Short code for the expression

h =2;
P = [1, 4, 5, 7];
a1 = (1-h)* 1 + h * P(:, 1);
a2 = (1-h) * a1 + h * P(:, 2);
a3 = (1-h) * a2 + h * P(:, 3);
a4 = (1-h) * a3 + h * P(:, 4);
disp(a1) = 1
disp(a2) = 7
disp(a3) = 3
disp(a4) = 11

2 comentarios

Tino
Tino el 1 de Mayo de 2019
Please I will appreciate the answer to this question
thanks in advance
Guillaume
Guillaume el 1 de Mayo de 2019
To what question? You haven't asked anything, just given us some code with no explanation nor context.

Iniciar sesión para comentar.

 Respuesta aceptada

madhan ravi
madhan ravi el 1 de Mayo de 2019
Editada: madhan ravi el 1 de Mayo de 2019

0 votos

a = num2cell( (1-h) * repmat( (1-h) * 1 + h * P(1), 1, 4) + h * P ) ;
[a1, a2, a3, a4] = deal( a{:} )

6 comentarios

Guillaume
Guillaume el 1 de Mayo de 2019
Note that deal is not needed in the above.
However, never create numbered variables. It's pointless, it makes the code more complex and has no advantages. Your original answer, which created a a vector was much better.
Another way of creating the a vector is with the filter function:
h = 2;
P = [1, 4, 5, 7];
a = filter(h, [1, h-1], P, -1)
madhan ravi
madhan ravi el 1 de Mayo de 2019
Yes true and thank you for showing an additional method.
Tino
Tino el 1 de Mayo de 2019
Hi Guillaume thanks for your swift response but if you check the code it does not follow sequence with different h values for instance
when h = 1
1
2
3
4
when h =2
1
3
3
5
when h =3
1
4
1
10
Guillaume
Guillaume el 1 de Mayo de 2019
You would have to adjust the 4th input of filter to match the starting a(0) of 1. Another option with filter:
a = filter(h, [1, h-1], [1/h, P]); %calculates a(0), ..., a(numel(P)), with a(0) forced to 1
a = a(2:end); %keep a(1), ..., a(numel(P))
I don't get why when h = 1, you'd get 1 2 3 4, since in that case, a(i) = P(i) and P is [1 4 5 7]
Tino
Tino el 1 de Mayo de 2019
Editada: Tino el 1 de Mayo de 2019
sorry the right figure is the one below. sorry for giving wrong computation
Can you help me with the short code to gives the following answers for h = any number for instance
when h = 1
1
4
5
7
when h = 2
1
7
7
3
when h = 3
1
10
1
13
thanks in advance
Guillaume
Guillaume el 1 de Mayo de 2019
In your question, you told us than when h = 2, a = [1 7 3 11] (for P = [1, 4, 5, 7]) and this is the answer that my code produces. I have no idea how you get your new answer. It doesn't match the formula you've posted.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Data Type Identification en Centro de ayuda y File Exchange.

Productos

Versión

R2019a

Etiquetas

Preguntada:

el 1 de Mayo de 2019

Comentada:

el 1 de Mayo de 2019

Community Treasure Hunt

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

Start Hunting!

Translated by