THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.

Transcription

1 THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (Regula-Falsi) Fied poit iteratio Newto Raphso Secat The root of a equatio is the value of that f() = 0. Roots are called the zeros of equatio. There are may fuctios for which the root caot be determied so easily. Methods for Determiig Roots Nocomputer methods.. Graphical method. Aalytical method Computer methods.. Bisectio method. False Positio method. Fied poit iteratio 4. Newto s method 5. Secat Method

2 NONCOMPUTER METHODS Separatio of roots To separate the roots of f()=0 is to divide the whole domai of permissible values ito itervals i each of which there is oly oe root. b a α a α b For the solutio of the roots, we will use,. Graphical method. Aalytical method GRAPHICAL METHOD OF SEPARATING ROOTS ST TECHNIQUE: It is easy to separate the roots if the graph of the fuctio y = f( ) is costructed Where α ad α are the roots of y = f ( ) =0

3 d TECHNIQUE : Give y = f( ) ad f( ) = r( ) s( ) =0, we ca write f ( ) = r( ) s( ) = 0 r( ) = s( ) The sketch the graph of y = r( ) ad y = s ( ), the poits of itersectio of the graph of these fuctios are the roots of the give equatio as follows root root Eample: Use graphical method to fid the iterval(s) i which the roots of the equatio = 0 are icluded. Where, = + let g( ) = ad s( ) = + Roots lies o [, ], [, 0] ad [, ] 0 NOTE: Graphical methods is ot very precise. It makes it possible to roughly determie the iterval.

4 ANALYTICAL METHOD OF SEPARATING OF ROOTS Theorem : Assume we wat to fid a root of equatio f()=0. Assume that f :R! R ad f is cotiuous. Let [a, b] R be such that f(a)f(b) < 0 The, by the itermediate value theorem, there eists f( ) = 0. [a, b] such that Now, we ca recommed the followig sequece of operatios to separate the roots usig the aalytic method.. Fid the critical values of f ( ), that is f ( ) = 0. Compute a table of sigs of the fuctio f() settig α equal to a) The critical values of the derivative or the values close them b) The boudaries of the iterval [ ab, ]. Determie the itervals at the edpoits of which the fuctio assumes values of opposite sigs. These itervals cotai oly oe root each i its iterior. Eample: Separate the roots of f( ) = 5 = 0 usig the aalytical method. Solutio: ( ) f( ) = 5, D:, ( ) 5 5 f ( ) = l 5= 0 = l = l l l 5 l 5 l l l ( ) = l =. 85 (. 85 close to ) l l f( ) = The roots of the equatio are i the iterval (-,0) ad (4,5) 4

5 Eample: Separate the roots of the equatio f( ) = + + = 0 D :(, ) f ( ) = + 6 = 0 the ( + ) = 0 = 0 ad = f( ) = Roots lie o (-,-), (-,-) ad (0,) COMPUTER METHODS BRACKETING METHODS(Or, two poit methods for fidig roots) Bracketig methods cosider the fact that a fuctio typically chages sig i the viciity of the root. I this process two iitial guesses for the root are required. These two iitial guesses are called mi (lower) (or =a) ad ma (upper) (=b). Defiitio: Assume that f ( ) is a cotiuous fuctio. Ay umber r for which f() r = 0 is called a root of the equatio f( ) = 0. Also we say that r is a zero of the fuctio f ( ). For eample, 5

6 No aswer (No root) Nice case (oe root) Oops!! (two roots!!) Three roots( Might work for a while!!) Two roots( Might work for a while!!) Discotiuous fuctio. Need special method 6

7 BISECTION METHOD I this sectio we develop our first bracketig method for fidig a zeros of cotiuous fuctio. f () = 0 Algorithm Step : The decisio step for this process of iterval halvig is first to choose the midpoit Where, f( a) f( b ) < 0 a+ b c =. () ad the to aalyze the three possibilities that might arise Step : a. If f ( a) ad f( c ) have opposite sigs ( ( ) ( ) f a f c < 0 ) a zero lies i [ ac, ] c b a Set b=c Therefore, set b=c ad retur equatio () 7

