minor integrative trapz assistance

       I try the following: 

c=3e8;
  eta=zeros(1,31);
  for xx=linspace(0,30,31);
      x=0:xx;
      if x==0
      upplimit=(28e9*3.14e7)./(1+(1+x).^2);
      y=(c*56e9*3.14e7*(1+x).^2)./(1+(1+x).^2);
      eta(1,1)=upplimit*y;
      else
      eta(1,xx+1)=trapz(upplimit,y);
      end
  end

for which I get the error message on the second instance of the 'for' loop:

Maximum variable size allowed by the program is exceeded.

I would be appreciative of a correction.

Answer 1

[Walter Roberson]

You have

x=0:xx;
if x==0

recall that an "if" statement is satisfied only if all of the values being tested are non-zero. You are 0:xx which is a vector if xx is not 0 so x == 0 would fail in that case, falling through to the else. Which might be what you want, only be accident.

In the case that xx is 0 then x will be the scalar 0 and (28e9*3.14e7)./(1+(1+x).^2); is going to become (28e9*3.14e7)/2 . And that is to be the increment between rows, a value on the order of 36e16. Increment between rows because when you use trapz(X,Y) the first parameter, X, is the increment.

Question: if x is not 0 then you do not change y, so you repeat the same trapz() over and over again. Are you sure that is what you want?

[Gloust]

Yes, it should be the following:

eta=zeros(1,31);
for xx=linspace(0,30,31);
    x=0:xx;
    if x==0
    upplimit=(28e9*3.14e7)./(1+(1+x).^2);
    y=(c*56e9*3.14e7*(1+x).^2)./(1+(1+x).^2);
    eta(1,1)=upplimit*y;
    else
    upplimit=(28e9*3.14e7)./(1+(1+x).^2);
    y=(c*56e9*3.14e7*(1+x).^2)./(1+(1+x).^2);
    eta(1,xx+1)=trapz(upplimit,y);
    end
end

When I give trapz a row vector in X, does it not take the difference between each adjacent pair and then multiply it by the corresponding y?