how to programme transition matrix with matlab

In dynamical system ( two dimensional ) we have transition matrix for a given A0, A1 and A2 ( are all matrices )

how we can programme this matrix with matlab ?

[Amon Festo Manirakiza]

good question

Answer 1

[Image Analyst]

It looks like they're starting with i and j of zero so you need to skip the first row. Did you try eye() and a simple for loop?

n = 7;
A0 = 1 * rand(n)
A1 = 10 * rand(n)
A2 = 100 * rand(n)
T = eye(n)
for i = 2 : n
	for j = 2 : n
		T(i, j) = ...
			A0 * T(i-1, j-1) + ...
			A1 * T(i, j-1) + ...
			A2 * T(i-1, j);
	end
end
T

Note that it fails because we're taking matrices (the right hand side of the equation) and trying to stuff a matrix into a single element (the element at row i and column j), which you can't do. I think you need to explain it better.

And regarding your tags, what does this have to do with image analysis or processing?

[azertazert azertazertazertazert]

I'm trying to plot a solution of two dimensional system in reality with matlab , and i=0.... +inf, j=0......+inf , but for exemple i and j are bounded i,=0...5, matlab give :

Assignment has more non-singleton rhs dimensions than non-singleton subscripts

when we trying the code ?

[Image Analyst]

Well you said that all were matrices so A0, A1, and A2 are matrices. And T(i,j) is one element out of a matrix T, so it's a scalar. Thus A1 * T(i, j) is a matrix times a scalar, which is a matrix. That's the middle term of your sum, and the other terms are also matrixes, so the sum of all 3 is a matrix. Now you're trying to stuff that matrix into a single element at T(I, j). That won't work. Are you sure that "A0, A1 and A2 ( are all matrices )"??? If the A were scalars, everything would work out fine.

[azertazert azertazertazertazert]

yes I'm sure A0 , A1 and A2 are all matrices

[Image Analyst]

Then what I said stands. What is on the right hand side of the equals sign is a matrix, and you can't stuff a whole matrix into one element of a different matrix, unless that different matrix is a cell array.

https://matlab.fandom.com/wiki/FAQ#What_is_a_cell_array.3F

[azertazert azertazertazertazert]

thank's for help, but until now I need this idea to my work and I can't do this programme with my self any help and thank's a lot?

[Image Analyst]

It would work if you replaced the A by A(i,j), like

n = 7;
A0 = 1 * rand(n)
A1 = 10 * rand(n)
A2 = 100 * rand(n)
T = eye(n)
for i = 2 : n
	for j = 2 : n
		T(i, j) = ...
			A0(i, j) * T(i-1, j-1) + ...
			A1(i, j) * T(i, j-1) + ...
			A2(i, j) * T(i-1, j);
	end
end
T

Do you think that's what you want?

[azertazert azertazertazertazert]

in this case we have A0 and A1 and A2 ares* the same case ( scalar coefficients not matrices) because the term T(i,j) is a matrix not coefficient ( scalar) that waht I have a problem

[Image Analyst]

I'm not sure what "ares*" means.

If T is the name of a variable that is a matrix, then T(i,j) is one single number - a scalar, NOT a matrix.

Perhaps if you gave some more context by showing the paper above and below the equations it might explain more.

[azertazert azertazertazertazert]

Hellow, my work based on the system

and it's solution for zero boundary condition is as follows x(t1,t2)=

that what how t plot this solution for A0 A1 A2 are latrices

[azertazert azertazertazertazert]

are is mistakes it mean all are matrices

[azertazert azertazertazertazert]

with i and j are bouned for exemple i=0..20 j=0....20 and \alpha =0.8 and \beta =0.2

[azertazert azertazertazertazert]

....?

[wassim bidi]

Excuse me ,i have a problem i can't ask

so, how do i solve this problem?

Error using plot

Vectors must be the same lengths

clc 
clear

sys1 = tf(1,[0.1   1])
t=0:0.1:10;
 step(sys1)
y=step(sys1);
plot(t,y)

[Image Analyst]

wassim, yes you CAN ask, and you just did. However you asked in the wrong place. You asked in azertazert's discussion instead of starting your own. Once you start your own, people will be able to tell you why t is not the same length as y, like ask you why you didn't define t as linspace(0, 10, length(y)).

