SOLUTIONS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS

Transcription

1 96 Egieerig Mathematics through Applicatios 5 SOLUTIONS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS aaaaa 5. INTRODUCTION The equatios of the form f() = 0 where f() is purely a polyomial i. e.g = 0 is called a algebraic equatio. But, if f() ivolves trigoometrical, arithmetic or epoetial terms i it, the it is called trascedetal equatio. E.g. e = 0 ad log 0. = 0. Basic Properties ad Observatios of a Algebraic Equatio ad its Roots: (i) If f () is eactly divisible by ( α), the α is a root of f(). (ii) Every algebraic equatio of th degree has ad oly real or imagiary roots. Coversely, if α, α,, α be the roots of the th degree equatio f() = 0, the f() = A( α )( a ) ( α ). (iii) If f () is cotiuous i the iterval [a, b] ad f(a), f (b) have differet sigs, the the equatio have at least oe root betwee = a ad = b (oftely kow as Itemediate Value Theorem.) (iv) I a equatio with real coefficiets, imagiary roots occur i cojugate pairs, i.e. if α + iβ is root of f() = 0, the α iβ is also its root. Similarly, if α+ β is a irratioal root of f() = 0, the α β is also its roots. Some Geeral Observatios o the roots of algebraic equatios are as follows: (i) The umber of positive roots of a algebraic equatio f() = 0 with real coefficiets ca ever eceed the umber of chages i sig from positive to egative ad egative to positive. E.g. Take = 0 Here the coefficiets are with sig + + i.e. they chage from positive to egative, egative to positive, positive to egative ad, hece the equatio cotais three positive roots. 96

2 Solutios of Algebraic ad Trascedetal Equatios 963 (ii) The umber of egative roots ca ot eceed the umber of chages of sig i f( ). E.g. f( ) = = 0; Here all the coefficiets are egative ad so there is o chage i sigs, meaig there by that the give equatio does ot cotai ay egative root. (Note: observatios (i) & (ii) are more commoly kow as Descrate s sig rule.) (iii) Every equatio of a odd degree has at least oe real root, whereas every equatio of a eve degree with last term i it if egative, has at least two real roots, oe positive ad the other egative. About Numerical Computatio of Roots: I Sciece ad egieerig ofte we require the root of algebraic, o-algebraic or trascedetal equatios ad that i a situatio whe o iformatio is available about the root, it is ot ecessarily be whole umber ad difficult to fid by covetioal Mathematics methods. I geeral to compute umerical value of the root, we wish to fid some approimate value of the root which satisfies our eed without much chage i its basic characters. For doig so, geerally we eed some rough estimate of the root to start with ad the iterate it for better approimatios. For that matter, we classify them i broadly i two types viz. closed ed ad ope ed methods. Uder closed ed method, as the ame it self, two iitial guesses are required to block the root. I these methods, root essetially coverges as we move closer ad closer to the root i each iteratio. That is why some time Bi-sectio ad Regula-Falsi Methods are closed bracketig or closed ed methods of iteratio. I some of the methods though we eed two iitial estimates, however, f() is ot required to chage sig betwee the estimates, that is why these are ot termed as bracketig methods ad call them ope ed, like Secat ad Newtos Raphso methods. Here, i this chapter, we discuss some of the methods amely Graphical, Iteratio Method, Bisectio Method, Regula-Falsi, Newto Raphso s, Horer s Method, Muller s Techique ad Li-Bairstow Method oe by oe. 5. GRAPHICAL METHOD A simple method for obtaiig the approimatio value = α (say) ear to the root or to estimate the root of the equatio f() = 0 is to make a plot of the fuctio ad observes where it crosses the -ais. This poit, which represets the value for which f () = 0, provides a rough approimatio of the root. More coveietly, we ca epress f() = 0 as f () = f () ad plot the graph of f () ad f () with respect to the same aes ad fid the abscissa of the poit of itersectio which is the root of the equatio f () = 0 E.g. ) Solve the equatio, f() = + a + b = 0 We draw the graph of y = f( ) = = 0, y = f( ) = ( a + b) ad the cosider their poit of itersectio. E.g. ) I order to fid the root of the equatio e si =, we rewrite it as e e = takig si = e ad the determiig the itersectio of y = e with y = e. Note: It is observed that graphical techiques are of limited practical value because they are ot precise. However, they ca be utilized to obtai rough estimates of roots.

3 964 Egieerig Mathematics through Applicatios Sigificace: Apart from providig rough estimates of the roots, graphical iterpretatios are importat tools for uderstadig the properties of the fuctios ad aticipatig the pitfalls of the umerical methods. Eample : Solve Graphically si = for a o-zero root. 4 Solutio: Let y = si; y = 4 TABLE 3.: VALUE OF y AND y FOR STARTING VALUES OF X (radia) y y f() = y y > < 0 Table 3. idicate that the root lies betwee ad. (radia) TABLE 3.: VALUES OF y AND y FOR VARIOUS SMALL INTERVALS y = si y = f() = y 4 y > > > > > > > > > > > < 0 Table 3. idicates the tred as there is a chage i sig i values of y for =.90 ad.00. (radia) y = si TABLE 3.3: y = f () = y 4 y > > > 0 Table 3.3 shows that =.93 is better approimatio. The real root of the equatio si = is.93. 4

4 Solutios of Algebraic ad Trascedetal Equatios 965 y () P = Poit of itersectio of two curves Root =.93. y y.0 4 P(.93, 0.94) d d y 0. c b a y Fig ITERATION METHOD This method is used fidig the root of oly those equatios f() = 0 () which are epressible as = φ() () The root of the equatio () are the same as the poit of itersectio of the straight lie y = ad y = φ(). For that, first we fid a iitial approimatio 0 of the required root ad the make it better value by replacig by 0 i the right had side of () i.e. = φ( 0 ) (3) Likewise a still better approimatio is computed by replacig = i the right had side of (3) i.e. = φ( ) (4) This procedure is cotiued, 3 =φ( ), 4 =φ( 3), =φ( ) (5) If this sequece of approimate values 0,,, coverges to a limit α, the α is take as the root of the equatio ()

5 966 Egieerig Mathematics through Applicatios Geometrical Sigificace: Draw the graph of y = ad y = φ () as show below i the figure 3.. Sice φ'() < for covergece, the iitial approimatio of the curve y = φ () must be less the 45 i the eighbourhood of 0. This fact has bee observed i costructig the graph. y P P P 3 Q3( 3, φ( )) Q(, φ( )) P 0 Q(, φ( 0)) y = φ( ) y = φ( 0 ) φ( ) φ( ) φ( 3) O A 0 A A A Fig. 5. For tracig the covergece of the iteratio process, draw the ordiates φ( 0 ). The from poit p 0, draw a lie parallel to OX util it itersects the lie y = at the poit Q (, φ ( 0 )). This poit Q is the geometric represetatio of the first iteratio = φ ( 0 ). The draw Q P, P Q, Q P, P Q 3 etc. as idicated by the arrows i geometry. The poits Q, Q, Q 3,, thus approaches to the poit of itersectio of the curves y = ad y = φ () as iteratio proceeds ad coverges to the desired root. Coditios for Covergece: This sequece of approimate roots does ot always coverge. To make its covergece sure, it have to satisfy certai coditios.. It be the iterval cotaiig the root = α, of the equatio = φ (), the φ() < for all I.. The iitial approimatio 0 for the root lies i I.