8 b. If f ( b) ad f ( c ) have opposite sigs ( ( ) ( ) (or f( a) f( c ) > 0 ) f b f c < 0 ) a zero lies i [ cb, ] b a c Set a=c Therefore, set a=c ad retur equatio () c. If f() c = 0, the the zero is c. Iterval halvig (bisectio) is a iterative procedure. The solutio is ot obtaied directly by a sigle calculatio. The iteratio are cotiued util the size of a iterval decreases below a prescribed tolerace ε, that is bi ai ε, or the value of f ( ) decreases a prescribed tolerace ε, that is f( ci ) ε, or both. 8

9 Eample: Let f ( ) = + a) Fid the locatio of the root(s) b) Fid a approimatio for the positive root usig Bisectio method with accuracy ε < 0 SOLUTION a) where = +, the f ( ) f ( ) = 0 + = 0 roots are = ( take = ) ad =.8 ( take =.5) Roots lies i (-, -.5), (-.5, 0) ad (,), positive root lie i (,) b) Use bisectio method to approimate positive root a c=(a+b)/ b f(a) f(c) f(a) f(c) ε = b a ε = f () c > < > > > < > > >

10 if ε = b a = is used or.74 if ε = f ( c) = is used Theorem (Bisectio Theorem) Assume that f [ ab, ] ad that there eist a umber r [ a, b] f ( a) ad f( b ) have opposite sigs, ad { } c =0 such that f() r = 0. If represets the sequece of midpoits geerated by the bisectio process a a... a... r... b... b b 0 0 ad a+, b+ = a, c ad a+, b+ = c, b (read page 54), the b a r c for = 0,, () c =0 Ad therefore the sequece { } coverges to the zero = r; that is limc = r () Proof : Sice both zero r ad the midpoit c lie i the iterval a, b, the distace betwee c ad r caot be greater tha half the width of this iterval (see fig. below) a r c b r c b a Thus, b a r c for all.(4) 0

11 Observe that the successive iterval width form the patter b0 a0 b a = b a b a b a = = the b b a 0 0 a = ( 5) Combiig (4) ad (5) result i b a r c 0 0 for all + The umber N of repeated bisectio eeded to guaratee that the Nth midpoit c is a approimatio to a zero ad has a error less tha the pre assiged value ε is Proof: Use b a + l( b a) l ε N = it l < ε the,. (6) b a l l ε l( b a) ( N ) l l ε + < + < l( b a) l ε N > l Therefore, the smallest value of N is N l( b a) l ε = it l Eample: The bisectio method is used to fid a zeros of f() i the iterval [,7]. How may times must this iterval be bisected to guaratee that the approimatio c has a accuracy of Give a =, b= 7 ad ε = 5 0 9

12 9 l( 7 ) l( 5 0 ) N = it = it(. ) take N = l If the give iterval 0 times bisected we obtai a accuracy Advatage: A global method: it always coverge o matter how far you start from the actual root. Disadvatage: It caot be used to fid roots whe the fuctio is taget is the ais ad does ot pass through the ais. For eample: It coverges slowly compared with other methods.

13 FALSE POSITION METHOD (REGULA FALSI) Aother popular algorithm is the method of false positio method. It was developed because the bisectio method coverges at a fairly slow speed. I the false positio method, the oliear fuctio f() is assumed to be liear fuctio g() i the iterval (a,b) as follows y f ( ) g( ) secat lie ( b, f ( b )) a c r o b ( a, f ( a )) False-poit method To fid the value of c, we write dow two versios of the slope m of the lie L=g() m = f( b) f( a) b a...( ) Where, the poits (a,f(a)) ad (b,f(b)) are used, ad 0 f( b) m = c b...( ) Where, the poits (c,0) ad (b,f(b)) are used. Equatig the slopes () ad (), we have

