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Lecture 3 Bisection method

Lecture 3 Bisection method. Download bisection02.m And ftest2.m From math.unm.edu/~plushnik/375. %Bisection method to find roots for function ftest2 istep=0;%set initial number of steps to zero a=0.1; %initial value for interval (a,b) b=2; %initial value for interval (a,b)

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Lecture 3 Bisection method

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  1. Lecture 3 Bisection method Download bisection02.m And ftest2.m From math.unm.edu/~plushnik/375

  2. %Bisection method to find roots for function ftest2 istep=0;%set initial number of steps to zero a=0.1; %initial value for interval (a,b) b=2; %initial value for interval (a,b) if sign(ftest2(a)*ftest2(b))>=0 error('sign(ftest2(a)*ftest2(b))>=0') end abserr=10^(-15); %stop criterion - desired absolute error while abs(ftest2((a+b)/2))>abserr c=(a+b)/2; %calculate midpoint istep=istep+1; fc=ftest2(c); if fc==0 break %if c is solution then exit end if (ftest2(a)*fc)<0 b=c; else a=c; end disp(['f(c)=',num2str(fc),' x=',num2str(c,10)]);%display value of function f(x) end disp(['number of steps for Bisection algorithm=',num2str(istep)]);

  3. %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;

  4. >> bisection02 f(c)=6.2162 x=1.05 f(c)=0.73319 x=0.575 f(c)=-0.69847 x=0.3375 f(c)=-0.053209 x=0.45625 f(c)=0.32019 x=0.515625 … f(c)=-2.2204e-015 x=0.465080868 f(c)=1.8652e-014 x=0.465080868 f(c)=7.9936e-015 x=0.465080868 f(c)=3.1086e-015 x=0.465080868 number of steps for Bisection algorithm=51

  5. It is often good idea to plot function first

  6. Inclass1 Modify bisection02.m and ftest2.m to find root of e^(-x)-x=0 at [0.2,1.5]

  7. Newton’s method Download newton02.m And ftest2.m From math.unm.edu/~plushnik/375

  8. %Newton's method to find roots for function ftest x0=0.5; %starting point abserr=10^(-15); %stop criterion - desired absolute error istep=0; x=x0; %set initial value of x to x0 %main loop to find root disp('Iterations by Newton Method'); while abs(ftest2(x))>abserr istep=istep+1; [f,fder]=ftest2(x); disp(['f(x)=',num2str(f),' x=',num2str(x,15)]);%display value of function f(x) x=x-f/fder; end [f,fder]=ftest2(x); disp(['f(x)=',num2str(f),' x=',num2str(x,15)]);%display value of function f(x) disp(['number of steps for Newton algorithm=',num2str(istep)]);

  9. >> newton02 Iterations by Newton Method f(x)=0.21828 x=0.5 f(x)=0.0061135 x=0.466087210490891 f(x)=5.1326e-006 x=0.46508171356867 f(x)=3.6251e-012 x=0.465080867976624 f(x)=-4.4409e-016 x=0.465080867976026 number of steps for Newton algorithm=4

  10. Inclass2 Modify newton02.m and ftest2.m to find root of e^(-x)-x=0 by Newton’s method starting at x=0.6

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

  12. %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)]);

  13. %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;

  14. >> 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 >>

  15. 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

  16. 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

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