lecture 06 solution of equations in one variable

8/15/2019 Lecture 06 Solution of Equations in One Variable

1/67

the most basic problems of numericalapproximation,

the root-finding problems

Solutions of Equations in One Variable

Chapter 2


2/67


3/67

#onlinear '0uations in /ne ariable

• Examples of nonlinear equation in one variable

•

Given a function , we are looking for a value ,s t ! a root of the equation, or a "ero of the

function # $he problem is calle% the root fin%ing

problem or the "ero fin%ing problem

sin&2 = x x

+2112111 2"&+ = x x x x x

3 x

f

( ) f x =

x ( ) f x


4/67

'xamples

• &onlinear equations can have an' number of solutions

2

" 2

1 has no solution

has one solution

&sin has t$o solutions

11 has three solutionssin has infinitel man solutions

x

x

e

e x

x x

x x x x

−

+ =

− =

− =

+ + − ==


5/67

'xistence4uni0ueness

• Existence an% uniqueness of solutions are much more

complicate% for nonlinear equations than for linear

equations• (f is continuous an%

then the interme%iate value theorem implies

there exists such that

f ))(sgn())(sgn( b f a f ≠

5 x .)( 5 = x f


6/67

(nterme%iate Value $heorem) (f an% is an'

number between an% , then there exists a

number in for which

6,7 baC f

c

K

.)( K c f =

)(a f )(b f

6,7 ba


7/67

*ssumption) is a continuous function %efine% on

, with an% of opposite signs(%ea) $he metho% calls for a repeate% halving of

subintervals of an%, at each step, locating the half

containing

Bisection Method

)(a f

f

)(b f 6,7 ba

p

6,7 ba


8/67


9/67

+epeat until

Set an% , let

(f then set

(f then , %one

-isection .etho%) /roce%ure

(f then

&ote) $he metho% can pro%uce a false root if is

%iscontinuous on

or

else set

,288 toleranceab k k 2k k

ba p +=

k p p =

bb =1aa =12

111

ba p

+=

, ,122

pbaa =)()( 11


10/67

(&/0$ en%points a, b tolerance $O1 maximumnumber of iterations & 3

O0$/0$ approximate solution p or message of failure

Step 4 set i546*5f!a#

Step 2 while ( & 3 %o Steps 789Step 7 set p5a:!b8a#;2 !compute p i #

6/5f!p#

≤


11/67

Step 2 while ( & 3 %o Steps 789

Step 7 set p5a:!b8a#;2 !compute p i #6/5f!p#

Step < (f 6/53 or !b8a#;2=$O1 then

O0$/0$ !p#

!proce%ure complete% successfull' #

Stop

Step > Set i5i:4

≤


12/67


13/67

4 4 3 2 3 4 > 2 7A>2 4 3 4 > 4 2> 84 AD9 A7 4 2> 4 > 4 7A> 3 49244< 4 2> 4 7A> 4 742> 83 < 7D> 4 742> 4 7A> 4 7 83 7>3D9 4 7 4 7A> 4 7>D7A> 83 3D9D7A> 4 7A> 4 79A4 A> 3 37279 4 7>D7A> 4 79A4 A> 4 7972 42> 83 3724>

D 4 7972 42> 4 79A4 A> 4 79>27 3 3333A243 4 7972 42> 4 79>27 4 79A 47 83 3493>44 4 79A 47 4 79>27 4 79 83 334D<

'xample has a root in F4,2Correct value) 4 79>273347

k k a k pk b )( k p f

1& 2" = x x


14/67

$heorem Suppose that an%

$he bisection metho% generates a sequence

approximating a "ero of with

&ote) !4# gives a boun% for the approximating error

!2# the number of iterations obtaine% from this boun%,

in man' cases, is much larger than the actualnumber of iterations require%

!7# assumes infinite8%igit arithmetic

∞=19: k k p

f

)()(


15/67

• Other stopping proce%ures can be applie%)

