I need to fix the code by using for loop to plot the relative error E in 2 norm versus n.

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%%%% Taylor ploynomials pn(x)
x=2:0.01:3;
f = 1./x;
p1=1/2.5;
p2= 1/2.5 -(4/25)*(x-2.5);
p3= 1/2.5 -(4/25)*(x-2.5) + (8/125)*(x-2.5).^2;
p4= 1/2.5 -(4/25)*(x-2.5) + (8/125)*(x-2.5).^2 -(16/625)*(x-2.5).^3;
E1=sqrt((f-p1).^2)/sqrt((f).^2)
E2=sqrt((f-p2).^2)/sqrt((f).^2)
E3=sqrt((f-p3).^2)/sqrt((f).^2)
E4=sqrt((f-p4).^2)/sqrt((f).^2)
n=[1 2 3 4]
E=[ E1 E2 E3 E4];
semilogy(n,E)

Answers (1)

Sivani Pentapati
Sivani Pentapati on 2 Sep 2021
Please refer to the below code snippet to calculate the l2 norm of error in iterative way. For more information, please refer to for loop in MATLAB documentation.
p(1,:)=1/2.5;
for i=2:4
p(i,:)= p(i-1,:)+ (4/25)*(2/5).^(i-2)*(-1).^(i-1)*(x-2.5).^(i-1);
end
E=sqrt((f-p).^2)/sqrt((f).^2);
n=1:4;
semilogy(n,E);

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