How to plot the error of two numerical methods on the same graph?

I'm trying to plot the error of two methods, but I got the following error

Index exceeds the number of array elements (3)

Also, How can I plot the error of the two methods on the same graph?

function xnew = newtonmethod(f,df,x0,tol,n)
%% Given data
f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
x0=1;
tol=0.0001;
n=50;
%% Newton code
disp('No Itr     Solution    Error  ')
Error=[];
for i=1:n
    xnew=x0-(f(x0)/df(x0));
    err=abs(xnew-x0);
    fprintf('%3i %11.4f  %11.4f %11.4f\n',i,x0,err);
    if (err<tol)
        break
    end
    x0=xnew;
    Error=[Error;err];
end
%% Graph

plot(1:i,Error(1:i),'r-','Linewidth',02)
xlabel('No of Iteration','Interpreter','latex','FontSize',12)
ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
title('Error Decay','Interpreter','latex','FontSize',12)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %% 2nd method
 function xnewh = Hmethod(f,df,ddf,x0,tol,n)
%% Given data
f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
ddf=@(x) -4.5*sin(x);

x0=1;
tol=0.0001;
n=50;
%% code
disp('No Itr     Solution    Errorh')

Errorh=[];
for i=1:n
    
    xnewh=x0- (2*f(x0).*df(x0)) ./ (2*(df(x0)).^2-ddf(x0).*f(x0));
    errh=abs(xnewh-x0);
    
    fprintf('%3i %11.4f  %11.4f\n',i,x0,errh);
    if (errh<tol)
        break
    end
    
    x0=xnewh;
    Errorh=[Errorh;errh];
end
%% Graph


plot(1:i,Errorh(1:i),'b-','Linewidth',02)
xlabel('No of Iteration','Interpreter','latex','FontSize',12)
ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
title('Error Decay','Interpreter','latex','FontSize',12)

%    

Answer 1

[Star Strider]

I cannot run your code because I do not have arguments for the functions. (I tweaked them to make them a bit more efficient.)

Plotting both results in the same axes is relativbely straightforward.

Example (using different functions) —

figure
plot1
hold on
plot2
hold off
grid

function plot1
plot((0:0.1:5), sin((0:0.1:5)*pi))
end

function plot2
plot((0:0.1:5), cos((0:0.1:5)*pi))
end



% 
% 
% function xnew = newtonmethod(f,df,x0,tol,n,Axh)
% %% Given data
% f=@(x) 8-4.5*(x-sin(x));
% df=@(x) -4.5*(1-cos(x));
% x0=1;
% tol=0.0001;
% n=50;
% %% Newton code
% disp('No Itr     Solution    Error  ')
% Error=zeros(1,n);
% for i=1:n
%     xnew=x0-(f(x0)/df(x0));
%     err=abs(xnew-x0);
%     fprintf('%3i %11.4f  %11.4f %11.4f\n',i,x0,err);
%     if (err<tol)
%         break
%     end
%     x0=xnew;
%     Error(i)=err;
% end
% %% Graph
%
% 
% plot(1:n,Error,'r-','Linewidth',02)
% xlabel('No of Iteration','Interpreter','latex','FontSize',12)
% ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
% title('Error Decay','Interpreter','latex','FontSize',12)
% end
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  %% 2nd method
%  function xnewh = Hmethod(f,df,ddf,x0,tol,n,Axh)
% %% Given data
% f=@(x) 8-4.5*(x-sin(x));
% df=@(x) -4.5*(1-cos(x));
% ddf=@(x) -4.5*sin(x);
% 
% x0=1;
% tol=0.0001;
% n=50;
% %% code
% disp('No Itr     Solution    Errorh')
% 
% Errorh=zeros(1,n);
% for i=1:n
% 
%     xnewh=x0- (2*f(x0).*df(x0)) ./ (2*(df(x0)).^2-ddf(x0).*f(x0));
%     errh=abs(xnewh-x0);
% 
%     fprintf('%3i %11.4f  %11.4f\n',i,x0,errh);
%     if (errh<tol)
%         break
%     end
% 
%     x0=xnewh;
%     Errorh(i)=errh;
% end
% %% Graph
% 
% 
% plot(1:n,Errorh,'b-','Linewidth',02)
% xlabel('No of Iteration','Interpreter','latex','FontSize',12)
% ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
% title('Error Decay','Interpreter','latex','FontSize',12)
%  end
% %    

.

[Omar B.]

I got an error when I plot the error

Please check my code

[Star Strider]

I cannot check it without the arguments.

Also, note the changes I made to your original code. They may solve that problem. (I commented-out your functions so they would not run and so I could provide an answer to the question you actually asked.)

[Omar B.]

What do you mean by arguments ?

I tried the following but I did not get the graph

f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
ddf=@(x) -4.5*sin(x);
x0=1;
tol=0.0001;
n=11;
Error = newtonmethod2(f,df,x0,tol,n);
Errorh = Hmethod2(f,df,ddf,x0,tol,n);
x=linspace(1,11);
figure
plot(x,Error,'r-','Linewidth',02)
hold on
plot(x,Errorh,'b-','Linewidth',02)
hold off
xlabel('No of Iteration','Interpreter','latex','FontSize',12)
ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)


   

[Star Strider]

I got the graph!

