Transcript Lecture 3 Bisection method Download And

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)
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)]);
%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;
>> 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
It is often good idea to plot function first
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ftest2
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Inclass1
Modify bisection02.m and ftest2.m to find root
of e^(-x)-x=0 at [0.2,1.5]
Newton’s method
Download
newton02.m
And
ftest2.m
From math.unm.edu/~plushnik/375
%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)]);
>> 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
Inclass2
Modify newton02.m and ftest2.m to find root
of e^(-x)-x=0 by Newton’s method starting at x=0.6
Secant method
Download
secant02.m
And
ftest2.m
From math.unm.edu/~plushnik/375
%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)]);
%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;
>> 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
>>
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
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