6 Solutios of Algebraic ad Trascedetal Equatios 967 Observatios:. As there could be several ways of rewritig the give equatio f() = 0 as = φ(), hece care should be take so that the coditio. φ () < for covergece is satisfied. Further, if φ () < the root coverges but oscillates about the eact value. e.g. illustratig the above fact: e.g.. The equatios 3 = 0 () ca be rewritte either = 3 + () ( 5 3 ) = + (3) 5 I () φ () = 3 ad φ ( ) 7.75 =.5 = > (4) where i (3), φ ( ) = (5 3 ) ad ( ) φ = < (5) = A root of the equatio () lies i the iterval (, ) ad hece we have evaluated φ ( ) = 0.35 <. Clearly equatio (5) will esure covergece, provided the iitial =.5 approimatio for the root is at =.5 choose i the iterval (, ).. Further, the covergece of a algorithm ca be categorized ito two types viz. local covergece if the startig poit 0 is sufficietly close to the root ad global covergece for ay startig value. E.g. illustrates e.g. the above facts. e.g. Take equatio f () = 3 = 0 () The root of this equatio are = 3 ad =. Case : We ca rewrite equatio () as = + 3 = g( ), say () Startig with 0 = 4.0, we get = = 3.0 = = = (3) As, α = Case : f() = 3 = 3 ca be rewritte as = (4) Startig with 0 = 4.0, we get =.500; 5 = 0.99; = =.08 3 = 0.375; 7 = 0.99; 4 =.63 8 =.003 (5) From above, we observe that { } sequece approimatio of root coverges to the root = α of f (), which is, but iteratio oscillates rather tha covergig mootoically. Case 3: Further f () = 3 3 = 0 ca be rewritte = (6) Startig with 0 = 4.0, we get = 6.5, = 9.635, 3 = 9.0 Which clearly goes o divergig, resultig i astray istead of the desired result. Hece, a care should be take while rewritig f() = 0 as = g().

7 968 Egieerig Mathematics through Applicatios Order of Covergece (Rate of Covergece): Ay method is said to have covergece of order p, if p is the largest positive real umber p such that ε+ k ε where k is a fiite positive costat, ad ε ad ε + are the errors i th ad ( + ) th iterated value of the root α of the equatio f() = 0 respectively. Whe p =, the covergece is said to be liear, whe p =, the covergece is said to be quadratic. Discussio o the Order of Covergece of Iteratio Method: Let α be the root of the equatio f() = 0 epressible as = φ(), the α = φ(α) () If ad + are the th ad ( + )th approimatio of the root α, the + α = φ( ) α () From () ad (). + = φ( ) (3) By mea value theorem of differetial calculus, φ( ) α = ( α)φ (θ), where < θ < α (4) Usig (4) i (3), + α = ( α)φ (θ) (5) Let p be the maimum value of φ (θ) i the iterval i.e. φ (θ) p for all i. The from (5), + α p α i.e. ε+ p ε (6) where ε ad ε + are errors i the th ad ( + )th iterated value of the root. Sice the ide of ε beig, the rate of covergece of the iteratio method + = φ( ) is liear. Observatios:. This method is particularly useful for fidig the real root of a equatio give i the form of a ifiite series.. Smaller the value of φ'(), the more rapidily it will coverge. Aitke s Method: Let,, + be the three successive approimatio to the root α of the equatio = φ (). The we kow that (α ) = k(α ) ad (α + ) = k(α ) (7) ( α ) ( O dividig the two, α + ) = ( α ) ( α ) Implyig ( + ) α + = + + But i the sequece of approimatios, = +

8 Solutios of Algebraic ad Trascedetal Equatios 969 = ( ) = ( + ) = + = + + = + Hece (7) ca be writte as ( ) α= + (8) + This yields successive approimatios to the root α ad eplais the term process. I umerical applicatio of the problem, the values of the followig quatities must be obtaied. + Eample : Fid the root of the equatio = cos + 3 correct to three decimal of places. Further, test its acceleratio of covergece: Aitke s process. Solutio: Re-write the equatio i the form = (cos + 3) () So that si φ () = (cos + 3) ad φ ( ) < < () π Hece, the iteratio method ca be applied to equatio () ad we start with 0 =. The successive iteratios are =.500, 4 =.56, 7 =.53, =.535, 5 =.5, 8 =.54; 3 =.58; 6 =.54; Clearly,.54 ca be cosidered as the solutio correct to three decimal places. Now, from the table of differeces, usig first three umbers of iterated values, fourth value is achieved. =.500 = =.535 = 0.05 = =.58 Ad ( ) ( 0.07) = =.58 = Eample 3: Use method of iteratio to fid a root of the equatio e = lyig betwee 0 ad correct to three decimal of places. Solutio: Rewrite the give equatio as = e () So that φ() = e ad φ () = e ()

9 970 Egieerig Mathematics through Applicatios Here φ () = for <, hece this method esures the coditio of covergece. Defie + = φ( ) ad take 0 =, we fid the successive iteratios are give by = , e = 0 e = = , = e = , = , 8 = = , 9 = , 6 = = , 0 = , 7 = = , = , 8 = = , = , So = i the desired root correct to three decimal of places. 5.4 BOLZANO OR BISECTION METHOD This method of solvig trascedetal equatios, cosists i locatig the roots of the equatio f() = 0 betwee two umbers say a ad b such that f() is cotiuous for a b ad f(a) ad f (b) are of opposite sigs so that the product f(a)f(b) < 0 i.e. the curve cuts the -ais betwee a ad b. The the desired root is approimately ( b, f( b)) a+ b =. If f ( ) = 0, the is a root of f() = 0. Otherwise the root lies betwee either 'a ad ' or ' ad b' fb ( ) y = f( ) accordig as f ( ) is positive or egative. The we bisect the iterval as before ad cotiue the process a 0 to improve the result to greater accuracy. 3 b From the figure 5.3, f( ) is positive, so that the ( af, ( a)) a+ root lies betwee a ad i.e. =. If f ( ) is Fig egative, the root lies betwee ad. The the et approimatio will be 3 = ad so o. The Bisectio Method is also called Biary Choppig 'or' Icremetal Search Method 'or' Iterval Halvig method. Observatios: (i) Though the bisectio method is ot rapid but it is simple ad reliable. (ii) Sice the ew iterval cotaiig the root is eactly half the legth of the previous oe i.e. the iterval legth is reduced by a factor of at each stage or operatio. Therefore, after repeated iteratio, the