14 f ( b) f ( a) 0 f ( b) = b a c b which easily solved for c to get c = b f( b)( b a) f( b) f( a)...( ) The three possibilities are the same as before (bisectio method). If f ( a) f ( c) < 0 the root lies i [ a, c]. If f ( a) f ( c) > 0 the root lies i [ c, b]. If f () c = 0 the = c Eample (False Positio Method) : Use False Positio Method to approimate 5 to decimal places ( i. e ) Solutio: First fid f ( ) where = = = that is f = ( ) 5 Use f = to locate the root(s) ( ) 5 f ( ) = = 0 = locate i [, ] a c= b f ( b)( b a) f ( b) f( a) b f(a) f(c) f(a)f(c) ε = f () c > > > >

15 Approimatio of the root = 5.59 with accuracy Eercise : Give + = l 0 Solve above equatio, usig a) Bisectio Method b) False Positio Method correct up to at least decimal places i.e ε = c) Compare the two methods accordig to the umber of iteratios performed. Eercise : Give e + = 0 Solve above equatio, usig d) Bisectio Method e) False Positio Method correct up to at least decimal places i.e ε = f) Compare the two methods accordig to the umber of iteratios performed. NOTE:. To compare betwee two methods you should start by the same iterval for both of them.. You must use same error calculatio for both of methods.. The termiatio criteria used i the Bisectio Method ( ε b a ) is ot useful for the False Positio method ad may result i ifiite loop. The best choice is to use ε f ( c ) to compare the two methods. 5

16 Ope Methods The ope methods are based o formulas that require oly a sigle startig value of or two startig values that do ot ecessarily bracket the root. Ope methods sometimes diverge or move away from the true root as the computatio progress. FIXED POINT ITERATION I this method, the equatio f( ) = 0 is rewritte i the form = g( )...( ) ad iterative procedure is adopted usig the relatio + = g( ) for = 0,,......( ) Where, a ew approimatio to root +, is foud usig the previous approimatio ( 0 deotes the iitial guess). The procedure is repeated util a covergece criterio satisfied. For eample ε ad / or f( ε...( ) + + Defiitio: Geometrically the fied poit of a fuctio y = g( ) are the poits of itersectio of y = g( ) ad y =. See figs below. (where, Fig. (a) ad (b) coverget fied poit iteratio ad Fig. (c) ad (d) diverget fied poit iteratio) 6

17 Defiitio: The iteratio + = g( ) for = 0,,... is called fied poit. Theorem: Let g be a cotiuously differetiable fuctio which maps the iterval I ito itself (I=[a,b]). Thus I g( ) I Suppose further that g ( ) <, I The a) g has a uique fied poit i I, α say b) for ay choice 0 I, the sequece + = g( ) coverges to α. The sequece 0,,,, will coverge to a root α of the equatio = g() provided a suitable startig value 0 is chose ad < g (α ) <. That is Fied poit iteratio coverges if g ( ) < 7

18 Eample: Use fied poit iteratio to fid the positive root of Solutio First locate the root(s) e = 0 with a accuracy ε = 0 f ( ) = e = 0 e = = l 069. ( take = ) f ( ) = e Positive root lies i [,] Fid g() a) e = 0 = ( e ) where g( ) = ( e ) g ( ) = e b) ad g ( ) > i [, ] ot satisfy theorem e = 0 e = + = l( + ) where g( ) = l( + ) g ( ) = ad g ( ) < i [, ] satisfy theorem + The fied poit iteratio + = g( ) + = l( + ) Choose ay poit betwee [,], let choose startig poit 0 = 5. ad start iteratio Iteratio : = l( + ) = l( (. 5) + ) = Successive error : = = residual error : f ( ) = f (. 869) =

