Numerical Solutio of Noliear Equatios. INTRODUCTION The most commo reallife problems are oliear ad are ot ameable to be hadled by aalytical methods to obtai solutios of a variety of mathematical problems. Iterative methods are the foremost amog the methods developed to obtai approimate solutios. The method of fidig a root of the oliear equatios of the form f () = 0 (.) where the fuctio f () may be algebraic, trascedetal or combiatio of both, plays a major role i the applicatios of mathematics as problems of such kid occur more frequetly i may scietific ad egieerig mathematical modelig. The followig equatios (i) 6 = 0 (ii) e cos = 0 (iii) si() e 5 = 0 ca be classified as follows. (i) The equatio 6 = 0 is a algebraic equatio of degree 6 havig oe root early at =.34743. (ii) The equatio e cos = 0 is a trascedetal equatio as it cotais trascedetal fuctios which has a root early about = 0.5775736. (iii) The equatio si () e 5 = 0 is a equatio combied of both algebraic ad trascedetal fuctios ad it has a root early about =.534530. If a umber a makes the fuctio of a scalar variable, i.e., f () to zero the a is called a zero or a root of the equatio f () = 0. To obtai this root, if oe obtais iterates {,,......} startig with a iitial guess 0, the this sequece of iterates coverge to the root wheever lim a = 0 (or) lim =a. Æ Æ I this chapter, we shall study the methods of obtaiig a approimate solutio of equatios of the form (.) ad discuss the importace of each of these methods comparig with oe aother whe required, through some eamples.
Numerical Solutio of Noliear Equatios 5 Y y = 0.8 0.6 0.4 0. y = e 0 0. 0.4 0.6 0.8 X Fig.. From the above plottig of the graphs, oe ca see that abscissa of the itersectio of two curves is aroud = 0.7. Therefore, the required approimate root of the equatio (..) is 0.7. Eample.: Obtai graphically the two real roots of the equatio 4 3 3 + = 0 Solutio: The give equatio is f () = 4 3 3 + = 0 Oe ca verify that Therefore, we have f (0) > 0, f () < 0, f () < 0, ad f (3) > 0 f (0) f () < 0, ad f () f (3) < 0 ad hece there eists roots i (0, ) ad (, 3). We ow rewrite the give equatio as 4 + = 3. 3 Let us plot the graphs for the curves y = ad y = Ê Á Ë 4 + ˆ 3 /3 o the same scale i the iterval (0, 3) with respect to the same ais formig the followig tabular values. Table. 0 0.5.5.5 3 y = 0 0.5.5.5 3 y = 3 4 + 3 0.69 0.7 0.87.6.78.37 3.0
5 Numerical Aalysis: Iterative Methods Y 3.5.5 0.5 y = 3 y = 4 + 3 0.5.5.5.3 Fig.. X From the above plottig of the curves, we have the abscissae of poits of itersectio of the curves are ear about 0.76 ad about.96. Threrefore, the required approimate roots of the give equatio i the itervals (0, ) ad (, 3) are = 0.76 ad =.96 respectively. Eample.3: Fid a root of the equatio e = 0 correct to 3 decimal places, by usig the bisectio method. Solutio: Let f () = e Sice f (0) < 0 ad f () > 0, there is a root i the iterval (0, ). We ow form the followig table to obtai the required root o takig a = 0 ad b =. Table.3 Sl. No. a b a+ b c = f (c). 0 0.5 0.065. 0.5 0.75 0.776 3. 0.5 0.75 0.65 0.0897 4. 0.5 0.65 0.565 0.0073 5. 0.565 0.65 0.5938 0.045 6. 0.565 0.5938 0.578 0.07 7. 0.565 0.578 0.5704 0.0050 8. 0.565 0.5704 0.5665 0.00 9. 0.5665 0.5704 0.5685 0.000 0. 0.5665 0.5685 0.5675 0.0006. 0.5665 0.5675 0.5670 0.000
Sice f (0.567) < 0.0005, the required root of the equatio e = 0 correct to three places of decimal is = 0.567. Note.: The above computed tabulated values are rouded to 4 decimals, though the root is required correct to 3 decimal places, i order to check whether the fuctio value is less tha 0.5 0 3 to obtai the root up to the desired accuracy. Remark.: If a is a real root of the equatio f () = 0 correct to N decimal places, the f (a) < 0.5 0 N i magitude. Eample.4: Obtai the smallest positive real root of the equatio e si = 0 by bisectio method, correct to 4 decimal places. Solutio: Let f () = e si Ê ˆ sice f (0) > 0, f (0.5) > 0 ad f () < 0, the lowest root lies i the iterval Á,. Now, Ë we form the followig table to obtai the root takig a = ad b = 0.5 Table.4 Sl. No. a b a+ b c =. 0.5 0.75 0.097. 0.75 0.5 0.65 0.04984 3. 0.65 0.5 0.565 0.03648 4. 0.65 0.565 0.59375 0.007 5. 0.59375 0.565 0.5783 0.0449 6. 0.59375 0.5783 0.58594 0.0036 7. 0.59375 0.58594 0.58985 0.008 8. 0.58985 0.58594 0.5879 0.00088 9. 0.58985 0.5879 0.58888 0.00047 0. 0.58888 0.5879 0.58839 0.000. 0.58888 0.58839 0.58864 0.0004. 0.58864 0.58839 0.5885 0.0000 Sice f (0.5885) = 0.0000 < 0.5 0 4, the root of the equatio e si = 0, correct to four decimal places is = 0.5885. Note.: As the width of the iterval is reduced by half at each step i the bisectio method, N bisectios will be required to have a legth of the iterval which cotais the root, is b a N. Ad, to have Numerical Solutio of Noliear Equatios 53 b a e (a small quatity as desired), oe ca easily deduce that N f (c)
54 Numerical Aalysis: Iterative Methods b a N l /l e Coditio (.) ca be verified for the eamples (.3) ad (.4) for e = 0.00. (.).4 ITERATION METHOD The iteratio method ivolves trasformig the equatio (.) ito the form = f() (.3) ad geeratig a sequece of approimatios,, 3,...,,... to a root of the equatio (.) from the scheme + = f( ) (.4) ( = 0,,,...) Choosig a proper iitial approimatio 0. We ow state ad prove the sufficiet coditio for the covergece of the method (.4). Theorem.: If I is the iterval i which * a root of the equatio f () = 0, lies ad if f () < for all i I the the iterative method (.4) will coverge to * provided 0 is properly chose i I. Proof: If * is a root of f () = 0, the from (.3), we have * = f(*) (.5) The scheme (.4) is give by + = f ( ) (.6) where + ad are the ( + ) th ad th approimatios to * respectively. Now, subtractig (.6) from (.5), we get + * = f( ) f(*) by mea value theorem, where < < *. If we let The (.7) ca be writte as = ( *) f () (.7) f ( ) K < for all i = 0,,,... (.8) + * K * (.9) K K * K K K * ad so o Proceedig i similar maer, fially we will be left with + * K + 0 *
For a large, the righthad side teds to zero. Thus, + coverges to *. Hece, the proof is complete. Note.3: The iterative method (.4) has a liear rate of covergece. Eample.5: Which of the followig forms will coverge to a root of the equatio Ê pˆ 3 cos = 0 lies i Á0, Ë. Numerical Solutio of Noliear Equatios 55 (a) = cos + (b) = ( + cos )/3 Solutio: Give f () = 3 cos = 0 (.5.) Ê pˆ which has a root i Á0,. Ë (a) The equatio (.5.) is writte as = cos + = f() So that f () = (si + ) Sice f () = (si + ) Ê pˆ for ay i Á0,, the iterative method will ot coverge to a root of (.5.) Ë with this form. (b) I this case, equatio (.5.) is writte as = ( + cos )/3 = f() Now, f () = si /3 si Sice f ( ) = < Ê pˆ for all i 0, 3 Á, the iterative scheme i this case will Ë Ê coverge to a root of the equatio (.5.) by choosig ay 0 i 0, pˆ Á Ë. Eample.6: Use iterative method to fid a root of the equatio + e = 0 up to 5 decimals writig it as (a) = e (b) = ( + + e ) Solutio: Give equatio is f () = + e = 0 (.6.) Ê 3ˆ Sice f Á < 0 Ë 5 ʈ ad f Á > 0 Ë, there eists a root i the iterval Ê 3, ˆ Á Ë 5.
