How does the following two codes differ? both are working, but they are yielding different answers

Hello,

Could you explain for me why I'm not getting the same answer from the following two codes??

function whatswrong(x)
x1= linspace(0,x);
n=length(x1);
    uy = -5/6.*(sing2(x1,0,4)-sing2(x1,5,4))...
        + 15/6.*sing2(x1,8,3) + 75*sing2(x1,7,2)...
        + 57/6.*x1^3 - 238.25.*x1;
plot(x1,uy,'--')
xlabel('x1')
     ylabel('uy'),
     title('displacement versus distance')

function check = sing2(x1,a,n)
    if x1 > a
        check = (x1 - a).^n;
    else
        check=0;
    end

------------------------------------------------------------------------------------------------------

function beam2(x)
x1= linspace(0,x);
n=length(x1);
for i=1:n
    uy(i) = -5/6.*(sing(x1(i),0,4)-sing(x1(i),5,4));
    uy(i) = uy(i) + 15/6.*sing(x1(i),8,3) + 75*sing(x1(i),7,2);
    uy(i) = uy(i) + 57/6.*x1(i)^3 - 238.25.*x1(i);
end
plot(x1,uy)
    function check = sing(x1,a,n)
    if x1 > a
        check = (x1 - a).^n;
    else
        check=0;
    end

many many thanks,

Answer 1

[Roger Stafford]

In the functions 'sing' and 'sing2' you have "if x1 > a". If x1 is a many-element vector, this only comes true what all elements are greater than a. Since you are presenting 'sing' with a scalar x1 one at a time and 'sing2' with a vector, you will get different results from them. You need to rewrite 'sing2' if you want the vectorized 'whatswrong' to work properly.

Also in 'whatswrong' you have written x1^3 where x1 is a vector and that should give you an error message, not a result. I assume you didn't actually run it that way if you got a result.

[B]

 "You need to rewrite 'sing2' if you want the vectorized 'whatswrong' to work properly." could you please explain this more?

[Roger Stafford]

It can be written in vectorized form as:

   function check = sing2(x1,a,n)
     check = (x1>a).*(x1-a).^n;
   end

If you prefer a for-loop, it could be:

   function check = sing2(x1,a,n)
     check = zeros(size(x1));
     for k = 1:length(x1);
       if x1(k) > a
         check(k) = (x1(k)-a)^n;
       end
     end
   end

Note that in this second method the condition after the 'if' gets applied one at a time and therefore gives you the same result as obtained with your use of 'sing'.

[B]

Thanks a lot :)