Question

3. Find the positive root of In(x²) = 0.7 20-points a) Using three iterations of the bisection method with initial guesses of Xi on method with initial guesses of x = 0.5 and Xu 2, and b) Using three iterations of the Secant method, with the same initial guesses as in .

Answer 1

Sol In(2) = 0.7 Inca?)-07=0 f)%3DIn(x)-0.7 9) given =0.5 ond t=9 fexs)70 bisechion methud fe21)20; first iteration X2= 1+ Xu D -5+2 *2= 1-25 fLb)= lnci-253)-0:7 fixx)=-0-254 fis) L0, SD the Root lies between 2=1-25 and Xu=2 itesation 1-25+2 2. 2. 3=1.625 f(73)= en(1-625?) -0-7 fug) = 0.211 fixg) 70, So the Reet lies between S=1-25 and X3=1-625 37 itesation X= X3+X2 Xy = 1-625+1-25 Root after the thid iteation Xy = 1.4375

b) fox) = Inlx)-07 2=05 72=2 fixg) = fix)=enco53)-0-1 en(2?)-0.7 +(xg)=0-6863 tix) = -2.0863 Secdnt method:- 1st itescetion foxs)=fi4). 2-0-5 23= 2- 0.6863 X D-6863+20863 23=1-6987 fix3) = In (162874-0-7 flg)= 0-2756 So the Roc lies betuleen X = D-5 and 리6287 fir3) 70 iteration :- X3-x, -{fix3)Lfrag)-fe4). Xy= X3- 1-6287-0.5 X4= 1.6287 - 0.2756 x D.2756+2.0863 Xy = 1-4970 fixu) = Inll-4972) -0-7 flz4) = 0-1069 So the Root lies betaseen X=0-5 and fizu) yo Xy=1-497

3d itesation Xy= Xy-}+Que) x tery)-tix) 1.497-0.5 - 497- D -1069 X 0.1069+2.0863 25=1:4484 %3D 5=1-4484 Root afder the third itration