56 Numerical Aalysis: Iterative Methods (a) Here, equatio (.6.) is writte as = e = f() Now, f () = e ad f () = e < for all i ( 0.6, 0.5) The iterative scheme for the solutio of (.6.) is + = e (.6.) ( = 0,,, ) If we start with 0 = 0.6, the from (.6.) we ca obtai the followig successive approimatios. Table.5 + + 0 0.5488 0.56779 0.577636 0.5673 0.564 3 0.56755 3 0.57050 4 0.56737 4 0.56537 5 0.56747 5 0.5685 6 0.5674 6 0.566530 7 0.56745 7 0.567490 8 0.5674 8 0.566946 9 0.56744 9 0.56755 0 0.56743 0 0.567080 0.56743 Hece, the required root of equatio (.6.) correct to five places of decimal is = 0.5674 (b) We ow solve (.6.) cosiderig the form = ( + + e ) = f() Here, f () = + + ( + ) e It ca be verified that f () = + + ( + ) e < for all i ( 0.6, 0.5) Now, takig 0 = 0.6 ad applyig the iterative scheme + = ( + + e ) (.6.3) ( = 0,,, ) the followig iterates are obtaied.
Numerical Solutio of Noliear Equatios 57 Table.6 0 3 4 5 + 0.56987 0.567375 0.56769 0.56746 0.56744 0.56743 Therefore, the required root correct to 5 decimal places is = 0.5674. Note.4: It may be oted that the scheme (.6.3) will ot coverge to the root if we start with 0 =.5. Remark.: Smaller the value of f (), larger the rate of covergece. Defiitio.: A sequece of approimatios { 0,,,...,,...} is said to coverge to a root * of a equatio f () = 0 with rate of covergece or the order of covergece p p if * K * + for some K > 0. (.0) Defiitio.: If p is the order of the method ad is the umber of fuctioal evaluatios per iteratio by a method, the the efficiecy ide of that method is p..5 ACCELERATION OF CONVERGENCE It is always possible to accelerate the covergece of the method of iteratio (.4) by itroducig a parameter a ad etrapolatig it as + = ( a) +af( ) (or) Ô + = +a( f( ) ) Ô ( = 0,, ) (.) The improvemet of covergece majorly depeds upo the choice of the relaatio factor a subject to costraits of the give problem. The followig are some of the methods which ehace the rate of covergece of the scheme (.4)..6 WEGSTEIN S METHOD I this method, the choice of a i the scheme (.) is take as a = f( ), D= D (.) The followig steps illustrate the workig procedure of this method. Step : Choose a iitial approimatio 0 to a root of f () = 0. Step : Assig Step 3: Calculate = f( )
58 Numerical Aalysis: Iterative Methods f( Step 4: Fid ) D= Step 5: Fid a= D Step 6: Calculate + = + a[f( ) ] Step 7: Stop the process if + is egligible up to the tolerace. Step 8: Assig + ad go to step 3. Note.5: Whe a = or D = 0, the above method reduces to the iterative method (.4)..7 AITKEN S D METHOD If the method (.4) coverges liearly ad if f (*) K < where * is a root of (.), the from (.9) approimately we have + * = K( *) * = K( *) Solvig the above equatios for * by elimiatig K from them, we obtai Replacig * by +, we have * = + = + ( + ) + + ( D ) + (.3) D Which is the Aitke s etrapolatio method, for a liearly coverget sequeces, yields approimatios to the root * with = f( 0 ) ad = f( ). Observatio.: It is see that the method (.) will accelerate the covergece of the method (.4) if we choose a proper a i (i) (0.5 < a < ) provided < f () < 0 (ii) (a > ) provided 0 < f () < Remark.3: By Theorem (.), oe ca easily obtai the coditio for covergece of the method (.) as: + a + af () < for all i I (.4) The coditio (.4) is same as < a + af () < 0 < (or) a< (.5) f ( ) for each.
