16Flowchart Loop without counter 0n Maplecodeforfxcosxx0xcosxsimpleiterprocxtollocalxoldxnewdifphitranscriptionofinputxoldxensuresthatloopstartsdiftol1iterative ID: 229201
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16Flowchart: INPUTex0,x = (x )nF0en0replaceOUTPUTxn Loop (without counter) 0n Maplecode(forf(x)=cos(x) x=0,(x)=cos(x)):simple_iter:=proc(x,tol)localxold,xnew,dif,phi;#transcriptionofinputxold:=x;#ensuresthatloopstartsdif:=tol+1;#iterativemapphi:=z-�cos(z);#iterationloopwhiledif�toldo#iterationxnew:=phi(xold);#errorestimatedif:=abs(xnew-xold);#transcriptionxold:=xnew;enddo;returnxnew;endproc:Problem:loopswithoutacountermayrunforever.b)Bisectionalgorithm 17Considera(continuous)functionfwheref(a)0andf(b)&#x-1.2;蝗0(orviceversa).Thenf(x)=0forsomex,axbbythein-termediatevaluetheorem.Howtocomputesuchavalue? ab Idea:Computethemidpointc=(a+b)=2.Iff(b)andf(c)havethesamesignthenxiscontainedin(thesmaller)interval[a;c].Iff(a)andf(c)havethesamesignthenxiscontainedin[c;b].Repeatthestepwiththesmallerintervaluntilitslengthisbelowtherequiredthreshold.Ineachstepthelengthoftheintervalshalves,i.e.,afternstepsthelengthoftheintervalisjb aj=2n.Thusthemethodalwaysconverges!Structureofthecode:Given:lowerlimita,upperlimitb,andthreshold".Checkwhetherf(a)andf(b)haveoppositesign.Computethemidpointc=(a+b)=2.Iff(a)andf(c)havethesamesignreplacethelowerlimitabyc(tocontinuewith[c;b]!),elsereplacetheupperlimitbbyc(tocontinuewith[a;c]!)Ifthelengthoftheintervaljb ajislargerthanthethresholdcontinuewithstep3,elsereturnthemidpoint.Flowchart: nob:=c 18Maplecode(forf(x)=cos(x) x=0):simple_bisection:=proc(a,b,tol)locallow,up,mid,f;f:=x-�cos(x)-x;#transcriptionofinputlow:=a;up:=b;#checkforsignchangeiff(up)*f(low)�0thenerror"f(a)f(b)�0";endif;#recursivebisectionwhileup-low�toldo#midpointmid:=(up+low)/2;#bisectionupdateiff(low)*f(mid)�0thenlow:=mid;elseup:=mid;endif;enddo;#outputreturn(up+low)/2;endproc:Example3.3:Supposeyoustartabisectionalgorithmwithinitialinterval[ 2;2]andthealgorithmreturns1=2.Howmanybisectionshavebeenperformed?The rstbisectionstepcomputesthemidpointvalue0.Asourresultiscontainedin[0;2]thealgorithmcontinueswiththeinterval[0;2]andthesecondbisectionstepcomputesforthemidpointthevalue1.Asourresultiscontainedin[0;1]thealgorithmreturnsthe 19midpointvalue1=2,i.e.,twobisectionstepshavebeenperformed.c)Newton-RaphsonmethodRemark:onTaylorseriesexpansionsf(x)=1Xk=0f(k)(x0) k!(x x0)k=f(x0)+f0(x0)(x x0)+f00(x0) 2(x x0)2+:::f(x)=f(x0)+f0(x0)(x x0)| {z }'(x);linearinx+R(x;x0)| {z }remainder(\small")Graphically: ff(x )0x1x0000 (x)=f(x )+f'(x )(x-x )j"first order approximation" Wewanttocomputexsuchthatf(x)=0.Supposeweknowsomeinitialguessx0(closetox).Replacefbythe rstorderapproximation'andsolve'(x)=0,i.e.,0=f(x0)+f0(x0)(x x0))x=x0 f(x0) f0(x0)=x1:Thusweobtainanimprovedvaluex1.Repeattheprocedureuntilthesequenceconverges,i.e.,xk+1=(xk)=xk f(xk) f0(xk):Iterativesolutionmethodwithsome\intelligent"iterativemap.(x)=x f(x)=f0(x)iscalledNewton-Raphsonmap.Propertiesof(seex3a)\Fixedpointcondition":if(x0;x1;x2;:::)converges,xn!xthenx=(x)=x f(x) f0(x))f(x)=0i.e.,thelimitisthedesiredsolution.