbecontain

. This

xis is

Thus

along

edbe


4/15

Figure

2:

Fa

OpenMe

The next

equation

twogues

FixedPoi

Thismeth

Whenav

f(x)=0.Th

Thisisco

lse

Position

Me

thods

three root

,andtypical

esthatcont

tIteration

odis

based

alueofxisf

ismethodfin

tinueduntil

thod.

www.wik

inding meth

lyonlyrequir

intheroot[

ethod

nchanging

t

undsuchth

dsanewval

g(xi)=x,mea

ipedia.org

ds used ar

eoneguessi

1].

heequation

tx=g(x),iti

eofxbased

ingthereis

known as

ncompariso

(x)=0to:

equivalentt

ontheoldv

rootatthe

pen metho

ntothebrac

ofindingas

aluesofx.Th

xvalueused

s. These m

ketingmeth

olutiontoth

isisseenas[

inthefuncti

thods depe

dswhichre

originaleq

1]:

n.

d on

uired

ation


5/15

Figure3:Fi

Newton

TheNewt

guess is

function.

Theform

Theproc

edPointMeth

aphsonMet

onRaphson

selected and

Thexinterc

ladescribin

ssoffinding

d.www.pathfi

od

methodhasi

a tangent l

ptoftheco

thisis[1]:

thetangentl

nder.scar.utoro

tsbasisinus

ine at that

putedtang

ineintercept

nto.ca

ingthetang

guess point

ntlineisfou

iscontinued

ntline(deri

is construct

nd,andthis

untilthetan

ative)ofthe

d using the

isthenewr

gentlineint

givenfuncti

derivative

otapproxim

rceptsther

on. A

f the

ation.

ot.


6/15

Figure4:N

SecantM

The seca

linesand

usesthe

compute

Theproc

wtonRaphson

ethod

tmethod is

their xinter

unctionsder

atangentlin

ssfromthis

Method.www.

very similar

cept toappr

ivativetoco

.Thismetho

pointonward

gilkalai.wordpr

to theNew

oximate the

structtheta

dusesthefo

isthesame

ess.com

onRaphson

root.Them

ngentline,

rmula[1]:

astheNewt

method in i

aindifferenc

hilethesec

nRaphson

tsapproach.

e is that the

ntmethodu

ethod

Bothuse ta

NewtonRa

sesadiffere

ngent

hson

ceto


7/15

Figure5:Se

Results

Firstthe

charge. T

cantMethod.

differentiale

hisisdoneb

ww.jwilson.co

quationdesc

:

.uga.edu

ribingthe transientresp

nsegivenabovemustbesolved for qor


8/15

Secondly, the resistor function is plotted to give a general idea of the shape and the root of the

equation.Theequationwasthis:

()=(2)cos(1/(/2)2)/0

Theplotdoneinmatlabisasfollows:

Figure6:PlotoftheResistorfunction

Fromthis

plot

itcan

be

seen

that

the

root

isin

the

range

of

550 2*epsilon)

'Calculate midpoint of domain

midpoint = (right + left) / 2

'Find f(midpoint)

If ((f(left) * f(midpoint)) < 0) Then

'Throw away right half

right = midpoint

ElseIf ((f(right) * f(midpoint)) < 0)

'Throw away left half

500 550 600 650 700 750 800 850 900 950 1000-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06


9/15

left = midpoint

Else

'Our midpoint is exactly on the root

Exit Do

End If

Loop

Return midpoint

TheresultingrootisR=605.8899after14iterationsofthecode.

FalsePositionMethod

Thepseudocodeusedforthismethodisasfollows[3]:

User inputs: a,b, max, epsilon

Initialize: iteration = 0

xold = b //to start out

Inside a loop with condition that iteration


10/15

WHILE

K :

Pold

Pterm

Pnew

Dg :Delta

RelErr

Dx

Slope :

Theresul

Newton

Thepseu

es=eo

xr=x0

iter =0

ea=999

% Begin

while

xro

xr=

ite

if

end

end

root=xr

Therooti

RelErr

= K+1

:=

:=

:= g(P

= Pnew:=

:

:= Pte

= Dg/Dx

ingrootisgi

aphsonMet

ocodeused

iteratio

a=

Pterm

Pnew

term)

-|

rm -

venas605.7

od

forthismet

loop

old)/f'(x

r-xrold)/

R=605.8841

Tol

Ptermg|

*Delta/(|

Pold

64with14i

odisasfoll

initial

initial

initial

% test

stopp

% update

old)

% update

% update

r)*100

% fin

after14iter

and

{

Pnew|+Sma

{g'(Pk)

erations.

ws[5]:

ze the ro

ze the it

ze the re

to see

ng criter

old value

% calcu

the iter

absolute

% calcula

l error

ations

K

{Increme

{Previou

{Curren

ompute

{Di{

ll) (

{

}

ot with s

eration c

lative er

f relati

ia

of root

late new

ation cou

relative

te relati


11/15

SecantMethod

Thepseudocodeisasfollows[6]:

Input xk, xkMinus1, xTol, maxiters

iters = 1

yk = (xk) (* invokes function f *)ykMinus1 = f(xkMinus1)

root = (xkMinus1*yk - xk*ykMinus1)/(yk - ykMinus1)

ykPlus1 = f(root)

While( (Abs(root - xk) > xTol) and (iters < maxiters) )

xkMinus1 = xk

ykMinus1 = yk

xk = root

yk = ykPlus1

root = (xkMinus1*yk - xk*ykMinus1)/(yk - yk Minus1)

ykPlus1 = f(root)

iters = iters + 1

Endofwhile

ThisgivesarootofR=605.8841after14iterationsofthecode

Summaryofresults(allafter14iterationsofthecode):

Method Bisection FalsePosition FixedPoint Newton Secant

RootValue 605.8899 605.8841 605.7264 605.8841 605.8841

Thistabledemonstratesthat,afterfourteen iterations,thefalseposition,newtonraphson,andsecant

root methods all yield results that are equivalent to four decimal places. For a fixed number of

iterations

the

newton

raphson

method

is

generally

considered

the

most

accurate

method

[1].