10 Solutios of Algebraic ad Trascedetal Equatios 97 ew iterval cotaiig the root be of legth ( b a ). Say after iteratios, the latest iterval is as small as give, the ( b a) or ( b a) or log( b a) log( ) log which give the umber of iteratios required to achieve the desired accuracy,. For eample, i particular the miimum umber of iteratios eeded for achievig a root i the iterval (0, ) for give error, are as follows: : : If (b-a) =, ad = 0.00, the it ca be see that 0. (iii) As by each iteratio the iterval is reduced by a factor of half meaig thereby that error is reduced by a factor of half. Or i other words, the errors i ( + ) th ad th iteratios are i the ratio of half i.e. ε+ =. Meaig there by, the covergece of this method is liear i.e.. ε Workig Rule: Step, Practically, choose the lower ed as 0 ( = a) ad upper ed as ( = b) to guess for the root so that the fuctio chages sig over the iterval. This ca be checked by esurig that f ( 0 ) f ( ) < Step : Make the followig root r is determied by r = Step 3: Make the followig evaluatios to determie i which subiterval the root lies. (a) If f ( 0 ) f ( ) < 0 root lies i the lower subiterval ad repeat. (b) If f ( 0 ) f ( ) > 0 the root lies i upper subiterval ad repeat the process. (c) If f ( 0 ) f ( ) = 0, the root equals ; termiate the computatio. Eample 4: Solve = 0 for the root betwee = ad = 4 by usig Bisectio Method. Solutio: Here f () = so that f() so cotiuous i 4. Further, f() = 9 f(4) = 9 ad f() f(4) < 0 so that a root lies betwee = ad = 4. a+ b + 4 Now = = = 3 ad f(3) = so that f() f(3) < Hece, = =.5 ad f(.5) = 5.87 For et iteratio, we see that f(3) f(.5) < 0 ad the root lies betwee.5 ad = =.75

11 97 Egieerig Mathematics through Applicatios Similarly, 4 =.875 ad 5 =.9375 ad the process is cotiued for better accuracy. Eample 5: Solve f() = 3 4 by the method of iterval halvig procedure. [KUK, MCA 004] Solutio: For give f () = 3 4, we fid two umbers a = ad b = 3 such that f(a) = f () = 4 ad f(b) = f (3) = 5 are of opposite sig Hece the root lies betwee a = ad b = 3. The for iteratio purpose, we follows as: A alterate way of epressig the problem i tabular form: Iteratio No. a b ( a+ b) = f(a) f(b) f() ve +ve ve (.8750) ve +ve +ve (.679) ve +ve +ve (0.3066) ve +ve ve ( ) ve +ve ve (0.0055) ve +ve +ve (0.488) ve +ve +ve (0.079) ve +ve +ve (0.039) ve +ve +ve (0.036) ve +ve ve (0.0040) Hece, the root correct to three decimal places is = REGULA FALSI OR FALSE POSITION METHOD This is the oldest method of computig the real root of a umerical equatio f () = 0 ad it is almost a replica of bisectio method. Draw the graph of the curve y = f () through = a ad = b such that f (a) ad f(b) are of opposite sigs. The = α is the abscissa of the poit C where the graph of y = f () meets the -ais. Let the chord AB meets the -ais at a poit whose abscissa is which is take as the first approimatio y Aa (, fa ( )) of the root. 0 b a To obtai the value of, cosider the equatio of the (, f( )) lie AB (two poit form) through (a, f (a)) as bellow: Bb (, fb ( )) y f(a) = m( a) Fig. 5.4 or y f() a = fb () fa () fb () fa () ( a) as m = ( b a) ( b a) But it meets the -ais at = a where y = 0 fb () fa () ( b a) i.e. 0 fa ( ) = ( a) or = a f() a () ( b a) fb () fa () Successive applicatio of this process gives,,,. Such poits leadig to a accurate value of (= α) as show i the figure 3.4. This iteratio process is kow as the

12 Solutios of Algebraic ad Trascedetal Equatios 973 method of false positio ad its rate if coversace equals.68, which is faster tha that of Bisectio Method. Some times this method is also kow as Method of Variable Secat. Geeral epressio represetig ( + ) th iterated value + is as follows: ( ) = f( ) + f ( ) f ( ) provide that at each step, f( ) f ( ) < 0. () Observatios: The method of false positio is based o the priciple that ay portio of a smooth curve is practically straight for a short distace ad therefore, it is legitimate to assume that the chage i f() is proportioal to the chage i over a short iterval as i the case of liear iterpolatio from logrithmic ad trigoometric tables. For it, assume that the graph of the curve y = f () is straight lie betwee the pts. (a, f(a)) ad (b, f (b)), these pts beig o the opposite sides of the -ais. Pit Falls of the False-Positio Method: Although false positio method is preferable amog all the bracketig methods, but there are certai fuctios o which it performs poor ad bisectio yields superior results, e.g. Take f () = 0. I this problem, after 5 iteratio, true error is reduced less tha % i bisectio method whereas i case of false positio after 5 iteratios results has bee reduced about 59% leavig very high error. Coditio for Covergece: Let { } be a sequece of values of obtaied by the above method ad let α be the eact root of the give equatio. The the method is said to be coverget if lim α = 0 Order of Covergece Let ε be the error i the th iterated value, so that = α + ε, + = α + ε +, O puttig these values i epressio ε f( α+ε) εf( α+ε ) ε + = f( α+ε ) f( α+ε ) ( ) = f( ), + f ( ) f ( ) we get = ε ε ε f( α ) +εf ( α ) + f ( α ) + ( ) ( ) ( )! ε f α +ε f α + f α +! ε ε f( ) ( ) ( ) ( ) α +εf α + f α + f α +ε f ( α ) + f ( α ) +!! f ( α) ε = ε ε + O ε f ( α) ( ) + Fid a umber k such that ε + = cε k, usig f(α) = 0 (3) (4)

13 974 Egieerig Mathematics through Applicatios k Ad therefore k c ε = c ε or Thus o usig values of ε ad ε i (3), ε = ε or ( ) k k k f ( α) ε k k + = c ε ε p, where p = f ( α) O equatig the powers of ε o both sides i (4), we get k = + or k ± 5 k = 0 or k =. k The positive value of k is + 5 =.68. Hece,.68 ε = cε + ε = c ε (5) (6) Eample 6. Use Regula-Falsi Method to fid a real root of the equatio log 0. = 0 correct to five places of decimal. [Bhopal, 00; Madras, 003; VTU, 004; Kottayam, 005; NIT Kurukshetra, 008] Solutio: Let f() = log 0. = 0 Now f() = log 0. = = ( ve) f(3) = 3log 0 3. = = (+ve) Thus, there is at least oe root of the give equatio which lies i the iterval (, 3). ( b a) Now = a f( a); for a =, b = 3 fb () fa () (3 ) = ( ) ( ) = + = = f( ) = f (.7) =.7 log 0.7. = = 0.07 Now, f( ) is ve ad f(b) is +ve, so the root lies i betwee.7 ad 3. Here for the et iteratio, a =.7 ad b = 3. ( b a) = a f() a fb () fa () (3.7) =.7 ( 0.07) = = ( 0.07) Now f( ) = f (.7400 =