-isection .etho%) &otes

• Hifficulties can arise (t is possible that

converges to "ero, while %iverges (t is also

possible that is close to "ero while

%iffers significantl' from

best, if noa%%itional knowle%ge

9: 1 N N p p

9: N p)( N p f N p

p

1

1

,

4 8 8 ,

8 ( ) 8

N N

N N N N

N

p p

p p p p

f p

ε

ε

ε

− <

− < ≠


16/67


17/67

• $o %etermine which subintervals ofcontains a root of , it is better to make use of

the signum function, which is %efine% as

gives the same result but avoi%s the possibilit' of

overflow or un%erflow in

$he test

6,7 k k ba

>

=<

=if 1

if

if 1

)sgn(

x

x

x

x

f

sgn( ( )) sgn( ( ))

in stead of ( ) ( )

k k

k k

f a f b

f a f b

>

>

)( k b f ( )5k f a


18/67

• use no magnitu%es of function values, but signsBisection Method 3 ;ummar

• given error tolerance of , requires

iterations regar%less of the function involve%

• at each iteration, length of interval containing the

solution is re%uce% b' half, length of interval after

iterations is , convergence rate is linear• one bit of accurac' is gaine% in the approximate

solution for each iteration

• is certain to converge, but slowl'

−tol

ab2logtol

k ab 24)( −

k


19/67


20/67

'xample "2,2)( 2 ≤ x x x g


21/67

$heorem)

a (f an% for all

then has a fixe% point in

b (f, in a%%ition, exists on an% a positive

constant exists with

then the fixe% point in

is unique

6,7 baC g 6,,7 ba x

6,7 ba

6,7)( ba x g

g

6,7 ba

)(< x g

1


22/67

'xample

minimum atmaximum at

11,"4)1()( 2 ≤ x x x g

11

,"42)(<

≤

=

x

x x g

)1,1("428)(


23/67


24/67

'xample has a root in F4,2

x x x x

x x g x

x x g x

x x g x

x x

x g x

x x x x g x

="1&

)( (e)

&1

)( (d)

24)1()( (c)

&1

)( (b)

1&)( (a)

2

2"

+

241

&

241""

241

2

2"

1

+−−

+

−

−

+

1& 2" = x x


25/67

Correct value) 4 79>273347 choose

!a# !b# !e#!c# !%#

"-+2" 1".1""-+2" -.12+"-+2" 2"-.12

"-+2" 1".1"-+22"-=.11+ "-+2" 1&.1"-+&1 -2.11

"-+2" 12.1"-&=>=21>.1?"-+2" 22.1"-+?1->"&.1="-+22??&2.1"-"==> &.1>"-+2" +>-.1"->=&-?-=.1-"-+22++?&.1"- ?&1?".1+

"-+2" 1".1"-+2-&>&=.1">+1> +2".1=e".1&"-+2" 1&.1"-&?+> 1+.1"&+&+=">&.1)-+.=(>.&-?""-+2-2 1+.1"->">-">2.1& 2+& = &.1??-?.2>"2.-2">""""""".1"&="??>2+.12=-?+">-=.1=1-+.=>+.1+.1+.1+.1+.1+.1

241−

−

k

+.1 = p


26/67

'xample has a root in F4,2

most powerful an%well8known

numerical metho%

x x x x x x g x

x x g x

x x g x

x x

x g x

x x x x g x

="1&)( (e)

&

1)( (d)

24)1()( (c)

&1

)( (b)

1&)( (a)

2

2"

+

241

&

241""

241

2

2"1

+−−

+

−

−

+

1& 2" = x x

)2,1( 18)(


27/67

6ixe%8/oint $heorem)

