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

Question

Omar B. 2024-2-16

0
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https://ww2.mathworks.cn/matlabcentral/answers/2082958-how-to-plot-the-error-of-two-numerical-methods-on-the-same-graph

评论： Star Strider 2024-2-16

采纳的回答： Star Strider

在 MATLAB Online 中打开

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)
%    

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Answer 1

Star Strider 2024-2-16

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2082958-how-to-plot-the-error-of-two-numerical-methods-on-the-same-graph#answer_1410448

在 MATLAB Online 中打开

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

% %

.

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Star Strider 2024-2-16

在 MATLAB Online 中打开

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. 2024-2-16

Thank you so much. Finally, I got it

Star Strider 2024-2-16

As always, my pleasure!

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Answer 2

Torsten 2024-2-16

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2082958-how-to-plot-the-error-of-two-numerical-methods-on-the-same-graph#answer_1410523

编辑：Torsten 2024-2-16

在 MATLAB Online 中打开

%% 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

%% 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

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Omar B. 2024-2-16

Thank you so much.

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

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