14 Solutios of Algebraic ad Trascedetal Equatios 975 Clearly, the root lies betwee (= a) =.7400 ad = (b) = 3. 3 (3.7400) =.7400 f(.7400) f(3) f(.7400) = = (+ve) Now, f (.70) is ve ad f(.7400) is + ve, so we ca proceed for further iteratios for more accuracy. But, here we observe that the two cosecutive values of are same up to 3 decimal place ad hece, is the correct values of up to 5 decimal places. Eample 7: Usig Regula Falsi Method, compute the real root of the equatio e = 0 correct up to three decimals places. [SVTU, 007; NIT Kurukshetra, 004, 007] Solutio: I f() = e ; for = f() = e = 0.78 (+ve) for = 0.5 f(0.5) = 0.5 e 5 =.7 ( ve) As the value of f() at = 0.5 is ve ad at =.0 it is +ve. Therefore, the root lies betwee = 0.5 ad =. I st Step: For 0 = 0.5 ad = ; ( ) = f( ) f ( ) ( 0) 0.5 = 0.5 (.7) 0.78 (.7) 0.5 = = = f( ) = f (0.8098) = e = 0.8 = d Step: Root lies betwee = ad = = f( ) 3 f ( ) f ( ) = ( 0.8) = = f( ) = f(0.8479) = e = 0.0( ve) 3 rd Step: Net root lies betwee 3 = ad = ( ) = f( ) f ( ) f ( 3) = ( 0.0) = = ( 0.0) f( 4 ) = e = 0.003

15 976 Egieerig Mathematics through Applicatios 4 th Step: Net value of i betwee = ad 4 = = f( ) f ( ) f ( 4) = = = Approimate root is correct to 4 decimal of place, Eample 8: Fid the equatio = 0 which lies betwee ad 3 by the method of false positio. Solutio: Let f() = () Now by puttig = ad = 3, we get f() = () 3 5() 7 = 9 ad f(3) = (3) 3 5(3) 7 = 5 Hece the root lies betwee ad 3 Take 0 = ad = = 0 f( 0) = ( 9) f ( ) f ( 0) 5 ( 9) = ( ) =.6486 () ad f(.6486) = (.6486) 3 5(.6486) 7 = = (3) Hece, the root of equatio lies betwee.6486 ad 3. For et iteratio, =.6486 ad = 3, so that =.4686 (.7547) 5 (.7547) =.6486 (.7547) =.6486 ( ) = ad f(.73564) = (.73564) 3 5(.73564) = 0.054( ve) (4) For et iteratio, root lies betwee 3 =.73564, = = (0.054) =.7356 ( ). 5 ( 0.054) (5) Ad f(.746) = (.746) 3 5(.746) 7 = 0.09 (6) For et iteratio, take 4 =.746, = = =.746 (0.0007) =.747, 5 (0.09)

16 Solutios of Algebraic ad Trascedetal Equatios 977 Ad f(.747) = (.747) 3 5(.747) 7 = (7) Hece agai root lies betwee.747 ad =.747 ( ) =.747 (0.0003) = ( ) Agai puttig =.7473 i equatio (), we get f(.7473) = (.7473) 3 5(.7473) 7 = Hece, the required root up to 4 decimal places is = (8) Eample 9: Obtai the root of the equatio = 00 correct to three decimal places, usig iteratio method. Give the root lies betwee 3 ad 4. Solutio: Give = 0 implyig log 0 = log 0 0 = () Let f() = log 0 f(3) = 3log 0 3 = ad f(4) = 3log 0 4 = () Usig iterative Regula-Falsi Formula, the st approimatio, ( b a) = a f(), a fb () fa () where a = 3 ad b = = = 3 + = ( ) (3) Here f(3.58) = 0.077( ve) ad f(4) = (+ve) (4) Clearly the root lies i betwee 3.58 ad 4. For d approimatio ( b a) = a f(), a where a = 3.58 ad b = 4.00 fb () fa () Roots of f() = = = = Eample 0: Fid the root of the equatio ta + tah = 0 correct to three sigificat figures, usig a iterative formula. Give that the root lies betwee ad 3. Solutio: Give f () = ta + tah f(a) = f() = =.0 () f(b) = f (3) = = () Usig iterative Regula-Falsi Formula, the first approimatio, ( b a) = a f(), a fb () fa () where a = 3 ad b = 4

17 978 Egieerig Mathematics through Applicatios = = 3 + = ( 5687) (3) Here f(3.58) = 0.077( ve) ad f(4) = (+ve) (4) Clearly the root lies i betwee 3.58 ad 4. For secod approimatio, ( b a) = a f(), a where a = 3.58 ad b = 4.00 fb () fa () Root of f() = = = = Eample 0: Fid the root of the equatio ta + tah = 0 correct to three sigificace figures, usig a iterative formula. Give that the root lies betwee ad 3. Solutio: Give f () = ta + tah f(a) = f () = =.0 () f(b) = f (3) = = () Applyig Regula-Falsi Method, the first approimatio is. ( b a) = a f() a fb () fa ().0 = (.0) = + = (.0).0735 (3) ad f(.59) = = , f() =.0 (4) Therefore, the root lies i betwee ad.59. For et iteratio, a =, b = = + (.0) = (5) ad f( ) = f (.45) = ta.45 + tah.45 = = 0.57; (6) For et iteratio, a = ad b = =.0 (.0) =.3987 (7).378 ad f(.3987) = = (+ve) (8) Hece, the root lies i betwee ad For 4 th approimatio, a = ad b =.3987 ( b a) = a f( a) =.0 (.0) =.3785 fb ( ) ( a).86 (9)

