Finding a variable value using no fzero etc.

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alexaa1989
alexaa1989 on 12 Aug 2014
Commented: alexaa1989 on 14 Aug 2014
Hi I have another problem that I cannot using fzero or solve or any method that uses initial answers because It can produce the exact answer
Y(t,j)=[0,0,1,0;1,0,0,0;0,0,0,1;0,1,0,0];
A(i,j)=[1,4,6,2;4,5,2,3;5,3,2,4];
now I have K(i,j) which needs to be calculated as follow
sigma(Y(1,j)*(K(1,j)-A(1,j)))=0
this part is simple I could easily calculate K(1,j) after having K(1,j) I need to use it in the following equation to calculate K(t,j)
sigma( sigma(Y(t,j)*(K(1,j)-(sigma(Y(t-1,f)*K(1,f)+A(1,j)))))=0
I calculated the answer K(1,3) = 6 from equation 1
from equation 2 we have K(1,1)=7, K(1,4)=9, K(1,2)=13
it can be easily calculated because it is gradually like this Y(K-A)=0 and we have y that is one and A
The problem is : how can I code the second equation that I would be able to calculate K(t,j) and use it in another equation? no fzero, solve etc. can anyone help?
  1 Comment
Matt J
Matt J on 12 Aug 2014
because It can produce the exact answer
Do you mean "can", or did you really mean "can't"? No solver will produce an exact answer. You always have inexact calculations in finite precision computers.

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Answers (1)

Matt J
Matt J on 12 Aug 2014
Your equations are linear, so you just have to build an appropriate matrix equation
P*X=Q
and solve as X=P\Q
  5 Comments
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Matt J
Matt J on 13 Aug 2014
There shouldn't be a fundamental difference. Both summations are equivalent to weighted sums of the K(i) with different sets of coefficients. Just put the coefficients into the rows of a matrix, one row per equation.
alexaa1989
alexaa1989 on 14 Aug 2014
the second summation have another K(1,f) in it
that makes everything different I have been trying to code it for a week now

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