How to display value of particular iteration in a loop

I am having a slight problem, I'd like to find a relative error without knowing the analytical solution. I have an equation where i need to find the value of F using composite mid rule when the relative error is less than 1%. Here is my code so far: The equation is F= integral (from 0 to 30) of (200*(z(n)/(5+z(n)))*exp((-2*z(n))/30))dz :

clc
clear all
a=0;
b=30;
s=1000;
dx=(b-a)/s;
z=zeros(1,s);
for n=1:s
z(n)=a+dx/2+(n-1)*dx;
F=0;
for n=1:s  
F=F+dx*(200*(z(n)/(5+z(n)))*exp((-2*z(n))/30));
end     
sprintf('%6.2f',F)

[Jan]

You forgot to ask a question. What is the problem with this code?

[Piotr Haciuk]

Finding solution of F using composite mid point rule with relative error less than 1%

[Piotr Haciuk]

here is the task: In a model of a racing sailboat, the total force, F, on the mast is expressed in the integral form - F= integral (from 0 to 30) of (200*(z(n)/(5+z(n)))*exp((-2*z(n))/30))dz Use the composite mid-point rule to approximate the value of F with a relative error of less than 1%. Assume that the analytic solution is not known when calculating the relative error. answers: 100 intervals: 1480.72; 200 intervals: 1480.61. Relative error much less than 1%]

[Matt J]

But you have code already. What is left to do? If the code is not working as you wish, in what way can we see that?

[Piotr Haciuk]

How can i find the relative error? the answers i wrote i should expect. When you change the value of s for 100 i get the answer 1480.72, for 200 i get the answer 1480.61, for s=1000 i get 1480.57, but when running it with s=1000 when i want to take the value for n=100 it doesnt give me same answer as for s=100

[Piotr Haciuk]

Also, at which iteration the relative error of the si=olution is less than 1%

Answer 1

[Alberto Mora]

Maybe you can use this piece of code inside the for loop:

   if n==desired_iteration
       disp('This is the desired iteration!');
       disp(['The result at ',num2str(n),' iteration is ',num2str(result)]);
   end

[Piotr Haciuk]

How to i implement it to a loop?

[Alberto Mora]

if true
  % code
endclc
clear all
a=0;
b=30;
s=1000;
dx=(b-a)/s;
z=zeros(1,s);
for n=1:s
   z(n)=a+dx/2+(n-1)*dx;
   F=0;
   for n=1:s  
     F=F+dx*(200*(z(n)/(5+z(n)))*exp((-2*z(n))/30));
     if n==desired_iteration
       disp('This is the desired iteration!');
       disp(['The result at ',num2str(n),' iteration is ',num2str(result)]);
     end
  end
end       
sprintf('%6.2f',F)

Obviously you need to use your variable instead of "desired_iteration", "result".

[Piotr Haciuk]

Okay what if i want to know at which iteration the relative arror is less than 1%?

[Torsten]

Choose n such that

K*(b-a)^3/(24*n^2) < 0.01

where

K = max_[x in [a;b]] |f''(x)|

http://math.cmu.edu/~mittal/Recitation_notes.pdf

Best wishes

Torsten.

[Piotr Haciuk]

Torsten, how to implement it to my code?

[Torsten]

As you can see from the answer provided in your textbook, the authors seem to define the relative error as

|F_(2*n)-F_n|/|F_n|

So simply calculate F for n and 2*n and evaluate the expression above. If it is less than 0.01, you are done. Otherwise you will have to increase n and repeat the procedure.