60 Numerical Aalysis: Iterative Methods Solutio: Give equatio is We rewrite this equatio as f () = si 6.8 = 0 = 6.8 + si = f() Here f () = cos Now, the iterative method for the solutio of give equatio is The Wegstei s iteratio is where 6.8 + si a=, D= D The Aitke s D process is + = 6.8 + si (.7.) ( = 0,, ) + = + a(6.8 + si ) (.7.) + = ( = 0,, ) + ( + ) + ( = 0,, ) + where = 6.8 + si ad + = 6.8 + si, whereas, the etrapolated iterative method with a=ais ˆ 6.8 si + = cos ( = 0,, ) (.7.3) (.7.4) The first 9 approimatios of these methods are tabulated hereuder takig 0 = 6 Table.7 Method (.7.) Method (.7.) Method (.7.3) + Method (.7.4) 6.00058450 6.00058450 6.00058450 6.0467506 6.004577 6.004577 6.004577 6.05500534 3 6.0068480 6.0476084 6.0470548 6.05503073 4 6.0006 6.04788597 6.04733649 6.05503073 5 6.00700060 6.05464679 6.04769 6 6.0037805 6.05466049 6.0550080
Numerical Solutio of Noliear Equatios 6 7 6.00363736 6.055073 6.05500883 8 6.00407889 6.0550337 6.0550096 9 6.00450386 6.0550335 6.05503073 Remark.5: The true root of the equatio of eample (.7) is 6.055030797 approimately. Comparig this value with the above tabulated approimatios, oe ca see that the etrapolated iterative method (.7.4) has a better rate of covergece for obtaiig the root of = 6.8 + si. Eample.8: Fid the largest root of the equatio 6 = 0 correct up to 8 places of decimal, startig with 0 = by usig: (i) Iterative method (ii) Etrapolated iterative method. Solutio: Give Which has largest root i the iterval < <. We rewrite (.8.) as Now, f ( ) = 6 5 ( + ) 6 < f() = 6 = 0 (.8.) 6 = + = f() [say] for all i (, ) The iterative method for solvig (.8.) is + = 6 + ( = 0,,, 3..) Ad the etrapolated iterative method i this case will be of the form where + = +a È 6 + Î ( = 0,,, 3 ) a = È 6 5 { ( ) /6 Í + Î } The computed results for obtaiig the root of equatio (i) are tabulated below:
6 Numerical Aalysis: Iterative Methods Table.8 Iterative Method Etrapolated Iterative Method a 0.000000000.000000000.00936955.07488330.438373.4055698.096873.3475697 3.3536648.09704065.347438 4.34769538.0940430.347438.347438 From the above tabulated values, we have the required largest root of the equatios (.8.) as =.34743 which is correct to 8 decimal places..9 METHOD OF FALSEPOSITION (OR) REGULAFALSI METHOD I this method, we eed to fid a sufficietly small iterval (a, b) i which the root of a equatio f () = 0 lies as i the case of bisectio method, so that the curve y = f () crosses the Xais i betwee = a ad = b ad also f (a) f (b) < 0. The falsepositio method is based o the priciple that ay small portio of a smooth curve is practically straight for a short distace. Takig this priciple ito cosideratio, we fit a straight lie passig through the poits P[a, f (a)] ad Q[b, f (b)] as Y y = f ( ) P [ a, f ( a)] = a = c = b Q [ b, f ( b)] X Fig..3
64 Numerical Aalysis: Iterative Methods Remark.7: Eve though the rate of covergece of the secat method is.6 approimately, it will fail to coverge whe f ( ) = f ( ) at ay stage. But, oce secat method coverges it will coverge more rapidly tha the RegulaFalsi method which is always guarateed to coverge. Eample.9: Solve the equatio give i eample (.8) by (i) RegulaFalsi method ad (ii) secat method to obtai the ith approimate 9 of the root with 0 = ad =. Solutio: Let f () = 6 = 0 So that f () = 6 > 0 & f () = < 0 therefore a root of the equatio lies i (, ). Takig 0 = = a, = = b, we applied the methods (.) ad (.) ad the successive approimatios up to 9 th approimate are tabulated hereuder. Table.9 C of (.) + of (.).06903.0690.030674754.90577768 3.0437660.765583 4.05534703.353550 5.0656678.3486808 6.07478340.3473646 7.088080.347438 8.0898338.347438 The required root is =.34743 which is correct to 8 decimals. Oe ca see that the more rapid covergece of secat method to the root, over falsepositio method. Eample.0: Use secat method to fid a root of the equatio + e = 0 correct to 8 places of decimal, by tabulatig the computed values. Solutio: Give f () = + e = 0 (.0.) Sice f ( ) < 0 ad f (0) > 0, there eists a root i the iterval (, 0). The secat method for the solutio of (.0.) is give by + = È ( ) ÍÎ f Í f( ) f( ) where f ( ) = + e ( =,, 3...) Let us take 0 = ad = 0. The f ( 0 ) = 0.630559 ad f ( ) =. Now, the followig table ehibits further approimatios.
Numerical Solutio of Noliear Equatios 65 Table.0 + f ( + ) 0.6699837 0.07083948 0.5784 0.00788873 3 0.5670080 0.000064583 4 0.5674338 0.000000059 5 0.5674390 0.00000000 6 0.567439 0.00000000 7 0.5674390 0.00000000 is Sice f ( 8 ) < 0.5 0 8, the desired root of the equatio correct to eight decimal places = 0.567439.. NEWTONRAPHSON (NR) METHOD To derive the NewtoRaphso method for fidig a root * of the equatio f () = 0, let 0 be a iitial approimatio ad let = 0 + h where h is a small quatity, be the first approimatio to *. If 0 is chose such that 0 lies i the eighbourhood of *, the will be early equal to *. Thus, f ( ) = 0 approimately. i.e., f ( 0 + h) = 0 Epadig the above fuctio by meas of Taylor s series, oe ca have h f ( 0) + hf ( 0) + f ( 0) +...! = 0 (.3) Sice h is small eough, we ca eglect the higher order powers of h startig from h owards. Thus, from (.3), we have which gives f( 0 ) h = f ( ) 0 f ( 0 ) + hf ( 0 ) = 0 Puttig this value of h i = 0 + h, f( 0 ) we obtai = 0. f ( 0 ) Now, for a better approimatio to *, if we let = + h be the secod approimatio ad proceedig i a similar maer as above, oe ca easily obtai f( ) = f ( )
.. Coditio for Covergece of the NewtoRaphso Method Comparig the iterative method (.4) with the NR method (.4), Oe ca have f( ) f ( ) = (.6) f ( ) We kow that the iterative method coverges by Theorem (.), if f () < for all i I From (.6), f () = f ( ) f( ) f ( ) f( ) f ( ) = f ( ) f ( ) Therefore, the coditio for covergece of the NR method is f( ) f ( ) < for all i I. [ f ( )] (.7) Observatio.: The covergece of NR method will be faster if f () is large eough, i.e., the graph of f () is early perpedicular to the ais. Observatio.3: By Theorem (.), we ca also have the covergece of NR method provided that the iitial approimatio 0 is chose sufficietly close to the root of f () = 0...3 NewtoRaphso Method has a Quadratic Covergece Let * be a root of f () = 0 such that f (*) = 0 ad let the small quatities e ad e + be the errors at th ad ( + ) th stages for the approimatios ad + respectively such that By (.8), the NR method (.4) takes the form e + + * = (or) e + = e = * ad e + = + * (.8) * e + * f( e + ) * f ( e + ) * f( +e) e * f ( +e ) Epadig the righthad side fuctio by Taylor s theorem, we have e + = e Numerical Solutio of Noliear Equatios 67 e ( ) +e ( ) + ( ) +... f ( ) +e f ( ) +... * * * f f f * * Takig f (*) = 0 ad eglectig the higher order terms of e startig from e 3 owards as e is small eough, we obtai