1et be such that for all

Suppose, in a%%ition, that exists on an% thata constant exist with

$hen for an' the sequence %efine% b'

converges to the unique fixe% point with error

boun%s

,1),( 1 ≥− n p g p nn

6,7 baC g 6.,7 ba x

),( ba

6,7)( ba x g

< g

),,( allfor ,8)(


28/67


29/67

1

1

1

1

1

1

1

1

)(

)()(

)(<

)(<

)(

−

−

−

−

−

−

−−

−−

≈

−−

≈

−

nn

n

nn

nn

n

n

n

nn

p p p f

p p p f p f

p f

p f

p f p p

p g


30/67

•

choose an initial approximation• let , for each

$heorem) 1et (fthen there exists a

&ewton8+aphson .etho%

p

)(<)(

1

11

−

−− −

n

nnn p f

p f p p 1n

).,( δ+ p p p

,)(


31/67

&ewton8+aphson .etho%) &otes

•

&ewton s formula can generall' be use% for an' pol'nomial or non8pol'nomial function

• 6irst approximations can be obtaine% b'

%rawing the graph or examining the function

• (n general, the closer the first approximations is to thereal root the faster the sequence converges to the root

• Gives problem if)($hen)(< = p f p f


32/67

&ewton8+aphson .etho%) &otes

•

Converges provi%e% a sufficientl' accurate initialapproximation is chosen

• (n a practical application, an initial approximation

chosen, the metho% will generall' convergesquickl' to the root, or it will be clear thatconvergence is unlikel'

• Extremel' powerful, but has maKor weakness) thenee% to know the %erivative of the function ateach iteration


33/67

let for each

choose an initial approximation

•

let

for each

Secant .etho%


choose an initial approximation p

1, p p

)()()()(

2121

1

1 −−

−− −− nnnn

n

nn p p p f p f p f

p p

2n

)()()(

)(<

21

211

−

−− −

−≈nn

nnn p p

p f p f p f

11

1

( ) ,


34/67

starting with twoinitialapproximations

in stea% of one asin the &ewton s

metho%

alwa's uses the

two newest points


35/67

Method of False osition (@egula Falsi)• choose an initial approximation such that

let

let

or

Continue Similarl'

1, p p

)()()(

)(1

1

112 p p p f p f

p f p p −

−

)()(!f 12 < p f p f )()()()( 12

12

12" p p p f p f

p f p p −−

)()( 1 < p f p f

)()()()(

22

2" p p p f p f p f

p p −− )()(!f 2 > p f p f


36/67


37/67

Aefinition ;uppose is a se0uence no$nto con*erge to Cero, and con*erges to a num

ber . !f a positi*e constant exists $ith

then $e sa that con*erges to $ith

rate of

for large n

&ot obvious how to use this %efinition to tell the rate ofconvergence of the sequence obtaine% using an iterativetechniqueL

'rror nal sis for !terati*e Methods

∞

= 19: nn∞

= 19: nn

K

)( nO

n n K ≤

1n n ∞

=


38/67

'rror nal sis for !terati*e Methods• 6or general iterative metho%s, %efine error at iteration

b' where is the approximate solutionan% is the true solution

• * sequence converges with rate if

for some finite non"ero constant

• 6or metho%s that maintain interval known to contain

the solution, rather than the specific approximatevalue for the solution, take error to be the length of

the interval that contains the true solution

k

5 x x e k k −5 x

k x

r C e

er

k

k

k =

∞

1lim

C


39/67


40/67

"?"1?2

12

+2

"1

11

11

1211

1=>>+.+1=12+.>>1=&2.11+-2+.1-

1-+--.&112+."+1+1=."12+.-&1=12+.>12+.1" 12+.11+.22

1.+1.+1)+.()+.(

9D :se0uence9:se0uence

eon*ergencEuadraticeon*ergencFinear

−

−

−

−

−

−

−

−

∞=∞=

××××××

×

nnnnnn

n p p


41/67


42/67


!a# !b# !e#!c# !%#

"-+2" 1".1""-+2" -.12+"-+2" 2"-.12

"-+2" 1".1"-+22"-=.11+ "-+2" 1&.1"-+&1 -2.11

"-+2" 12.1"-&=>=21>.1?"-+2" 22.1"-+?1->"&.1="-+22??&2.1"-"==> &.1>"-+2" +>-.1"->=&-?-=.1- "-+22++?&.1"- ?&1?".1+