18 Solutios of Algebraic ad Trascedetal Equatios 979 Here f(.3785) = = Hece, f() = 0 has the root, = SECANT METHOD OR CHORD METHOD This method is a improvemet over the False Positio Method as it does ot require the coditio of f (a) f(b) < 0. Here i this method also the graph of the curve is approimated by the secat chord (straight lie) but at each iteratio step, two recet most approimatios to the root are used to fid et approimatio with the coditio that the iterval must cotai the root. Iitially takig two poits a ad b, we write the equatio of the chord joiig these poits as: fb () fa () y f() a = ( a) b a The the abscissa of the poit where it crosses the -ais (y = 0) is give by ( b a) = a f() a fb () fa () which is a approimatio to the root. The geeral formula for successive approimatios is therefore give by ( ) = f( ), + f ( ) f ( ) Observatios: This method fails if at ay stage of iteratio f ( ) = f ( + ) which shows that it ot ecessarily coverges. Oce Secat coverges, it covergeces faster tha Regula-Falsi with covergece rate of.68. Eample : Fid the root of the equatio e = cos usig the Secat method correct to four decimal places. Solutio: Here e = cos for iitial approimatios a = 0, b = gives, f(0) =, f() = cos e =.7798 ( b a) The = b f( b) = (.7798) = fb ( ) fa ( ) ad f( ) = For et iteratio cosider b ad (beig the recet most iterated values), so that ( b ) ( ) = f( ) = (0.5987) = ( ) ( ) fb f O y Fig. 5.5 Bb, ( fb ( ) D (, f ( ) = a 3 4 α = b E F A C(, f( )) ( a, f( a)) 4 3

19 980 Egieerig Mathematics through Applicatios ad f( ) = For et iteratio cosider ad (beig the recet most iterated values), so that ( ) ( ) = f( ) = (0.0354) = ( ) ( ) ) 3 f f Repeatig this process, the successive iteratio are 4 = , 5 = , 6 = Hece, the root is Observatios: Here we see that up to d iteratios, values are same by False-Positio method as well as by Secat Method, but after 3 rd iteratio oward values are defiitely achieved faster i Secat Method tha that of False Positio Method as here blockig of the root is ot required ad we work recet most values. 5.7 NEWTON METHOD OR NEWTON-RAPHSON METHOD [KUK, 007, 008] This method of successive approimatio is due to Eglish Mathematicia ad Physicist, Sir Issac Newto (64-77) is also called fied poit method. This techique is very useful for fidig the root of the equatio of the form f () = 0, where is real ad f() is a easily differetial fuctio. To derive the formula for computig the root by this method, let 0 deote the approimate value of the desired root ad let h deote the correctio which must be applied to 0 to give the eact value of the root, so that = 0 + h. The equatio, f() = 0 the becomes f( 0 + h) = 0 () Epadig by Taylor s theorem, we get h 0 = f( 0 + h) = f( 0) + hf ( 0) + f ( 0 +θh), 0 θ ; ()! Now if h is relatively small, we may eglect the term cotaiig h ad get the simple relatio, f ( 0) hf ( 0) 0 implyig f ( 0) h f ( 0) (3) So the st iterated value of i.e. 0 h 0 f ( 0) f ( 0) (4) The et improved value of the root is the = f ( ) f ( ) f ( ) Ad so o, + = f ( ) (5) Observatios: (i) This method is geerally used to improve the results obtaied by other methods. It is applicable to solutios of both algebraic ad trascedetal havig real as well for comple roots.

20 Solutios of Algebraic ad Trascedetal Equatios 98 (ii) Newto formula coverges provided the iitial approimatio 0 is chose of sufficietly close to the root, otherwise, it may lead to a astray. Thus, a proper choice of iitial guess is very importat for covergece of this method. For h to be small, f () has to be large. Because, whe f () is large, the the graph of f() while crossig the -ais will be early vertical ad the value is foud rapidly. I other cases, whe f () is small, the correctio may take more umber of iteratios or eve the method may fail leadig to a edless cycle. (iii) If the equatio has a pair of double roots i the eighbourhood of, the Newto Raphso method is f ( ) modified as + = which also shows its quadratic covergece. f ( ) (vi) Divisio by zero may occurs if f () is zero. (v) I Newto-Raphso method, f ( ) has to be evaluated at each iteratio step, istead if we use f (0) at f ( ) each iteratio step, the + = f (0) is called Vo Mises Method. Geometrical Sigificace: Let 0 be a poit ear the eact α of the equatio, f() = 0. See the figure 5.6. the taget at A 0 ( 0, f( 0 ) is y f ( 0 ) = f ( 0 )( 0 ) f ( 0) It cuts the -ais at = 0, f ( 0) (sice at =, y = 0) which is the first approimatio of the root α. If A is the poit correspodig to o the curve, o cotiuig, the taget at A will cut the -ais at which is earer to the root α, a secod approimatio to the root. This way, by repeatig, we approaches the root quite rapidly. Clearly, here we replace the part of the curve betwee the poit A 0 ad the -ais by meas of the taget at A 0. Coditio for Covergece: f ( ) From the formula, + = f ( ) it is evidet that + is a fuctio of [KUK, 007] f ( ) i.e. + =φ ( ) = f ( ) Fig. 5.6 f () ff () () I geeral, φ () = implyig φ () = f ( ) f ( ) Sice the iteratio method ( 5.3) coverges if φ () <. Newto formula coverges if f () f () < f () is close iterval. Hece assumig f (), f (), f () to be cotiuous, we ca select a small iterval i the viciity of α i which the above coditio is satisfied. Order of Covergece: Let α be the eact root of the equatio f() = 0 ad let ad + be the th ad ( + ) th approimatios of the root with ε ad ε + as the correspodig errors i them, so that [KUK, 007, 08, 09] y A A A0( 0 f( 0)) 0 = α 0 y= f( )

21 98 Egieerig Mathematics through Applicatios Now by formula, + = i.e. α+ε + =α+ε ε = α + ε ad + = α + ε + f ( ) f ( ) f( α+ε) f ( α+ε) + ε ε f( α ) + εf ( α ) + f ( α ) + = ε!, Taylor's epasio f ( α ) + ε f ( α ) + f ( α ) +! ε ε f ( α ) + f ( α ) + =ε!, ( α ) = 0 ε f ( α ) + ε f ( α ) + f ( α ) +! ( Q f ) = ε f ( α), f ( α ) + εf ( α) o eglectig 3rd ad higher power of ε = ε f ( α), f ( α) f ( α ) + ε f ( α) This is a quadratic covergece, if ε ε f ( α) f ( α) f ( α) f ( α) = + ε = ε f ( α) f ( α) f ( α) f ( α) ε f ( α) f ( α), where ε for very small ε f ( α) f ( α) f ( α ) <. f ( α) Here it is cleat that error at each subsequet step is proportioal to the square of the error i the previous step. Or i other words, error is corrected to double decimal places at every iteratio at least if the factor f ( α) f ( α) is ot too large. e.g. if error is of order 0.0 after st iteratio, meas it will reduce to after d iteratio ad hece as such the covergece is quadratic. Special Forms i Particular Cases: (i) + = + N to compute the square root of N i.e. N