68 Numerical Aalysis: Iterative Methods e + = * * * e * e f ( ) +e f ( ) +... e f ( ) f ( ) +... f ( ) +e f ( ) * * which give us * È f ( ) e + = e Í * Î f ( ) (.9) e + µ e This relatio shows that the subsequet error is proportioal to the square of the previous error ad hece the NR method has a quadratic or secod order covergece ad its efficiecy ide is.44...4 NewtoRaphso Method Algorithm Step : Choose a iitial approimatio 0 to obtai the root of f () = 0. Step : Calculate f ( 0 ) ad f ( 0 ). Step 3: Evaluate f( 0 ) = f ( ) 0 0 Step 4: If 0 is egligible up to the tolerace stop the process. Otherwise, go to step 5 Step 5: Icrease the subscripts of s by oe uit i the above 3 steps ad go to step. Eample.: Fid a root of the equatio cos e = 0 by Newto s method. Solutio: Let f () = cos e = 0 The, f () = [si + ( + ) e ] Sice f (0) > 0 ad f () < 0, there is a root i the iterval (0, ). Let us choose 0 = 0.5 The, the NR method (.4) yields For = 0, we obtai + = = = cos e + si + ( + ) e ( = 0,,.) 0.5 cos(0.5) (0.5) e 0.5 + si(0.5) + (0.5 + ) e 0.053 0.5 +.95507 = 0.5806 0.5
70 Numerical Aalysis: Iterative Methods Eample.3: Develop NR method for fidig Square root of a umber M Reciprocal of a umber M ad fid 3 ad correct to 5 decimals. 3 Solutio: (a) The square root of a umber M is a root of the equatio M = 0. Let f () = M The, f () = From (.4), we have + = M (or) + = Ê Á Ë ( = 0,,...) Mˆ + is the Newto s formula for obtaiig square root of a umber M. (.3.) Sice 3 = 9 < 3 < 6 = 4, we take 0 = 3.5 to obtai 3. From (.3.), we have for = 0; = Ê 3ˆ Á0 + Ë 0 = 3.6074 for =; = Ê 3ˆ Á + Ë = 3.60555 for = ; we obtai 3 = 3.60555 Sice = 3, the square root of 3 correct to 5 decimals is 3.6055. (b) The reciprocal of a umber M is the root of Let f( )= M The, f ( )= M =0
Numerical Solutio of Noliear Equatios 7 From (.4), we obtai + = Ê ˆ Á M Ë Ê ˆ Á Ë (or) + = ( M ) (.3.) ( = 0,,,...) is the NR formula for gettig reciprocal of a umber M. Sice 0.05 = < < = 0., we choose 0 = 0.075 to obtai reciprocal of 3. 0 3 0 for = 0, (.3.) gives us = 0 ( M 0 ) = 0.07688 for = ; = 0.07688( 3 0.07688) = 0.0769 for = ; 3 = 0.0769 Therefore, the required reciprocal of 3 correct to 5 places of decimal is 0.0769.. SOME VARIANTS OF NEWTONRAPHSON METHOD I this sectio, we shall discuss some of variats of the NR method ad make ote of the importace of these methods over the NR method i the specific cases whe it requires, cosiderig some eamples for the compariso... Modified NewtoRaphso Method (or) VoMises Method This method cosists of fidig the slope at iitial iteratio ad obtaiig these by usig the formula + = f( ) f ( 0 ) (.30) ( = 0,, ) It ca be see from the formula (.30) that the slopes are ot calculated at each ad every iteratio as i NR method but are obtaied by drawig the parallel lies to f ( 0 ) from the poits o the curve as show i the followig figure.
7 Numerical Aalysis: Iterative Methods Y y = f ( ) (, y ) 0 0 Lies parallel to taget Taget of f( ) at ( 0, y0) 0 X Fig..6 Note.8: The VoMises method is useful i the cases whe the evaluatio of f () at every iteratio requires a large amout of computatioal time ad i such cases this modified NR method takes less computatioal time tha the NR method... Higher Order Newto s Method (or) Halley s Method I sectio (.), we have obtaied the value of h to get = 0 + h by eglectig the higher powers of h startig from the secod degree term owards as h = f( 0 ) (.3) f ( ) If we eglect the terms i (.3) from h 3 powers owards, we get Now, h = = 0 f( 0 ) h f ( 0) + f ( 0) f( ) f( 0) f ( 0) f ( 0 ) f ( ) 0 by (.3) (.3) I geeral, = + = 0 f( ) f( 0) f ( 0) f ( 0 ) f ( ) f( ) f( ) f ( ) f ( ) f ( ) ( = 0,, ) which is the Halley s method havig third order covergece. 0 (.33)
74 Numerical Aalysis: Iterative Methods From (.30), we have + = 3 3 (.4.) ( = 0,,,...) Let 0 =.5. The, f ( 0 ) = 5.75 The followig approimatios are obtaied by applyig the VoMises formula (.4.). Table. =.34786087 9 =.3478378 =.3303644 0 =.3478066 3 =.36446 =.3477985 4 =.3508457 =.3477964 5 =.348558 3 =.3477959 6 =.3474389 4 =.3477958 7 =.347468 5 =.3477957 8 =.3479587 6 =.3477957 Sice 5 = 6, the required root of equatios (.4.) is.347795, which is correct to 8 decimal places. Eample.5: Solve the problem give i eample (.4) by usig Halley s method. Solutio: Give f () = 3 = 0 (.5.) The f () = 3 ad f () = 6 The Halley s method (.33) for the solutio of (.5.) is f( ) f ( ) + = f( ) f ( ) f ( ) ( = 0,,,...) Usig this formula with 0 =.5, we obtaied the followig approimatios =.37539, =.3477968 3 =.3477957, 4 =.3477957 Sice 3 = 4, the desired root of the equatio is =.3477957. (.5.) Eample.6: Solve 3 0 = 0 to obtai a root by usig etrapolated NR method. Solutio: Give f () = 3 0 So f () = 3 ad f () = 6
The etrapolated NR method (.39) i this case takes the form + = 3 ( 0) ( 0) (6 ) (3 ) 3 (3 ) ( = 0,,, 3...) Sice f () < 0 ad f (3) > 0, there is root i the iterval (, 3). Let us choose 0 =.5 The, the above formula gives us the followig successive approimatios. =.309794563, =.308907398 3 =.308907397, 4 =.308907397 Therefore, the root of the equatio is 3 =.308907397 which is correct to 0 places of decimal as error reduces cubically i this method..3 NEWTONRAPHSON METHOD FOR MULTIPLE ROOTS (OR) GENERALIZED NEWTONRAPHSON (GNR) METHOD If * is a root of the equatio f () = 0 with multiplicity m > the f () ca be epressed as f () = ( *) m g() (.40) where g () is bouded ad g (*) π 0 ad f (*) = f (*) = f (*) =... = f (m ) (*) = 0 & f (m) (*) π 0 The geeralized NR method i the case of (.34) is defied by + = f( ) ( ) m f which is the NR method for multiple roots havig secod order covergece..4 GENERALIZED EXTRAPOLATED NEWTONRAPHSON (GENR) METHOD (.4) If h is a root of (.) with multiplicity m, the the geeralized NewtoRaphso method is defied as f( ) + = m (.4) f ( ) ( = 0,, ) Numerical Solutio of Noliear Equatios 75 We develop geeralized etrapolated NewtoRaphso (GENR) method by itroducig computatioal parameters a of the form + = ( a ) + a * + (.43) where * + is + of the geeralized NR method (.4).
