Creating a matrix with for loop

2018 4 27
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更新时间：2018 4 28
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I need to create a matrix that increases or decreases in size with the change in variable n. The matrix should be having only one column and the number of rows need to change with the n value.
The inputs are n, (dx=L/n), L. The matrix that needs to be created should be [0;dx/2;(dx/2+dx);((dx/2+dx)+dx);(((dx/2+dx)+dx)+dx);.....;0.5].
The fourth value of the matrix is simply the third value +dx, this should go on until it reaches 0.5. I can't seem to work out how to make a for loop for this problem.
The number of values in between 0 and 0.5 should also be equal to the n value, so if the n value is 8 the number of values in between should also be 8.

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Stephen23
Stephen23 2018-4-27
编辑：Stephen23 2018-4-27

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"It should definitely be a for loop. I don't think such a matrix can be produced using for loop"
Using colon and concatenation, without any loop:
[0,dx/2:dx:L,L].'
Try it yourself, I don't see any difference from what you requested (shown in vector A):
>> n = 8;
>> L = 0.5;
>> dx = L/n;
>> A = [0;dx/2;dx/2+dx;dx/2+2*dx;dx/2+3*dx;dx/2+4*dx;dx/2+5*dx;dx/2+6*dx;dx/2+7*dx;0.5];
>> B = [0,dx/2:dx:L,L].';
>> [A,B]
ans =
0.00000 0.00000
0.03125 0.03125
0.09375 0.09375
0.15625 0.15625
0.21875 0.21875
0.28125 0.28125
0.34375 0.34375
0.40625 0.40625
0.46875 0.46875
0.50000 0.50000
>> A-B
ans =
0
0
0
0
0
0
0
0
0
0
Why do you need to use a loop?

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Harin Nelumdeniya
Harin Nelumdeniya 2018-4-27
编辑：Harin Nelumdeniya 2018-4-27
Hey Stephen, Thanks for that but i need the values in between to change according to n if n is 5 the values in between should also be 5. A numerical example would be [0;0.05;0.15;0.25;0.35;0.45;0.5], this matrix has five values in between 0 and 5 so n=5. If i set n=10 then i should get 10 values in between 0 and 5. The inputs for the numerical matrix is L=0.5, n=5 and dx=L/n.
Stephen23
Stephen23 2018-4-27
编辑：Stephen23 2018-4-27
It works when I try it:
>> n = 5;
>> L = 0.5;
>> dx = L/n;
>> [0,dx/2:dx:L,L].'
ans =
0.00000
0.05000
0.15000
0.25000
0.35000
0.45000
0.50000
This gives five values between 0 and L, just as requested. And this gives ten, just as requested:
>> n = 10;
>> L = 0.5;
>> dx = L/n;
>> [0,dx/2:dx:L,L].'
ans =
0.00000
0.02500
0.07500
0.12500
0.17500
0.22500
0.27500
0.32500
0.37500
0.42500
0.47500
0.50000
I still don't see why you think that you need a loop.
Harin Nelumdeniya
Harin Nelumdeniya 2018-4-28
Thanks Stephen, I thought it couldn't be done without using a loop, but i thought wrong.

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