[Kent Millard]

@wassim bidi Hi Wassim. If you're still interested in asking your question, please click the 'Ask' link beneath the blue bar or use this link to start a new question thread. Best - Kent

Answer 2

[Walter Roberson]

n = 7;
A0 = 1 * rand(n);
A1 = 10 * rand(n);
A2 = 100 * rand(n);
T(1,1:n) = {eye(n)};
T(1:n,1) = {eye(n)};

for i = 2 : n
	for j = 2 : n
		T{i, j} = ...
			A0 * T{i-1, j-1} + ...
			A1 * T{i, j-1} + ...
			A2 * T{i-1, j};
	end
end
T

T = 7x7 cell array
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}
    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}    {7×7 double}

[azertazert azertazertazertazert]

matlab v 2015 give

T

Conversion to double from cell is not possible.

[Image Analyst]

You forgot to give the ENTIRE error message, which includes line numbers and the actual line of code.

The code Walter posted runs without error. I just tried it. How did you alter it?

[azertazert azertazertazertazert]

Image Analysit In the previous (answer) and question I show the system and the solution that I want plot , the code of Walter runs without error.

[Image Analyst]

Put it in a script/editor window, not on the command line.

Also put

clear all

because evidently you have another variable called T that is a double so doing

T(1, 1:n) = {eye(n)};

won't work because you're trying to put a cell into a double. Clearing everything will get rid of your prior T and let it work.

[azertazert azertazertazertazert]

Thank , it work now like you and Walter say, but the last question how I can eca;uate T(1,2) for exemple matlab show me {7×7 double} I want the value of this matrix

[Image Analyst]

I can tell you how to do that, but I'd rather you read the FAQ and learn how to do that:

https://matlab.fandom.com/wiki/FAQ#What_is_a_cell_array.3F

[azertazert azertazertazertazert]

Thank you a gain, I used cell2mat(T([4 5])) to extract the value of T(i,j) , Now when I'm tryin a noter exemple where i,j=2.....25 and we have small matrix for example A0=[1 2;3 4] we have a proble of dimension

[Image Analyst]

OK, but you should have read the FAQ. If you had, you would know that you can simply use braces to extract the contents of a cell. This is how most MATLAB programmers would do it:

cellContents = T{1,2};

rather than call the cell2mat() function. The FAQ does not even mention cell2mat().

[azertazert azertazertazertazert]

yes it work now, but If we wante change for simple example I wante chosse dimesnion of our matrix =2 and i=2:25 for example

clear all
n=2
A0 = [0.5 1 ;2 3];
A1 = [4 1 ;0.8 -2];
A2 = [3.4 1.2 ;-0.4 1];
T(1,1:n) = {eye(n)};
T(1:n,1) = {eye(n)};
for i = 2 : 25
	for j = 2 : 25
		T{i, j} = ...
			A0 * T{i-1, j-1} + ...
			A1 * T{i, j-1} + ...
			A2 * T{i-1, j};
	end
end
T

we will have a problem of Index exceeds matrix dimensions.

[Image Analyst]

Of course. Because if A is a matrix and you're multiplying it by some matrix inside the cell of T, the number of columns of A must match the number of rows of whatever is inside that cell of T. Evidently it doesn't match.

[azertazert azertazertazertazert]

any solution of this situation ????? if we have this case I want dim(A0)=2 or 3 for example and "i" and "j" are bouded max=25

[Image Analyst]

That's what we've been trying to tell you. The original equation does not make sense if the A are matrices. The only suggestion I had to fix it was to index A with (i, j) so that you are using JUST ONE ELEMENT of A instead of the whole matrix.