76 Numerical Aalysis: Iterative Methods Now the GENR method for multiple roots of equatio (.) ca be defied as + = f( ) ( ) ma f.4. Covergece Criteria of GENR Method (.44) As it is well kow that ay iterative method of the form + = f( ) coverges if f ( ) < for all i I. Hece, the method (.44) coverges uder the coditio where m = ma + ma w < for all i I (.45) w = f( ) f ( ) [ f ( )] The fuctio f () i the immediate eighbourhood of = h, ca be writte as f () = k ( h) m where k k(h) is effectively costat. The, f () = k m( h) m (.46) We thus have f( ) f ( ) [ f ( )] f () = k m(m ) ( h) m = m m k km m m ( h) ( )( h) = km h m m [ ( ) ] (.47) We eed to fid a real value of a for each iteratio, which miimizes m of (.45). Sice w of (.46) is positive ad real for all, as oted i the Remark.4 i the immediate viciity of h, we have i geeral m m f( ) f ( ) w [ f ( )] (.48) ( = 0,, ) The process of miimizig m of (.45) keepig i view of (.47) with respect to a, gives the optimal choice for a as ma = fi a = Êm ff ˆ + Ë Á m [ f ] m+ mw ( = 0,, ) (.49)
With this optimal choice of a, m of (.45) takes the form Numerical Solutio of Noliear Equatios 77 m = w m + m m+ mw m+ mw = m+ mw m+ mw m+ mw = m( w ) + m( w ) (.50) Which is always less tha uity as log as w of (.46) less tha oe i magitude ad hece the covergece of the GENR method (.44) is assured. Note.9: It ca be see the GENR method (.44) with the choice of a of (.49) i the case whe m = takes the form + = f( ) f( ) f ( ) f ( ) [ f ( ) ] ( = 0,, ) (.5) which is the cubic covergece higher order Newto s method for fidig a simple root of the equatio (.)..4. Rate of Covergece of Geeralized Etrapolated NewtoRaphso Method Let h be the multiple root of the equatio f () = 0 ad e +, e be errors whe +, are the ( + ) th ad th approimates. The, + = h + e +, = h + e (.5) Substitutig these values of + ad i (.44), we have from (.49) f e + = e m ff f m+ m f [ ] ff = e Êm + ˆ [ Á f ] ff Ë m
Numerical Solutio of Noliear Equatios 79 È iv 3 Ê f ˆ Ê f ˆ (ii) Whe m = e + µe Í 4 Á Ë f 4 Á Ë f Í Î È iv v 3 Ê f ˆ Ê f ˆ (iii) Whe m = 3 e + µe Í 59 Á Ë f 60 Á Ë f Í Î I geeral, oe ca have e + µ e 3 k where ( m+ ) ( m+ ) Ê f ˆ Ê f ˆ k = ká k m m Ë f Á Ë f where k ad k are costats. Hece, the GENR method has a cubic rate of covergece. Eample.7: Fid the double root of the equatio 3 + = 0 startig with 0 = 0 by GNR method. Solutio: Let f () = 3 3 + The, f () = 3 3 For m =, the GNR method give as + = 3 3 + 3( ) ( = 0,,, 3,...) Startig with 0 = 0, we obtai the followig approimatios. =.3333333, =.058730 3 =.0000407, 4 =.0000407 Therefore, the required double root of the equatio is.0000407 which is almost close to the actual double root = of the give equatio. Eample.8: Solve eample.7 by usig the GENR method. Solutio: Let f () = 3 3 + The, f () = 3 3 ad f () = 6 The GENR method (.44) i this case takes the form + = 4 f( ) f ( ) 3 f ( ) f( ) f ( ) (.8.)