There is also an error in the ‘newtonmethod’ fprintf call that I corrected. (The format string had too many fields.) I also added a legend call.

f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
ddf=@(x) -4.5*sin(x);
x0=1;
tol=0.0001;
n=11;

figure
Error = newtonmethod(f,df,x0,tol,n);

No Itr     Solution    Error  
  1        1.0000      3.5224
  2        4.5224      3.1347
  3        1.3877      1.6791
  4        3.0668      0.6080
  5        2.4588      0.0282
  6        2.4306      0.0001
  7        2.4305      0.0000

hold on
Errorh = Hmethod(f,df,ddf,x0,tol,n);

No Itr     Solution    Errorh
  1        1.0000      0.8339
  2        1.8339      0.5654
  3        2.3993      0.0312
  4        2.4305      0.0000

hold off
grid
legend('newtonmethod','Hmethod','Location','best')

function xnew = newtonmethod(f,df,x0,tol,n,Axh)
%% Given data
f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
x0=1;
tol=0.0001;
n=50;
%% Newton code
disp('No Itr     Solution    Error  ')
Error=zeros(1,n);
for i=1:n
    xnew=x0-(f(x0)/df(x0));
    err=abs(xnew-x0);
    fprintf('%3i   %11.4f %11.4f\n',i,x0,err);
    if (err<tol)
        break
    end
    x0=xnew;
    Error(i)=err;
end
%% Graph


plot(1:n,Error,'r-','Linewidth',02)
xlabel('No of Iteration','Interpreter','latex','FontSize',12)
ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
title('Error Decay','Interpreter','latex','FontSize',12)
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %% 2nd method
 function xnewh = Hmethod(f,df,ddf,x0,tol,n,Axh)
%% Given data
f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
ddf=@(x) -4.5*sin(x);

x0=1;
tol=0.0001;
n=50;
%% code
disp('No Itr     Solution    Errorh')

Errorh=zeros(1,n);
for i=1:n

    xnewh=x0- (2*f(x0).*df(x0)) ./ (2*(df(x0)).^2-ddf(x0).*f(x0));
    errh=abs(xnewh-x0);

    fprintf('%3i   %11.4f %11.4f\n',i,x0,errh);
    if (errh<tol)
        break
    end

    x0=xnewh;
    Errorh(i)=errh;
end
%% Graph


plot(1:n,Errorh,'b-','Linewidth',02)
xlabel('No of Iteration','Interpreter','latex','FontSize',12)
ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
title('Error Decay','Interpreter','latex','FontSize',12)
 end

.

[Omar B.]

Thank you so much. Finally, I got it

[Star Strider]

As always, my pleasure!

Answer 2

[Torsten]

%% Given data
f=@(x) 8-4.5*(x-sin(x));
df=@(x) -4.5*(1-cos(x));
x0=1;
tol=0.0001;
n=50;
[x_newton,i_newton,Error_newton]=newtonmethod(f,df,x0,tol,n);

No Itr     Solution    Error  
  1      1.0000      3.5224
  2      4.5224      3.1347
  3      1.3877      1.6791
  4      3.0668      0.6080
  5      2.4588      0.0282
  6      2.4306      0.0001
  7      2.4305      0.0000

ddf=@(x) -4.5*sin(x);
[x_Hmethod,i_Hmethod,Error_Hmethod]=Hmethod(f,df,ddf,x0,tol,n);

No Itr     Solution    Errorh
  1      1.0000       0.8339
  2      1.8339       0.5654
  3      2.3993       0.0312
  4      2.4305       0.0000

%Plot results
hold on
plot(1:i_newton,Error_newton,'r-','Linewidth',02)
plot(1:i_Hmethod,Error_Hmethod,'b-','Linewidth',02)
hold off
xlabel('No of Iteration','Interpreter','latex','FontSize',12)
ylabel('Error=$|x_{n+1}-n_n|$','Interpreter','latex','FontSize',12)
title('Error Decay','Interpreter','latex','FontSize',12)

function [xnew,i,Error] = newtonmethod(f,df,x0,tol,n)
  %% Newton code
  disp('No Itr     Solution    Error  ')
  Error=[];
  for i=1:n
    xnew=x0-(f(x0)/df(x0));
    err=abs(xnew-x0);
    Error=[Error;err];
    fprintf('%3i %11.4f %11.4f\n',i,x0,err);
    if (err<tol)
        break
    end
    x0=xnew;
  end
end
%% 2nd method
function [xnewh,i,Errorh] = Hmethod(f,df,ddf,x0,tol,n)
  %% code
  disp('No Itr     Solution    Errorh')
  Errorh=[];
  for i=1:n   
    xnewh=x0- (2*f(x0).*df(x0)) ./ (2*(df(x0)).^2-ddf(x0).*f(x0));
    errh=abs(xnewh-x0);
    Errorh=[Errorh;errh];
    fprintf('%3i %11.4f  %11.4f\n',i,x0,errh);
    if (errh<tol)
        break
    end
    x0=xnewh;
  end
end

[Omar B.]

Thank you so much.