19 Iteratio : = l( + ) = l( (. 869) + ) =. 775 Successive error : = = residual error : f ( ) = f (. 775) = 0. 7 Iteratio : = l( + ) = l( (. 775) + ) =. 969 Successive error : = = 0. 0 residual error : f ( ) = f (. 969) = Iteratio 4: = l( + ) = l( (. 969) + ) = Successive error : = = residual error : f ( ) = f (. 788) = Iteratio 5: = l( + ) = l( (. 788) + ) = Successive error : = = < ε = 0 ( STOP) 5 residual error : f ( ) = f (. 69) = ( DO NOT STOP if Re sidual error used) Iteratio 6: = l( + ) = l( (. 69) + ) = Successive error : = = < ε = residual error : f ( ) = f (. 66) = ε = 0 6 Approimate root. 69 (whe successive error is used) Approimate root. 66 (whe residual error is used) Eample: Let f( ) = 9

20 If we write f( ) = r( ) h( ), where r( ) = ad h( ) = + which of the followig iterative methods will coverge to the positive root a) h ( ) r ( ) + = b) h ( ) = r ( + ) Also defie the regio through which the iterative method is coverget, apply steps of the iterative method. Solutio:First locate the roots f ( ) = e Positive root lie i [,] because f( ) f ( ) < 0 a) h ( + ) = r ( ) that is = = therefore, g ( ) = ad g ( ) = where g ( ) = > i [, ] That is iteratio ot coverge to fied poit b) h ( ) = r ( + ) that is + = + + = + g ( ) = + g ( ) = whereg ( ) = < i [, ] + + That is iteratio coverges to fied poit Fied poit iteratio is + = + choose 0 = 5. Iteratio : = + = (. 5) + = Iteratio : = + = (. 4494) + = Iteratio : = + = (. 4878) + =. 404 After iteratio 0

21 . 404 with successive error : = residual error : f ( ) = NEWTON METHOD (NEWTON RAPHSON METHOD) To obtai a iteratio with rapid covergece to the solutio of the equatio f( ) = 0. We seek a rearragemet of that equatio satisfyig = g( ) f( ) =0 with a additioal property gs () = 0. DERIVATION OF THE METHOD Most widely used method. Based o Taylor series epasio: Δ f( + ) = f( ) + f ( ) Δ + f ( ) + OΔ! The root is the value of whe f( ) = 0 Rearragig, 0 = f( ) + f( )( ) f( ) = f ( ) ANOTHER DERIVATION OF NEWTON METHOD Newto-Raphso formula

22 f ( ) f ( ) f( ) Slope of the taget lie f( ) 0 f( ) m = where m = f ( ) f ( ) = + + f( ) f( ) + = ad Newto method + = f ( ) f ( ) For Newto-Raphso method f ( ) f ( ) f ( ) g ( ) = ad g ( ) = f ( ) f ( ) ( ) For covergece f( ) f ( ) g ( ) = < for all I ( f ( ) ) Eample: Show that Newto s İterative Method to fid the th root of the umber C is give by C i = ( ) + i + i Apply iteratios of the above form to fid the approimatio to the root of the followig equatio f( ) = 6 what is the accuracy of the last iterative value of? Solutio

23 where f ( ) = C the f ( ) = f( i) ( i C) i i C C Newto Method : i+ = i = i = + i ( + = ) i + f ( i) i i i i i = Where f ( ) 6 locate the root [ ] ( eercise) root lie i 5,6 choose 0 = 5. 5 where = ad C = 6 6 ITERATION : = (5.5) = ITERATION : = (5.44) = ITERATION : = (5.496) = after iteratio ε = f (5.440) = Eample : Use Newto Method to fid the upper root α of the followig equatio with a accuracy ε = 0 Solutio First locate the root(s) 7+ = 0 7 f( ) = 7+ the f ( ) = 7 = 0 = ± ±. 575 take =± 5. which is close to ±

24 f( ) = Roots lie i (-,-), (0,) ad (,) upper root α (, ) choose 0 = Newto iteratio for give fuctio + = 7 +, = = = = =. 55 = = =. 4.. = residual error : f ( ) = < 0 therefore, Multiple Roots A multiple root (double, triple, etc.) occurs where the fuctio is taget to the ais 4

