1 / 16

Lecture 5 Fixed point iteration

Lecture 5 Fixed point iteration. Download fixedpoint.m From math.unm.edu/~plushnik/375. %fixedpoint.m - solution of nonlinear equation by fixed point iterations function [x,n, xn] = fixedpoint(f, x0, tol, nmax) % find the root of equation x=f(x) by fixed point method; % input:

ailish
Download Presentation

Lecture 5 Fixed point iteration

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 5 Fixed point iteration Download fixedpoint.m From math.unm.edu/~plushnik/375

  2. %fixedpoint.m - solution of nonlinear equation by fixed point iterations function [x,n, xn] = fixedpoint(f, x0, tol, nmax) % find the root of equation x=f(x) by fixed point method; % input: % f - inline function % x0 - initial guess % tol - exit condition f(x) < tol % nmax - maximum number of iterations % output: % x - the approximation for the root % n - number of iterations % xn - vector of apporximations x(iter) % compute function at initial guesses f0 = f(x0); n = 0; % begin iterations while ((abs(f0-x0) > tol) && (n < nmax)) x0 = f0; f0 = f(x0); disp(['Error: f0-x0=',num2str(f0-x0)]); if f0 == x0 % x0 is a root, done break; end n = n+1; xn(n) = x0; end if n==nmax disp('warning: maximum iterations reached without conversion'); end x=x0; disp(['Number of iterations: n = ',num2str(n)]); end

  3. Enter inline functions: >> g1=inline('1-x^3'); >> g2=inline('(1-x)^(1/3)'); >> g3=inline('(1+2*x^3)/(1+3*x^2)');

  4. >>fixedpoint(g1,0.5,10^(-8),10); Error: f0-x0=-0.54492 Error: f0-x0=0.63396 Error: f0-x0=-0.85998 Error: f0-x0=0.89482 Error: f0-x0=-0.9955 Error: f0-x0=0.99662 Error: f0-x0=-1 Error: f0-x0=1 Error: f0-x0=-1 Error: f0-x0=1 warning: maximum iterations reached without conversion Number of iterations: n = 10

  5. >> fixedpoint(g2,0.5,10^(-8),100); Error: f0-x0=-0.20282 Error: f0-x0=0.15148 Error: f0-x0=-0.10605 Error: f0-x0=0.077491 Error: f0-x0=-0.054795 Error: f0-x0=0.039626 Error: f0-x0=-0.028184 Error: f0-x0=0.020281 Error: f0-x0=-0.01447 Error: f0-x0=0.010387 Error: f0-x0=-0.0074232 Error: f0-x0=0.0053217 Error: f0-x0=-0.0038066 Error: f0-x0=0.0027272 Error: f0-x0=-0.0019517 Error: f0-x0=0.0013978 Error: f0-x0=-0.0010005 Error: f0-x0=0.00071648 Error: f0-x0=-0.00051291 Error: f0-x0=0.00036726 Error: f0-x0=-0.00026293 Error: f0-x0=0.00018826 Error: f0-x0=-0.00013478 Error: f0-x0=9.6501e-005 Error: f0-x0=-6.9091e-005 Error: f0-x0=4.9467e-005 Error: f0-x0=-3.5416e-005 Error: f0-x0=2.5357e-005 Error: f0-x0=-1.8155e-005 Error: f0-x0=1.2998e-005 Error: f0-x0=-9.3063e-006 Error: f0-x0=6.663e-006 Error: f0-x0=-4.7705e-006 Error: f0-x0=3.4155e-006 Error: f0-x0=-2.4454e-006 Error: f0-x0=1.7508e-006 Error: f0-x0=-1.2535e-006 Error: f0-x0=8.9748e-007 Error: f0-x0=-6.4257e-007 Error: f0-x0=4.6006e-007 Error: f0-x0=-3.2938e-007 Error: f0-x0=2.3583e-007 Error: f0-x0=-1.6885e-007 Error: f0-x0=1.2089e-007 Error: f0-x0=-8.6551e-008 Error: f0-x0=6.1968e-008 Error: f0-x0=-4.4367e-008 Error: f0-x0=3.1765e-008 Error: f0-x0=-2.2743e-008 Error: f0-x0=1.6283e-008 Error: f0-x0=-1.1658e-008 Error: f0-x0=8.3468e-009 Number of iterations: n = 52

  6. >> fixedpoint(g3,0.5,10^(-8),100); Error: f0-x0=-0.031106 Error: f0-x0=-0.0008513 Error: f0-x0=-6.1948e-007 Error: f0-x0=-3.2774e-013 Number of iterations: n = 4 >>

