Limit of a Cell

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?

Answer 1

[Image Analyst]

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.

[fert]

That is correct, thank you. But for example can I do "while t goes 200" then?

[Image Analyst]

Sure, as long as x1c has at least 200 rows.

[fert]

Yes, it has. But when I attempt to do that, in this way:

tn=1
for N=1:6591
for t=1:200
    sumx1c{tn}=(((abs(x1c{t,1}(N)-x1c{1,1}(N))).^2))/(6591*200*6);
    tn=tn+1;
end
end

sumx1cc=cell2mat(sumx1c);

syms t sumx1cc
limit(sumx1cc,t,200)

it doesn't give any number.

[Image Analyst]

I don't have any optimization toolboxes so I don't know what the limit() function does. Why are you having sumx1c be a cell array when you immediately cast it to a double after the loop. It can be a double right from the start.

We can't try your code conveniently unless you supply us with x1c.

[Image Analyst]

Alright, I made code to create some x1c, made slight modifications to make sumx1c a double array and get rid of unneeded second index, and then ran it, and it ran fine.

% Create some random numbers for x1c
numElements = 6591;
for k = 1 : 300
  x1c{k} = rand(1, numElements);
end

% Now do the code.
tn=1
for N=1:6591
  for t=1:200
    sumx1c(tn)=(((abs(x1c{t}(N)-x1c{1}(N))).^2))/(6591*200*6);
    tn=tn+1;
  end
end
% Compute the sum array
sumx1cc=sum(sumx1c)

% syms t sumx1cc
% limit(sumx1cc,t,200)

[fert]

Thank you so much for your effort, and your time. But I want the take the limit, not the sum?

I just want to do this operation:

http://imgur.com/Lp4HMJ6

[Image Analyst]

You could have saved some wasted effort if you have posted that first. Apparently there are different x functions: an x1, an x2, an x3, etc. So it's best to make x a 2D array with N rows, and like a million or so columns (or some big number). So, to get started we need to know what the various x functions are. What are they? Do you have some equation for the x's?

[fert]

The N is the number of atoms, the t is the time, and the x is the displacement. The x(disp) is not an equation; they have been given as scalar numbers. I suppose my "for loop" meet with the sum in the equation, but as you state, in order to take the limit, I need an equation depends on t in the equation; though I find a matrix contains of the numbers...

I don't have an equation for x's; but the values.

[Image Analyst]

You have x "sub i" as a function of t. In other words, an array x1 as a function of t, and x2 as a function of t, and so on. Can you give me code for creating those? Can we just say N=3 and create 3 random x's, like

x1 = rand(1, 200);
x2 = rand(1, 200);
x3 = rand(1, 200);

If so, we can then get started. Or would you rather give me the code for the x1, x2, and x3?

[fert]

First of all, thank you so much! Absolutely I can give:

This is "x1c":

6591x3 double

6591x3 double

6591x3 double

.
.
.
(200 of them which respect to "t=time")

This is what one of the 6591*3 double includes:

4.80604400000000  5.71385900000000  39.6054360000000
4.21084900000000  5.39276800000000  40.3420780000000
5.10345200000000  4.93471100000000  39.0536430000000
6.80840100000000  4.58005700000000  37.6887870000000
.                       .                       .                
.                       .                       .
.                       .                       .

[fert]

By the way, a correction, i states the N, not the "t".

[Image Analyst]

So, x1c{1} is x1, and x1c{2} is x2, and x1c{3} = x3? Is that correct?

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.

Anyway, I still can't run anything because you still haven't given your data. Please attach your x1c as a .mat file or Excel workbook or something.

[fert]

In formula it says, for every 6591 atoms, find the difference between positions for different times relative to the time=1.

Because there are three directions for x(i j k).

File exceeds the 5Mb.

[Image Analyst]

So clip it to make a smaller one. While you're at it, can you answer my two questions in my last reply?

[fert]

x1c{1,1} is 6591 different atoms' position, in i j k directions for time=1.

I will try to zip...

[fert]

I couldn't make smaller the file for 5 Mb, thus I uploaded here:

http://www.filedropper.com/matlab

[Image Analyst]

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]

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.