How to display value of particular iteration in a loop

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Piotr Haciuk
Piotr Haciuk 2018-8-14
编辑： Stephen23 2018-8-14
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)
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Piotr Haciuk
Piotr Haciuk 2018-8-14
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
Piotr Haciuk 2018-8-14
Also, at which iteration the relative error of the si=olution is less than 1%

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回答（1 个）

Alberto Mora
Alberto Mora 2018-8-14
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
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Torsten
Torsten 2018-8-14
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.

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