80 Numerical Aalysis: Iterative Methods For the give problem, (.8.) gives us + = 4( 3 + ) (3 3) 7( ) ( 3 ) 3 3 + (.8.) ( = 0,,,...) If we choose 0 = 0, the oe ca have the followig successive iterates from (.8.) = 0.8888889, = 0.999937, 3 =.0000000 Therefore, the double root of the equatio is =. Eample.9: Fid the triple root of 7 5 + 7 4 + 36 3 + 8 + 9 + = 0 correct to si places of decimal, usig the GENR method. Solutio: Give f () = 7 5 + 7 4 + 36 3 + 8 + 9 + fi f () = 35 4 + 08 3 + 08 + 56 + 9, f () = 540 3 + 34 + 6 + 56 The GENR method (.44) i this case takes the form + = For the give problem, (.9.) gives us 6 f( ) f ( ) 4[ f ( )] 3 f( ) f ( ) (.9.) 5 4 3 4 3 6(7 + 7 + 36 + 8 + 9+ ) (35 + 08 + 08 + 56+ 9) + = 5 4 3 È(7 + 7 + 36 + 8 + 9+ ) 4 3 4(35 + 08 + 08 + 56+ 9) 3 Í 3 ÍÎ (540 + 34 + 6+ 56) ( = 0,,,...) (.9.) Usig this formula (.9.) with 0 =, we obtaied the followig approimatios = 0.3465347, = 0.333338, 3 = 0.333334. Sice the differece betwee d ad 3 rd approimatios are less tha 0.5 0 6, the required solutio is = 0.33333.5 TWOSTEP ITERATIVE METHODS FOR SOLVING NONLINEAR EQUATIONS I this sectio, we discuss a twostep iterative methods for solvig oliear equatios of the form (.)..5. Accelerated Iterative Method Let a be the eact root of the equatio (.) i a ope iterval D i which f () is cotiuous ad has well defied first derivative ad let be the th approimate to the root of (.), the
8 Numerical Aalysis: Iterative Methods Takig a i (.53) as ( + ) th approimate to the root, from (.53) ad (.60), the twostep accelerated iterative method ca be defied as follows. (i) Calculate f( ) y = (.6) f ( ) (ii) Obtai È Í f( ) Í + = Í f ( ) Í Ê 4 f( y ) ˆ Í+ Í Á f( ) Î Ë ( = 0,,...) (.63) This method is free from the secod derivative ad requires two fuctioal evaluatios ad oe of its first derivatives. The efficiecy ide of this method is 3 4..5. Covergece Criteria of Accelerated Iterative Method If a is the root ad is the th approimate to the root, the epadig f ( ) about a usig Taylor s epasio, we have f ( ) = where = e e e e ( ) ( ) ( ) ( ) ( ) ( ) ( )! 3! 4! 5! 3 4 5 iv v 6 f a + f a e + f a + f a + f a + f a + Oe iv v È f ( a) f ( a) 3 f ( a) 4 f ( a) 5 6 f ( a ) Íe + e + e + e + e + Oe ( ) Î! f ( a) 3! f ( a) 4! f ( a) 5! f ( a) f ( a ) È Îe + c e + c e + c e + c e + O( e ) (.64) 3 4 5 6 = 3 4 5 c j j f ( a) =, ( j =, 3, 4 ) j! f ( a) f ( ) = f ( a )[+ c e + 3c e + 4c e + 5 c e + O( e )] (.65) 3 4 5 ad, 3 4 5 Now, f( ) f e c e (c c ) e (3c 7c c 4 c ) e O( e ) 3 3 4 5 3 4 3 ( ) = + + From (.53), (.6) ad (.66), we have 3 3 4 5 3 4 3 (.66) y =a+ c e + (c c ) e + (3c 7c c + 4 c ) e + O( e ) (.67) f( y) = f ( a)[ ce (c c) e (7cc 5c 3 c) e + Oe ( ) (.68) 3 3 4 5 3 3 4
84 Numerical Aalysis: Iterative Methods Thus, È f( ) f( y) Í Ê ˆ + 4 f ( ) Í Á Ë f( ) ÍÎ [ e c e (c c ) e (3c 7c c + 4 c ) e + O( e )] = [ ( ) (6 4 3 ) ( )] 3 3 4 5 3 4 3 3 3 4 ce + c c3 e + cc3 c c4 e + O e = e c e (c c ) e (3c 7c c + 4 c ) e + O( e ) 3 3 4 5 3 4 3 [ { c e + (c c ) e + (4c + 3c 6 c c ) e + O( e )] 3 3 4 3 4 3 3 3 4 5 = [ e ce (c3 c) e (3c4 7cc3 + 4 c) e + O( e)] 3 3 [ + ce + (c3 c) e + (4c+ 3c46 cc3) e + ( c e + (c c ) e + c (c c ) e + c (4c + 3c 6 c c ) e + ( c e + c (c c ) e + c (c c ) e 4 3 3 4 3 3 4 3 3 3 4 4 3 3 + ce 4 4 ] = e c e (c c ) e (3c 7c c + 4 c ) e + O( e ) [ + + ( ) + (4 + 3 6 + 4 4 + ) + ( )] 3 3 4 5 3 4 3 3 3 3 3 4 ce c3 c e c c4 cc3 cc3 c c e O e = [ e + c e + c e c e + ( c + 3c 6 c c ) e + + + 3 3 3 4 3 4 3 3 3 4 3 4 ce ce ( cc3 c) e (c3 c) e c(c3 c) e 3 4 5 (3c4 7cc 3 4 c) e Oe ( )] = [ e + ( c c ) e + (c c c + c c ) e + (c c 4c + c + 3c 3c + 7 ) + (e )] 3 3 3 3 3 3 3 4 4 4 5 cc 3 cc 3 cc 3 cc 3 e O = e + ( cc c) e + Oe ( ) (.73) 3 4 5 3 From (.53), (.63) ad (.73), we have the rate of covergece of the method (.63) is four..5.3 Variat of Acccelerated Iterative Method By epadig f( y ) appearig i the deomiator of the method (.63), we obtai 4 f ( ) + = Ï Ô f( ) Ô Ì 3 f ( ) Ô ÔÏ f( y) Ê f( y) ˆ Ê f( y) ˆ Ô Ô Ô Ì +... f( ) Á f( ) + Á + f( ) Ô Ô Ô Ë Ë Ó Ó Ô Ô Ô Ô (.74)