It

is

also

worthnotingthedifferencebetweenthemethodstohelpfurtheranalyzetheresults.

Method Input Converge ConvergeRate

Bisection Upper,lower Yes Linear

FalsePosition Upper,lower Yes Better

FixedPoint Anyvalue Usually Better

Newton Anyvalue Usually UsuallyBest

Secant Anytwovalues Usually Better

Fromcomparingtheprevioustwotablesitcanbedeterminedthatforarelativelysimpleequationwith

a derivative that can be taken and that has only one root, the false position is the best bracketing

method, and the newtonraphsonmethod is the best openmethod. While the secant and newton

methodsgavethesameresult,foramorecomplexfunctionthenewtonmethodwillmostlikelyfindthe

rootmorequicklythanthesecantmethod.


12/15

Conclusion

When comparing the various methods available to find the root of a function, there is no definite

method

that

stands

out

above

the

rest.

For

most

general

equations

it

seems

that

the

newton

method

wouldbethepreferedoptioninthatitconvergesthequickestandonlyoneguesshastobemadethat

canbeaboveorbelowtheroot.However ifthefunctionsderivative istaxingtofind,thismethodmay

notbethebest,andthesecantmethodmayworkbetter. Thebracketingmethodsholdonesignificant

advantage inthattheyalwaysconvergeontheroot.As longasthetwoguessesbrackettheroot,the

rootcanandwillbefoundwiththesemethods. Howeverthisisalsothedisadvantagewithbracketing

methods. Ifagraphingutility isnotavailable itmaybedifficult topicka rangewhere the root falls

makingthebracketingmethodsuseless. Whileacomputercanrunalloftheseprogramsefficiently, it

requiresaknowledgeablepersontounderstandwhichofthesemethodsbestfitstherequirementsof

theproblem.

References

[1]Chapra,Steven.NumericalMethodsforEngineers.Pg130150.2010

[2]Wikipedia.BisectionMethod.2010

[3]www.physics.arizona.edu.FalsePositionMethod.2010

[4]www.physics.arizona.edu.FixedPointMethod.2010

[5]www.cs.purdue.edu.NewtonRaphsonMethod.2010

[6]www.vortex.bd.psu.edu.TheSecantMethod.2010

[7]Wikipedia.NumericalAnalysis.2010

AppendixMatlabCode

Functioncode

function y = fcn(x)

y = exp(-x.*(.05)./(2.*5)).*(cos(sqrt((1./(5.*10.^-4))-

((x./(2.*5)).^2).*.05)))-.01;

FunctionandDerivativecode(forusewithNewtonMethod)


13/15

function [y, deriv] = fcn_nr(x)y = exp(-x.*(.05)./(2.*5)).*(cos(sqrt((1./(5.*10.^-4))-

((x./(2.*5)).^2).*.05)))-.01;deriv=exp(-0.005.*x)*((-0.005*cos(sqrt(2000 - 0.0005.*x.^2)) +

(0.0005.*x.*sin(sqrt(2000 - 0.0005*x.^2)))/sqrt(2000 - 0.0005* x.^2)))

Bisection

Code

functiony=bisection(f,a,b,TOL)

sfa=sign(feval(f,a));

Nmax=floor(log((ba)/TOL)/log(2.0))+1

fori=1:Nmax

p=(a+b)/2.0;

sfp=sign(feval(f,p));

if((ba)


14/15

rel_error = 0;

elseif test < 0xu = xr;

elsexl = xr;

end

end

xr

FixedPointCode

function y=op(maxerror, guess, it)count = 0;

error = 1;

maxit=itx=guesswhile (error > maxerror) & (count < maxit)

count = count + 1xnew=feval('fcn',x)error = abs((xnew - x)/xnew) * 100x=xnew

end

NewtonRaphsonCode

function y=newrap(itermax, errmax, guess)

x = guess;

iter = 0;

error = 1;

while error > errmax & iter < itermax

iter = iter + 1;[f fprime] = fcn_nr(x);xnew = x - f / fprime;

error = abs((xnew - x)/xnew) * 100;

x = xnew;

end

SecantCode

function y=secan(guess1, guess2, itermax, errmax)


15/15

x1 = guess1;x2 = guess2;

iter = 0;

error = 1;

while error > errmax & iter < itermax

iter = iter + 1;f1 = fcn(x1);f2 = fcn(x2);

x3 = x2 - f2*(x2-x1) / (f2 - f1);

error = abs((x3 - x2)/x3) * 100;

x1 = x2;

x2 = x3;endx3

numerical methods for use in root discovery

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