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Answered
How to find the maximum of mu by solving eqn
I think you are after eqn = ((W_bar - mu)/sigma) - lambda*sum(normpdf(lambda*(W-mu)/sigma)./normcdf(lambda*(W-mu)/sigma)) ...
How to find the maximum of mu by solving eqn
I think you are after eqn = ((W_bar - mu)/sigma) - lambda*sum(normpdf(lambda*(W-mu)/sigma)./normcdf(lambda*(W-mu)/sigma)) ...
8 years ago | 1
| accepted
Answered
How can I determine the time and corresponding values of two functions when the value one function is 20% larger than the other one with a FOR oder WHILE loop?
I would do the following: First, you want x such that Ps(x)=Pu(x)+0.2*Pu(x). So, define an <https://www.mathworks.com/help/matl...
How can I determine the time and corresponding values of two functions when the value one function is 20% larger than the other one with a FOR oder WHILE loop?
I would do the following: First, you want x such that Ps(x)=Pu(x)+0.2*Pu(x). So, define an <https://www.mathworks.com/help/matl...
8 years ago | 0
Answered
Create rolling-window matrix from vector
n=3;m=5;repmat(1:n,m,1)+repmat((0:m-1)',1,n) gives what you want. A more complete version is: n=5;m=100;o=3;repmat(1:n,ceil(m...
Create rolling-window matrix from vector
n=3;m=5;repmat(1:n,m,1)+repmat((0:m-1)',1,n) gives what you want. A more complete version is: n=5;m=100;o=3;repmat(1:n,ceil(m...
8 years ago | 0
Answered
Matlab and memory use
xx= rand(500,500); h=whos('xx'); h.bytes is what you want then. This is an estimate, as there might be some compressions i...
Matlab and memory use
xx= rand(500,500); h=whos('xx'); h.bytes is what you want then. This is an estimate, as there might be some compressions i...
8 years ago | 0
