Limit of a Cell

2 次查看（过去 30 天）
fert
fert 2016-2-27
评论： fert 2016-2-28
I have a "x1c" cell, and with the following code:
tn=1
for N=1:6591
for t=1:200
k=(((abs(x1c{t,1}(N)-x1c{1,1}(N))).^2))/(6591*200*6)
tn=tn+1;
end
end
I would like to take the "limit of the "k" while t goes infinity"; is it possible?

请先登录，再回答此问题。

采纳的回答

Image Analyst
Image Analyst 2016-2-27
No because you have x1c{t,1} and if t is infinite, then you'd need infinite number of cells in your cell array, which you don't have enough memory for.
  18 个评论
显示 16更早的评论
Image Analyst
Image Analyst 2016-2-28
Perhaps I'll try something later if you you answer my two questions in my earlier reply (third and last chance or I'm moving on).
fert
fert 2016-2-28
Oh I am so sorry.
So, x1c{1} is x1, and x1c{2} is x2, and x1c{3} = x3? Is that correct?
It is not correct. x1c{1} is a cell which includes 6591 atoms' i j k position directions.
Okay, but then why does x1, x2, and x3 have 3 columns instead of 1? According to the formula, x sub n takes one index and a 2D array would take 2 indexes.
Since it contains both i j k directions.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by