"-+2" 1".1"-+2-&>&=.1">+1> +2".1=e".1&"-+2" 1&.1"-&?+> 1+.1"&+&+=">&.1)-+.=(>.&-?""-+2-2 1+.1"->">-">2.1& 2+& = &.1??-?.2>"2.-2">""""""".1"&="??>2+.12=-?+">-=.1=1-+.=>+.1+.1+.1+.1+.1+.1

241−

−

k

+.1 = p


43/67


44/67

$heorem)

1et be the solution of the equation

Suppose that an% iscontinuous with

on an open interval containing

$hen there exists a numbersuch that, for , the sequence

%efine% b' converges at

least qua%raticall' to .oreover, for sufficientl' large values of ,

,

p

I

)( x g x =


45/67

&ewton s metho% is qua%raticall' convergent whenit converges

6ixe%8/oint metho% is generall' linear convergent


46/67


Secant .etho%

Generall', experience %ifficult' if

Secant metho% is superlinearl' convergent

)(<

)(

1

1

1 −

−

−−

n

n

nn p f

p f p p 1n

)()()()(

2121

11 −

−

−− −− nnnn

nnn p p p f p f

p f p p 2n

)($hen)(< = p f p f


47/67


48/67

'xample

1)(


49/67

then

also has a "ero at Jowever is a simple "ero

of

&ewton s metho% can then be applie% to

to give the mo%ifie% &ewton s metho%

(f is a "ero of of multiplicit' an%

)(


50/67

(n practice, multiple roots can cause serious roun%8off problems since the %enominator consists of the

%ifference of two numbers that are both close to 3

$heoreticall', the onl' %rawback is the a%%itional

calculation of an% the more laborious

proce%ure of calculating the iterates

)(


51/67

ccelerating on*ergence

• *itken s metho% can be use% to accelerate theconvergence of a sequence which is linearl'convergent , regar%less of its origin or application

•

*itken s metho% is base% on the assumption thatthe sequence %efine% b'

converges more rapi%l' to than %oes the originalsequence

• (t is rare to have qua%ratic convergence

2∆

nnn

nnnn

p p p

p p p p

+

−−+

+

12

21

2

)(G

2

∆

p

∞=19G: nn p

∞=19: nn p


52/67

$hen the sequence %efine% as

converges to faster than in the sense that

$heorem) Suppose that is a sequence that

converges linearl' to the limit an% that

1lim 1 <−−

∞ p p p p

n

n

n

∞=19G: nn p

p

∞=19: nn p

p

Glim =

−−

∞ p p p p

n

n

n

∞

=19: nn p

nnn

nnnn p p p

p p p p

+−−

+

+

12

21

2)(G


53/67

Hefinition) 6or a given sequence , the forwar%

%ifference is %efine% b'

Jigher powers of the operator are %efine%

recursivel' b'

(n particular,

for ,1 ≥+ n p p p nnn

∞=19: nn p

n p

2∆

2for ),( 1 ≥− k p p nk nk

21

1 2 1

( ) ( )

( ) ( ) 2n n n n

n n n n n

p p p p

p p p p p

∆ = ∆ ∆ = ∆ −

= ∆ − ∆ = − +


54/67


55/67

;teffensen%s method

n

nnnnn

nnnn p p p p p p p p p p 22

12

2

1 )(2 )(G ∆∆−+−− ++

2)2(

1)2(

2

)2()2(1

1)2(

1)1(

1)1(

2

)1()1(1

)1()(1

)(2

)()(1

)(

G)()(+

GG)(&)("

GG)(2)(1

p p g p p g p p p p p g p

p g p p p p p g p

p g p

p p

=

=

=

=

=

=

=


56/67


!a# !b# !e#!c# !%#

"-+2" 1".1""-+2" -.12+"-+2" 2"-.12

"-+2" 1".1"-+22"-=.11+"-+2" 1&.1"-+&1 -2.11"-+2" 12.1"-&=>=21>.1?"-+2" 22.1"-+?1->"&.1="-+22??&2.1"-"==> &.1>"-+2" +>-.1"->=&-?-=.1- "-+22++?&.1"- ?&1?".1+

