I'm trying to make a Simpsons Rule Function Myself

I've had a few attempts at this question and have developed my answer further each time. If people wouldn't mind giving feedback or correcting my code that'd be great.

This is the question:

and the code I have written to answer this is, there is more to th question but this is just my code for the first bit:

function [y] =MPR_Asgn4_Q4a_26024405(f,a,b,n)

h=(b-a)/n;
x=zeros(1,n+1);
x(1)=a;
x(n+1)=b;
s1=0;
s2=0;
s3=0;

for k = 2:n
    x(k)=a+h*(1i-1);
end

for k=1:n/2
    s1= s1 + f(x(2*1i-1));
end

for k=1:n/2
    s2=s2+f(x(2*1i));  
end

for k= 1:n/2
    s3 = s3 +f(x(2*1i+1));
end
    
y = (h/3)*(s1+(4*s2)+s3);
disp(y);
end

[Omer Yasin Birey]

Hi Jack,

I tried to understand your code and the formula together. However, firstly I believe your "i" usages are quite wrong. If you put "i" or "j" without multiplication sign (*) that gives complexity to the number. However, if Im not wrong there is no calculations in complex domain here. Secondly, you are using loops without using the loop indexes maybe you meant "k" instead of "i"?

[Walter Roberson]

Please do not close questions that have answers.

Answer 1

[Omer Yasin Birey]

Hi Jack,

I fixed your code as much as I can. You can use the code below:

function [y] =MPR_Asgn4_Q4a_26024405(f,a,b,n)
    x = linspace(a,b,n);
    x(1)=a;
    x(n+1)=b;
    h=(b-a)/n;
    s1=0;
    s2=0;
    s3=0;
    %First if statements is to check if f is an anonymous function  
    %or an array.
    
    if length(f) >  1%array state
        %if it is a array your sub intervals will be the length of array
        %So you have to update it.
        n = length(f)-1;
        h=(b-a)/n;

        for k=3:2:length(f)-2
            s1= s1 + 2*f(k);
        end


        for k=2:2:length(f)
              s2=s2+f(k);
        end

        s3 = f(length(f));

        y = (h/3)*(f(1)+(2*s1+(4*s2)+s3));
    else%anonymous function state
        
        for k=3:2:length(f)-2
            s1= s1 + 2*f(x(k));
        end


        for k=3:2:length(f)
              s2=s2+f(x(k));
        end


        s3 = f(length(f));
        y = (h/3)*(f(x(1))+(2*s1+(4*s2)+s3));
    end
    disp(y);
end