I am having problems in anonymous function (y=@(x)(...))

Let me re-word my querry.

Let's say there is a square matrix of v=[1 2 3 4 5 6 78 9 10; 11 12 13 14 15 16 17 18 19 20;21 22 23 24 25 26 27 28 29 30;31 32 33 34 35 36 37 38 39 40;41 42 43 44 45........ 99 100] and a is a variable that prompt the user to input. It is easy to obtain a square matrix of size a by a of the original matrix v (left hand corner) with the use of w=v(1:a,1:a), of course a is less than length(v).

However, I would like to know how to do the same for anonymous functions. v in this case will be a column matrix a quantities of "@(x)(B.*[x x^2 x^3 ......]);", where [x x^2 x^3 ......] is a row matrix with a elements.

Here is an example:

Let's say v={@(x)([x x^2 x^3 x^4]; @(x)([x^2 x^3 x^4 x^5]); @(x)([x^3 x^4 x^5 x^6]); @(x)([x^4 x^5 x^6 x^7])};

if a = 3, the new v would be v={@(x)([x x^2 x^3]; @(x)([x^2 x^3 x^4]); @(x)([x^3 x^4 x^5])};

[Kevin Phung]

are you missing sum()'s?

[Angus Wong]

Oh, there supposed to be sum() there, my copy and paste skills should improve. Let me change that. But the program still doesnt work.

[Kevin Phung]

ok, just wanted to make sure. Also, what is x supposed to be? I think maybe it'll be easier to construct the matrix if x was an input to your function fun, instead of creating so many anonymous functions.

[Angus Wong]

x is a variable.

The script is something like this:

clear; close all; clear;
dat=load(data);
leadingcoe=coef(dat);
yo=fun(leadingcoe);
k=input('Input a number for x');
for a=1:k
    disp(yo{a}(k));
end

This is a simplified version.

Answer 1

[Kevin Phung]

let me know if this is what youre trying to do with your anon functions:

function y = fun(a,x)
y = sum(triu(repmat(flip(x.^(1:a)),numel(a),1)),2)

%uncomment this if you want to see how the matrix looks like before summing the rows
%y = triu(repmat(flip(x.^(1:a)),numel(a),1)) 
end

[Angus Wong]

Emmm. Sorry. No, this is not I am looking for. Let's say a=[1 2 3 4 5];

I am expecting the output to be:

y={@(x)(sum(a.*[x x^2 x^3 x^4 x^5]));

@(x)(sum(a.*[0 x x^2 x^3 x^4]));

@(x)(sum(a.*[0 0 x x^2 x^3]));

@(x)(sum(a.*[0 0 0 x x^2]));

@(x)(sum(a.*[0 0 0 0 x]))};

After this, in the main script, I then define a numerical value to the variable x (let's say 6) and display the result. I expect the final output from the main script to be [44790;7464;1242;204;30].

44790 comes from 1*6+2*6^2+3*6^3+4^6^4+5*6^5.

I then am able to change the value assigned to x and get a new vector.

Hence, my x is not defined as a numerical value inside the anonymous function, but a variable.

Basically my intended program steps are: import the data set, determine the number of columns using size(,2), running the anonymous function (fun) with the leading coefficients as the sole input and output a set of anonymous functions based on that, able to set a numerical value for the variable x and output the results for the anonymous functions based on the vzalue of x as a column vector.

[Angus Wong]

This function is intended to output an anonymous function.

[Angus Wong]

But thanks a lot for your hard work!