22 Solutios of Algebraic ad Trascedetal Equatios 983 (ii) N + = + 3 N (iii) N ( ) = k + k k to compute the kth root of N i.e. k N (iv) [ N] + = to compute the root of reciprocal of N i.e. N (v) + = + N N Proofs: (i) Let N or = N so that f() = N ad f () = The, f ( ) N N + = = = + f ( ), N = 0,, 3, (ii) Let ( ) 3 3 = N 3 or = N N = 0 so that f() = 3 N ad f () = 3 The, = = = = + N = f ( ) N 3 + N N +, 0,,, f ( ) (iii) k k k = N or = N N = 0 so that f() = k N ad f () = k k (iv) Let k k k f( ) N k + N The, ( ) k N + = = = = k k k + k f ( ) k k k = or N 0 N = so that f () = Nad f() =, the by Newto's Formula N f ( ) N + = = = + f ( ) = + [ N ] = [ N ] (v) Let = or 0 N N = so that

23 984 Egieerig Mathematics through Applicatios f () = ad the by Newto's formula N f ( ) N + = = = + f ( ) N Geeralized Newto Raphso Method For Multiple Roots: f ( ) If a root α of the equatio f () = 0 is repeated m times, the + = m which is f ( ) called a geeral Newto's formula ad is reduces to Newto-Raphso formula, whe m =. Observatios:. If α is oe of the root of the equatio f() = 0 with m times repetitio, the α will be (m ) times repeated root of the equatio f () = 0, (m ) times repeated root of the equatio f () = 0 ad so o. Further, if the iitial approimatio 0 is sufficietly close to the root, the the epressios, f ( 0) f ( 0) f ( 0) 0 m, 0 ( m ), 0 ( m ), will have the same value. f ( 0) f ( 0) f ( 0). Geeralized Newto formula has a secod order covergece for multiple roots also. f ( ) Eample : Show that the geeralized Newto formula + =, f ( ) gives a quadratic covergece whe f() = 0 has a pair of double roots i the eighborhood of =. Solutio: Let = α is a double ear =, the f() = 0, f () = 0 f( α+ε) Now, ε + = ε, f ( α+ε) where ε ad ε + are the errors i the ad + respectively. 3 ε ε f ( α ) + f ( α ) +! 3! =ε ε εf ( α ) + f ( α ) +! ε f( ) f ( ) α + ε α 3! =ε, Neglectig terms with high powers of ε ε f ( α ) + f ( α) ε f ( α) f ( α) = ε 6 ε ( ) ( ) 6 f ( ) f f α α + α

24 Solutios of Algebraic ad Trascedetal Equatios 985 Which clearly shows that ε+ ε ad hece a secod order covergece. 5.8 NEWTON RAPHSON EXTENDED FORMULA: CHEBYSHEV METHOD We kow that o epadig f () i the eighborhood of 0 by Taylor's series, 0 = f ( ) = f ( + ) = f ( ) + ( ) f ( ) to the first approimatio Hece the Ist approimatio to the root at = is give by f ( 0) 0 = f ( ) () 0 Agai epadig f( ) by Taylor's series to the secod approimatio, we get f ( ) = f ( 0 + 0) = f ( 0) + ( 0) f ( 0) + ( 0) f ( 0) Sice is a approimatio to the root, therefore, f( ) = 0 meas f ( 0) + ( 0) f ( 0) + ( 0) f ( 0) = 0 () Implyig ( 0) f ( 0) = f( 0) ( 0) f ( 0) O dividig throughout by f ( 0 ) ad the usig () f ( 0) f ( 0) f ( 0) ( 0) =, f ( ) f ( ) f ( ) (3) or f ( 0) f ( 0) f ( 0) = 0 f ( ) f ( ) f ( ) This is kow as Chebyshev's Formula of 3 rd Order. (4) Eample : Compute the real root of log 0. = 0 upto four decimal places. [Burdwa, 003] Solutio: For give f () = log 0., fid f() = log 0. =. f() = log 0. = f (3) = 3 log 0 3. = 0.33 So the root lies betwee ad 3 sice for =, f() is egative where as for = 3, f() is positive. Take 0 = 3, f( 0 ) = 0.33, f ( ) = log + log e = log 3 + log e = = 0.93 Ad (Usig, log a = log a e log e )

25 986 Egieerig Mathematics through Applicatios Now by Newto Raphso Method for 0 = 3, f ( 0) 0.33 = 0 = 3 =.746 f ( ) For et iteratio, fid f( ) at =.746 i.e. f( ) =.746 log 0 (.746).0 = = Ad f ( ) = log log 0 e = = f ( ) = f ( ) = = So.7407 is the real root of the give equatio correct upto four decimal places. Eample 4: Usig Newto-Raphso Method, compute the real root of the followig equatio correct to four decimal places, = 8. Solutio: Give = 8 ( 8) = 0 = f( ) so that f ( ) = To start with, take = 5, 6 ad fid f(5) = 5 8 = 3 ad f(6) = 36 8 = 8 As value of f() at = 5 is ve, where as at = 6 it is + ve, therefore, the root lies betwee 5 ad 6, ad ear to 5. st Approimatio: For 0 = 5, f( 0 ) = 3, f ( 0 ) = 0 f ( 0) 3 = 0 = 5 = = f ( ) 0 0 d Approimatio: For = 5.30, f( ) = 8 = (5.30) 8 = 0.09, f ( ) = 5.30 = 0.60 = f ( ) f ( ) = 0.60 = 3 rd Approimatio: For = 5.095, f( ) = (5.95) 8 = , f ( ) = 5.95 = = f ( ) f ( ) = = 3 Which is correct upto 4 decimals. Alteratively usig the special form for fidig the value of the square root of umber, = = +, i = 0,,. i i+ i i i 5+ 6 Now to start with 0 = = = 5.5 i

26 Solutios of Algebraic ad Trascedetal Equatios 987 Usig the above formula for = = ( ) = + = + = = ( ) = + = + = ad f() = f(5.95) = 8 = (5.95) 8 = À 0 at = Hece, 5.95 is the correct value upto 4 decimal of places. ASSIGNMENT. Usig the iteratio method, fid the root of the equatio = + si correct to four decimal places. [MDU, 00]. Fid the real root of log 0 = 7 correct to four decimal places usig iteratio method. [MDU, 00] 3. Usig the method of iteratio, fid the real root of the equatio cos = 3 correct to three decimal of places. Verify the result so achieved by Aitki's Method Fid the real root of the equatio = [MDU, 007] 5. Usig Bisectio method, fid the approimate root of the equatio si =, that lies betwee = ad =.5, correct to the third decimal place. [VTU, 003] 6. Usig Bisectio method fid the root of the equatio, (i) cos = 0 [Madras, 003] (ii) + log 0 = Usig the Regula-Falsi method, solve correct to four decimal places (i) log 0 = [MDU,003] (ii) = 0 8. Fid the root of the equatio e = cos, usig the Regula-Falsi method correct to four decimal places. [Hit: f(0) =.7798] [KUK, 008; NIT Kurukshetra, 008] 9. Use the method of False Positio, to fid the fourth root of 3 correct to three decimal places. [Hit: Take 4 3 = 0] 0. Use Regula Falsi method to compute real root of the equatio = 0 if the root lies betwee ad 3. [NIT Jaladhar, 006; 007] N. Fid the cube of 4, usig Newto-Raphso method, Hit : + = 3 + [Madras, 006; NIT Kurukshetra, 003]. Use Newto's method to fid the smallest root of the equatio e si = to four places of decimal. [MDU Rohtak, 007]