[azertazert azertazertazertazert]

yes I understood your idea, but there is some researchers studie this classe of system and they give the plot of solution for the considered system and A0 A1 A2 are matrices but how did they do this I don't Know , for this reason i ask questions

[Walter Roberson]

If you want to iterate i and j to 25, then you must initialize T(1:1:25) and T(1:25,1)

n=2

n = 2

maxiter = 25;
A0 = [0.5 1 ;2 3];
A1 = [4 1 ;0.8 -2];
A2 = [3.4 1.2 ;-0.4 1];
T(1,1:maxiter) = {eye(n)};
T(2:maxiter,1) = {eye(n)};
for i = 2 : maxiter
	for j = 2 : maxiter
		T{i, j} = ...
			A0 * T{i-1, j-1} + ...
			A1 * T{i, j-1} + ...
			A2 * T{i-1, j};
	end
end
T

T = 25x25 cell array
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}
    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}    {2×2 double}

[azertazert azertazertazertazert]

Thanks WAlter and Image Analyst now I can see some result that I need I will try other exemple and I will tele yous the results,

again thak you so much.

[azertazert azertazertazertazert]

tha last question If I want plot the surface " x "

clc
maxiter = 25;
A0 = [0.5 1 ;2 3];
A1 = [4 1 ;0.8 -2];
A2 = [3.4 1.2 ;-0.4 1];
T(1,1:maxiter) = {eye(2)};
T(2:maxiter,1) = {eye(2)};
for i = 2 : maxiter
	for j = 2 : maxiter
		T{i, j} = ...
			A0 * T{i-1, j-1} + ...
			A1 * T{i, j-1} + ...
			A2 * T{i-1, j};
	end
end
T;
x=[0 0]';
B=[1 4]';

syms t1 t2
for i = 2 : maxiter
	for j = 2 : maxiter
		x=x+T{i,j}*B*t1^i*t2^j;
	end
end

x depend on variable t1 t2 any solution to plot directly because It's long expression ?

[Walter Roberson]

subplot(2,1,1)
fsurf(x(1), 0.9*[-1 1 -1 1])
xlabel('t1')
ylabel('t2')
title('first x')

subplot(2,1,2)
fsurf(x(2), 0.9*[-1 1 -1 1])
xlabel('t1')
ylabel('t2')
title('second x')

[azertazert azertazertazertazert]

thank walter, excuse me for being late in a book the example

clc
maxiter = 25;
A0 = [-0.1 0 ;0.1 -0.05];
A1 = [-0.01 0.1; 0.1 -0.05];
A2 = [-0.05 0;0.1 -0.01];
T(1,1:maxiter) = {eye(2)};
T(2:maxiter,1) = {eye(2)};
for i = 2 : maxiter
	for j = 2 : maxiter
		T{i, j} = ...
			A0 * T{i-1, j-1} + ...
			A1 * T{i, j-1} + ...
			A2 * T{i-1, j};
	end
end
T;
x=[0 0]';
B=[0.1 0.1]';
syms t1 t2
alpha= 0.7;
beta=0.9;
for i = 2 : maxiter
	for j = 2 : maxiter
	x=x+T{i-1,j-1}*B*((t1^(i*alpha))/gamma(i*alpha+1))*((t2^(j*beta))/gamma(j*beta+1));
	end
end


fsurf(x(1), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('first x')

the plot is shown

but in this code , my matlab don show me the plot a white page only and this message

Undefined function or variable 'fsurf'.

Error in Transition_matrix (line 35)

fsurf(x(1), 0.9*[-1 1 -1 1])

bu If Iuse matlab online It show me this plot !!!

[Image Analyst]

fsurf() was introduced in R2016a. Do you have a version older than that?

[azertazert azertazertazertazert]

and If I use In my matlab 2015 this code

ezsurf(x(2), [0 20 0 20])

xlabel('t1')

ylabel('t2')

title('first x')

it show

Perhaps we have a mistakes In the code because only T(0,0)= eye(2) in the first picture that contain the expression of T(i,j) not a cel .?

[azertazert azertazertazertazert]

I have only matlab 2015, the command ezurf work but is note the same graphe ?

