Effacer les filtres
Effacer les filtres

Finding the last value of a difference equation

2 vues (au cours des 30 derniers jours)
Nick Vasilakis
Nick Vasilakis le 24 Jan 2022
Modifié(e) : Yongjian Feng le 27 Jan 2022
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
  2 commentaires
KALYAN ACHARJYA
KALYAN ACHARJYA le 24 Jan 2022
Note: Matlab allows the positive indices only, Y(1), Y(2) etc
Nick Vasilakis
Nick Vasilakis le 24 Jan 2022
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!

Connectez-vous pour commenter.

Réponses (2)

Yongjian Feng
Yongjian Feng le 24 Jan 2022
Use:
y(end) = -3
  5 commentaires
Nick Vasilakis
Nick Vasilakis le 26 Jan 2022
Modifié(e) : Nick Vasilakis le 26 Jan 2022
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
Yongjian Feng le 27 Jan 2022
Modifié(e) : Yongjian Feng le 27 Jan 2022
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)

Connectez-vous pour commenter.


Torsten
Torsten le 24 Jan 2022
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
  6 commentaires
Nick Vasilakis
Nick Vasilakis le 26 Jan 2022
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
Torsten le 26 Jan 2022
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.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Produits


Version

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by