Homework 2 solutions

Transcription

1 Sectio 2.1: Ex 1,3,6,11; AP 1 Sectio 2.2: Ex 3,4,9,12,14 Homework 2 solutios 1. Determie i ech uctio hs uique ixed poit o the speciied itervl. gx = 1 x 2 /4 o [0,1]. g x = -x/2, so g is cotiuous d decresig o [0,1]. g0 = 1 d g1 = ¾. Thereore gx is i [0,1] or ll x i [0,1]. Furthermore, g ' x x / 2 1/ 2 1 or x i [0,1]. The ixed poit theorem thus implies the existece o uique ixed poit i [0,1]. gx = 2 -x o [0,1]. g x = -log2 2 -x, so g is cotiuous d decresig o [0,1]. g0=1 d g1=1/2. Thereore gx is i [0,1] or ll x i [0,1]. Ad sice g is decresig, there is uique ixed poit i [0,1]. c gx = 1/x o [0.5, 5.2]. gx = x is 1/x = x, or x 2 = 1, which hs sigle solutio x = 1 i the itervl [0.5, 5.2]. 3. Grph the uctio with the lie y=x, costruct cowe digrm, d determie i the ixed poit itertio coverges. gx = 6 + x 1/2, P = 3, d p o = 7. gx = 1 + 2/x, P = 2, d p o = 4. c gx = x 2 /3, P = 3, d p o = 3.5. d gx = -x 2 + 2x + 2, P = 2, d p o = 2.5. The digrms re show elow. The itertio coverges mootoiclly. The itertio exhiits oscilltig covergece. c The itertio does ot coverge it diverges to iiity. d The itertio does ot coverge.

2 6. Suppose gx d g x re cotiuous o,; p o, p 1, p 2 re i,, d p 1 =gp o d p 2 =gp 1. Also ssume tht there exists costt K such tht g x < K. Show tht p 2 - p 1 < K p 1 -p o. Hit: Use the me vlue theorem. We hve p 2 -p 1 = gp 1 gp o = g c o p 1 p o or some c o etwee p o d p 1 y the me vlue theorem Sice p o d p 1 re i,, c o must lso e i,, d thereore g c o <K. Thereore, p 2 -p 1 = g c o p 1 p o = g c o p 1 p o < K p 1 -p o. 11. For ixed-poit itertio, discuss why it is dvtge to hve g P = 0. The itertio stisies p +1 p < K p p -1, where K is oud o the solute vlue o the derivtive i the regio o the ixed poit. Thus, the smller g P, the smller will K e i the iequlity ove, d the closer will successive itertes e. This mes tht the itertio will coverge ster i the derivtive is smller. The smllest derivtive is 0, so itertio roud ixed poit with zero derivtive will coverge rpidly.

3 ALGORITHMS AND PROGRAMS 1. Use the ixed poit itertio lgorithm to pproximte the ixed poits i y o the ollowig uctios to 12 deciml plces. Produce grph o ech uctio d the lie y=x tht clerly shows y ixed poits. gx = x 5 3x 3 2x The itertio does ot coverge. There re 3 ustle ixed poits gx = cossix x* = c gx = x 2 six d gx = x x-cosx Cowe digrms re elow. MATLAB code ollows. The itertio does ot coverge The itertio will to coverge to the ixed poit x*=1 i the iitil coditio is i the itervl [.6, 1.2]. There is other ustle ixed poit t pproximtely

4 MATLAB code %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Computes itertes d cowe digrm or give uctio % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % PARAMETERS % ic - iitil coditio % um - umer o itertios um=20; % DEFINE FUNCTION USED IN ITERATIONS =@xcossix; ic=-1; =-1;=2;c=-.1;d=1.1; s=sprit'cossix'; %%%%%%%%%% ITERATIONS %%%%%%%%%%%%% % x = ic; % or i=1:um % xi+1 = xi; % ed % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% err = sxi+1-xi/sxi; dispsprit'fixed poit t x = %.12\Reltive error: %.3e',xi+1,err % <<< Plot results >>> igure2; cl suplot2,1,1 hold o; plot0:um,x,'r.','liewidth',4; plot0:um, x,'' xis[0 um mi0,mxmix,-10, mx1,mxx] xlel'','otsize',20; ylel'x_','otsize',20 titlesprit'itertios o x_{+1}=gx_; gx=%s',s,'otsize',16 suplot2,1,2 xx=lispce,,100; hold o; plotxx,xx,'r'; plotxx,xx,'k' plot[x1 x1],[0 x2],'' plot[x1 x2],[x2 x2],'' or i=2:um plot[xi xi],[xi xi+1],'' plot[xi xi+1],[xi+1 xi+1],'' ed plotxx,xx,'r'; plotic,0,'c*' xis[ c d] xlel'x_','otsize',20 ylel'x_{+1}','otsize',20 titlesprit'cowe digrm; gx=%s',s,'otsize',16

5 Sectio For ech uctio, id itervl [,] so tht d hve dieret sigs. x = e x 2 x -3 > 0, 0 < 0, 3 > 0; thus roots lie i the itervls [-3,0] d [0,3]. x = cosx + 1 x /4 > 0 d /2 < 0; thus root lies i the itervl [ /4, /2]. c x = lx 5 + x 3 < 0 d 5 > 0; thus root lies i the itervl [3,5] d x = x 2 10x > 0, 5 < 0, 7 > 0; thus roots lie i the itervls [3,5] d [5,7]. 4. Strt with [ o, o ] d use the lse positio method to compute c o, c 1, c 2, d c 3. x = e x 2 x = 0, [ o, o ] = [-2.4, -1.6] c o = , c 1 = , c 2 = , c 3 = Wht will hppe i the isectio method is used with the uctio x = 1/x-2 d the itervl is [3,7]? 37 = 1/5 > 0, so the lgorithm would tell us tht there is o root i [3,7], which is true. the itervl is [1,7]? 17 = -1/5 < 0, ut there is o root i 1,7. The method ils, ecuse the uctio is ot cotiuous i [1,7]. 12. Show tht the ormul c is lgericlly equivlet to c. We dd the terms: c

6 14. The polyomil x = x-1 3 x-2x-3 hs 3 zeros: x=1 o multiplicity 3, x=2, d x=3, ech o multiplicity 1. I o d o re y two rel umers such tht o < 1 d o > 3, the o o <0. Thus, o the itervl [ o, o ] the isectio method will coverge to oe o the three zeros. I o <1 d o >3 re selected so tht c = + /2 is ot equl to 1, 2, or 3 or y >= 1, the the isectio method will ever coverge to which zeros? Why? It c t coverge to x=2, uless it hits c =2 exctly extremely ulikely, or the ollowig reso. Excludig the cses where c hits 1 or 3 exctly, i o <1, d o >3, the the midpoit c o will e either less th 1, greter th 3, or i the itervl 1,3. I the irst d secod cses, 1 =c o or 1 =c o, d we ck to the origil cse o <1, o >3. Assume, thereore, tht the itertio hs proceeded util c is i 1,3, d relel this poit c o. I, 1 < c o < 3, the there re three possiilities: i 1 < c o < 2, i which cse c o >0, so 1 = c o, d x=2 is outside o the rcket, ii 2 < c o < 3, i which cse c o <0, so 1 = c o, d gi, x=2 is outside o the rcket, or iii c o = 2. I either cse, oce itertio gets poit i the itervl 1,3, it rckets i o either 1 or 3, ut ever 2, uless it hppes to hit o 2 exctly.