[Walter Roberson]

The online version you plotted x(1), the ezsurf you plotted x(2). The plots are different

[azertazert azertazertazertazert]

WAlter yes this ezsurf is the same as fsurf ??

and the last question why this graphe is note the same as in the originale paper

may be there a simple mistakes in the programme in the beginig

because it must T(0,0)=eye(2) or in this case T(1,1) =eye(2) not all tha cell

T(1,1:maxiter) = {eye(2)};

T(2:maxiter,1) = {eye(2)}; ??????

[Walter Roberson]

I did identify a problem in the initialization but I have not had time to write it up yet... been dealing with hardware problems

[Walter Roberson]

maxiter = 25;

A0 = [-0.1 0 ;0.1 -0.05];
A1 = [-0.01 0.1; 0.1 -0.05];
A2 = [-0.05 0;0.1 -0.01];

%corrected initialization
T{1,1} = eye(2);
for i = 2 : maxiter
    T{i,1} = A2*T{i-1,1};
end
for j = 2 : maxiter
    T{1,j} = A1*T{1,j-1};
end

%do the work
for i = 2 : maxiter
	for j = 2 : maxiter
		T{i, j} = ...
			A0 * T{i-1, j-1} + ...
			A1 * T{i, j-1} + ...
			A2 * T{i-1, j};
	end
end

x=[0 0]';
B=[0.1 0.1]';
syms t1 t2
alpha= 0.7;
beta=0.9;
for i = 2 : maxiter
	for j = 2 : maxiter
	x=x+T{i-1,j-1}*B*((t1^(i*alpha))/gamma(i*alpha+1))*((t2^(j*beta))/gamma(j*beta+1));
	end
end


fsurf(x(1), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('first x')

fsurf(x(2), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('second x')

[azertazert azertazertazertazert]

Thank you walter and Paul for helping me, Now It should be work but we have to find this graph

[Paul]

Can you post a screen capture of the equation for x from the source, like you did for T(i,j)?

[Walter Roberson]

Here is an implementation of T as a recursive function, to ensure that the proper thing is being calculated.

Note that in this implementation, what is to be passed to T is the zero-based indices exactly as in the definition of .

But please be sure to check the bounds of iteration of i and j, keeping in mind that for this purpose it is fine to pass 0 to T. In particular, check what the powers of t1 should be in the variable ti, and the powers of t2 should be in the variable tj, and how those power relate to the (zero based) indices to pass into T.

A0 = [-0.1 0 ;0.1 -0.05];
A1 = [-0.01 0.1; 0.1 -0.05];
A2 = [-0.05 0;0.1 -0.01];

maxiter = 25;

x=[0 0]';
B=[0.1 0.1]';
syms t1 t2
alpha = 0.7;
beta = 0.9;
for i = 1 : maxiter
    ti = B * ((t1^(i*alpha))/gamma(i*alpha+1));
    for j = 1 : maxiter
        tb = T(i-1, j-1, A0, A1, A2);
        tj = ((t2^(j*beta))/gamma(j*beta+1));
        x = x + tb * ti * tj;
    end
end

fsurf(x(1), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('first x')

fsurf(x(2), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('second x')

function Tij = T(i, j, A0, A1, A2)
   persistent tij Z
   if isempty(tij);
       tij = {eye(size(A0))};
       Z = zeros(size(A0));
   end
   if i < 0 || j < 0
       Tij = Z;
   elseif i + 1 <= size(tij,1) && j+1 <= size(tij,2) && ~isempty(tij{i+1,j+1})
       %i, j, size(tij)
       Tij = tij{i+1,j+1};
   else
       ta0 = T(i-1, j-1, A0, A1, A2);
       ta1 = T(i, j-1, A0, A1, A2);
       ta2 = T(i-1, j, A0, A1, A2);
       Tij = A0 * ta0 + A1 * ta1 + A2 * ta2;
       tij{i+1, j+1} = Tij;
   end
end

[azertazert azertazertazertazert]

firs the system is:

the solution is defined as follow

for a bounded i and j by 25 we have

for example :

the solution is given in the previous picture and T(i,j)

the plot of solution is

and Thank a gain Paul

[azertazert azertazertazertazert]

Walter Roberson thak you thak you thank you it work perfectly 100 % ,!!!