Numerical Solutio of Noliear Equatios 85 Ad, the above further yields + = 3 f( ) È f( y ) Ê f( y ) ˆ Ê f( y ) ˆ Í 5... Î + + + + f ( ) f( ) Á Ë f( ) Á Ë f( ) Í (.75) Cosiderig up to the first ad secod degree terms of the above epressio lyig withi the brackets of the formula (.75), the twostep variat of accelerated iterative method is defied by (i) Calculate y (ii) Obtai f( ) = f ( ) f( ) È f( y ) Ê f( y ) ˆ Í Î + = + + f ( ) f( ) Á Ë f( ) Í ( = 0,,...).5.4 Covergece Criteria of Variat of Accelerated Iterative Method As doe i sectio (.5.), oe ca easily obtai the error relatio as which gives us 3 4 5 a + e + = a+ e [ e + ( cc 3 5 c) e + O(e ) +...] (.76) e + µ e 4 (.77) Therefore, the method (.76) has fourth order covergece. Note.0: If we eglect the secod degree term lyig withi the brackets of the variat of accelerated iterative method, oe will be left with the formula + = which has cubic rate of covergece. f( ) È f( y ) Í + f ( ) Î f( ).5.5 TwoStep Etrapolated NewtoRaphso (ENR) Method (.78) Let a be the eact root of the equatio (.) i a ope iterval D i which f () is cotiuous ad has well defied first derivative ad let be the th approimate to the root of (.) If we let y = + h (.79) where h = f( ) f ( )
Numerical Solutio of Noliear Equatios 87 iv v È f ( a) f ( a) 3 f ( a) 4 f ( a) 5 6 = f ( a ) Íe + e + e + e + e + Oe ( ) Î! f ( a) 3! f ( a) 4! f ( a) 5! f ( a) = f ( a ) È Îe + ce + ce + ce + ce + Oe ( ) (.85) 3 4 5 6 3 4 5 f ( ) = f ( a ) È Î+ ce + 3ce + 4ce + 5 ce + Oe ( ) (.86) 3 4 5 ad 3 4 5 j f ( a) c j = j! f ( a) where (j =, 3, 4 ) Now, from (.80) ad (.8), we obtai ad ad f( ) e c e (c c ) e (3c 7c c 4 c ) e O( e ) 3 3 4 5 = + + 3 + 4 3 + + f ( ) From (.83), (.84) ad (.87), we have y =a + c e + (c c ) e + (3c 7c c + 4 c ) e + O( e ) 3 3 4 5 3 4 3 (.87) 3 3 4 5 f (y ) = f ( a)[ ce (c c3) e (7cc 3 5c 3 c4) e + Oe ( ) (.88) From (.84) ad (.88), we have 3 3 4 5 f ( ) f (y ) = f ( a ) [ e + ( c c3) e + (7cc 3 5c c4) e + Oe ( )] (.89) f( ) f( ) f( y ) Fially, With (.9), (.8) yields 3 4 5 = È Î + ce + c3e + c4e + c5e + O( e) [ e + ( c c) e + (7cc 5c c) e + Oe ( )] 3 3 4 5 3 3 4 = + c e + (c c ) e + (3c + 3c 6 c c ) e + O( e )] (.90) 3 3 4 3 4 3 È f( ) f( ) Í Î f( ) f( y) f ( ) e + = c e 3 + O(e 4 ) = e ce Oe ( ) 3 4 Which shows that the method (.8) has third order covergece..5.7 TwoStep Derivative Free Etrapolated Newto s Method (.9) As it is kow that the backward differece approimatio for the first derivative for f () at is f () ª f ( ) f ( h ) (.9) h
Numerical Solutio of Noliear Equatios 89 ad, f ( f ( )) = ( c+ c) e + ( cc 3 cc + c) e 3 3 + ( cc + 3cc + cc 4cc c + c) e 3 3 3 3 + ( c + c( c + cc ) 5cc 5cc 4 3 4 3 (.98) Subtractig (.96) from (.98), we get f ( ) f ( f ( )) + 6cc 4cc + cc + 6ccc 3 ccc) e + Oe ( ) 3 4 4 5 4 4 4 3 3 ce + ( cc + 3 cc) e + ( cc 3cc cc + 4cc + c ) e (.99) = 3 3 3 3 3 + ( c( c + cc ) + 5cc + 5cc 6cc 3 4 3 4 + 4cc cc 6ccc + 3 ccc) e + Oe ( ) 3 4 4 5 4 4 3 3 [ f( )] = [ ce + ce + ce + ce + ce + Oe ( )] 3 4 5 6 Also 3 4 5 = ce + cce + ( c + cc) e + (cc + cc ) e (.00) From (.99) ad (.00), we get 3 4 5 3 3 4 [ f( )] ce + cce + ( c + cc) e + (cc + cc ) e + Oe ( ) 3 4 5 6 3 3 4 = 3 3 + + + 3 3 + 3 + f( ) f( f( )) { ce ( cc 3 cc) e ( cc 3cc cc 4cc c ) e c 3 + ( c( c + cc) + 5cc + 5cc 6cc + 4cc 4 cc 6ccc + 3c cc ) e 4 + Oe ( 5 )} 3 3 4 3 4 4 4 3 ( c + cc ) (cc + cc ) 3 3 3 4 4 e + e + e + e c c c 3 cc cc cc 3 cc 3 cc + cc 3 + c e c c 3 3 4 3 4 4 4 cc 4 6ccc 3 3 ccc 3) 3 e c = Ï ( + 3 ) ( 3 4 ) Ô + + eô Ô Ô Ô È( c( c + cc ) + 5cc + 5cc 6cc + 4cc Ô Ì Í Ô Í + + + Oe ( Í 4 ) Ô Ô Ô Ô Í Ô ÔÓ ÍÎ Ô = c È c cc cc cc ( + 3) 3 ( 3+ 4) 4 Íe + e + e + e Î c c c 3 È Ï( cc + 3 cc ) ( cc 3 3cc 3 cc + 4cc 3 + c) Í Ì e + e c c Í Ó Í 3 ( cc 3 cc ) ( cc + 3 cc )( cc 3 3cc 3 cc + 4cc 3 + c) Í + 3 + e + e 4 4 ÍÎ c c
Numerical Solutio of Noliear Equatios 9 4 5 3 3 È Êc ˆ (3cc 3 cc 3cc cc 3 + cc ) 3 + Á c e e Oe ( 4 ) c + ÍÎ Ë c with (.05), (.95) ad (.96) oe ca have È c Ê ˆ 3 e + + a = e +a e + e c Á c ÍÎ Ë which yields (.05) (.06) e + µ e 3 which shows the method (.94) has cubic covergece. Eample.0: Fid a root of the equatio 3 + 4 0 = 0 which is ear about = by usig the twostep accelerated iterative method tabulatig all computatios. Solutio: Let f () = 3 + 4 0 the f () = 3 + 8 The twostep accelerated iteratio method is give by where + = y = È Í f( ) Í Í f ( ) Í Ê 4 f( y ) ˆ Í+ Í Á f( ) Î Ë ( = 0,,.) + 4 0 3 8 3 + Takig the iitial guess 0 as uity, the computatios for obtaiig the solutio of the give equatio usig the above method are tabulated below: Table.3 f ( ) f ( ) y f (y ) f( y ) 4 f ( ) 0 5.45454545.5409534.3567.3645053 0.09630 6.506667.365307 0.45406 0 5.0044.365300 0.35003 0 6.53399.365300 0 From the above tabulated results, we thus have the required root of the equatio as =.365300.