27 988 Egieerig Mathematics through Applicatios 3. The bacteria cocetratio i reservoirs as C = 4e t + e 0.t, usig Newto Raphso method, calculate the time required for the bacteria cocetratio to be 0.5. [MDU, 007] [Hit: As here f(6) = , f(7) = , C = 0.5, take 0 = 6.80] 4. Fid by Newto's method, the root of the equatio cos = e. [Madras 003; VTU, 003] 5. Fid the square root of 0 correct to three decimal of places by usig recurrece formula + 0 =. + [NIT Allahabad, 006] 5.9 HORNER S METHOD This method cosists i fidig of both ratioal ad irratioal roots of the polyomial equatio figure by figure first fidig the iteger part ad the decimal part to ay desired decimal place of accuracy. The step by step procedure for dimiutio is give below:. Here by trial, we fid itegers 'a ad b' such that f(a) ad f(b) are of opposite sig for cotiuous f(), thus bracketig the root. Let the bracketed root be a.d d d 3 d 4 where d, d, d 3, d 4 etc. are the digits i the decimal part.. Dimiish the root of the equatio f() = 0 by 'a', to get ew equatio f () = 0 with root 0.d d d 3 d 4 3. Multiply the root of the equatio f () = 0 by 0, so that ew equatio g() = 0 is obtaied with oe of the root as d. d d 3 d 4 which lies betwee 0 ad Now d is foud such that g(d ) ad g(d + ) are of opposite sig. The the root of the equatio g() = 0 is dimished by d ad the resultig equatio g () = 0 with root 0.d d 3 d 4 5. Multiply the root of the equatio g () = 0 by 0 so that the ew equatio h() = 0 is obtaied with oe of the root as d. d 3 d 4 which lies betwee 0 ad 0. d is obtaied by trial as before. By cotiuig this process, the digits d 3, d 4 are successively obtaied. Observatios: This method is best for fidig the real roots of a polyomial equatio but ca be applied for a trascedetal equatio. Eample 5: Fid by Horer's Method the root of the equatio = 0 that lies betwee ad to three decimal places. Solutio: Let the root of equatio f() = = 0 be. d d d 3 d 4 Dimiish the root of the above equatio by, so that the root of the trasformed equatio is 0.d d d 3 d

28 Solutios of Algebraic ad Trascedetal Equatios 989 Thus, the trasformed equatio f () = has a root 0. d d d 3 d 4 lyig betwee 0 ad. Now, multiply the root of f () = 0 by 0 so that the chaged equatio g() = = 0 has oe of its roots as d. d d 3 d 4 Fid g(8) < 0 ad g(9) > 0 so that d = 8. Now dimiish the root of the above equatio by Thus the trasformed equatio g () = = 0 has a root 0.d d 3 d 4 lyig betwee 0 ad. Multiply the root of g () = 0 by 0. The ew equatio h() = = 0 has its root as d. d 3 d 4 lyig betwee ad 0. Fid p(8) < 0 ad p(9) > 0 so that this root is 8. d 4 Now dimiish the root of the equatio h() = 0 by The trasformed equatio h () = = 0 has its root 0.d 3 d 4 betwee 0 ad. Multiply the root of this equatio by 0. The ew equatio, p() = = 0 has oe of the root as d 3. d 4 lyig betwee 0 ad 0. Fid p(9) < 0 ad p(9) > 0 so that this root is 8.d 4 Now dimiish the root of the above equatio by Hece, the root of the equatio f() = = 0 correct to 4 decimal of place is MULLER'S METHOD I this method first we assume three approimatios,, as the root of the equatio f() = 0. The et better approimatio + is obtaied as the root of this equatio f() = 0 by approimatig the polyomial f() as a secod degree parabola passig through these three poits (, y ), (, y ), (, y ).

29 990 Egieerig Mathematics through Applicatios Let p() = A( ) + B( ) + y () be the desired parabola passig through these f () poits, so that y = A( ) + B( ) + y () y = A( ) + B( ) + y (3) f () O simplificatio, we get two simultaeous equatios i ukows A ad B, y i y i y i y y = A( ) + B( ) i + y y = A( ) 0 + B( ) i i i More precisely, write the above equatios as: Fig. 5.7 d = Ah + Bh where h = ; d = y y (4), d = Ah + Bh, where h = ; d = y y (5) From (4) ad (5), we get hd hd ad hd A = B = hd hh ( h h) hh ( h h) With these values of A ad B, the quadratic equatio (), p() = A( ) + B( ) + y = 0 gives the et approimate root at = + as (6) B± B 4Ay ( + ) = A B± B 4Ay B± B 4Ay y = = A (8) B± B 4Ay B± B 4Ay I (8), the sig i the deomiator is so chose that it is largest i magitude meaig there by that the differece betwee + ad is very small. See the subsequet eample as a illustratio of the method. Observatios: This method of iteratio coverges quadratically almost for all iitial approimatios. I case, if o better approimatios are kow, we ca take =, = 0 ad =. It ca also be used for fidig comple roots. Eample 6: Fid the root of the equatio 3 5 = 0 which lies betwee ad 3 by Muller's method. Solutio: Le =, = ad = 3 so that y = 6, y = ad y = 6 The h = =, d = y y = 7; h = =, d = y y = ; So that hd hd 6, hd A = = B = hd = 3 hh( h h) hh( h h)

