Finding the last value of a difference equation

Question

Nick Vasilakis 2022-1-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1634940-finding-the-last-value-of-a-difference-equation

编辑： Yongjian Feng 2022-1-27

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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

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KALYAN ACHARJYA 2022-1-24

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

Nick Vasilakis 2022-1-24

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!

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Answer 1

Yongjian Feng 2022-1-24

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Use:

y(end) = -3

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Nick Vasilakis 2022-1-26

编辑：Nick Vasilakis 2022-1-26

在 MATLAB Online 中打开

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 2022-1-27

编辑：Yongjian Feng 2022-1-27

在 MATLAB Online 中打开

You need to call this function with an A value.

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

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

my_matlab_function(100)

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Answer 2

Torsten 2022-1-24

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https://ww2.mathworks.cn/matlabcentral/answers/1634940-finding-the-last-value-of-a-difference-equation#answer_880635

在 MATLAB Online 中打开

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

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Nick Vasilakis 2022-1-26

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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