"-+2" 1".1"-+2-&>&=.1">+1> +2".1=e".1&"-+2" 1&.1"-&?+> 1+.1"&+&+=">&.1)-+.=(>.&-?""-+2-2 1+.1"->">-">2.1& 2+& = &.1??-?.2>"2.-2">""""""".1"&="??>2+.12=-?+">-=.1=1-+.=>+.1+.1+.1+.1+.1+.1

241−

−

k

+.1 = p


57/67


!%# Steffensen s metho%

*bout the same accurac' as the

&ewton s metho%

"-+2" 1".1"-+2" 1&.1"-+2" 12.1"-+2" 22.1"-+22??&2.1"-+2" +>-.1"-+22++?&.1"-+2-&>&=.1

"-&?+> 1+.1"->">-">2.1"&="??>2+.1

+.1

+.1 = p

"->">-">2.1"&="??>2+.1

+.1

"-+2" +=".1"-+22++"&.1"-+2-+22&.1

"-+2" 1".1


58/67

orollar 3 (f is a pol'nomial of %egree


59/67

orollar 3 1et an% be pol'nomials of

%egree at most (f with , are

%istinct numbers with

then for all values of

with real or complex coefficients, then there exists

unique constants , possibl' complex, an%

unique positive integers such that

k x x x ,,,

21 n

)( x P )( x Q

nk >

,,,2,1 for ),()( k i x Q x P i i

)()( x Q x P = x

k x x x ,,, 21

1n)( x P

k mmm ,,, 21

1 2

1 2( ) ( ) ( ) ( ) .k mm m

n k P x a x x x x x x − − −L

1

1 andk

i i

m=

=


60/67


61/67

Ihen &ewton8+aphson metho% is use% to fin% an

approximate "ero of a pol'nomial, nee% to evaluate

an% at specifie% values

can be evalueate% in a similar manner

)( x P

)(< x P

)()(<

)(


62/67

$o overcome this %ifficult')

begin with a complex initial value or

.Mller s metho%

One problem with appl'ing the Secant , 6alse /osition , or

&ewton s metho% to pol'nomials is the possibilit' of the

pol'nomial having complex roots even when all thecoefficients are real numbers

$his is because, all subsequent approximations will also be

real numbers if the initial approximation is a real number

$h ) (f i l " f l i li i '


63/67

$heorem) (f is a complex "ero of multiplicit'

of the pol'nomial with real coefficients, then

is also a "ero of multiplicit' of the

pol'nomial , an% is a factor of

the pol'nomial

* secon% %egree pol'nomial is use% to fit three pointsin the vicinit' of the root instea% of two points as inthe Secant metho%

.uller s metho% is an interpolation metho% that usesqua%ratic interpolation rather than linear

m )( x P

).( x P

bi a z −

)( x P

m

mbaax x )2( 222 +

z a bi


64/67


65/67

2c−


66/67

Solving for the "eros of the qua%ratic allows the metho% tofin% complex pairs of roots -oth real "eros an% complex"eros can be foun%

Given three previous guesses for the root

" 2 "2

" 2 2

2

" 1 2 "

&

2I therefore , t$o possibilities of

&

2;eclect ,

sgn( ) &

$hich is the closest Cero of ( ) to .

/nce is determined, reinitialiCe using , ,

to obtain .

c p p p

b b ac

c p p

b b b ac

P x p

p p p p

p

− =± −

= −+ −


67/67

Jome$or &•

#umerical nal sis, ?th edition• age 1 E. 11 use

Bisection Method, #e$ton%s Method

;ecant Method, Method of False osition

Muller%s Method

ompare the results, chec the rates of con*ergence

Aue on Monda , #o*ember ?, 2 1+ =3 am

lecture 06 solution of equations in one variable

Documents