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Letitia Dawson
5 years ago
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7 function [xc,yc,errx]=newtonldr(fun,funpr,xl,tol,kmax,pcon) Input. fun function (inline function or m-file function) funpr derivative function (inline or m-file) xl starting estimate o tol allowable tolerance in computed zero kmax maximum number of iterations o pcon=expected order of convergence, p $ Output: xc approximation to the root fun(xc)=0 o yc =fun (xc) errx = approximation to error in xc x(1)=xl;err(1)=0.;papp(1)=0.;capp(1)=0.;cap2(1)=0.; y(1)=feval (fun,x(1)) ; ypr(1)=feval(funpr,x(1)); for k = 2:kmax x(k)=x(k-1)-y(k-1)/ypr(k-1); err(k)=abs(y(k-1)/ypr(k-1)); if k<4 papp(k)=0.;capp(k)=0.;cap2(k)=0.; if k==3 cap2(3)=abs(err(k)/(err(k-1))^pcon); end %if k==3 else papp(k)=(log(err(k))-log(err(k-1)))/(log(err(k-1))-log(err(k-2))); capp(k)=abs(err(k)/(err(k-lj)^papp(k)); cap2(k)=abs(err(k)/(err(k-1))^pcon); end oif k<4 y (k) =feval (fun, x (k)) ; if abs(x(k)-x(k-1)) < tol disp('newton method has converged'); break; end ypr(k)=feval(funpr,x(k)); iter=k; end if liter >= kmax) disp(' zero not found to desired tolerance'); end n=length(x); if imag(x(1)) =0 imag(y(1)) =0 disp(' n x y err Pa1P ) else disp(' n x y err papp Capp cwithpcon') end oif for k=1:n; gout = [k' x' y' err' papp']; odisp(out) if imag(x(l))= 0 ( imag(y(1)) = 0 fprintf('%3.of o10.6f o10.6f %-13.5g o-13.5g o o8.4f\n',k,real(x(k)),imag(x(k)),real(y(k)),,imag(y(k)),err(k),papp(k)); else fprintf('%3.of o10.6f o-13.5g o-13.5g %8.4f o8.4f %8.4f\n',k,x(k),y(k),err(k),papp(k),capp(k),cap2(k)); end oif end %for Page 1 of

8 >o output for problem 6 >[xc yc errx]=newtonldr("pr6fun","pr6fpr",o,l.e-11,20,2) Newton method has converged n x y err papp capp cwithpcon e e e e xc = yc = 0 errx = e-012 >% >o output for problem 3b >[xc yc errx]=newtonldr("pr3fun","pr3fpr",l,l.e-11,20,2) Newton method has converged n x y err papp capp cwithpcon e e e e e xc = yc = e-015 errx = e-012 >o >% output for problem 4a >[xc yc errx]=newtonldr("pr4fun","pr4fpr",0,5.e-9,26,1) Newton method has converged n x y err papp Capp cwithpcon e e e e e e e e e e e e e e e e e Page 1 of

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10 0 output for problem 6 using pr6ftaylor.m to evaluate f(x)=integral 0 to x e^(t^2) dt with a Taylor polynomial >[xc yc errx]=newtonldr("pr6ftaylor","pr6fpr",o,l.e-11,20,2) Newton method has converged n x err papp cape e e e e xc = yc 0 errx = e-012 cwithpco function [y ier nfun err]=pr6ftaylor(x) [y ier nfun err]=quad("pr6fpr",o,x); This integrates the function integral 0 to x e^(t^2) dt using a Taylor polynomial. ier=0 is returned iff the integration was successful nfun=how many terms were used in the Taylor polynomial err = an estimate of the error in the solution sum=x; dfact=l.; term=x; nmax=900; efact=exp(x.^2);ier=0;k=0;err=1; if x < 0 efact=l; end oif while err > l.e-15 k=k+l; dfact=dfact+2; term=term*x^2/k; sum=sum+term/dfact; err=abs(efact*term*x^2/(k+l)/(dfact+2)); if k > nmax ier=1; break; end oif end while nfun=k; if ier==1; disp 'more terms are needed in the Taylor polynomial' x y=sum nfun err break; end %if y=sum-3; return

A L A BA M A L A W R E V IE W

A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N