Using anonymous functions with vectors of unknown number of elements and getting the coefficients associated with the variables

Hello everyone,

I hope that everyone is doing good.

I am aware that the description or summary of my question is quite not clear, so I will explain here what I mean exactly by the title, and feel free to ask any question if you feel things are still not clear.

So I have the following example, from the book (Kernel Adaptive Filtering: A Comprehensive Introduction):

and basically, what I am trying to implement (1.29), such that u will be a (1 x n) vector given by the user, and then c will be (1 x n) arbitrary vector with the values [c1 c2 c3 c4 ... cn]. So obviously, I do not know the number of elements in c unless I know the number of elements of the given vector, u.

So that is my first concern, how I can identify a function with a vector of unknown variables, such that the number of variables is equal to the number of elements in the given vector.

Second, after implementing this function, I want to find the coefficients of c1, c2, c1c2, such that Ii put the coefficients in an array.

Can anyone help me with this? I have been looking for some helpful functions in matlab, but I did not find any.

Thank you

Ahmed AR

[Torsten]

Please show the code how you want to supply the u-vector.

[Ahmed Radhi]

So basically, through some inputs, the user will declare the dimension of the vector

% n_u = input("input dimension n_u: ");
% n_s = input("state dimension n_s: ");
% n_y = input("output dimension n_y: ");

Based on the inputs, I will have a randomly initialized vector of size (1 x n_s) as shown in the following code:

% Assume n_s = 0
n_s = 2;
u = randi([-9999, 9999], 1, n_s)

u = 1×2

6113 1041

So yah, that is how we initially supply the vector.

Answer 1

[Torsten]

Then

n_s = 2;
u = randi([-9999, 9999], 1, n_s)

u = 1×2

-5089 -4994

c = sym('c',size(u)); 
kappa = expand((1+u*c.')^2)

kappa = 

coefficients = coeffs(kappa)

coefficients = 

[Ahmed Radhi]

Thank you very much, I really appreciate that.

Can you please explain the line

"kappa = expand((1+u*c.')^2)"

I didn't really understand how you formulated the variables as (u*c.')

Thank you

[Torsten]

Maybe you understand better if u is also symbolic:

c = sym('c',[1 2]);
u = sym('u',size(c));
kappa = (1+u*c.')^2

kappa = 

kappa = expand(kappa)

kappa = 

coefficients = coeffs(kappa,c)

coefficients = 

[Ahmed Radhi]

Thank you very much.

I understand now, it seems that it is a way that you use to formulate the dot product, I use the "dot()" function instead.

One more concern please, is there a way that I can go through the array and square root the coefficient in front of u? So basically after taking the coefficients with respect to c, I want to take square root of the coefficients with respect to u.

Thank you

[Torsten]

You mean

c = sym('c',[1 2]);
u = sym('u',size(c));
kappa = (1+u*c.')^2 

kappa = 

kappa = expand(kappa)

kappa = 

coefficients = coeffs(kappa,c)

coefficients = 

coefficients = sqrt(coefficients)

coefficients = 

? What if the coefficients are negative like in your example case ?

[Ahmed Radhi]

No, that's not really what I meant.

Now, as you showed in the last example, where we have vector u and vector c.

As I previously mentioned, we will find the coefficients with respect to vector c, which will be initiated as variables (c1, c2, ...).

Then, I want to go through the result array, and take the square root of the coefficients only with respect to u, such that

()

I hope it is clear now.

[Torsten]

c = sym('c',[1 2]);
u = sym('u',size(c));
kappa = (1+u*c.')^2 

kappa = 

kappa = expand(kappa)

kappa = 

cc = coeffs(kappa,c)

cc = 

cu = subs(cc,u,[1 1])

cu = 

cusqrt = sqrt(cu)

cusqrt = 

cc = cc./cu.*cusqrt

cc = 

[Ahmed Radhi]

Thank you very much

I really appreciate it

[Ahmed Radhi]

Sorry but one more concern.

Looking at the result that I want to reach, which is also from the book (Kernel Adaptive Filtering: A Comprehensive Introduction):

we find that the elements of the vector are arranged in a specific way, which is different than the way matlab arranges them (we can find that in your example above).

So can we arrange them in the same way or something? Becuase that is a bit weird.

[Torsten]

I don't see the rule of the ordering - neither in your vectors phi(u) and phi(c) nor in the MATLAB ordering.

But I think you cannot inluence it. If it were the case, it should be mentionned in the documentation of "coeffs".

Look it up.

[Ahmed Radhi]

Okay. Thank you.