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How do I create such matrix ? (please look at the thread for further details)

1 Ansicht (letzte 30 Tage)
Derick Wong
Derick Wong am 19 Dez. 2013
Geschlossen: MATLAB Answer Bot am 20 Aug. 2021
Hi,
Let say I have a matrix [ X1 X2 X3 .... Xm]
I need to use a method base on X1, X2 and X3 to get my X4, X5 till Xm.
The method in order to find X4 is X4=(X1+X2+X3)/3. Once X4 is calculated, we use the X4 to calculate X5 which now the it will turn out to be X5=(X2+X3+X4)/3 and find X6 using the found X5 and X4, X6=(X3+X4+X5)/3 and so on until we find Xm.
My question is, how do we come out with such matrix ?

Antworten (3)

Walter Roberson
Walter Roberson am 19 Dez. 2013
X = rand(1,3);
for K = 4 : m
X(K) = mean(X(K-3:K-1));
end
  2 Kommentare
Derick Wong
Derick Wong am 19 Dez. 2013
That is useful,thanks. But what if now that there is an occasion where a matrix eg. [ 1 2 3 4 5 ; 1 2 3 4 5 ; 1 2 3 4 5] is to be the variable X u declared ? Given that matrix, I will have to recalculate X4 and X5 which is 4 and 5 in this case for all rows.
Walter Roberson
Walter Roberson am 21 Dez. 2013
Do you mean the case where X4 and X5 have already been found, so you want to continue on from X6 ? But 4 is not (1 + 2 + 3)/3 ?
If it is the question of how to do this for several rows simultaneously, then
X = rand(2,3); %example 2 rows
for K = 4 : m
X(:,K) = mean(X(:,K-3:K-1),2);
end

Andrei Bobrov
Andrei Bobrov am 19 Dez. 2013
Bearbeitet: Andrei Bobrov am 19 Dez. 2013
X = randi(25,1,10);
n = 3;
X = X(:);
X = [X(1:n);conv2(X(1:end-n+2),ones(n,1)/n,'valid')];
on Deric's comment
X = randi(1500,93,343);
n = 3;
X = [X(:,1:n), conv2(X(:,1:end-n+2),ones(1,n)/n,'valid')];
  2 Kommentare
Derick Wong
Derick Wong am 19 Dez. 2013
Hi,
May I ask what if that the matrix has 343 columns and 93 rows ?
Andrei Bobrov
Andrei Bobrov am 19 Dez. 2013
see in my answer code after row with "on Deric ..."

Roger Stafford
Roger Stafford am 21 Dez. 2013
In case it is of interest to you, Derick, here is an explicit formula for individual elements of your vector X in terms of its first three elements. That is, it doesn't involve iteration - one can find the n-th element without evaluating others.
Let x1, x2, and x3 be the first three elements.
a = (x1+2*x2+3*x3)/6;
b = (-x1+4*x2-3*x3)/6;
c = (-2*x1-x2+3*x3)/3/sqrt(2);
t = atan2(sqrt(2),-1);
X(n) = a+3^(-(n-2)/2)*(b*cos((n-2)*t)+c*sin((n-2)*t));
This shows that X consists of rather widely-spaced points in an exponentially decaying sine function.

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