small trapz clarification needed
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I would appreciate some clarification.
I attempt the following:
zz=zeros(1,31);
for xx=linspace(0,30,31);
x=0:xx;
y=(c*56e9*3.14e7*(1+x).^2)./(1+(1+x).^2);
zz(1,xx+1)=trapz(x,y);
end
I get the following errata description:
??? Error using ==> colon
Maximum variable size allowed by the program is exceeded.
Error in ==> trapz at 43
perm = [dim:max(ndims(y),dim) 1:dim-1];
Error in ==> Cs_kSZ at 14
eta(1,xx+1)=trapz(x,y);
What is the particular problem?
1 个评论
the cyclist
2014-3-5
I deleted the duplicate of this question that you posted earlier.
回答（1 个）
the cyclist
2014-3-5
编辑：the cyclist
2014-3-5
[FYI, I put in c=1 to get your code to work at all.]
In the first iteration of your for loop, you are calling trapz() with these inputs:
trapz(0,8.79e17)
which is I suppose you mean to be zero (because you are "integrating" over a single point). However, MATLAB does not handle that case very well, possibly due to floating point error.
I think you might get what you want by skipping that first evaluation, using this modification of your code:
zz=zeros(1,31);
for xx=linspace(1,30,30);
x=0:xx;
y=(c*56e9*3.14e7*(1+x).^2)./(1+(1+x).^2);
zz(1,xx+1)=trapz(x,y);
end
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2014-3-5
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