Question

I wrote the code for the secant algorithm , and now I have the function :

f(x) = 2x^3 - 4x^2 + 3x , with two initial points : x0 = -1 , x1 = 2 .

My question is how can I plot the function that I wrote ,i.e. secant , with the function above , f , and the results below , in one graph ?

Is it even possible to do that ?

The results that I got after I used the secant algorithm , are :

    v =
   -4.0000
    2.2069
    2.3699
    2.6617
    2.5683
    2.5804

Those are 6 iterations that I used on my secant algorithm , for the given x0 & x1 above .

I’d appreciate if you can explain .

EDIT :

This is the code that I used in order to get the results :

[z,n,v]=secant([-1,2],10^(-5),10^(-5),10)

for the prototype :

function [x,n,v]=secant(X,d,e,N)

% x0 is the first point
% x1 is the end point 
% d is the tolerance 
% e is the requested precision 
% N is the number of iterations 

Thanks .

Answer 1

I quickly threw this together, it illustrates the powerful anonymous function
and it shows you how to plot the results of the secant function (the same way as on wikipedia: http://en.wikipedia.org/wiki/File:Secant_method.svg)

What I don’t understand however is why your secant function has both a tolerance and a requested precision as an input; I would think the tolerance is the result of the secant algorithm..

function [  ] = tmp1(  )

    f=@(x) x.^2;
    [xend,n,v]=secant(f,-4,3,1e-4,50);
    fprintf('after %d iterations reached final x_end = %g, f(x_end) = %g\n',n,xend,f(xend))


    figure;hold on;
    xtmp = linspace(min(v),max(v),250);
    plot(xtmp,f(xtmp),'r'); % plot the function itself
    line([v(1:end-2) v(3:end)]',[f(v(1:end-2)) zeros(n+1,1)]','Color','b','Marker','.','MarkerEdgeColor','b'); % plot the secant lines
    plot(v,f(v),'.','MarkerEdgeColor','r')% plot the intermediate points of the secant algorithm
    line([v(3:end) v(3:end)]',[zeros(n+1,1) f(v(3:end))]','Color','k','LineStyle','--'); % vertical lines

    ylim([-4 max(f(xtmp))]); % set y axis limits for nice plotting view algorithm

end

function [xnew,n,v]=secant(f, x0,x1,e,N)
% x0 is the first point
% x_end is the end point
% e is the requested precision
% N is the number of iterations

v=zeros(N+2,1);
v(1)=x0;
v(2)=x1;

for n=0:N-1
    xnew = x1 - f(x1) * (x1-x0)/(f(x1)-f(x0));
    v(3+n) = xnew;
    if abs(f(xnew)) <e
        break;
    else
        x0=x1;
        x1=xnew;
    end
end
v(n+4:end)=[];

end