30 Solutios of Algebraic ad Trascedetal Equatios 99 Ad the quadratic becomes 6( + 3) + 3( + 3) + 6 = 0 (From (), Multer s Method) The et approimatio of the root + to the desired root is 3 ( + 3) =, Sice B is positive or =.0868 This process ca ow be repeated with three approimatios,,.0868 respectively for still better value. 5. LIN-BAIRSTOW METHOD Here i this method, by a iteratio process we resolve a th degree polyomial f() ito quadratic factors correspodig to a pair of real or comple roots. For purpose of illustratio of the method, here we take a fourth degree polyomial, however, the cotetio ca be eteded to a polyomial of th degree. Let the polyomial be a a 3 + a + a 3 + a 4. Whe this polyomial will be divided by ( + p + q), the quotiet will be a secod degree polyomial of the form (b 0 + b + b ) ad the remaider will be a first degree epressio of the form (R + S) Thus a a 3 + a + a 3 + a 4 = ( + p + q)(b 0 + b + b ) + (R + S) () Here values of R ad S will deped o p ad q ad hece they are take fuctios of p ad q. If p ad q are so chose that R ad S vaish (i.e. o remaider), the ( + p + q) will be factor of the give polyomial. Meaig thereby, the problem reduces to fidig pad q such that R(p, q) = 0 ad S(p, q) = 0 () Let (p 0 q 0 ) be a iitial approimatio ad (p 0 + p 0, q 0 + q 0 ) be the actual solutio of the equatio (). The R(p 0 + p 0, q 0 + q 0 ) = 0 ad S(p 0 + p 0, q 0 + q 0 ) = 0 (3) By Taylor's series, from (3) Rp ( 0, q0) + ( Rp p0 + Rq q0) = 0 (app.), where R R Rp = ad Rq =! p q i.e. (R p p 0 + R q q 0 ) = R(p 0, q 0 ) (4) Similarly (S p p 0 + S q q 0 ) = S(p 0, q 0 ) (5) Solvig equatios (4) ad (5) SR p0 = RS q p q RS q RS q p RSp SRp ad q0 = RS RS p q q p (6) O repeatig like terms i (), a 0 = b 0, b 0 = a 0, a = b 0 p + b, b = a pb 0 a = b 0 q + b p + b, b = a pb qb 0, a 3 = b q + b p + R, implyig b 3 = a 3 pb qb, (7) a 4 = b q + S, b 4 = a 4 pb 3 qb, where we have assumed that R = b 3, ad S = b 4 + pb 3 (8) The above equatios ca be represeted by recurrece formula

31 99 Egieerig Mathematics through Applicatios b α = a α pb α qb α, where b = 0 ad b = 0; α = 0,,, 3, 4 (9) bα Differetiatig (9) partially w.r.t. p ad deotig by c p α, we get c α = ( pc α + b α ) q( c α 3 ) i.e. c α = b α pc α qc α 3 (0) bα Differetiatig (9) partially w.r to q ad deotig q by c α, we get c α = p( c α 3 ) ( qc α 4 + b α ) i.e. c α = b α pc α 3 qc α 4 () The equatios (0) ad () ca be represeted by a sigle recurrece relatio c α = b α pc α qc α, k = 0,, 3; where c 0 = b 0 ad c = 0 = c () Now b p b p 3 α Rp = = c Q = cα, b b S p b c pc b p p 4 3 p = = ; b q b q 3 k 4 3 Rq = = c Q = cα, q b b S = + p = c pc q q O usig the values of R, S, R p, R q, S p, S q i (6), we get c ( b + pb ) + b ( c + pc ) bc bc p = = ( ) ( ) ( ) c c + pc c c3 + pc b3 c c c3 b3 b ( c pc + b ) + c ( b + pb ) bc b( c b) q = = c c( c3 b3) c c( c3 b3) The better approimatios for p ad q are the obtaied as p = p 0 + p 0 ad q = q 0 + q 0 (3) Still better approimatios for p ad q are obtaied by iterative procedure similar to sythetic divisio method for computig b α s ad c α s as below. a 0 a a a 3 a 4 p pb 0 pb pb pb 3 q qb 0 qb qb b 0 b b b 3 b 4 p pc 0 pc pc q qc 0 qc c 0 c c c 3 Eample 7: Etract the quadratic factor of the form + p + q from the polyomial usig Bairstow method ad assumig that p 0 =.5 ad q 0 = p 0 = q 0 =

32 Solutios of Algebraic ad Trascedetal Equatios 993 (b 0 ) 4.5(b ).5(b ).875(b 3 ).85(b 4 ) p 0 = q 0 = (c ) 8.75(c ).5(c 3 ) bc bc p = = = = ( ) (8.75) 6( ) c c c3 b3 bc b( c b) q = = = ( ) c c c 3 b 3 p = p 0 + p 0 = =.954 q = q 0 + q 0 = = p = q = (b 3 ) 4.368(b 4 ) p = q = (c ) 9.568(c ) 3.47(c 3 ) p = = = (9.568) q = = = p = p + p = =.08 q = q + q = = p = q = (b 3 ) (b 4 ) p = q = (c ) 6.438(c ) (c 3 ) p = = = 0.08 (6.438) q = = = p 3 = p + p = = q 3 = q + q = =.9997 The required quadratic factor is , while the actual factor is + +.

33 994 Egieerig Mathematics through Applicatios Assigmet. Apply Muller's method to obtai the root of the equatio e = 0, which lies betwee 0 ad. [Hit: 0 =, = 0, = ]. Solve the equatio = 0 by the Muller's method. [Hit: 0 = 0, =, = ; h = ] 3. Usig Muller's method, fid the root of the equatio 3 5 = 0, which lies betwee ad. [NIT Allahabad 007] 4. Apply Muller's method to obtai the real root of the equatio 3 = 0, which lies betwee ad. [MDU, 006] 5. Fid by Horer's method, the positive root of = 0 correct to three decimal place. 6. Usig Horer's method, fid the root of the equatio = 0, which lies betwee ad correct to three decimal place. 7. Fid the cube root of 30 correct to 3 decimal place, usig Horer's method. 8. Usig Barstow's method, obtai the quadratic factors of the equatio = 0 with p, q =.4,.4 (perform two iteratios). [NIT Jaladhar, 007] 5. COMPARATIVE STUDY The bisectio method is the most reliable oe as here we shall certaily reach the root. But it is very slow sice oly oe biary digit precisio is achieved i each iteratio. If evaluatio of f () is ot difficult, the Newto-Raphso Method is recommeded. The Newto-Raphso Method is cosidered to be the fastest method. But this method approaches to the root oly whe the iitial guess is chose sufficietly ear to the root. Hece, its covergece must be verified usig bisectio method oce i every 0 iteratio. If evaluatio of f () is difficult, Secat Method (Chords Method) is recommeded. It gives better results tha Regula-Falsi Method, but this may ot coverge to the root some time. Regula-Falsi Method always coverges. For locatig comple roots, Newto's Method, Muller's Method ad Li-Bairstow Methods are recommeded. However, Bairstow is best amog all. If all the roots of the give equatio are required, the Li-Bairstow Method is quite useful. For fidig the real roots of a polyomial, Horer's ad Graeffe's Root Squarig Method are composed. But, if the roots are real ad distict, the Graeffe's Root Squarig Method is quite useful.

34 Solutios of Algebraic ad Trascedetal Equatios 995 ANSWERS Assigmet. =.5, =.49749, 3 =.4973, 4 = = , 5 = 0.607, 6 = = , = , 3 = , 4 = , 5 = (i) (ii).9 7. (i).9 (ii) Assigmet. 3 = 0.54; 4 = 0.577; 5 = = ; 4 = ; 5 =