Finding the last value of a difference equation

Hello!

So I've been assigned to solve this equation:

where A is a positive nymber and 0<=n<= N-1, while also N=30

Furthermore I have an initial value of y(-1)=3. I'm looking to find the last value of this equation (so for n=29).

Last, I want my function to be called back by using this code:

result = my_matlab_function(A,N);
disp(['A= ' num2str(A) 'Result=' num2str(result)])

This is what I've managed to write so far, yet MATLAB won't accept it:

function res = my_matlab_function(A,n)

assume(A,'positive');
n:0:1:29;
y(-1)=3;
y(n)=(1/2)*(y(n-1)+(A^2)/(y(n-1)));

result = my_matlab_function(A,N);
disp(['A= ' num2str(A) 'Result=' num2str(result)])
end

[KALYAN ACHARJYA]

Note: Matlab allows the positive indices only, Y(1), Y(2) etc

[Nick Vasilakis]

Ok, but i want for the first value of n (meaning n=0), and given that y(-1)=3 to begin the calculations. So I've got to declare it somewhere!

Answer 1

[Yongjian Feng]

Use:

y(end) = -3

[Nick Vasilakis]

I want as an initial value y(-1)=3 not y(-1)=-3

[Yongjian Feng]

Then

y(end) = 3

[Yongjian Feng]

You actually want this:

function my_matlab_function(A)
    if A < 0
        % you don't really need this. You use A^2 later, it doesn't really
        % matter A < 0 or not
        disp('A must be positive');
        return;
    end
    % initialize the first one in the list
    y(1) = 3;
    % you want apply the expression from n = -1 to n = 29
    % but matlab has to start with i = 0. So it is i from 1 to 31
    
    N = 31;
    for i=2:N
        % each y(i) depends on y(i-1)
        y(i)=0.5*(y(i-1)+((A^2)/y(i-1)));    
    end
    
    % plot it if you want to see how it goes through different n
    % plot(1:N, y);
    
    disp(['A= ' num2str(A) '; Result=' num2str(y(N))])
end

[Nick Vasilakis]

It returns:

Not enough input arguments.

Error in my_matlab_function (line 2)
    if A < 0

And using Matlab Grader to solve my function, it returns:

Unrecognized function or variable 'A'.

[Yongjian Feng]

You need to call this function with an A value.

1. Save the function as my_matlab_function.m. It is a function that takes an input argument A.

function res = my_matlab_function(A)
    if A < 0
        % you don't really need this. You use A^2 later, it doesn't really
        % matter A < 0 or not
        disp('A must be positive');
        return;
    end
    % initialize the first one in the list
    y(1) = 3;
    % you want apply the expression from n = -1 to n = 29
    % but matlab has to start with i = 0. So it is i from 1 to 31
    
    N = 31;
    for i=2:N
        % each y(i) depends on y(i-1)
        y(i)=0.5*(y(i-1)+((A^2)/y(i-1)));    
    end
    
    % plot it if you want to see how it goes through different n
    % plot(1:N, y);
    
    disp(['A= ' num2str(A) '; Result=' num2str(y(N))]);
    res = y(N);
end

1. From command line, call this function with an A value. Something like:

my_matlab_function(100)

Answer 2

[Torsten]

Then iterate 2-times more starting with n=1 instead of n=-1.

function res = my_matlab_function(A,N)
  y = zeros(N+1,1)
  y(1) = 3;
  for i = 1:N
    y(i+1) = 0.5*(y(i) + A^2/y(i))
  end
  res = y(end)
end

[Nick Vasilakis]

It returns:"Unrecognized function or variable 'A'."

[Torsten]

And did you call my_matlab_function as

function main
  A = 4;
  N = 10;
  res = my_matlab_function(A,N)
end

function res = my_matlab_function(A,N)
  y = zeros(N+1,1);
  y(1) = 3;
  for i = 1:N
    y(i+1) = 0.5*(y(i) + A^2/y(i));
  end
  res = y(end);
end

?

[Nick Vasilakis]

I'm looking to find the value of A and the result of the equation for n=29, given that I have an initial value of y(-1)=3. I can't declare them and use them as I wish

[Torsten]

You can't "find" A because you have no degree of freedom in the difference equation.

Maybe you want to prescribe the last value of y (say yend) after 30 iterations and determine A such that y(end) = yend ?

If you want the expression of y(30) dependent on A, you must use symbolic computations.

[Nick Vasilakis]

I've managed to create a new code, but I don't know how to implement that variable A is a positive variable

function res = my_matlab_function(A,N)

%ορισμός της αρχικής συνθήκης y(-1)=3
y_1=3;
assume (A,'positive');
% υπολογισμός της πρώτης τιμής χωρίς την χρήση του βρόχου
y(0)=0.5*(y_1+((A^2)/y_1));

%υπολογισμός των υπόλοιπων τιμών

for i=1:29
    
    y(i)=0.5*(y(i-1)+((A^2)/y(i-1)));
    
    result = my_matlab_function(A,N);
disp(['A= ' num2str(A) 'Result=' num2str(result)])
end

[Torsten]

Well, maybe I was not clear in my answer: A has to be prescribed, it can't be deduced.

At least if you don't modify your question somehow.

And again: Arrays in Matlab start with index 1. 0 gives an error message.