25 Eamples of Multiple Roots Problems with multiple roots The fuctio does ot chage sig at eve multiple roots (i.e., M =, 4, 6, ) f () goes to zero - eed to put a zero check for f() i program slower covergece (liear istead of quadratic) of Newto-Raphso ad secat methods for multiple roots. 5

26 ORDER OF A ROOT Assume that f ( ) ad its derivatives f ( ), f ( ),..., f ( ) are defied ad cotiuous o a iterval about = p. We say that f ( ) has a root of order M at = p if ad oly if ( M ) ( M ) ( M) f( p) = 0, f ( p) = 0, f ( p) = 0,..., f ( p) = 0 ad f ( p) 0 A root of order M= is ofte called a simple root if M> it is called multiple root. Note: For o-simple root if we use Newto Method we get first order covergece but for secod order covergece we use acceleratio Newto Method. Whe the multiplicity of the root is kow f( ) + = M f ( ) Eample: Determie the multiplicity of the root α = 05. for the followig oliear equatio = 0 perform iteratios of Newto s method ad the accelerated Newto s method ad compare the accuracy. Solutio: Multiplicity of the root α = f( ) = f ( 05. ) = 0 f ( ) = f ( 05. ) = 0 f ( ) = f ( 05. ) = 0 f ( ) = 4+ 5 f ( 05. ) = 7 0 That is M = Fid locatio of the root (EXERCISE) where α =. [, ] 05 0 choose 0 = 07. Newto s Method 6

27 f( ) + = f ( ) Iteratio : Iteratio : f ( 07. ) = 07. = 064. f ( 07. ) f ( 0. 64) = = f ( 064. ) Successive error : = = Residual error : f ( ) =. 7 0 Accelerated Newto s Method Mf ( ) + = where M= f ( ) Iteratio : Iteratio : f ( 0. 7) = 0. 7 = f ( 07. ) f ( ) = = f ( ) Successive error : = = Residual error : f ( ) = 5 0 SECANT METHOD Whe the derivative fuctio f ( ), is uavailable or prohibitively costly to evaluate, a alterative to Newto s Method is required. The preferred alterative is the Secat Method. 7

28 f ( ) Secat lie f( ) Newto formula : f( ) + = f ( ) From above figure, slope of the secat lie is f( ) f( ) f ( ) = substitute i Newto method f( ) f( ) + = + = SECANTMETHOD f( ) f( ) f( ) f( ) Notice that this is very similar to the false positio method i form Still requires two iitial estimates But it does't bracket the root at all times - there is o sig test Eample : Fid the commo poit betwee f( ) = ad f( ) = usig secat method for iteratios. What is the achieved accuracy for the last iterative value. Solutio : Root lie i [-,-] (EXERCISE) 8

29 Choose = 8. ad =. 0 Iteratio : Iteratio : Iteratio : f( )( 0) = =. 6 f( ) f( ) 0 f( )( ) = =. 698 f( ) f( ) f( )( ) = = f( ) f( ) 4 With error ε = f ( ) = Eercises: Q-) Costruct the coverget fied poit iteratio to fid the lowest root of the followig equatio with a accuracy ε = 0. 9

30 l + 7 8= 0 Q-) Use the method of fied poit to solve the equatio e + = 0 with a accuracy ε = Q-) Fid the smallest root of the fuctio f ( ) = si cos over [, 0π] usig Newto s method with a accuracy ε = Q-4) Estimate usig Newto s method with a accuracy ε 000. Q-5) Fid a approimatio to the root of e ( ) = 0, usig False positio method for steps. Q-6) Give e + = 0 a) Separate the roots usig aalytical method. b) Approimate the largest root of the above equatio with a accuracy of ε 00. usig i) Fied poit iteratio ii) Newto s Method commets the result 0