[Walter Roberson]

I was right in my previous concern: you do want to start the powers with 0.

A0 = [-0.1 0 ;0.1 -0.05];
A1 = [-0.01 0.1; 0.1 -0.05];
A2 = [-0.05 0;0.1 -0.01];

maxiter = 25;

x=[0 0]';
B=[0.1 0.1]';
syms t1 t2
alpha = 0.7;
beta = 0.9;
for i = 0 : maxiter
    ti = B * ((t1^(i*alpha))/gamma(i*alpha+1));
    for j = 0 : maxiter
        tb = T(i-1, j-1, A0, A1, A2);
        tj = ((t2^(j*beta))/gamma(j*beta+1));
        x = x + tb * ti * tj;
    end
end

fsurf(x(1), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('first x')

fsurf(x(2), [0 20 0 20])
xlabel('t1')
ylabel('t2')
title('second x')

function Tij = T(i, j, A0, A1, A2)
   persistent tij Z
   if isempty(tij);
       tij = {eye(size(A0))};
       Z = zeros(size(A0));
   end
   if i < 0 || j < 0
       Tij = Z;
   elseif i + 1 <= size(tij,1) && j+1 <= size(tij,2) && ~isempty(tij{i+1,j+1})
       %i, j, size(tij)
       Tij = tij{i+1,j+1};
   else
       ta0 = T(i-1, j-1, A0, A1, A2);
       ta1 = T(i, j-1, A0, A1, A2);
       ta2 = T(i-1, j, A0, A1, A2);
       Tij = A0 * ta0 + A1 * ta1 + A2 * ta2;
       tij{i+1, j+1} = Tij;
   end
end

[azertazert azertazertazertazert]

yes walter you are right about the power should be begin from zero

Answer 3

[azertazert azertazertazertazert]

Thank' Walter I will wait when you have time to write to me , thank you again

[azertazert azertazertazertazert]

Hellow, WALter If You can help me for the last time

I a similar problem that I want how to plot and obtain the same solution that exist :

1-------- the system:

2- The solution is as follow

with T(i,j) is :

fo example

the plot must be :

but I'm havn't this graphIn my plot.

The modified programme is:

function Tij = T(i, j, A1, A2)

persistent tij Z

if isempty(tij);

tij = {eye(size(A1))};

Z = zeros(size(A1));

end

if i < 0 || j < 0

Tij = Z;

elseif i + 1 <= size(tij,1) && j+1 <= size(tij,2) && ~isempty(tij{i+1,j+1})

%i, j, size(tij)

Tij = tij{i+1,j+1};

else

ta1 = T(i, j-1, A1, A2);

ta2 = T(i-1, j, A1, A2);

Tij = A1 * ta1 + A2 * ta2;

tij{i+1, j+1} = Tij;

end

end

clear all

clc

%exmple1

A1 = [-0.1 0; 0.1 -0.05];

A2 = [-0.01 0.1;0.1 -0.05];

maxiter = 25;

x=[0 0]';

%exmple1

B1=[-0.05 0.01]';

B2=[-0.1 -0.1]';

syms t1 t2

alpha = 0.7;

beta = 0.9;

for i = 1 : maxiter

ti = ((t1^(i*alpha))/gamma(i*alpha+1));

ti1 = ((t1^((i+1)*alpha))/gamma((i+1)*alpha+1));

for j = 1 : maxiter

tb1 = T(i, j, A1, A2);

tb2 = T(i-1, j, A1, A2);

tb3 = T(i, j-1, A1, A2);

tj = ((t2^(j*beta))/gamma(j*beta+1));

tj1 = ((t2^((j+1)*beta))/gamma((j+1)*beta+1));

x = x + (-tb1*B1*ti*tj1 -tb1*B2*ti1*tj)+ (tb3*B1+tb2*B2)*ti*tj;

end

end

ezsurf(x(1), [0 20 0 20])

xlabel('t1')

ylabel('t2')

title('first x')