Short code for the expression
显示 更早的评论
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
采纳的回答
madhan ravi
2019-5-1
编辑：madhan ravi
2019-5-1
a = num2cell( (1-h) * repmat( (1-h) * 1 + h * P(1), 1, 4) + h * P ) ;
[a1, a2, a3, a4] = deal( a{:} )
6 个评论
Guillaume
2019-5-1
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
2019-5-1
Yes true and thank you for showing an additional method.
Tino
2019-5-1
Guillaume
2019-5-1
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]
Guillaume
2019-5-1
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.
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