  7. Secant method Download secant02.m And ftest2.m From math.unm.edu/~plushnik/375

  8. %Secant method to find roots for function ftest2 x0=0.1; x1=2.0;%starting points abserr=10^(-14); %stop criterion - desired absolute error istep=0; xn1=x0; %set initial value of x to x0 xn=x1; %main loop to find root disp('Iterations by Secant Method'); while abs(ftest2(xn))>abserr istep=istep+1; fn=ftest2(xn); fn1=ftest2(xn1); disp(['f(x)=',num2str(fn),' xn=',num2str(xn,15)]);%display value of function f(x) xtmp=xn-(xn-xn1)*fn/(fn-fn1); xn1=xn; xn=xtmp; end f=ftest2(xn); disp(['f(x)=',num2str(fn),' xn=',num2str(xn,15)]);%display value of function f(x) disp(['number of steps for Secant algorithm=',num2str(istep)]);

  9. %test function is defined at fourth line; %derivative of function is defined at firth line function [f,fderivative]=ftest2(x) f=exp(2*x)+x-3; fderivative=2*exp(2*x)+1;

  10. >> secant02 Iterations by Secant Method f(x)=53.5982 xn=2 f(x)=-1.4715 xn=0.157697583825433 f(x)=-1.2804 xn=0.206925256821038 f(x)=0.46299 xn=0.536842578960542 f(x)=-0.094954 xn=0.449229649271443 f(x)=-0.0057052 xn=0.464140200867443 f(x)=7.5808e-005 xn=0.465093357175321 f(x)=-5.9571e-008 xn=0.465080858161814 f(x)=-6.2172e-013 xn=0.465080867975924 f(x)=-6.2172e-013 xn=0.465080867976027 number of steps for Secant algorithm=9 >>

  11. Inclass3 Modify secant02.m and ftest2.m to find root of e^(-x)-x=0 by secant method starting at x=0.2 and x=1.5

  12. Answer to inclass3 >> secant02 Iterations by Secant Method f(x)=-1.2769 xn=1.5 f(x)=-0.088702 xn=0.624324608254261 f(x)=0.012856 xn=0.558951914931113 f(x)=-0.00013183 xn=0.567227412711665 f(x)=-1.9564e-007 xn=0.567143415251049 f(x)=2.9781e-012 xn=0.567143290407884 f(x)=2.9781e-012 xn=0.567143290409784 number of steps for Secant algorithm=6

  13. Matlab function: fzero • X = FZERO(FUN,X0), • X0 a scalar: Attempts to find a zero of the function FUN near X0. • FUN is a function handle. • The value X returned by FZERO is near a point • where FUN changes sign (if FUN is continuous), or NaN, if the srootfinding • search fails. • X = FZERO(FUN,X0), • X0 a 2-vector. Assumes that FUN(X0(1)) • and FUN(X0(2)) differ in sign, insuring a root. • X = FZERO(FUN,X0,OPTIONS). • Solves the equation with default optimization parameters replaced by values • in the string OPTIONS, an argument created with the OPTIMSET function.

  14. Example of fzero use >> options = optimset('disp', 'iter', 'tolx', 1.e-15); >> fzero(@ftest2,[0.1 2],options) Func-count x f(x) Procedure 2 0.1 -1.6786 initial 3 0.157698 -1.4715 interpolation 4 0.556708 0.601452 interpolation 5 0.440938 -0.143633 interpolation 6 0.463256 -0.0110609 interpolation 7 0.465084 2.0255e-005 interpolation 8 0.465081 -3.08857e-008 interpolation 9 0.465081 -8.61533e-014 interpolation 10 0.465081 0 interpolation Zero found in the interval [0.1, 2] ans = 0.4651 >>

  15. Inclass4 Modify ftest2.m to find root of e^(-x)-x=0 by Brent’s method starting at x=0.2 and x=1.5

  16. Answer to inclass4 >> fzero(@ftest2b,[0.2 1.5],options) Func-count x f(x) Procedure 2 0.2 0.618731 initial 3 0.624325 -0.0887015 interpolation 4 0.571121 -0.0062285 interpolation 5 0.567143 1.13316e-006 interpolation 6 0.567143 -8.15018e-010 interpolation 7 0.567143 -1.11022e-016 interpolation 8 0.567143 -1.11022e-016 interpolation Zero found in the interval [0.2, 1.5] ans = 0.5671

More Related