9 Numerical Aalysis: Iterative Methods Eample.: Obtai a root of the equatio e 3 + = 0 tabulatig all computatios, by usig the twostep variat of accelerated iterative method, takig 0 =. Solutio: Let f () = e 3 + The, f () = e 3 The variat of accelerated iterative method is give by where y e = e 3 + 3 + = f( ) È f( y ) Ê f( y ) ˆ Í Î + + f ( ) f( ) Á Ë f( ) Í ( = 0,, ) Takig 0 = ad applyig the above scheme, the successive approimatios are tabulated below: Table.4 f ( ) f ( ) y f (y ) + 0.78883 3.78883 0.68944 0.0430736 0.56990 0.56990 0.044 0 3.77905 0.575306 0.030958 0 6 0.575309 0.575309 0.888784 0 5 3.7786704 0.575309 0.444089 0 5 0.575309 From the above table, we have the required root = 0.575309. Eample.: Solve si () + = 0 with 0 = 5, by usig (i) Twostep accelerated iterative method (.63) (ii) Variat of twostep accelerated iterative method (.76). Tabulate all computatios. Solutio: (i) with 0 = 5, the computatios of accelerated iterative method are give below: Table.5 f ( ) f ( ) y f (y ) f( y ) 4 f ( ) 0 5 3.080464 0.5440.803774 6.79658885 0.7789465
94 Numerical Aalysis: Iterative Methods EXERCISES. Fid graphically the positive root of the equatio 3 6 3 = 0. Usig graphical method, fid a approimate root of the equatio = si. 3. Fid the positive real root of the equatio log 0 =. usig bisectio method. 4. By usig bisectio method, fid a approimate root of the equatio si = that lies betwee = ad =.5. Carry out computatios to obtai the root correct to 3 decimals. 5. Fid by the iteratio method, the root ear 3.8 of the equatio log 0 = 7 correct to four decimal places. 6. Solve = + ta by iteratio method. 7. By the Wegstei s method, fid the root of the equatio 3 5 = 0 that is ear about =, correct to 3 decimal places. 8. Fid a real root of cos = 3 by Wegstei s method correct to three decimal places. 9. Fid a root of cos e = 0 correct to 3 decimal places by Aitke s D method, with 0 = 0. 0. Perform Aitke s D process to obtai the root of the equatio + = si + after 3 approimatios to the root are available which are obtaied by iteratio method takig =. How may Aitke D process iteratios are required to obtai the root correct to 6 decimals.. Fid the root of the equatio si = + 3 betwee ad by etrapolated iteratio method, correct to 3 decimal places.. Fid the real root of + log = 0 by etrapolated iteratio method, correct to 5 decimals places. 3. Solve 3 = + 5 for a positive root by etrapolated iteratio method. 4. Fid the real root of the equatio 3 + si e = 0 by the method of falsepositio method correct to four decimal places. 5. Fid all real zeros of the fuctio f () = 3 4 3 by RegulaFalsi method. 6. A real root of the equatio f () = 3 5 + = 0 lies i the iterval (0, ). Perform four iteratios of the secat method. 7. Computer the root of the equatio e = i the iterval (0, ) usig the secat method correct to three decimal places. 8. Fid a root of e si = usig NR method correct to 3 decimal places. 9. Usig the NR method, fid the root of the equatio e = 3 that lies betwee 0 ad, correct to 6 decimal places. 0. Fid all real zeros of f () = 3 9 + by NR method.
96 Numerical Aalysis: Iterative Methods (iv) Fid the positive root of si + = 0 correct to three decimal places. (v) Fid the real root of the equatio e 4 = 0 correct to three decimals. (vi) Fid the real root of the equatio cos e = 0 correct to three decimals. (vii) Fid a real root of the equatio 3 + si e = 0 correct to four decimal places. (viii) Fid a real root of the equatio e 3 + = 0 correct to three decimal places. (i) Fid the approimate root of 4 0 = 0 give that the root lies betwee.8 ad correct to four decimals. () Fid the real root of the equatio 3 = cos + correct to four decimal places. (i) Fid root of the equatio e si + 3 cos + 5 = 0 correct to four decimal places, give that root lies betwee 0 ad. (ii) Fid to four places of decimal, the smallest root of the equatio e = si. (iii) Fid the root of the equatio si = + 3 betwee ad correct to 3 places of decimal. (iv) Fid a real root of + log = 0 correct to 5 decimals. (v) Fid real root of the equatio cos = 0 correct to three decimal places. (vi) Compute the root of the equatio e three decimal places. = i the iterval (0, ) correct to (vii) ʈ Compute the root of the equatio si Á Ë correct to four decimals. = 0 i the iterval [, ] ad (viii) Solve the cubic 3 6 + 8 + 0.8 = 0 usig =.5 & = correct up to four decimals. (i) Obtai a root of the equatio. 0.5 3 = 0 usig iitial values ad.5. () The equatio 4 5 3 + 76 79 = 0 has two roots close to =, fid these roots to four decimal places. (i) Fid the double root of 3 + = 0 close to 0.8. (ii) Obtai a root, correct to three decimal places for equatio 3 + + + 7 = 0. (iii) Compute the positive root of 3 8 = 0 correct to two decimal places. (iv) Fid a real root of the equatio e si cos si + 8 = 0 correct to three decimal places. (v) Fid a real root of 4 4 9 = 0 correct to 4 decimal places. (vi) Fid the root of the equatio e = cos correct to four decimal places. (vii) Fid all real zeros of the fuctio f () = 3 + 3 4 correct to four decimals (viii) Fid the smallest positive root of e = 0 correct to three decimal places. (i) Fid the root of the equatio log +.5 = 0 startig from =, =.50 compute up to 3 decimals. () Fid the fourth root of 3 correct to three decimal places.
98 Numerical Aalysis: Iterative Methods 3. 0.33334467 3..365300 33..0764783 34. 4 35. Equatio Iitial guess No. of iteratios take by * f () = 0 0 Method.8 Method.94 Root () e cos = 0 0 5 4 0.5775736368458 () e si = 0 0. 5 5 0.588537439886 (3) e.5 ta = 0 7 4 4 4.06977739964 (i) 0.739 (ii).0 (iii) 0.6 (iv).404 (v) 3.574 (vi) 0.55 (vii) 0.3604 (viii) 0.57 (i).8555 () 0.607 (i).076 (ii) 0.5885 (iii).49 (iv).5574 (v) 0.739 (vi).49 (vii).9337 (viii).00 (i).965 ().40,.7684 (i) (ii).05 (iii).33 (iv) 3.437 or 4.6 (v).6875 (vi) 0.577 (vii).8794, 0.657, 0.53 (viii) 0.567 (i).359 ().378 (i) 6.8506 (ii) 3.340,.697, 5.9437 (iii) 0.856 (iv).8 (v).54 (vi).33 (vii).73 (viii).395 (i).4 (l).933 (li) 3.30 (lii) 0.69 (liii) 3. (liv).93