Calculating the contributing terms of a summation equation
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I am trying to calculate the value of a variable in a summation term. Could any of you share some leads about the coding technique in Matlab? My equation looks like:
,
in which I know the values of Y, and z. I am looking for an idea and methodology to calculate the values of the variable .
12 件のコメント
David Goodmanson
2020 年 7 月 15 日
HI Bharath,
I am aware that Q (used to be called Y) and y have the same length, and that q and x have the same length, which may well be diffferent than the length of Q&y. Given the equatiion you posted on July 4, I believe the answer I posted is apt.
回答 (2 件)
KALYAN ACHARJYA
2020 年 6 月 21 日
編集済み: KALYAN ACHARJYA
2020 年 6 月 21 日
Please try any issue let me know
David Goodmanson
2020 年 7 月 5 日
編集済み: David Goodmanson
2020 年 7 月 5 日
Hi Bharath,
suppose the sum involving q is done over a dummy index j instead of i. It appears that Y and y are vectors of the same length so the expression is
Yi = sum{j=1,N} qj*xj*yi/(1+xj^2*yi^2)
or in matrix notation
Y = M*q where M(i,j) = xj*yi/(1+xj^2*yi^2)
You don't say how the length of Y and y compare to N, but if that length is >= N, then a least squares solution for q is just
q = M\Y.
2 件のコメント
David Goodmanson
2020 年 7 月 6 日
編集済み: David Goodmanson
2020 年 7 月 6 日
I will assume here that vectors Y and y have the same length as each other (if not then I don't understand the equation at all). But assuming Y and y are the same length, the key is that the entire expression is linear in q, so that a minimal q can be determined with standard linear algebra.
Assume that Y and y have length A. Then as I mentioned before, this can be put into matrix multiplication form,
Y = M*q
where Y is Ax1, q is Nx1 and M is AxN. This expression has A equations and N unknowns which are the elements of q.
If A=N there is an exact solution for q (assuming that matix M is nonsingular, which seems likely). If A>N there are more equations than unknowns and q = M\Y solves for q in the least squares sense. If A< N there are more unknowns than equations and q is not determined completely but can vary somewhat.
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