CS321. Numerical Analysis and Computing

Transcription

1 CS Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY September 8 5

2 What is the Root May physical system ca be writte i the orm o a equatio This equatio represets the relatioship betwee the depedet variables ad idepedet variables The root o a oliear equatio is the solutio o that equatio It ca also be said to be the solutio o a oliear system Most oliear equatios are too complicated to have a aalytical solutio I practice we are more iterested i idig some umerical solutios eplicit umbers approimate solutios

3 Roots Zeroso a Fuctio

4 Roots Zeroso a Fuctio 4

5 Roots o a Fuctios Let be a uctio that has values o opposite sigs at the two eds o a give iterval [ab] with a < b i.e. a b <. I is cotiuous o [ab] the there eists a umber c i [ab] such that c c is called a root o uctio Eample. The uctio has a root i the iterval []. It has two roots i the iterval [-5] Remark: roots are ot ecessarily uique i a give iterval Need some root idig algorithms or geeral uctios 5

6 A Root o a Fuctio 6

7 A Fuctio with Four Roots 7

8 Bisectio Method Give a iterval [ab] ad a cotiuous uctio i a b < the must have a root i [ab]. How to id it? We suppose a > ad b < b a b a Step. Compute the midpoit c stop i is small ad take c as the root Step. Evaluate c i c a root is oud Step. I c the either c > or c < Step 4. I c < a root must be i [ac] Step 5. Let b c b c go to Step. 8

9 Bisectio Process 9

10 Bisectio Process Fid a root o the uctio

11 Bisectio Process

12 Covergece Aalysis Let r be a root o i the iterval [a b ]. Let c be the midpoit the I we use the bisectio algorithm we compute ad have a b c a b c the Sice the iterval legth is halved at each step we have Hece which is the maimum error i we take c as a approimate to the root r b a a b c r a b c r a b a b a b a b c r

13 Liear Covergece A sequece { } has liear covergece to a limit i there eists a costat C i the iterval [ such that By recursio we have Or equivaletly a liear covergece satisies For some positive umber A C C The bisectio algorithm has a liear covergece rate with C ½ ad A b -a / C C C AC <

14 Stoppig Criterio What is our goal? Whe to stop? How may iteratios? Our goal is to id r Є [ab] such that r With the bisectio algorithm we geerate a sequece such that r c < ε or some prescribed umber ε > i.e. we id a poit c iside the iterval [ab] that is very close to the root r. We the use c as a approimate to r It is ot guarateed however that c is very close to 4

15 How May Iteratios I we wat the approimate root c is close to the true root r i.e. we wat c < ε The the umber o bisectio steps satisies b a < ε Or log b a logε > log r Eample. Fid a root i [67] up to machie sigle precisio a. b. so r must have a biary orm r.***. We have a total o 4 bits 5 is already ied. The accuracy will be up to aother 9 bits which is betwee -9 ad -. We choose є -. Sice b a we eed > yieldig 5

16 Newto s Method Give a uctio ad a poit i we kow the derivative o at we ca costruct a liear uctio that passes through with a slope as l ' Sice l is close to at i is close to r we ca use the root o l as a approimate to r the root o ' may ot be close to r eough we repeat the procedure to id Uder certai coditios { } coverges to r ' 6

17 Newto s Method 7

18 From Taylor Series I but is close to r we may assume that they dier by h i.e. h r or h r Usig Taylor series epasios h h ' " Igorig the higher order terms we have Or h ' h ' Sice h does ot satisy h we use h ' as a approimate to r ad repeat the process 8

19 First Few Approimatios 9

20 Fast Covergece Fid a root or the ollowig uctio startig at 4 4 ' Each iteratio gais double digits o accuracy ad decreases quadratically to

21 Eample

22 Covergece Aalysis Let the uctio have cotiuous irst ad secod derivatives ad ad r be a simple root o with r. I is suicietly close to r the Newto s method coverges to r quadratically. r c r I diers rom r by at most oe uit i the kth decimal place i.e. The or c we have r r The umber o correct decimal digits doubles ater aother iteratio k k

23 Covergece Proo Let e r. Newto s method geerates a sequece { } such that Usig Taylor s epasio there eists a poit betwee ad r or which It ollows that ' ' ' ' e e r r e " ' e e e r ξ " ' e e ξ

24 4 Covergece Proo Cot. We thus have Deie a upper boud We ca choose δ small so that This is to guaratee that is close to r withi a distace o δ ' " e e ξ ' mi " ma > δ δ δ δ c r r ξ δ r ad δ r e

25 For very small δ > we have Covergece Proo Cot. With ρ δcδ < i δ is small eough thereore is also close to r withi a distace o δ. By recursio i is close to r the e Sice ρ < this is to say e " ξ e c δ e ' δ c δ e r e ρ e ρ e e δ ρ e ρ e ρ e lim e as 5

26 Weakess o Newto s Method Newto s method coverges ast oly whe is chose close to r. I practice there might also be a umber o problems. eeds derivative value ad availability. startig poit must be close to r. lose quadratic covergece i multiple root 4. iterates may ru away ot i covergece domai 5. lat spot with 6. cyclig iterates aroud r 6

27 Problems o Newto s Method 7

28 Newto s Method Cyclig 8

29 9 Systems o Noliear Equatios Newto s method is really useul or idig zero o a system o oliear equatios Writte i vector orm as Where We have k J k is the Jacobia matri T T ] [' k k k k

30 A Equatio Eample Usig Taylor epasio Let be a approimate solutio ad the computed correctio be. Hece h h h h h h i i i i i T T h h h h h ' h

31 Eample Cot. The Jacobia matri is It ollows that Hece the ew iterate is I practice we solve the Jacobia matri i So that J ] -[' h ] [' ]h [J k k k h k k k

32 Secat Method I Newto s method We eed to evaluate ad at each iteratio We ca approimate the derivative at by Thus the secat method geerates iterates Oly oe uctioal evaluatio at each iteratio ' '

33 Secat Method

34 Commets Secat method eeds two iterates to start with we ca use bisectio method to geerate the secod iterate Secat method does ot eed to kow the derivative o I - is small the computatio may lose sigiicat digits ad becomes ustable The covergece rate o the secat method is superliear e C e With α 5.6. Its covergece rate is betwee that o the bisectio method ad the Newto s method α

35 Hybrid Approaches I practice hybrid methods are usually used For eample we ca use Bisectio Method to geerate iitial iterates that are close to the root so that Newto s method ca be applied Whe the evaluatio o derivative is epesive the Secat Method should be used to replace the Newto s method The trade-os betwee Bisectio Method ad Newto s Method are robustess ad ast covergece The Secat Method is supposed to come betwee these two methods to take the advatages o both ad avoid the disadvatages o either 5